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Tipler volume 2 resolução 6 edição

1. This document contains conceptual problems and their solutions related to measurement and vectors in physics. It covers topics like SI base units, significant figures, dimensions of physical quantities, and vector operations. 2. Conceptual questions are answered by determining the relevant concepts, relationships between quantities, and applying definitions. Numerical problems are solved by substituting values into appropriate equations. 3. Estimation problems involve approximating quantities using given values and reasonable assumptions. Diagrams are used to illustrate vector concepts and solutions.

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1 Chapter 1 Measurement and Vectors Conceptual Problems 1 • [SSM] Which of the following is not one of the base quantities in the SI system? (a) mass, (b) length, (c) energy, (d) time, (e) All of the above are base quantities. Determine the Concept The base quantities in the SI system include mass, length, and time. Force is not a base quantity. )(c is correct. 2 • In doing a calculation, you end up with m/s in the numerator and m/s2 in the denominator. What are your final units? (a) m2 /s3 , (b) 1/s, (c) s3 /m2 , (d) s, (e) m/s. Picture the Problem We can express and simplify the ratio of m/s to m/s2 to determine the final units. Express and simplify the ratio of m/s to m/s2 : s sm sm s m s m 2 2 = ⋅ ⋅ = and )(d is correct. 3 • The prefix giga means (a) 103 , (b) 106 , (c) 109 , (d) 1012 , (e) 1015 . Determine the Concept Consulting Table 1-1 we note that the prefix giga means 109 . )(c is correct. 4 • The prefix mega means (a) 10–9 , (b) 10–6 , (c) 10–3 , (d) 106 , (e) 109 . Determine the Concept Consulting Table 1-1 we note that the prefix mega means 106 . )(d is correct. 5 • [SSM] Show that there are 30.48 cm per foot. How many centimeters are there in one mile? Picture the Problem We can use the facts that there are 2.540 centimeters in 1 inch and 12 inches in 1 foot to show that there are 30.48 cm per ft. We can then use the fact that there are 5280 feet in 1 mile to find the number of centimeters in one mile.
Chapter 12 Multiply 2.540 cm/in by 12 in/ft to find the number of cm per ft: cm/ft48.30 ft in 12 in cm 540.2 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Multiply 30.48 cm/ft by 5280 ft/mi to find the number of centimeters in one mile: cm/mi10609.1 mi ft 5280 ft cm 48.30 5 ×=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Remarks: Because there are exactly 2.54 cm in 1 in and exactly 12 inches in 1 ft, we are justified in reporting four significant figures in these results. 6 • The number 0.000 513 0 has significant figures. (a) one, (b) three, (c) four, (d) seven, (e) eight. Determine the Concept Counting from left to right and ignoring zeros to the left of the first nonzero digit, the last significant figure is the first digit that is in doubt. Applying this criterion, the three zeros after the decimal point are not significant figures, but the last zero is significant. Hence, there are four significant figures in this number. )(c is correct. 7 • The number 23.0040 has significant figures. (a) two, (b) three, (c) four, (d) five, (e) six. Determine the Concept Counting from left to right, the last significant figure is the first digit that is in doubt. Applying this criterion, there are six significant figures in this number. )(e is correct. 8 • Force has dimensions of mass times acceleration. Acceleration has dimensions of speed divided by time. Pressure is defined as force divided by area. What are the dimensions of pressure? Express pressure in terms of the SI base units kilogram, meter and second. Determine the Concept We can use the definitions of force and pressure, together with the dimensions of mass, acceleration, and length, to find the dimensions of pressure. We can express pressure in terms of the SI base units by substituting the base units for mass, acceleration, and length in the definition of pressure.
Measurement and Vectors 3 Use the definition of pressure and the dimensions of force and area to obtain: [ ] [ ] [ ] 22 2 LT M L T ML === A F P Express pressure in terms of the SI base units to obtain: 22 2 2 sm kg m s m kg m N ⋅ = ⋅ = 9 • True or false: Two quantities must have the same dimensions in order to be multiplied. False. For example, the distance traveled by an object is the product of its speed (length/time) multiplied by its time of travel (time). 10 • A vector has a negative x component and a positive y component. Its angle measured counterclockwise from the positive x axis is (a) between zero and 90 degrees. (b) between 90 and 180 degrees. (c) More than 180 degrees. Determine the Concept Because a vector with a negative x-component and a positive y-component is in the second quadrant, its angle is between 90 and 180 degrees. ( ) correct.isb 11 • [SSM] A vector A r points in the +x direction. Show graphically at least three choices for a vector B r such that AB rr + points in the +y direction. Determine the Concept The figure shows a vector A r pointing in the positive x direction and three unlabeled possibilities for vector .B r Note that the choices for B r start at the end of vector A r rather than at its initial point. Note further that this configuration could be in any quadrant of the reference system shown. x y A r choicesSeveralB r 12 • A vector A r points in the +y direction. Show graphically at least three choices for a vector B r such that AB rr − points in the + x direction.
Chapter 14 Determine the Concept Let the +x direction be to the right and the +y direction be upward. The figure shows the vector A r − pointing in the −y direction and three unlabeled possibilities for vector .B r Note that the choices for B r start at the end of vector A r − rather than at its initial point. A r − x y choicesSeveralB r 13 • [SSM] Is it possible for three equal magnitude vectors to add to zero? If so, sketch a graphical answer. If not, explain why not. Determine the Concept In order for the three equal magnitude vectors to add to zero, the sum of the three vectors must form a triangle. The equilateral triangle shown to the right satisfies this condition for the vectors A r , B r , and C r for which it is true that A = B = C, whereas .0=++ CBA rrr A r B r C r Estimation and Approximation 14 • The angle subtended by the moon’s diameter at a point on Earth is about 0.524º (Fig. 1-2). Use this and the fact that the moon is about 384 Mm away to find the diameter of the moon. HINT: The angle can be determined from the diameter of the moon and the distance to the moon. Picture the Problem Let θ represent the angle subtended by the moon’s diameter, D represent the diameter of the moon, and rm the distance to the moon. Because θ is small, we can approximate it by θ ≈ D/rm where θ is in radian measure. We can solve this relationship for the diameter of the moon. Express the moon’s diameter D in terms of the angle it subtends at Earth θ and the Earth-moon distance rm: mrD θ= Substitute numerical values and evaluate D: ( ) m1051.3 Mm384 360 rad2 524.0 6 ×= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° ×°= π D
Measurement and Vectors 5 15 • [SSM] Some good estimates about the human body can be made if it is assumed that we are made mostly of water. The mass of a water molecule is 29.9 ×10−27 kg. If the mass of a person is 60 kg, estimate the number of water molecules in that person. Picture the Problem We can estimate the number of water molecules in a person whose mass is 60 kg by dividing this mass by the mass of a single water molecule. Letting N represent the number of water molecules in a person of mass mhuman body, express N in terms of mhuman body and the mass of a water molecule mwater molecule: moleculewater bodyhuman m m N = Substitute numerical values and evaluate N: molecules100.2 molecule kg 109.29 kg60 27 27 ×= × = − N 16 •• In 1989, IBM scientists figured out how to move atoms with a scanning tunneling microscope (STM). One of the first STM pictures seen by the general public was of the letters IBM spelled with xenon atoms on a nickel surface. The letters IBM were 15 xenon atoms across. If the space between the centers of adjacent xenon atoms is 5 nm (5 ×10−9 m), estimate how many times could ″IBM″ could be written across this 8.5 inch page. Picture the Problem We can estimate the number of times N that ″IBM″ could be written across this 8.5-inch page by dividing the width w of the page by the distance d required by each writing of ″IBM.″ Express N in terms of the width w of the page and the distance d required by each writing of ″IBM″: d w N = Express d in terms of the separation s of the centers of adjacent xenon atoms and the number n of xenon atoms in each writing of ″IBM″: snd = Substitute for d in the expression for N to obtain: sn w N =
Chapter 16 Substitute numerical values and evaluate N: ( ) ( ) 6 9 103 15 nm m10 nm5 in cm 540.2in5.8 ×= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − N 17 •• There is an environmental debate over the use of cloth versus disposable diapers. (a) If we assume that between birth and 2.5 y of age, a child uses 3 diapers per day, estimate the total number of disposable diapers used in the United States per year. (b) Estimate the total landfill volume due to these diapers, assuming that 1000 kg of waste fills about 1 m3 of landfill volume. (c) How many square miles of landfill area at an average height of 10 m is needed for the disposal of diapers each year? Picture the Problem We’ll assume a population of 300 million and a life expectancy of 76 y. We’ll also assume that a diaper has a volume of about half a liter. In (c) we’ll assume the disposal site is a rectangular hole in the ground and use the formula for the volume of such an opening to estimate the surface area required. (a) Express the total number N of disposable diapers used in the United States per year in terms of the number of children n in diapers and the number of diapers D used by each child per year: nDN = (1) Use the estimated daily consumption and the number of days in a year to estimate the number of diapers D required per child per year: yilddiapers/ch101.1 y d24.365 dchild diapers3 3 ⋅×≈ × ⋅ =D Use the assumed life expectancy to estimate the number of children n in diapers yearly: ( ) children10 children10300 y76 y5.2 7 6 ≈ ×⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =n Substitute numerical values in equation (1) to obtain: ( ) diapers101.1 y diapers 103children10 10 37 ×≈ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×=N
Measurement and Vectors 7 (b) Express the required landfill volume V in terms of the volume of diapers to be buried: diaperoneNVV = Substitute numerical values and evaluate V: ( ) 36 33 10 m105.5 L m10 diaper L5.0 diapers101.1 ×≈⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ××= − V (c) Express the required volume in terms of the volume of a rectangular parallelepiped: AhV = ⇒ h V A = Substitute numerical values evaluate A: 25 36 m105.5 m10 m105.5 ×= × =A Use a conversion factor (see Appendix A) to express this area in square miles: 2 2 2 25 mi2.0 km1 mi3861.0 m105.5 ≈ ××=A 18 •• (a) Estimate the number of gallons of gasoline used per day by automobiles in the United States and the total amount of money spent on it. (b) If 19.4 gal of gasoline can be made from one barrel of crude oil, estimate the total number of barrels of oil imported into the United States per year to make gasoline. How many barrels per day is this? Picture the Problem The population of the United States is roughly 3 × 108 people. Assuming that the average family has four people, with an average of two cars per family, there are about 1.5 × 108 cars in the United States. If we double that number to include trucks, cabs, etc., we have 3 × 108 vehicles. Let’s assume that each vehicle uses, on average, 14 gallons of gasoline per week and that the United States imports half its oil. (a) Find the daily consumption of gasoline G: ( )( ) gal/d106 gal/d2vehicles103 8 8 ×= ×=G Assuming a price per gallon P = $3.00, find the daily cost C of gasoline: ( )( ) dollars/dbillion2 d/1018$ gal/00.3$gal/d106 8 8 ≈ ×= ×== GPC
Chapter 18 (b) Relate the number of barrels N of crude oil imported annually to the yearly consumption of gasoline Y and the number of gallons of gasoline n that can be made from one barrel of crude oil: n tfG n fY N Δ == where f is the fraction of the oil that is imported. Substitute numerical values and estimate N: ( ) y barrels 106 barrel gallons 19.4 y d 24.365 d gallons 1065.0 9 8 ×≈ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × =N Convert barrels/y to barrels/d to obtain: d barrels 102 d365.24 y1 y barrels 106 7 9 ×≈ ××=N 19 •• [SSM] A megabyte (MB) is a unit of computer memory storage. A CD has a storage capacity of 700 MB and can store approximately 70 min of high-quality music. (a) If a typical song is 5 min long, how many megabytes are required for each song? (b) If a page of printed text takes approximately 5 kilobytes, estimate the number of novels that could be saved on a CD. Picture the Problem We can set up a proportion to relate the storage capacity of a CD to its playing time, the length of a typical song, and the storage capacity required for each song. In (b) we can relate the number of novels that can be stored on a CD to the number of megabytes required per novel and the storage capacity of the CD. (a) Set up a proportion relating the ratio of the number of megabytes on a CD to its playing time to the ratio of the number of megabytes N required for each song: min5min70 MB700 N = Solve this proportion for N to obtain: ( ) MB50min5 min70 MB700 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =N
Measurement and Vectors 9 (b) Letting n represent the number of megabytes per novel, express the number of novels novelsN that can be stored on a CD in terms of the storage capacity of the CD: n N MB700 novels = Assuming that a typical page in a novel requires 5 kB of memory, express n in terms of the number of pages p in a typical novel: pn ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = page kB 5 Substitute for n in the expression for novelsN to obtain: p N ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = page kB 5 MB700 novels Assuming that a typical novel has 200 pages: novels107 novel pages 200 page kB 5 MB kB10 MB700 2 3 novels ×= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × =N Units 20 • Express the following quantities using the prefixes listed in Table 1-1 and the unit abbreviations listed in the table Abbreviations for Units. For example, 10,000 meters = 10 km. (a) 1,000,000 watts, (b) 0.002 gram, (c) 3 × 10–6 meter, (d) 30,000 seconds. Picture the Problem We can use the metric prefixes listed in Table 1-1 and the abbreviations on page EP-1 to express each of these quantities. (a) MW1watts10watts000,000,1 6 == (c) m3meter103 6 μ=× − (b) mg2g102gram002.0 3 =×= − (d) ks30s1003seconds000,30 3 =×= 21 • Write each of the following without using prefixes: (a) 40 μW, (b) 4 ns, (c) 3 MW, (d) 25 km.
Chapter 110 Picture the Problem We can use the definitions of the metric prefixes listed in Table 1-1 to express each of these quantities without prefixes. (a) W0.000040W1040W40 6 =×= − μ (c) W000,000,3W103MW3 6 =×= (b) s40.00000000s104ns4 9 =×= − (d) m000,52m1025km25 3 =×= 22 • Write the following (which are not SI units) using prefixes (but not their abbreviations). For example, 103 meters = 1 kilometer: (a) 10–12 boo, (b) 109 low, (c) 10–6 phone, (d) 10–18 boy, (e) 106 phone, (f) 10–9 goat, (g) 1012 bull. Picture the Problem We can use the definitions of the metric prefixes listed in Table 1-1 to express each of these quantities without abbreviations. (a) picoboo1boo10 12 =− (e) megaphone1phone106 = (b) gigalow1low109 = (f) nanogoat1goat10 9 =− (c) microphone1phone10 6 =− (g) terabull1bull1012 = (d) attoboy1boy10 18 =− 23 •• [SSM] In the following equations, the distance x is in meters, the time t is in seconds, and the velocity v is in meters per second. What are the SI units of the constants C1 and C2? (a) x = C1 + C2t, (b) x = 1 2 C1t2 , (c) v2 = 2C1x, (d) x = C1 cos C2t, (e) v2 = 2C1v – (C2x)2 . Picture the Problem We can determine the SI units of each term on the right- hand side of the equations from the units of the physical quantity on the left-hand side. (a) Because x is in meters, C1 and C2t must be in meters: m/sinism;inis 21 CC (b) Because x is in meters, ½C1t2 must be in meters: 2 1 m/sinisC
Measurement and Vectors 11 (c) Because v2 is in m2 /s2 , 2C1x must be in m2 /s2 : 2 1 m/sinisC (d) The argument of a trigonometric function must be dimensionless; i.e. without units. Therefore, because x is in meters: 1 21 sinism;inis − CC (e) All of the terms in the expression must have the same units. Therefore, because v is in m/s: 1 21 sinism/s;inis − CC 24 •• If x is in feet, t is in milliseconds, and v is in feet per second, what are the units of the constants C1 and C2 in each part of Problem 23? Picture the Problem We can determine the US customary units of each term on the right-hand side of the equations from the units of the physical quantity on the left-hand side. (a) Because x is in feet, C1 and C2t must be in feet: ft/msinisft;inis 21 CC (b) Because x is in feet, 2 12 1 tC must be in feet: ( )2 1 msft/inisC (c) Because v2 is in ft2 /(ms)2 , 2C1x must be in ft2 /s2 : ( )2 1 msft/inisC (d) The argument of a trigonometric function must be dimensionless; that is, without units. Therefore, because x is in feet: ( ) 1 21 msinisft;inis − CC (e) The argument of an exponential function must be dimensionless; that is, without units. Therefore, because v is in ft/s: ( ) 1 21 msinisft/ms;inis − CC Conversion of Units
Chapter 112 25 • From the original definition of the meter in terms of the distance along a meridian from the equator to the North Pole, find in meters (a) the circumference of Earth and (b) the radius of Earth. (c) Convert your answers for (a) and (b) from meters into miles. Picture the Problem We can use the formula for the circumference of a circle to find the radius of Earth and the conversion factor 1 mi = 1.609 km to convert distances in meters into distances in miles. (a) The Pole-Equator distance is one-fourth of the circumference: m1044 7 equator-pole ×== DC (b) The formula for the circumference of a circle is: RDC ππ 2== ⇒ π2 C R = Substitute numerical values and evaluate R: m106 m1037.6 2 m104 6 6 7 ×= ×= × = π R (c) Use the conversion factors 1 km = 1000 m and 1 mi = 1.609 km to express C in mi: mi102 km1.609 mi1 m10 km1 m104 4 3 7 ×= ×××=C Use the conversion factors 1 km = 1000 m and 1 mi = 1.61 km to express R in mi: mi104 km1.609 mi1 m10 km1 m1037.6 3 3 6 ×= ×××=R 26 • The speed of sound in air is 343 m/s at normal room temperature. What is the speed of a supersonic plane that travels at twice the speed of sound? Give your answer in kilometers per hour and miles per hour. Picture the Problem We can use the conversion factor 1 mi = 1.61 km to convert speeds in km/h into mi/h. Find the speed of the plane in km/h: ( ) km/h102.47 h s 3600 m10 km1 s m 686 m/s686m/s3432 3 3 ×= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ==v
Measurement and Vectors 13 Convert v into mi/h: mi/h101.53 km1.609 mi1 h km 102.47 3 3 ×= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×=v 27 • A basketball player is 6 ft 10 1 2 in tall. What is his height in centimeters? Picture the Problem We’ll first express his height in inches and then use the conversion factor 1 in = 2.540 cm. Express the player’s height in inches: in82.5in10.5 ft in12 ft6 =+×=h Convert h into cm: cm210 in cm2.540 in2.58 =×=h 28 • Complete the following: (a) 100 km/h = mi/h, (b) 60 cm = in, (c) 100 yd = m. Picture the Problem We can use the conversion factors 1 mi = 1.609 km, 1 in = 2.540 cm, and 1 m = 1.094 yd to perform these conversions. (a) Convert h km 100 to : h mi mi/h62.2 km1.609 mi1 h km 100 h km 100 = ×= (b) Convert 60 cm to in: in24 in23.6 cm2.540 in1 cm60cm60 = =×= (c) Convert 100 yd to m: m91.4 yd1.094 m1 yd100yd100 = ×= 29 • The main span of the Golden Gate Bridge is 4200 ft. Express this distance in kilometers. Picture the Problem We can use the conversion factor 1.609 km = 5280 ft to convert the length of the main span of the Golden Gate Bridge into kilometers.
Chapter 114 Convert 4200 ft into km by multiplying by 1 in the form ft5280 km609.1 : km1.28 ft5280 km1.609 ft4200ft4200 = ×= 30 • Find the conversion factor to convert from miles per hour into kilometers per hour. Picture the Problem Let v be the speed of an object in mi/h. We can use the conversion factor 1 mi = 1.609 km to convert this speed to km/h. Multiply v mi/h by 1.609 km/mi to convert v to km/h: km/h61.1 mi km1.609 h mi h mi vvv =×= 31 • Complete the following: (a) 1.296 × 105 km/h2 = km/(h⋅s), (b) 1.296 × 105 km/h2 = m/s2 , (c) 60 mi/h = ft/s, (d) 60 mi/h = m/s. Picture the Problem Use the conversion factors 1 h = 3600 s, 1.609 km = 1 mi, and 1 mi = 5280 ft to make these conversions. (a) skm/h36.00 s3600 h1 h km 10296.1 h km 10296.1 2 5 2 5 ⋅=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×=× (b) 2 32 2 5 2 5 m/s10.00 km m10 s3600 h1 h km 10296.1 h km 10296.1 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×=× (c) ft/s88 s3600 h1 mi1 ft5280 h mi 60 h mi 60 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = (d) m/s27m/s8.26 s3600 h1 km m10 mi1 km1.609 h mi 60 h mi 60 3 ==⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = 32 • There are 640 acres in a square mile. How many square meters are there in one acre? Picture the Problem We can use the conversion factor given in the problem statement and the fact that 1 mi = 1.609 km to express the number of square meters in one acre. Note that, because there are exactly 640 acres in a square mile, 640 acres has as many significant figures as we may wish to associate with it.
Measurement and Vectors 15 Multiply by 1 twice, properly chosen, to convert one acre into square miles, and then into square meters: ( ) 2 22 m4045 mi m1609 acres640 mi1 acre1acre1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 33 •• [SSM] You are a delivery person for the Fresh Aqua Spring Water Company. Your truck carries 4 pallets. Each pallet carries 60 cases of water. Each case of water has 24 one-liter bottles. You are to deliver 10 cases of water to each convenience store along your route. The dolly you use to carry the water into the stores has a weight limit of 250 lb. (a) If a milliliter of water has a mass of 1 g, and a kilogram has a weight of 2.2 lb, what is the weight, in pounds, of all the water in your truck? (b) How many full cases of water can you carry on the cart? Picture the Problem The weight of the water in the truck is the product of the volume of the water and its weight density of 2.2 lb/L. (a) Relate the weight w of the water on the truck to its volume V and weight density (weight per unit volume) D: DVw = Find the volume V of the water: L5760 ) case L 24() pallet cases (60pallets)4( = =V Substitute numerical values for D and L and evaluate w: ( ) lb103.1 lb10267.1L5760 L lb 2.2 4 4 ×= ×=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =w (b) Express the number of cases of water in terms of the weight limit of the cart and the weight of each case of water: waterofcaseeachofweight carttheoflimitweight =N Substitute numerical values and evaluate N: cases7.4 case L 24 L lb 2.2 lb250 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =N cases.4carrycanYou
Chapter 116 34 •• A right circular cylinder has a diameter of 6.8 in and a height of 2.0 ft. What is the volume of the cylinder in (a) cubic feet, (b) cubic meters, (c) liters? Picture the Problem The volume of a right circular cylinder is the area of its base multiplied by its height. Let d represent the diameter and h the height of the right circular cylinder; use conversion factors to express the volume V in the given units. (a) Express the volume of the cylinder: hdV 2 4 1 π= Substitute numerical values and evaluate V: ( ) ( ) ( ) ( ) 33 2 2 4 1 2 4 1 ft50.0ft504.0 in12 ft1 ft2.0in6.8 ft2.0in6.8 == ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = = π πV (b) Use the fact that 1 m = 3.281 ft to convert the volume in cubic feet into cubic meters: ( ) 3 3 3 3 m0.014 m0.0143 ft3.281 m1 ft0.504 = =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =V (c) Because 1 L = 10−3 m3 : ( ) L14 m10 L1 0.0143m 33 3 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − V 35 •• [SSM] In the following, x is in meters, t is in seconds, v is in meters per second, and the acceleration a is in meters per second squared. Find the SI units of each combination: (a) v2 /x, (b) ax , (c) 1 2 at2 . Picture the Problem We can treat the SI units as though they are algebraic quantities to simplify each of these combinations of physical quantities and constants. (a) Express and simplify the units of v2 /x: ( ) 22 22 s m sm m m sm = ⋅ = (b) Express and simplify the units of ax : ss m/s m 2 2 ==
Measurement and Vectors 17 (c) Noting that the constant factor 2 1 has no units, express and simplify the units of 2 2 1 at : ( ) ms s m 2 2 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Dimensions of Physical Quantities 36 • What are the dimensions of the constants in each part of Problem 23? Picture the Problem We can use the facts that each term in an equation must have the same dimensions and that the arguments of a trigonometric or exponential function must be dimensionless to determine the dimensions of the constants. (a) x = C1 + C2 t T T L LL (d) x = C1 cos C2 t T T 1 LL (b) 2 12 1 tCx = 2 2 T T L L (e) v2 = 2 C1 v − (C2)2 x2 2 2 L T 1 T L T L T L ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ (c) xCv 1 2 2= L T L T L 22 2 37 • The law of radioactive decay is ( ) t eNtN λ− = 0 , where N0 is the number of radioactive nuclei at t = 0, N(t) is the number remaining at time t, and λ is a quantity known as the decay constant. What is the dimension of λ? Picture the Problem Because the exponent of the exponential function must be dimensionless, the dimension of λ must be .T 1− 38 •• The SI unit of force, the kilogram-meter per second squared (kg⋅m/s2 ) is called the newton (N). Find the dimensions and the SI units of the constant G in Newton’s law of gravitation F = Gm1m2/r2 .
Chapter 118 Picture the Problem We can solve Newton’s law of gravitation for G and substitute the dimensions of the variables. Treating them as algebraic quantities will allow us to express the dimensions in their simplest form. Finally, we can substitute the SI units for the dimensions to find the units of G. Solve Newton’s law of gravitation for G to obtain: 21 2 mm Fr G = Substitute the dimensions of the variables: [ ] 2 3 2 2 2 MT L M L T ML = × =G Use the SI units for L, M, and T to obtain: 2 3 skg m ⋅ 39 •• The magnitude of the force (F) that a spring exerts when it is stretched a distance x from its unstressed length is governed by Hooke’s law, F = kx. (a) What are the dimensions of the force constant, k? (b) What are the dimensions and SI units of the quantity kx2 ? Picture the Problem The dimensions of mass and velocity are M and L/T, respectively. We note from Table 1-2 that the dimensions of force are ML/T2 . (a) We know, from Hooke’s law, that: x F k = Write the corresponding dimensional equation: [ ] [ ] [ ]x F k = Substitute the dimensions of F and x and simplify to obtain: [ ] 2 2 T M L T L M ==k (b) Substitute the dimensions of k and x2 and simplify to obtain: [ ] 2 2 2 2 2 T ML L T M ==kx Substitute the units of kx2 to obtain: 2 2 s mkg⋅
Measurement and Vectors 19 40 •• Show that the product of mass, acceleration, and speed has the dimensions of power. Picture the Problem We note from Table 1-2 that the dimensions of power are ML2 /T3 . The dimensions of mass, acceleration, and speed are M, L/T2 , and L/T respectively. Express the dimensions of mav: [ ] 3 2 2 T ML T L T L M =××=mav From Table 1-2: [ ] 3 2 T ML =P Comparing these results, we see that the product of mass, acceleration, and speed has the dimensions of power. 41 •• [SSM] The momentum of an object is the product of its velocity and mass. Show that momentum has the dimensions of force multiplied by time. Picture the Problem The dimensions of mass and velocity are M and L/T, respectively. We note from Table 1-2 that the dimensions of force are ML/T2 . Express the dimensions of momentum: [ ] T ML T L M =×=mv From Table 1-2: [ ] 2 T ML =F Express the dimensions of force multiplied by time: [ ] T ML T T ML 2 =×=Ft Comparing these results, we see that momentum has the dimensions of force multiplied by time. 42 •• What combination of force and one other physical quantity has the dimensions of power? Picture the Problem Let X represent the physical quantity of interest. Then we can express the dimensional relationship between F, X, and P and solve this relationship for the dimensions of X.
Chapter 120 Express the relationship of X to force and power dimensionally: [ ][ ] [ ]PXF = Solve for [ ]X : [ ] [ ] [ ]F P X = Substitute the dimensions of force and power and simplify to obtain: [ ] T L T ML T ML 2 3 2 ==X Because the dimensions of velocity are L/T, we can conclude that: [ ] [ ][ ]vFP = Remarks: While it is true that P = Fv, dimensional analysis does not reveal the presence of dimensionless constants. For example, if πFvP = , the analysis shown above would fail to establish the factor of π. 43 •• [SSM] When an object falls through air, there is a drag force that depends on the product of the cross sectional area of the object and the square of its velocity, that is, Fair = CAv2 , where C is a constant. Determine the dimensions of C. Picture the Problem We can find the dimensions of C by solving the drag force equation for C and substituting the dimensions of force, area, and velocity. Solve the drag force equation for the constant C: 2 air Av F C = Express this equation dimensionally: [ ] [ ] [ ][ ]2 air vA F C = Substitute the dimensions of force, area, and velocity and simplify to obtain: [ ] 32 2 2 L M T L L T ML = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =C 44 •• Kepler’s third law relates the period of a planet to its orbital radius r, the constant G in Newton’s law of gravitation (F = Gm1m2/r2 ), and the mass of the sun Ms. What combination of these factors gives the correct dimensions for the period of a planet?
Measurement and Vectors 21 Picture the Problem We can express the period of a planet as the product of the factors r, G, and MS (each raised to a power) and then perform dimensional analysis to determine the values of the exponents. Express the period T of a planet as the product of cba MGr Sand,, : cba MGCrT S= (1) where C is a dimensionless constant. Solve the law of gravitation for the constant G: 21 2 mm Fr G = Express this equation dimensionally: [ ] [ ][ ] [ ][ ]21 2 mm rF G = Substitute the dimensions of F, r, and m: [ ] ( ) 2 3 2 2 MT L MM L T ML = × × =G Noting that the dimension of time is represented by the same letter as is the period of a planet, substitute the dimensions in equation (1) to obtain: ( ) ( )c b a T M MT L L 2 3 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = Introduce the product of M 0 and L0 in the left hand side of the equation and simplify to obtain: bbabc 23100 TLMTLM −+− = Equating the exponents on the two sides of the equation yields: 0 = c – b, 0 = a + 3b, and 1 = –2b Solve these equations simultaneously to obtain: 2 1 2 1 2 3 and,, −=−== cba Substitute for a, b, and c in equation (1) and simplify to obtain: 23 S 21 S 2123 r GM C MGCrT == −− Scientific Notation and Significant Figures 45 • [SSM] Express as a decimal number without using powers of 10 notation: (a) 3 × 104 , (b) 6.2 × 10–3 , (c) 4 × 10–6 , (d) 2.17 × 105 .
Chapter 122 Picture the Problem We can use the rules governing scientific notation to express each of these numbers as a decimal number. (a) 000,30103 4 =× (c) 000004.0104 6 =× − (b) 0062.0102.6 3 =× − (d) 000,2171017.2 5 =× 46 • Write the following in scientific notation: (a) 1345100 m = ____ km, (b) 12340. kW = ____ MW, (c) 54.32 ps = ____ s, (d) 3.0 m = ____ mm Picture the Problem We can use the rules governing scientific notation to express each of these numbers in scientific notation. (a) km103451.1 m103451.1m1345100 3 6 ×= ×= (c) s105.432 s1054.32ps32.54 11 12 − − ×= ×= (b) MW101.2340 W101.2340 kW101.2340kW.12340 1 7 4 ×= ×= ×= (d) mm100.3 m mm10 m3.0m0.3 3 3 ×= ×= 47 • [SSM] Calculate the following, round off to the correct number of significant figures, and express your result in scientific notation: (a) (1.14)(9.99 × 104 ), (b) (2.78 × 10–8 ) – (5.31 × 10–9 ), (c) 12π /(4.56 × 10–3 ), (d) 27.6 + (5.99 × 102 ). Picture the Problem Apply the general rules concerning the multiplication, division, addition, and subtraction of measurements to evaluate each of the given expressions. (a) The number of significant figures in each factor is three; therefore the result has three significant figures: ( )( ) 54 1014.11099.914.1 ×=×
Measurement and Vectors 23 (b) Express both terms with the same power of 10. Because the first measurement has only two digits after the decimal point, the result can have only two digits after the decimal point: ( ) ( ) ( ) 8 8 98 1025.2 10531.078.2 1031.51078.2 − − −− ×= ×−= ×−× (c) We’ll assume that 12 is exact. Hence, the answer will have three significant figures: 3 3 1027.8 1056.4 12 ×= × − π (d) Proceed as in (b): ( ) 2 2 1027.6 627 5996.271099.56.27 ×= = +=×+ 48 • Calculate the following, round off to the correct number of significant figures, and express your result in scientific notation: (a) (200.9)(569.3), (b) (0.000000513)(62.3 × 107 ), (c) 28 401 + (5.78 × 104 ), (d) 63.25/(4.17 × 10–3 ). Picture the Problem Apply the general rules concerning the multiplication, division, addition, and subtraction of measurements to evaluate each of the given expressions. (a) Note that both factors have four significant figures. ( )( ) 5 10144.13.5699.200 ×= (b) Express the first factor in scientific notation and note that both factors have three significant figures. ( )( ) ( )( ) 2777 1020.3103.621013.5103.62000000513.0 ×=××=× − (c) Express both terms in scientific notation and note that the second has only three significant figures. Hence the result will have only three significant figures. ( ) ( ) ( ) ( ) 4 4 444 1062.8 1078.5841.2 1078.510841.21078.528401 ×= ×+= ×+×=×+
Chapter 124 (d) Because the divisor has three significant figures, the result will have three significant figures. 4 3 1052.1 1017.4 25.63 ×= × − 49 • [SSM] A cell membrane has a thickness of 7.0 nm. How many cell membranes would it take to make a stack 1.0 in high? Picture the Problem Let N represent the required number of membranes and express N in terms of the thickness of each cell membrane. Express N in terms of the thickness of a single membrane: nm7.0 in1.0 =N Convert the units into SI units and simplify to obtain: 66 9 106.31063.3 m10 nm1 cm100 m1 in cm2.540 nm7.0 in1.0 ×=×=×××= − N 50 •• A circular hole of radius 8.470 ×10−1 cm must be cut into the front panel of a display unit. The tolerance is 1.0 × 10−3 cm, which means the actual hole cannot differ by more than this much from the desired radius. If the actual hole is larger than the desired radius by the allowed tolerance, what is the difference between the actual area and the desired area of the hole? Picture the Problem Let r0 represent the larger radius and r the desired radius of the hole. We can find the difference between the actual and the desired area of the hole by subtracting the smaller area from the larger area. Express the difference between the two areas in terms of r and r0: ( )22 0 22 0Δ rrrrA −=−= πππ Factoring 22 0 rr − to obtain: ( )( )rrrrA +−= 00Δ π Substitute numerical values and evaluate ΔA: ( ) ( )[ ] 23 3113 cm103.5 cm100.1cm10470.8cm10470.8cm100.1Δ − −−−− ×= ×+×+××= πA 51 •• [SSM] A square peg must be made to fit through a square hole. If you have a square peg that has an edge length of 42.9 mm, and the square hole has an edge length of 43.2 mm, (a) what is the area of the space available when
Measurement and Vectors 25 the peg is in the hole? (b) If the peg is made rectangular by removing 0.10 mm of material from one side, what is the area available now? Picture the Problem Let sh represent the side of the square hole and sp the side of the square peg. We can find the area of the space available when the peg is in the hole by subtracting the area of the peg from the area of the hole. (a) Express the difference between the two areas in terms of sh and sp: 2 p 2 hΔ ssA −= Substitute numerical values and evaluate ΔA: ( ) ( ) 2 22 mm62 mm9.42mm2.43Δ ≈ −=A (b) Express the difference between the area of the square hole and the rectangular peg in terms of sh ,sp and the new length of the peg lp: pp 2 hΔ lssA' −= Substitute numerical values and evaluate ΔA′: ( ) ( )( ) 22 mm03mm10.0mm9.42mm9.42mm2.43Δ ≈−−=A' Vectors and Their Properties 52 • A vector 7.00 units long and a vector 5.50 units long are added. Their sum is a vector 10.0 units long. (a) Show graphically at least one way that this can be accomplished. (b) Using your sketch in Part (a), determine the angle between the original two vectors. Picture the Problem Let A r be the vector whose length is 7.00 units and let B r be the vector whose length is 5.50 units. BAC rrr += is the sum of A r and .B r The fact that their sum is 10.0 long tells us that the vectors are not collinear. We can use the of the vectors to find the angle between A r and .B r (a) A graphical representation of vectors A r , B r and C r is shown below. θ is the angle between A r and .B r
Chapter 126 α θ 00.7=A r 50.5= B r 0.10=C r y x (b) The components of the vectors are related as follows: xxx CBA =+ and yyy CBA =+ Substituting for the components gives: ( ) ( ) αθ cos0.10cos50.500.7 =+ and ( ) ( ) αθ sin0.10sin50.5 = Squaring and adding these equations yields: ( ) ( ) ( ) ( )[ ]2222 cos50.500.7sin50.5cos100sin100 θθαα ++=+ or ( )( ) ( ) ( )[ ]2222 cos50.500.7sin50.5cossin100 θθαα ++=+ Because 1cossin 22 =+ αα : ( ) ( )[ ]222 cos50.500.7sin50.5100 θθ ++= Solve this equation for θ (you can (1) use the same trigonometric identity used in the previous step to eliminate either sin2 θ or cos2 θ in favor of the other and then solve the resulting equation or (2) use your graphing calculator’s SOLVER program) to obtain: °= 4.74θ Remarks: You could also solve Part (b) of this problem by using the law of cosines. 53 • [SSM] Determine the x and y components of the following three vectors in the xy plane. (a) A 10-m displacement vector that makes an angle of 30° clockwise from the +y direction. (b) A 25-m/s velocity vector that makes an
Measurement and Vectors 27 angle of −40° counterclockwise from the −x direction. (c) A 40-lb force vector that makes an angle of 120° counterclockwise from the −y direction. Picture the Problem The x and y components of these vectors are their projections onto the x and y axes. Note that the components are calculated using the angle each vector makes with the +x axis. (a) Sketch the displacement vector (call it A r ) and note that it makes an angle of 60° with the +x axis: A r x y °30 Ay Ax Find the x and y components of A r : ( ) m0.560cosm10 =°=xA and ( ) m.7860sinm10 =°=yA (b) Sketch the velocity vector (call it v r ) and note that it makes an angle of 220° with the +x axis: v r °40 y xxv yv Find the x and y components of v r : ( ) m/s19220cosm/s52 −=°=xv and ( ) m/s16220sinm/s52 −=°=yv
Chapter 128 (c) Sketch the force vector (call it F r ) and note that it makes an angle of 30° with the +x axis: x °120 F r y Find the x and y components of F r : ( ) lb3530coslb04 =°=xF and ( ) lb0230sinlb04 =°=yF 54 • Rewrite the following vectors in terms of their magnitude and angle (counterclockwise from the + x direction). (a) A displacement vector with an x component of +8.5 m and a y component of −5.5 m. (b) A velocity vector with an x component of −75 m/s and a y component of +35 m/s. (c) A force vector with a magnitude of 50 lb that is in the third quadrant with an x component whose magnitude is 40 lb. Picture the Problem We can use the Pythagorean Theorem to find magnitudes of these vectors from their x and y components and trigonometry to find their direction angles. (a) Sketch the components of the displacement vector (call it A r ) and show their resultant: y x A r m5.8=xA m5.5−=yA θ Use the Pythagorean Theorem to find the magnitude of A r : ( ) ( ) m10 m5.5m5.8 2222 = −+=+= yx AAA Find the direction angle of A r : °=°−= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = −− 32733 m5.8 m5.5 tantan 11 x y A A θ
Measurement and Vectors 29 (b) Sketch the components of the velocity vector (call it v r ) and show their resultant: v r x y α m/s35=yv m/s75−=xv θ Use the Pythagorean Theorem to find the magnitude of v r and trigonometry to find its direction angle: ( ) ( ) m/s38 m/s53m/s75 2222 = +−=+= yx vvv °=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = −− 25 m/s57 m/s53 tantan 11 x y v v α and °=°−°=−°= 15525180180 αθ (c) Sketch the force vector (call itF r ) and its x-component and show Fy: x y α lb40−=xF y F rlb 50 = F θ Use trigonometry to find α and, hence, θ : °=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = −− 37 lb50 lb40 coscos 11 F Fx α °=°+°=+°= 21737180180 αθ Use the relationship between a vector and its vertical component to find Fy: ( ) lb30 217sinlb50sin −= °== θFFy Remarks: In Part (c) we could have used the Pythagorean Theorem to find the magnitude of Fy. 55 • You walk 100 m in a straight line on a horizontal plane. If this walk took you 50 m east, what are your possible north or south movements? What are the possible angles that your walk made with respect to due east?
Chapter 130 Picture the Problem There are two directions that you could have walked that are consistent with your having walked 50 m to the east. One possibility is walking in the north-east direction and the other is walking in the south-east direction. We can use the Pythagorean Theorem to find the distance you walked either north or south and then use trigonometry to find the two angles that correspond to your having walked in either of these directions. Letting A represent the magnitude of your displacement on the walk, use the Pythagorean Theorem to relate A to its horizontal (Ax) and vertical (Ay) components: 22 xy AAA −= Substitute numerical values and evaluate Ay: ( ) ( ) m87m50m100 22 ±=−=yA The plus-and-minus-signs mean that you could have gone 87 m north or 87 m south. Use trigonometry to relate the directions you could have walked to the distance you walked and its easterly component: °±=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − 60 m100 m50 cos 1 θ The plus-and-minus-signs mean that you could have walked 60° north of east or 60° south of east. 56 • The final destination of your journey is 300 m due east of your starting point. The first leg of this journey is the walk described in Problem 55, and the second leg is also a walk along a single straight-line path. Estimate graphically the length and heading for the second leg of your journey. Picture the Problem Let A r be the vector whose length is 100 m and whose direction is 60° N of E. Let B r be the vector whose direction and magnitude we are to determine and assume that you initially walked in a direction north of east. The graphical representation that we can use to estimate these quantities is shown below. Knowing that the magnitude of A r is 100 m, use any convenient scale to determine the length of B r and a protractor to determine the value of θ.
Measurement and Vectors 31 m 100 = A r °60 θ B r 250 m50 m 86.6 m E N The magnitude of the second leg of your journey is about 260 m at an angle of approximately 20° S of E. If you had initially walked into the southeast, the magnitude of the second leg of your journey would still be about 260 m but its direction would be approximately 20° N of E. Remarks: If you use the Pythagorean Theorem and right-triangle trigonometry, you’ll find that the length of the second leg of your journey is 265 m and that θ = 19° S of E. 57 •• Given the following vectors: r A = 3.4ˆi + 4.7 ˆj, r B = −7.7( )ˆi + 3.2 ˆj , andr C = 5.4ˆi + −9.1( )ˆj . (a) Find the vector r D, in unit vector notation, such thatr D + 2 r A − 3 r C + 4 r B = 0. (b) Express your answer in Part (a) in terms of magnitude and angle with the +x direction. Picture the Problem We can find the vector D r by solving the equation 0432 =+−+ BCAD rrrr for D r and then substituting for the vectors .and, CBA rrr In (b) we can use the components of D r to find its magnitude and direction. (a) Solve the vector equation that gives the condition that must satisfied for D r : BCAD rrrr 432 −+−= Substitute for BCA rrr and, and simplify to obtain: ( ) ( ) ( ) ( ) ( ) jiji ji jijijiD ˆ50ˆ40ˆ5.49ˆ2.40 ˆ8.123.274.9ˆ8.302.168.6 ˆ2.3ˆ7.74ˆ1.9ˆ4.53ˆ7.4ˆ4.32 −=−= −−−+++−= +−−−++−= r (b) Use the Pythagorean Theorem to relate the magnitude of D r to its components Dx and Dy: 22 yx DDD +=
Chapter 132 Substitute numerical values and evaluate D: ( ) ( ) 8.635.492.40 22 =−+=D Use trigonometry to express and evaluate the angle θ : °−= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = −− 51 2.40 5.49 tantan 11 x y D D θ where the minus sign means that D r is in the 4th quadrant. 58 •• Given the following force vectors: r A is 25 lb at an angle of 30° clockwise from the +x axis, and r B is 42 lb at an angle of 50° clockwise from the +y axis. (a) Make a sketch and visually estimate the magnitude and angle of the vector r C such that 2 r A + r C − r B results in a vector with a magnitude of 35 lb pointing in the +x direction. (b) Repeat the calculation in Part (a) using the method of components and compare your result to the estimate in (a). Picture the Problem A diagram showing the condition that iBCA ˆ352 =−+ rrr and from which one can scale the values of C and θ is shown below. In (b) we can use the two scalar equations corresponding to this vector equation to check our graphical results. (a) A diagram showing the conditions imposed by iBCA ˆ352 =−+ rrr approximately to scale is shown to the right. The magnitude of C r is approximately lb57 and the angle θ is approximately .68° B r B r − A r 2 C r °50 θ °50 °30 x y iˆ35 (b) Express the condition relating the vectors ,A r ,B r and :C r iBCA ˆ352 =−+ rrr The corresponding scalar equations are: 352 =−+ xxx BCA and 02 =−+ yyy BCA
Measurement and Vectors 33 Solve these equations for Cx and Cy to obtain: xxx BAC +−= 235 and yyy BAC +−= 2 Substitute for the x and y components of A r and B r to obtain: ( )[ ] ( ) lb9.23 40coslb42 330coslb252lb35 = °+ °−=xC and ( )[ ] ( ) lb52 40sinlb42330sinlb252 = °+°−=yC Use the Pythagorean Theorem to relate the magnitude of C r to its components: 22 yx CCC += Substitute numerical values and evaluate C: ( ) ( ) lb57lb0.52lb9.23 22 =+=C Use trigonometry to find the direction of C r : °=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = −− 65 lb9.23 lb0.52 tantan 11 x y C C θ Remarks: The analytical results for C and θ are in excellent agreement with the values determined graphically. 59 •• [SSM] Calculate the unit vector (in terms of iˆ and jˆ ) in the direction opposite to the direction of each of the vectors in Problem 57. Picture the Problem The unit vector in the direction opposite to the direction of a vector is found by taking the negative of the unit vector in the direction of the vector. The unit vector in the direction of a given vector is found by dividing the given vector by its magnitude. The unit vector in the direction of A r is given by: 22 ˆ yx AA + == A A A A r r r Substitute for ,A r Ax, and Ay and evaluate :ˆA ( ) ( ) ji ji A ˆ81.0ˆ59.0 7.44.3 ˆ7.4ˆ4.3ˆ 22 += + + =
Chapter 134 The unit vector in the direction opposite to that of A r is given by: jiA ˆ81.0ˆ59.0ˆ −−=− The unit vector in the direction of B r is given by: 22 ˆ yx BB + == B B B B r r r Substitute for B r , Bx, and By and evaluate :ˆB ( ) ( ) ji ji B ˆ38.0ˆ92.0 2.37.7 ˆ2.3ˆ7.7ˆ 22 +−= +− +− = The unit vector in the direction opposite to that of B r is given by: jiB ˆ38.0ˆ92.0ˆ −=− The unit vector in the direction of C r is given by: 22 ˆ yx CC + == C C C C r r r Substitute for C r , Cx, and Cy and evaluate :ˆC ( ) ( ) ji ji C ˆ86.0ˆ51.0 1.94.5 ˆ1.9ˆ4.5ˆ 22 −= −+ − = The unit vector in the direction opposite that of C r is given by: jiC ˆ86.0ˆ51.0ˆ +−=− 60 •• Unit vectors iˆ and jˆ are directed east and north, respectively. Calculate the unit vector (in terms of iˆ and jˆ ) in the following directions. (a) northeast, (b) 70° clockwise from the −y axis, (c) southwest. Picture the Problem The unit vector in a given direction is a vector pointing in that direction whose magnitude is 1. (a) The unit vector in the northeast direction is given by: ji jiu ˆ707.0ˆ707.0 ˆ45sin)1(ˆ45cos)1(ˆNE += °+°= (b) The unit vector 70° clockwise from the −y axis is given by: ji jiu ˆ342.0ˆ940.0 ˆ200sin)1(ˆ200cos)1(ˆ −−= °+°=
Measurement and Vectors 35 (c) The unit vector in the southwest direction is given by: ji jiu ˆ707.0ˆ707.0 ˆ225sin)1(ˆ225cos)1(ˆSW −−= °+°= Remarks: One can confirm that a given vector is, in fact, a unit vector by checking its magnitude. General Problems 61 • [SSM] The Apollo trips to the moon in the 1960's and 1970's typically took 3 days to travel the Earth-moon distance once they left Earth orbit. Estimate the spacecraft's average speed in kilometers per hour, miles per hour, and meters per second. Picture the Problem Average speed is defined to be the distance traveled divided by the elapsed time. The Earth-moon distance and the distance and time conversion factors can be found on the inside-front cover of the text. We’ll assume that 3 days means exactly three days. Express the average speed of Apollo as it travels to the moon: timeelapsed traveleddistance av =v Substitute numerical values to obtain: d3 mi102.39 5 av × =v Use the fact that there are 24 h in 1 d to convert 3 d into hours: mi/h1032.3 mi/h10319.3 d h24 d3 mi102.39 3 3 5 av ×= ×= × × =v Use the fact that 1 mi is equal to 1.609 km to convert the spacecraft’s average speed to km/h: km/h1034.5 h km 10340.5 mi km 609.1 h mi 10319.3 333 av ×=×=××=v
Chapter 136 Use the facts that there are 3600 s in 1 h and 1000 m in 1 km to convert the spacecraft’s average speed to m/s: m/s1049.1m/s10485.1 s3600 h1 km m10 h km 10340.5 33 3 6 av ×=×=×××=v Remarks: An alternative to multiplying by 103 m/km in the last step is to replace the metric prefix ″k″ in ″km″by 103 . 62 • On many of the roads in Canada the speed limit is 100 km/h. What is this speed limit in miles per hour? Picture the Problem We can use the conversion factor 1 mi = 1.609 km to convert 100 km/h into mi/h. Multiply 100 km/h by 1 mi/1.609 km to obtain: mi/h2.62 km1.609 mi1 h km 100 h km 100 = ×= 63 • If you could count $1.00 per second, how many years would it take to count 1.00 billion dollars? Picture the Problem We can use a series of conversion factors to convert 1 billion seconds into years. Multiply 1 billion seconds by the appropriate conversion factors to convert into years: y31.7 days365.24 y1 h24 day1 s3600 h1 s10s10 99 =×××= 64 • (a) The speed of light in vacuum is 186,000 mi/s = 3.00 × 108 m/s. Use this fact to find the number of kilometers in a mile. (b) The weight of 1.00 ft3 of water is 62.4 lb, and 1.00 ft = 30.5 cm. Use this and the fact that 1.00 cm3 of water has a mass of 1.00 g to find the weight in pounds of a 1.00-kg mass. Picture the Problem In both the examples cited we can equate expressions for the physical quantities, expressed in different units, and then divide both sides of the equation by one of the expressions to obtain the desired conversion factor.
Measurement and Vectors 37 (a) Divide both sides of the equation expressing the speed of light in the two systems of measurement by 186,000 mi/s to obtain: km/mi61.1 m10 km1 mi m 1061.1 m/mi1061.1 mi/h101.86 m/s103.00 1 3 3 3 5 8 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×= ×= × × = (b) Find the volume of 1.00 kg of water: Volume of 1.00 kg = 103 g is 103 cm3 Express 103 cm3 in ft3 : ( ) 3 3 3 ft0.03525 cm5.03 ft1.00 cm10 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ Relate the weight of 1.00 ft3 of water to the volume occupied by 1.00 kg of water: 33 ft lb 62.4 ft0.03525 kg1.00 = Divide both sides of the equation by the left-hand side to obtain: lb/kg20.2 ft0.03525 kg1.00 ft lb 62.4 1 3 3 == 65 • The mass of one uranium atom is 4.0 × 10–26 kg. How many uranium atoms are there in 8.0 g of pure uranium? Picture the Problem We can use the given information to equate the ratios of the number of uranium atoms in 8 g of pure uranium and of 1 atom to its mass. Express the proportion relating the number of uranium atoms NU in 8.0 g of pure uranium to the mass of 1 atom: kg104.0 atom1 g0.8 26 U − × = N Solve for and evaluate NU: 23 263U 100.2 kg104.0 atom1 g10 kg1 g0.8 ×= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×= − N
Chapter 138 66 •• During a thunderstorm, a total of 1.4 in of rain falls. How much water falls on one acre of land? (1 mi2 = 640 acres.) Express your answer in (a) cubic inches, (b) cubic feet, (c) cubic meters, and (d) kilograms. Note that the density of water is 1000 kg/m3 . Picture the Problem Assuming that the water is distributed uniformly over the one acre of land, its volume is the product of the area over which it is distributed and its depth. The mass of the water is the product of its density and volume. The required conversion factors can be found in the front material of the text. (a) Express the volume V of water in terms of its depth h and the area A over which it falls: AhV = Substitute numerical values and evaluate V: ( ) ( ) 36 36 222 in108.8 in10782.8in4.1 ft in12 mi ft5280 acres640 mi1 acre1 ×= ×=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =V (b) Convert in3 into ft3 : 3333 3 36 ft101.5ft10082.5 in12 ft1 in10782.8 ×=×= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ××=V (c) Convert ft3 into m3 : 32 3 333 33 m104.1 m102439.1 cm100 m1 in cm540.2 ft in12 ft10082.5 ×= ×=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ××=V (d) The mass of the water is the product of its density ρ and volume V: Vm ρ= Substitute numerical values and evaluate m: ( ) kg104.1 m10439.1 m kg 10 5 32 3 3 ×= ×⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =m
Measurement and Vectors 39 67 •• An iron nucleus has a radius of 5.4 × 10–15 m and a mass of 9.3 × 10–26 kg. (a) What is its mass per unit volume in kg/ m3 ? (b) If Earth had the same mass per unit volume, what would be its radius? (The mass of Earth is 5.98 × 1024 kg.) Picture the Problem The mass per unit volume of an object is its density. (a) The density ρ of an object is its mass m per unit volume V: V m =ρ Assuming the iron nucleus to be spherical, its volume as a function of its radius r is given by: 3 3 4 rV π= Substitute for V and simplify to obtain: 3 3 4 3 3 4 r m r m ππ ρ == (1) Substitute numerical values and evaluate ρ: ( ) ( ) 317 317 315 26 kg/m104.1 kg/m10410.1 m104.54 kg103.93 ×= ×= × × = − − π ρ (b) Solve equation (1) for r to obtain: 3 4 3 πρ m r = Substitute numerical values and evaluate r: ( ) m102.2 m kg 10410.14 kg1098.53 2 3 3 17 24 ×= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × = π r or about 200 m! 68 •• The Canadian Norman Wells Oil Pipeline extends from Norman Wells, Northwest Territories, to Zama, Alberta. The 8.68 ×105 -m-long pipeline has an inside diameter of 12 in and can be supplied with oil at 35 L/s. (a) What is the volume of oil in the pipeline if it is full at some instant in time? (b) How long would it take to fill the pipeline with oil if it is initially empty?
Chapter 140 Picture the Problem The volume of a cylinder is the product of its cross- sectional area and its length. The time required to fill the pipeline with oil is the ratio of its volume to the flow rate R of the oil. We’ll assume that the pipe has a diameter of exactly 12 in. (a) Express the volume V of the cylindrical pipe in terms of its radius r and its length L: LrV 2 π= Substitute numerical values and evaluate V: ( ) 345 22 m103.6m1068.8 in m102.540 in6 ×=×⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × ×= - V π (b) Express the time Δt to fill the pipe in terms of its volume V and the flow rate R of the oil: R V t =Δ Substitute numerical values and evaluate Δt: s108.1 L m10 s L 35 m103.6 Δ 6 33 34 ×= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = − t or about 21 days! 69 •• The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the sun, namely 1.496 × 1011 m. The parsec is the radius of a circle for which a central angle of 1 s intercepts an arc of length 1 AU. The light-year is the distance that light travels in 1 y. (a) How many parsecs are there in one astronomical unit? (b) How many meters are in a parsec? (c) How many meters in a light-year? (d) How many astronomical units in a light-year? (e) How many light-years in a parsec? Picture the Problem We can use the relationship between an angle θ, measured in radians, subtended at the center of a circle, the radius R of the circle, and the length L of the arc to answer these questions concerning the astronomical units of measure. We’ll take the speed of light to be 2.998 × 108 m/s. (a) Relate the angle θ subtended by an arc of length S to the distance R: R S =θ ⇒ θRS = (1)
Measurement and Vectors 41 Substitute numerical values and evaluate S: ( )( ) parsec10848.4 360 rad2 min60 1 s60 min1 s1parsec1 6− ×=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ °⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ° ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = π S (b) Solving equation (1) for R yields: θ S R = Substitute numerical values and evaluate R: ( ) m10086.3 360 rad2 min60 1 s60 min1 s1 m10496.1 16 11 ×= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ °⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ° ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × = π R (c) The distance D light travels in a given interval of time Δt is given by: tcD Δ= Substitute numerical values and evaluate D: m10461.9 min s 60 h min 60 d h 24 y d 365.24 s m 102.998 158 ×=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×=D (d) Use the definition of 1 AU and the result from Part (c) to obtain: ( ) AU106.324 m101.496 AU1 m109.461y1 4 11 15 ×= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × ×=⋅c (e) Combine the results of Parts (b) and (c) to obtain: ( ) y262.3 m109.461 y1 m10086.3parsec1 15 16 ⋅=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × ⋅ ×= c c 70 •• If the average density of the universe is at least 6 × 10–27 kg/m3 , then the universe will eventually stop expanding and begin contracting. (a) How many electrons are needed in each cubic meter to produce the critical density? (b) How many protons per cubic meter would produce the critical density? (me = 9.11 × 10–31 kg; mp = 1.67 × 10–27 kg.)
Chapter 142 Picture the Problem Let Ne and Np represent the number of electrons and the number of protons, respectively and ρ the critical average density of the universe. We can relate these quantities to the masses of the electron and proton using the definition of density. (a) Using its definition, relate the required density ρ to the electron density Ne/V: V mN V m ee ==ρ ⇒ e e mV N ρ = (1) Substitute numerical values and evaluate Ne/V: 33 33 31 327 e melectrons/107 melectrons/10586.6 nkg/electro109.11 kg/m106 ×≈ ×= × × = − − V N (b) Express and evaluate the ratio of the masses of an electron and a proton: 4 27 31 p e 10455.5 kg101.67 kg109.11 − − − ×= × × = m m Rewrite equation (1) in terms of protons: p p mV N ρ = (2) Divide equation (2) by equation (1) to obtain: p e e p m m V N V N = or ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = V N m m V N e p ep Substitute numerical values and use the result from Part (a) to evaluate Np/V: ( )( ) 3334p protons/m4melectrons/10586.610455.5 ≈××= − V N 71 ••• You are an astronaut doing physics experiments on the moon. You are interested in the experimental relationship between distance fallen, y, and time elapsed, t, of falling objects dropped from rest. You have taken some data for a falling penny, which is represented in the table below. You expect that a general relationship between distance y and time t is y = BtC , where B and C are constants to be determined experimentally. To accomplish this, create a log-log plot of the data: (a) graph log(y) vs. log(t), with log(y) the ordinate variable and log(t) the abscissa variable. (b) Show that if you take the log of each side of your equation, you get log(y) = log(B) + Clog(t). (c) By comparing this linear relationship to the graph of the data, estimate the values of B and C. (d) If you
Measurement and Vectors 43 drop a penny, how long should it take to fall 1.0 m? (e) In the next chapter, we will show that the expected relationship between y and t is ,2 2 1 aty = where a is the acceleration of the object. What is the acceleration of objects dropped on the moon? y (m) 10 20 30 40 50 t (s) 3.5 5.2 6.0 7.3 7.9 Picture the Problem We can plot log y versus log t and find the slope of the best- fit line to determine the exponent C. The value of B can be determined from the intercept of this graph. Once we know C and B, we can solve C Bty = for t as a function of y and use this result to determine the time required for an object to fall a given distance on the surface of the moon. (a) The following graph of log y versus log t was created using a spreadsheet program. The equation shown on the graph was obtained using Excel’s ″Add Trendline″ function. (Excel’s ″Add Trendline″ function uses regression analysis to generate the trendline.) log y = 1.9637log t - 0.0762 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 0.50 0.60 0.70 0.80 0.90 1.00 log t logy (b) Taking the logarithm of both sides of the equation C Bty = yields: ( ) tCB tBBty CC loglog loglogloglog += +== (c) Note that this result is of the form: mXbY += where Y = log y, b = log B, m = C, and X = log t
Chapter 144 From the regression analysis (trendline) we have: log B = −0.076 Solving for B yields: 2076.0 m/s84.010 == − B where we have inferred the units from tho given for C Bty = . Also, from the regression analysis we have: 0.296.1 ≈=C (d) Solve C Bty = for t to obtain: C B y t 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Substitute numerical values and evaluate t to determine how long it would take a penny to fall 1.0 m: s1.1 s m 84.0 m0.1 2 1 2 ≈ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ =t (e) Substituting for B and C in C Bty = yields: 2 2 s m 84.0 ty ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Compare this equation to 2 2 1 aty = to obtain: 22 1 s m 84.0=a and 22 s m 7.1 s m 84.02 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =a Remarks: One could use a graphing calculator to obtain the results in Parts (a) and (c). 72 ••• A particular company’s stock prices vary with the market and with the company’s type of business, and can be very unpredictable, but people often try to look for mathematical patterns where they may not belong. Corning is a materials-engineering company located in upstate New York. Below is a table of the price of Corning stock on August 3, for every 5 years from 1981 to 2001. Assume that the price follows a power law: price (in $) = BtC where t is expressed in years. (a) Evaluate the constants B and C. (b) According to the power law, what should the price of Corning stock have been on August 3, 2000? (It was actually $82.83!)
Measurement and Vectors 45 Price (dollars) 2.10 4.19 9.14 10.82 16.85 Years since 1980 1 6 11 16 21 Picture the Problem We can plot log P versus log t and find the slope of the best-fit line to determine the exponent C. The value of B can be determined from the intercept of this graph. Once we know C and B, we can use C BtP = to predict the price of Corning stock as a function of time. (a) The following graph of log P versus log was created using a spreadsheet program. The equation shown on the graph was obtained using Excel’s ″Add Trendline″ function. (Excel’s ″Add Trendline″ function uses regression analysis to generate the trendline.) log P = 0.658log t + 0.2614 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 log t logP Taking the logarithm of both sides of the equation C BtP = yields: ( ) tCB tBBtP CC loglog loglogloglog += +== Note that this result is of the form: mXbY += where Y = log P, b = log B, m = C, and X = log t From the regression analysis (trendline) we have: log B = 0.2614⇒ .831$10 2614.0 ==B Also, from the regression analysis we have: 658.0=C (b) Substituting for B and C in C BtP = yields: ( ) 658.0 83.1$ tP =
Chapter 146 Evaluate P(20 y) to obtain: ( ) ( )( ) 14.13$2083.1$y20 658.0 ==P Remarks: One could use a graphing calculator to obtain these results. 73 ••• [SSM] The Super-Kamiokande neutrino detector in Japan is a large transparent cylinder filled with ultra pure water. The height of the cylinder is 41.4 m and the diameter is 39.3 m. Calculate the mass of the water in the cylinder. Does this match the claim posted on the official Super-K Web site that the detector uses 50000 tons of water? Picture the Problem We can use the definition of density to relate the mass of the water in the cylinder to its volume and the formula for the volume of a cylinder to express the volume of water used in the detector’s cylinder. To convert our answer in kg to lb, we can use the fact that 1 kg weighs about 2.205 lb. Relate the mass of water contained in the cylinder to its density and volume: Vm ρ= Express the volume of a cylinder in terms of its diameter d and height h: hdhAV 2 base 4 π == Substitute in the expression for m to obtain: hdm 2 4 π ρ= Substitute numerical values and evaluate m: ( ) ( ) ( ) kg10022.5 m4.41m3.39 4 kg/m10 7 233 ×= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = π m Convert 5.02 × 107 kg to tons: ton104.55 lb2000 ton1 kg lb2.205 kg10022.5 3 7 ×= ×××=m The 50,000-ton claim is conservative. The actual weight is closer to 55,000 tons. 74 ••• You and a friend are out hiking across a large flat plain and decide to determine the height of a distant mountain peak, and also the horizontal distance from you to the peak Figure 1-19). In order to do this, you stand in one spot and determine that the sightline to the top of the peak is inclined at 7.5° above the horizontal. You also make note of the heading to the peak at that point: 13° east of north. You stand at the original position, and your friend hikes due west for 1.5 km. He then sights the peak and determines that its sightline has a heading of 15° east of north. How far is the mountain from your position, and how high is its summit above your position?
Measurement and Vectors 47 Picture the Problem Vector A r lies in the plane of the plain and locates the base or the peak relative to you. Vector B r also lies in the plane of the plain and locates the base of the peak relative to your friend when he/she has walked 1.5 km to the west. We can use the geometry of the diagram and the E-W components of the vectors d r , B r and A r to find the distance to the mountain from your position. Once we know the distance A, we can use a trigonometric relationship to find the height of the peak above your position. °15 °13 β α W E N d = 1.5 km A rB r Your positionYour friend's position d d r Peak Referring to the diagram, note that: hB =βsin and hA =αsin Equating these expressions for h gives: αβ sinsin AB = ⇒ AB β α sin sin = Adding the E-W components of the vectors d r , B r and A r yields: αβ coscoskm5.1 AB −= Substitute for B and simplify to obtain: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= −= α β α αβ β α cos tan sin coscos sin sin km5.1 A AA Solving for A yields: α β α cos tan sin km5.1 − =A
Chapter 148 Substitute numerical values and evaluate A: km42 km52.41 77cos 75tan 77sin km5.1 = = °− ° ° =A Referring to the following diagram, we note that: h A°5.7 ( ) km5.5 5.7tankm52.41 5.7tan = °= °= Ah Remarks: One can also solve this problem using the law of sines. 75 ••• The table below gives the periods T and orbit radii r for the motions of four satellites orbiting a dense, heavy asteroid. (a) These data can be fitted by the formula n CrT = . Find the values of the constants C and n. (b) A fifth satellite is discovered to have a period of 6.20 y. Find the radius for the orbit of this satellite, which fits the same formula. Period T, y 0.44 1.61 3.88 7.89 Radius r, Gm 0.088 0.208 0.374 0.600 Picture the Problem We can plot log T versus log r and find the slope of the best-fit line to determine the exponent n. We can then use any of the ordered pairs to evaluate C. Once we know n and C, we can solve n CrT = for r as a function of T. (a) Take the logarithm (we’ll arbitrarily use base 10) of both sides of n CrT = and simplify to obtain: ( ) ( ) Crn rCCrT nn loglog loglogloglog += +== Note that this equation is of the form bmxy += . Hence a graph of log T vs. log r should be linear with a slope of n and a log T -intercept log C.
Measurement and Vectors 49 The following graph of log T versus log r was created using a spreadsheet program. The equation shown on the graph was obtained using Excel’s ″Add Trendline″ function. (Excel’s ″Add Trendline″ function uses regression analysis to generate the trendline.) log T = 1.5036log r + 1.2311 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 -1.1 -1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 log r logT From the regression analysis we note that: 50.1=n , ( ) 232311.1 Gmy/0.1710 ==C , and ( )( ) 50.123 Gmy/0.17 rT = (1) (b) Solve equation (1) for the radius of the planet’s orbit: ( ) 32 23 Gm/y0.17 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = T r Substitute numerical values and evaluate r: ( ) Gm510.0 Gmy/0.17 y20.6 32 23 =⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =r 76 ••• The period T of a simple pendulum depends on the length L of the pendulum and the acceleration of gravity g (dimensions L/T2 ). (a) Find a simple combination of L and g that has the dimensions of time. (b) Check the dependence of the period T on the length L by measuring the period (time for a complete swing back and forth) of a pendulum for two different values of L. (c) The correct formula relating T to L and g involves a constant that is a multiple of π , and cannot be obtained by the dimensional analysis of Part (a). It can be found by
Chapter 150 experiment as in Part (b) if g is known. Using the value g = 9.81 m/s2 and your experimental results from Part (b), find the formula relating T to L and g. Picture the Problem We can express the relationship between the period T of the pendulum, its length L, and the acceleration of gravity g as ba gCLT = and perform dimensional analysis to find the values of a and b and, hence, the function relating these variables. Once we’ve performed the experiment called for in Part (b), we can determine an experimental value for C. (a) Express T as the product of L and g raised to powers a and b: ba gCLT = (1) where C is a dimensionless constant. Write this equation in dimensional form: [ ] [ ] [ ]ba gLT = Substituting the dimensions of the physical quantities yields: b a ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = 2 T L LT Because L does not appear on the left-hand side of the equation, we can write this equation as: bba 210 TLTL −+ = Equate the exponents to obtain: 12and0 =−=+ bba Solve these equations simultaneously to find a and b: 2 1 2 1 and −== ba Substitute in equation (1) to obtain: g L CgCLT == − 2121 (2) (b) If you use pendulums of lengths 1.0 m and 0.50 m; the periods should be about: ( ) ( ) s4.1m50.0 and s0.2m0.1 = = T T (c) Solving equation (2) for C yields: L g TC =
Measurement and Vectors 51 Evaluate C with L = 1.0 m and T = 2.0 s: ( ) π226.6 m1.0 m/s9.81 s2.0 2 ≈==C Substitute in equation (2) to obtain: g L T π2= 77 ••• A sled at rest is suddenly pulled in three horizontal directions at the same time but it goes nowhere. Paul pulls to the northeast with a force of 50 lb. Johnny pulls at an angle of 35° south of due west with a force of 65 lb. Connie pulls with a force to be determined. (a) Express the boys' two forces in terms of the usual unit vectors (b) Determine the third force (from Connie), expressing it first in component form and then as a magnitude and angle (direction). Picture the Problem A diagram showing the forces exerted by Paul, Johnny, and Connie is shown below. Once we’ve expressed the forces exerted by Paul and Johnny in vector form we can use them to find the force exerted by Connie. N S EW θ°45 °35 Connie F r Paul F r Johnny F r lb 50 lb 56 (a) The force that Paul exerts is: ( )[ ] ( )[ ] ( ) ( ) ( ) ( ) ji jijiF ˆlb35ˆlb35 ˆlb4.35ˆlb4.35ˆ45sinlb50ˆ45coslb50Paul += +=°+°= r The force that Johnny exerts is: ( )[ ] ( )[ ] ( ) ( ) ( ) ( ) ji jijiF ˆlb37ˆlb53 ˆlb3.37ˆlb53.2ˆ215sinlb56ˆ215coslb56Johnny −+−= −−=°+°= r
Chapter 152 The sum of the forces exerted by Paul and Johnny is: ( ) ( ) ( ) ( ) ( ) ( ) ji jijiFF ˆlb9.1ˆlb8.71 ˆlb3.37ˆlb53.2ˆlb4.35ˆlb4.35JohnnyPaul −−= −−+=+ rr (b) The condition that the three forces must satisfy is: 0ConnieJohnnyPaul =++ FFF rrr Solving for ConnieF r yields: ( )JohnnyPaulConnie FFF rrr +−= Substitute for JohnnyPaul FF rr + to obtain: ( ) ( )[ ] ( ) ( ) ji jiF ˆlb.91ˆlb81 ˆlb9.1ˆlb8.71Connie += −−−= r The magnitude of ConnieF r is: ( ) ( ) lb18lb9.1lb8.17 22 Connie =+=F The direction that the force exerted by Connie acts is given by: EofN1.6 lb17.8 lb9.1 tan 1 °=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − θ 78 ••• You spot a plane that is 1.50 km North, 2.5 km East and at an altitude of 5.0 km above your position. (a) How far from you is the plane? (b) At what angle from due north (in the horizontal plane) are you looking? (c) Determine the plane's position vector (from your location) in terms of the unit vectors, letting iˆ be toward the east direction, jˆ be toward the north direction, and kˆ be vertically upward. (d) At what elevation angle (above the horizontal plane of Earth) is the airplane? Picture the Problem A diagram showing the given information is shown below. We can use the Pythagorean Theorem, trigonometry, and vector algebra to find the distance, angles, and expression called for in the problem statement.
Measurement and Vectors 53 θ φ r h d km5.2 km 5.1 N E hereareYou Plane (a) Use the Pythagorean Theorem to express d in terms of h and l: 22 hd += l Substitute numerical values and evaluate d: ( ) ( ) ( ) km8.5 km0.5km5.1km5.2 222 = ++=d (b) Use trigonometry to evaluate the angle from due north at which you are looking at the plane: NofE59 km1.5 km5.2 tan 1 °=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − θ (c) We can use the coordinates of the plane relative to your position to express the vector d r : ( ) ( ) ( )kjid ˆkm.05ˆkm5.1ˆkm5.2 ++= r (d) Express the elevation angle φ in terms of h and l: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − l h1 tanφ Substitute numerical values and evaluate φ: ( ) ( ) horizontheabove60 km5.1km5.2 km0.5 tan 22 1 °= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = − φ
Chapter 154
55 Chapter 2 Motion in One Dimension Conceptual Problems 1 • What is the average velocity over the ″round trip″ of an object that is launched straight up from the ground and falls straight back down to the ground? Determine the Concept The "average velocity" is being requested as opposed to "average speed". The average velocity is defined as the change in position or displacement divided by the change in time. t y v Δ Δ =av The change in position for any "round trip" is zero by definition. So the average velocity for any round trip must also be zero. 0 Δ 0 Δ Δ av === tt y v 2 • An object thrown straight up falls back and is caught at the same place it is launched from. Its time of flight is T, its maximum height is H. Neglect air resistance. The correct expression for its average speed for the entire flight is (a) H/T, (b) 0, (c) H/(2T), and (d) 2H/T. Determine the Concept The important concept here is that "average speed" is being requested as opposed to "average velocity". Under all circumstances, including constant acceleration, the definition of the average speed is the ratio of the total distance traveled (H + H) to the elapsed time, in this case 2H/T. )(d is correct. Remarks: Because this motion involves a round trip, if the question asked for "average velocity," the answer would be zero. 3 • Using the information in the previous question, what is its average speed just for the first half of the trip? What is its average velocity for the second half of the trip? (Answer in terms of H and T.) Determine the Concept Under all circumstances, including constant acceleration, the definition of the average speed is the ratio of the total distance traveled to the elapsed time. The average velocity, on the other hand, is the ratio of the displacement to the elapsed time.
Chapter 256 The average speed for the first half of the trip is the height to which the object rises divided by one-half its time of flight: T H T H v 2 2 1half1stav, == The average velocity for the second half of the trip is the distance the object falls divided by one-half its time of flight: T H T H 2 vel 2 1half2ndav, −= − = Remarks: We could also say that the average velocity for the second half of the trip is −2H/T. 4 • Give an everyday example of one-dimensional motion where (a) the velocity is westward and the acceleration is eastward, and (b) the velocity is northward and the acceleration is northward. Determine the Concept The important concept here is that a = dv/dt, where a is the acceleration and v is the velocity. Thus, the acceleration is positive if dv is positive; the acceleration is negative if dv is negative. (a) An example of one-dimensional motion where the velocity is westward and acceleration is eastward is a car traveling westward and slowing down. (b)An example of one-dimensional motion where the velocity is northward and the acceleration is northward is a car traveling northward and speeding up. 5 • [SSM] Stand in the center of a large room. Call the direction to your right ″positive,″ and the direction to your left ″negative.″ Walk across the room along a straight line, using a constant acceleration to quickly reach a steady speed along a straight line in the negative direction. After reaching this steady speed, keep your velocity negative but make your acceleration positive. (a) Describe how your speed varied as you walked. (b) Sketch a graph of x versus t for your motion. Assume you started at x = 0. (c) Directly under the graph of Part (b), sketch a graph of vx versus t. Determine the Concept The important concept is that when both the acceleration and the velocity are in the same direction, the speed increases. On the other hand, when the acceleration and the velocity are in opposite directions, the speed decreases. (a) Your speed increased from zero, stayed constant for a while, and then decreased.
Motion in One Dimension 57 (b) A graph of your position as a function of time is shown to the right. Note that the slope starts out equal to zero, becomes more negative as the speed increases, remains constant while your speed is constant, and becomes less negative as your speed decreases. (c) The graph of v(t) consists of a straight line with negative slope (your acceleration is constant and negative) starting at (0,0), then a flat line for a while (your acceleration is zero), and finally an approximately straight line with a positive slope heading to v = 0. x t t xv 6 • True/false: The displacement always equals the product of the average velocity and the time interval. Explain your choice. Determine the Concept True. We can use the definition of average velocity to express the displacement Δx as Δx = vavΔt. Note that, if the acceleration is constant, the average velocity is also given by vav = (vi + vf)/2. 7 • Is the statement ″for an object’s velocity to remain constant, its acceleration must remain zero″ true or false? Explain your choice. Determine the Concept True. Acceleration is the slope of the velocity versus time curve, a = dv/dt; while velocity is the slope of the position versus time curve, v = dx/dt. The speed of an object is the magnitude of its velocity. Zero acceleration implies that the velocity is constant. If the velocity is constant (including zero), the speed must also be constant. 8 • Draw careful graphs of the position and velocity and acceleration over the time period 0 ≤ t ≤ 30 s for a cart that, in succession, has the following motion. The cart is moving at the constant speed of 5.0 m/s in the +x direction. It passes by the origin at t = 0.0 s. It continues on at 5.0 m/s for 5.0 s, after which it gains speed at the constant rate of 0.50 m/s each second for 10.0 s. After gaining speed for 10.0 s, the cart loses speed at the constant rate of 0.50 m/s for the next 15.0 s. Determine the Concept Velocity is the slope of the position versus time curve and acceleration is the slope of the velocity versus time curve. The following graphs were plotted using a spreadsheet program.
Chapter 258 0 20 40 60 80 100 120 140 160 0 5 10 15 20 25 30 t , s x,m 0 1 2 3 4 5 6 7 8 9 10 0 5 10 15 20 25 30 t , s v,m/s -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0 5 10 15 20 25 30 t, s a,m/s2
Motion in One Dimension 59 9 • True/false; Average velocity always equals one-half the sum of the initial and final velocities. Explain your choice. Determine the Concept False. The average velocity is defined (for any acceleration) as the change in position (the displacement) divided by the change in time txv ΔΔ=av . It is always valid. If the acceleration remains constant the average velocity is also given by 2 fi av vv v + = Consider an engine piston moving up and down as an example of non-constant velocity. For one complete cycle, vf = vi and xi = xf so vav = Δx/Δt is zero. The formula involving the mean of vf and vi cannot be applied because the acceleration is not constant, and yields an incorrect nonzero value of vi. 10 • Identical twin brothers standing on a horizontal bridge each throw a rock straight down into the water below. They throw rocks at exactly the same time, but one hits the water before the other. How can this be? Explain what they did differently. Ignore any effects due to air resistance. Determine the Concept This can occur if the rocks have different initial speeds. Ignoring air resistance, the acceleration is constant. Choose a coordinate system in which the origin is at the point of release and upward is the positive direction. From the constant-acceleration equation 2 2 1 00 attvyy ++= we see that the only way two objects can have the same acceleration (–g in this case) and cover the same distance, Δy = y – y0, in different times would be if the initial velocities of the two rocks were different. Actually, the answer would be the same whether or not the acceleration is constant. It is just easier to see for the special case of constant acceleration. 11 •• [SSM] Dr. Josiah S. Carberry stands at the top of the Sears Tower in Chicago. Wanting to emulate Galileo, and ignoring the safety of the pedestrians below, he drops a bowling ball from the top of the tower. One second later, he drops a second bowling ball. While the balls are in the air, does their separation (a) increase over time, (b) decrease, (c) stay the same? Ignore any effects due to air resistance. Determine the Concept Neglecting air resistance, the balls are in free fall, each with the same free-fall acceleration, which is a constant. At the time the second ball is released, the first ball is already moving. Thus, during any time interval their velocities will increase by exactly the same amount. What can be said about the speeds of the two balls? The first ball will always be moving faster than the second ball. This being the case, what happens to the separation of the two balls while they are both falling? Their separation increases. )(a is correct.
Chapter 260 12 •• Which of the position-versus-time curves in Figure 2-28 best shows the motion of an object (a) with positive acceleration, (b) with constant positive velocity, (c) that is always at rest, and (d) with negative acceleration? (There may be more than one correct answer for each part of the problem.) Determine the Concept The slope of an x(t) curve at any point in time represents the speed at that instant. The way the slope changes as time increases gives the sign of the acceleration. If the slope becomes less negative or more positive as time increases (as you move to the right on the time axis), then the acceleration is positive. If the slope becomes less positive or more negative, then the acceleration is negative. The slope of the slope of an x(t) curve at any point in time represents the acceleration at that instant. (a) The correct answer is .)(d The slope of curve (d) is positive and increasing. Therefore the velocity and acceleration are positive. We would need more information to conclude that a is constant. (b) The correct answer is .)(b The slope of curve (b) is positive and constant. Therefore the velocity is positive and constant. (c) The correct answer is .)(e The slope of curve (e) is zero. Therefore, the velocity and acceleration are zero and the object remains at the same position. (d) The correct answers are ( ) .and)( ca The slope of curve (a) is negative and becomes more negative as time increases. Therefore the velocity is negative and the acceleration is negative. The slope of curve (c) is positive and decreasing. Therefore the velocity is positive and the acceleration is negative. 13 •• [SSM] Which of the velocity-versus-time curves in figure 2-29 best describes the motion of an object (a) with constant positive acceleration, (b) with positive acceleration that is decreasing with time, (c) with positive acceleration that is increasing with time, and (d) with no acceleration? (There may be more than one correct answer for each part of the problem.) Determine the Concept The slope of a v(t) curve at any point in time represents the acceleration at that instant. (a) The correct answer is ( ) .b The slope of curve (b) is constant and positive. Therefore the acceleration is constant and positive. (b) The correct answer is .)(c The slope of curve (c) is positive and decreasing with time. Therefore the acceleration is positive and decreasing with time.
Motion in One Dimension 61 (c) The correct answer is .)(d The slope of curve (d) is positive and increasing with time. Therefore the acceleration is positive and increasing with time. (d) The correct answer is .)(e The slope of curve (e) is zero. Therefore the velocity is constant and the acceleration is zero. 14 •• The diagram in Figure 2-30 tracks the location of an object moving in a straight line along the x axis. Assume that the object is at the origin at t = 0. Of the five times shown, which time (or times) represents when the object is (a) farthest from the origin, (b) at rest for an instant, (c) in the midst of being at rest for awhile, and (d) moving away from the origin? Determine the Concept Because this graph is of distance-versus-time we can use its displacement from the time axis to draw conclusions about how far the object is from the origin. We can also use the instantaneous slope of the graph to decide whether the object is at rest and whether it is moving toward or away from the origin. (a) The correct answer is .B Because the object’s initial position is at x = 0, point B represents the instant that the object is farthest from x = 0. (b) The correct answers are D.andB Because the slope of the graph is zero at points B and D, the velocity of the object is zero and it is momentarily at rest at these points. (c) The correct answer is E . Because the graph is a horizontal line with zero slope, the object remains at rest at the same position (its velocity is zero). (d) The correct answer is A. Because the slope of the graph is positive at point A, the velocity of the object is positive and it is moving away from the origin. 15 •• [SSM] An object moves along a straight line. Its position versus time graph is shown in Figure 2-30. At which time or times is its (a) speed at a minimum, (b) acceleration positive, and (c) velocity negative? Determine the Concept Because this graph is of distance-versus-time we can use its instantaneous slope to describe the object’s speed, velocity, and acceleration. (a) The minimum speed is at EandD,B, , where it is zero. In the one- dimensional motion shown in the figure, the velocity is a minimum when the slope of a position-versus-time plot goes to zero (i.e., the curve becomes horizontal). At these points the velocity is zero and, therefore, the speed is zero.
Chapter 262 (b) The acceleration is positive at points D.andA Because the slope of the graph is increasing at these points, the velocity of the object is increasing and its acceleration is positive. (c) The velocity is negative at point C. Because the slope of the graph is negative at point C, the velocity of the object is negative. 16 •• For each of the four graphs of x versus t in Figure 2-31 answer the following questions. (a) Is the velocity at time t2 greater than, less than, or equal to the velocity at time t1? (b) Is the speed at time t2 greater than, less than, or equal to the speed at time t1? Determine the Concept In one-dimensional motion, the velocity is the slope of a position-versus-time plot and can be either positive or negative. On the other hand, the speed is the magnitude of the velocity and can only be positive. We’ll usev to denote velocity and the word ″speed″ for how fast the object is moving. (a) curve a: ( ) ( )12 tvtv < curve b: ( ) ( )12 tvtv = curve c: ( ) ( )12 tvtv > curve d: ( ) ( )12 tvtv < (b) curve a: ( ) ( )12 speedspeed tt < curve b: ( ) ( )12 speedspeed tt = curve c: ( ) ( )12 speedspeed tt < curve d: ( ) ( )12 speedspeed tt > 17 •• True/false: Explain your reasoning for each answer. If the answer is true, give an example. (a) If the acceleration of an object is always zero, then it cannot be moving. (b) If the acceleration of an object is always zero, then its x-versus-t curve must be a straight line. (c) If the acceleration of an object is nonzero at an instant, it may be momentarily at rest at that instant. Explain your reasoning for each answer. If an answer is true, give an example. (a) False. An object moving in a straight line with constant speed has zero acceleration. (b) True. If the acceleration of the object is zero, then its speed must be constant. The graph of x-versus-t for an object moving with constant speed is a straight line. (c) True. A ball thrown upward is momentarily at rest when it is at the top of its trajectory. Its acceleration, however, is non-zero at this instant. Its value is the same as it was just before it came to rest and after it has started its descent.
Motion in One Dimension 63 18 •• A hard-thrown tennis ball is moving horizontally when it bangs into a vertical concrete wall at perpendicular incidence. The ball rebounds straight back off the wall. Neglect any effects due to gravity for the small time interval described here. Assume that towards the wall is the +x direction. What are the directions of its velocity and acceleration (a) just before hitting the wall, (b) at maximum impact, and (c) just after leaving the wall. Determine the Concept The tennis ball will be moving with constant velocity immediately before and after its collision with the concrete wall. It will be accelerated during the duration of its collision with the wall. (a) Just before hitting the wall the velocity of the ball is in the +x direction and, because its velocity is constant, its acceleration is zero. (b) At maximum impact, the ball is reversing direction and its velocity is zero. Its acceleration is in the −x direction. (c) Just after leaving the wall, the velocity of the ball is in the −x direction and constant. Because its velocity is constant, its acceleration is zero. 19 •• [SSM] A ball is thrown straight up. Neglect any effects due to air resistance. (a) What is the velocity of the ball at the top of its flight? (b) What is its acceleration at that point? (c) What is different about the velocity and acceleration at the top of the flight if instead the ball impacts a horizontal ceiling very hard and then returns. Determine the Concept In the absence of air resistance, the ball will experience a constant acceleration and the graph of its position as a function of time will be parabolic. In the graphs to the right, a coordinate system was chosen in which the origin is at the point of release and the upward direction is the +y direction. The top graph shows the position of the ball as a function of time and the bottom graph shows the velocity of a ball as a function of time. y v t t (a) 0flightoftop =v (b) The acceleration of the ball is the same at every point of its trajectory, including the point at which v = 0 (at the top of its flight). Hence ga −= flight oftop .
Chapter 264 (c) If the ball impacts a horizontal ceiling very hard and then returns, its velocity at the top of its flight is still zero and its acceleration is still downward but greater than g in magnitude. 20 •• An object that is launched straight up from the ground, reaches a maximum height H, and falls straight back down to the ground, hitting it T seconds after launch. Neglect any effects due to air resistance. (a) Express the average speed for the entire trip as a function of H and T. (b) Express the average speed for the same interval of time as a function of the initial launch speed v0. Picture the Problem The average speed is being requested as opposed to average velocity. We can use the definition of average speed as distance traveled divided by the elapsed time and the expression for the average speed of an object when it is experiencing constant acceleration to express vav in terms of v0. (a) The average speed is defined as the total distance traveled divided by the change in time: timetotal traveleddistancetotal av =v Substitute for the total distance traveled and the total time and simplify to obtain: T H T HH v 2 av = + = (b) The average speed for the upward flight of the object is given by: T Hv v 2 1 0 upav, 2 0 = + = ⇒ 04 1 v T H = The average speed for the same interval of time as a function of the initial launch speed v0 is twice the average speed during the upward portion of the flight: ( ) 02 1 04 1 upav,av 22 vvvv === Because 00 ≠v , the average speed is not zero. Remarks: 1) Because this motion involves a roundtrip, if the question asked for ″average velocity″, the answer would be zero. 2) Another easy way to obtain this result is take the absolute value of the velocity of the object to obtain a graph of its speed as a function of time. A simple geometric argument leads to the result we obtained above. 21 •• A small lead ball is thrown directly upward. Neglect any effects due to air resistance. True or false: (a) The magnitude of its acceleration decreases on the way up. (b) The direction of its acceleration on its way down is opposite to the direction of its acceleration on its way up. (c) The direction of its velocity on its way down is opposite to the direction of its velocity on its way up. Determine the Concept For free fall, the acceleration is the same (g) throughout the entire flight.
Motion in One Dimension 65 (a) False. The velocity of the ball decreases at a steady rate. This means that the acceleration of the ball is constant. (b) False. The velocity of the ball decreases at a steady rate (g) throughout its entire flight. (c) True. On the way up the velocity vector points upward and on the way down it points downward. 22 •• At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the roof. Air resistance is negligible. During the descent of B to the ground, the distance between the two objects (a) is proportional to t, (b) is proportional to t2 , (c) decreases, (d) remains 10 m throughout. Determine the Concept Both objects experience the same constant acceleration. Choose a coordinate system in which downward is the positive direction and use a constant-acceleration equation to express the position of each object as a function of time. Using constant-acceleration equations, express the positions of both objects as functions of time: 2 2 1 0A0,A gttvxx ++= and 2 2 1 0B,0B gttvxx ++= where v0 = 0. Express the separation of the two objects by evaluating xB − xA: m10A.0B,0AB =−=− xxxx )(d is correct. 23 •• You are driving a Porsche that accelerates uniformly from 80.5 km/h (50 mi/h) at t = 0.00 to 113 km/h (70 mi/h) at t = 9.00 s. (a) Which graph in Figure 2-32 best describes the velocity of your car? (b) Sketch a position-versus- time graph showing the location of your car during these nine seconds, assuming we let its position x be zero at t = 0. Determine the Concept Because the Porsche accelerates uniformly, we need to look for a graph that represents constant acceleration. (a) Because the Porsche has a constant acceleration that is positive (the velocity is increasing), we must look for a velocity-versus-time curve with a positive constant slope and a nonzero intercept. Such a graph is shown in ( ).c
Chapter 266 (b) Use the data given in the problem statement to determine that the acceleration of the Porsche is 1.00 m/s2 and that its initial speed is 22.4 m/s. The equation describing the position of the car as a function of time is ( ) ( ) .m/s00.1m/s4.22 22 2 12 ttx += The following graph of this equation was plotted using a spreadsheet program. 0 50 100 150 200 250 0 1 2 3 4 5 6 7 8 9 t , s x,m 24 •• A small heavy object is dropped from rest and falls a distance D in a time T. After it has fallen for a time 2T, what will be its (a) fall distance from its initial location in terms of D, (b) its speed in terms of D and t, and (c) its acceleration? (Neglect air resistance.) Picture the Problem In the absence of air resistance, the object experiences constant acceleration. Choose a coordinate system in which the downward direction is positive and use the constant-acceleration equation to describe its motion. (a) Relate the distance D that the object, released from rest, falls in time t: ( ) 2 2 1 gtDtx == (1) Evaluate x(2t) to obtain: ( ) ( ) 22 2 1 222 gttgtx == (2) Dividing equation (2) by equation (1) and simplifying yields: ( ) 4 22 2 2 1 2 == gt gt D tx ⇒ ( ) Dtx 42 =
Motion in One Dimension 67 (b) Express the speed of the object as a function of time: gtvv += 0 or, because v0 = 0, gtv = (3) Solving equation (1) for g yields: 2 2 t x g = Substitute for g in equation (3) to obtain: t x t t x v 22 2 == (4) Evaluating equation (4) at time 2t and simplifying yields: ( ) ( ) t D t tx tv 4 2 22 2 == (c) The acceleration of the object is independent of time (that is, it is constant) and is equal to .g 25 •• In a race, at an instant when two horses are running right next to each other and in the same direction (the +x direction), horse A's instantaneous velocity and acceleration are +10 m/s and +2.0 m/s2 respectively, and horse B's instantaneous velocity and acceleration are +12 m/s and –1.0 m/s2 respectively. Which horse is passing the other at this instant? Explain. Determine the Concept The information about the horses’ accelerations is irrelevant to the determination of which horse is passing the other at this instant. The horse running with the greater instantaneous velocity will by passing the slower horse. Hence A.passingisB The accelerations are relevant to the determination of which horse will be in the lead at some later time. 26 •• True or false: (a) The equation 2 2 1 00 tatvxx xx +=− is always valid for all particle motion in one dimension. (b) If the velocity at a given instant is zero, the acceleration at that instant must also be zero. (c) The equation Δx = vavΔt holds for all particle motion in one dimension. Determine the Concept As long as the acceleration remains constant the following constant-acceleration equations hold. If the acceleration is not constant, they do not, in general, give correct results except by coincidence. 2 2 1 00 attvxx ++= atvv += 0 xavv Δ+= 22 0 2 2 fi av vv v + = (a) False. This statement is true if and only if the acceleration is constant.
Chapter 268 (b) False. Consider a rock thrown straight up into the air. At the "top" of its flight, the velocity is zero but it is changing (otherwise the velocity would remain zero and the rock would hover); therefore the acceleration is not zero. (c) True. The definition of average velocity, txv ΔΔ=av , requires that this always be true. 27 •• If an object is moving in a straight line at constant acceleration, its instantaneous velocity halfway through any time interval is (a) greater than its average velocity, (b) less than its average velocity, (c) equal to its average velocity, (d) half its average velocity, (e) twice its average velocity. Determine the Concept Because the acceleration of the object is constant, the constant-acceleration equations can be used to describe its motion. The special expression for average velocity for constant acceleration is 2 fi av vv v + = . ( )c is correct. 28 •• A turtle, seeing his owner put some fresh lettuce on the opposite side of his terrarium, begins to accelerate (at a constant rate) from rest at time t = 0, heading directly toward the food. Let t1 be the time at which the turtle has covered half the distance to his lunch. Derive an expression for the ratio of t2 to t1, where t2 is the time at which the turtle reaches the lettuce. Picture the Problem We are asked, essentially, to determine the time t2, at which a displacement, Δx, is twice what it was at an earlier time, t1. The turtle is crawling with constant acceleration so we can use the constant-acceleration equation ( )2 2 1 0 ΔΔΔ tatvx xx += to describe the turtle’s displacement as a function of time. Express the displacement Δt of the turtle at the end of a time interval Δt: ( )2 2 1 0 ΔΔΔ tatvx xx += or, because v0x = 0, ( )2 2 1 ΔΔ tax x= For the two time intervals: 2 12 1 1Δ tax x= and 2 22 1 2Δ tax x= Express the ratio of Δx2 to Δx1 to obtain: 2 1 2 2 2 12 1 2 22 1 1 2 Δ Δ t t ta ta x x x x == (1) We’re given that: 2 Δ Δ 1 2 = x x
Motion in One Dimension 69 Substitute in equation (1) and simplify to obtain: 22 1 2 2 = t t ⇒ 2 1 2 = t t 29 •• [SSM] The positions of two cars in parallel lanes of a straight stretch of highway are plotted as functions of time in the Figure 2-33.Take positive values of x as being to the right of the origin. Qualitatively answer the following: (a) Are the two cars ever side by side? If so, indicate that time (those times) on the axis. (b) Are they always traveling in the same direction, or are they moving in opposite directions for some of the time? If so, when? (c) Are they ever traveling at the same velocity? If so, when? (d) When are the two cars the farthest apart? (e) Sketch (no numbers) the velocity versus time curve for each car. Determine the Concept Given the positions of the two cars as a function of time, we can use the intersections of the curves and their slopes to answer these questions. (a) The positions of cars A and B are the same at two places where the graphs cross. x t Cars are side by side Car A Car B 1 9 (s) (b) When the slopes of the curves have opposite signs, the velocities of the cars are oppositely directed. Thus, after approximately 7 s, car A is moving leftward while car B is moving rightward. Before t = 7 s, the two cars are traveling in the same direction. (c) The two cars have the same velocity when their curves have the same slopes. This occurs at about 6 s. (d) The time at which the two cars are farthest apart is roughly 6 s as indicated by the place at which, vertically, the two curves are farthest part.
Chapter 270 (e) Car A Car B v t 30 •• A car driving at constant velocity passes the origin at time t = 0. At that instant, a truck, at rest at the origin, begins to accelerate uniformly from rest. Figure 2-34 shows a qualitative plot of the velocities of truck and car as functions of time. Compare their displacements (from the origin), velocities, and accelerations at the instant that their curves intersect. Determine the Concept The graph is a plot of velocity versus time. Thus, where the two curves cross, the truck and car are, at that instant, moving with equal velocities. The slope of a velocity versus time curve is equal to the instantaneous acceleration – thus, since the curve that represents the truck’s velocity has a positive slope, and the car’s curve has zero slope, the truck is accelerating at a higher rate than the car. Finally, the displacements of the two cars are determined by calculating the areas under the curves. In this instance, the curve representing the truck’s velocity as a function of time encloses a triangular area that is exactly half that of the curve representing the car’s velocity. Thus, at the instant represented by the point where the curves intersect, the truck has gone half as far as has the car. 31 •• Reginald is out for a morning jog, and during the course of his run on a straight track, has a velocity that depends upon time as shown in Figure 2-35. That is, he begins at rest, and ends at rest, peaking at a maximum velocity vmax at an arbitrary time tmax. A second runner, Josie, runs throughout the time interval t = 0 to t = tf at a constant speed vR, so that each has the same displacement during the time interval. Note: tf is NOT twice tmax, but represents an arbitrary time. What is the relationship between vJ and vmax? Determine the Concept In this problem we are presented with curves representing the velocity as a function of time for both runners. The area under each curve represents the displacement for each runner and we are told that Josie and Reginald each have the same displacement during the time interval of length tf. Since this is the case, we can find the relationship between vR and vmax by equating the areas under the two curves. Express the condition on the displacement of the two runners: JR ΔΔ xx = (1)
Motion in One Dimension 71 Josie runs at a constant velocity v for the whole of the time interval. Express her displacement ΔxJ: fJJΔ tvx = Reginald has a different velocity profile, one which results in a triangle of height vmax and length tf. Express his displacement ΔxJ: fmax2 1 RΔ tvx = Substitute for ΔxR and ΔxJ in equation (1) and simplify to obtain: == fJfmax2 1 tvtv ⇒ max2 1 J vv = 32 •• Which graph (or graphs), if any, of v versus t in Figure 2-36 best describes the motion of a particle with (a) positive velocity and increasing speed, (b) positive velocity and zero acceleration, (c) constant non-zero acceleration, and (d) a speed decrease? Determine the Concept The velocity of the particle is positive if the curve is above the v = 0 line (the t axis), and the acceleration is positive if the curve has a positive slope. The speed of the particle is the magnitude of its velocity. (a) Graph ( )c describes the motion of a particle with positive velocity and increasing speed because v(t) is above the t axis and has a positive slope. (b) Graph ( )a describes the motion of a particle with positive velocity and zero acceleration because v(t) is above the t axis and its slope is zero. (c) Graphs ( ) ( ) ( )and, edc describe the motion of a particle with constant non- zero acceleration because v(t) is linear and has a non-zero slope. (d) Graph ( )e describes the motion of a particle with a speed decrease because it shows the speed of the particle decreasing with time. 33 •• Which graph (or graphs), if any, of vx versus t in Figure 2-36 best describes the motion of a particle with (a) negative velocity and increasing speed, (b) negative velocity and zero acceleration, (c) variable acceleration, and (d) increasing speed? Determine the Concept The velocity of the particle is positive if the curve is above the v = 0 line (the t axis), and the acceleration is positive if the curve has a positive slope. The speed of the particle is the magnitude of its velocity.
Chapter 272 (a) Graph ( )d describes the motion of a particle with negative velocity and increasing speed because v(t) is below the t axis and has a negative slope. (b) Graph ( )b describes the motion of a particle with negative velocity and zero acceleration because v(t) is below the t axis and its slope is zero. (c) None of these graphs describe the motion of a particle with a variable acceleration because v(t) is linear. (d) Graphs ( ) ( )and dc describe the motion of a particle with an increasing speed because they show the speed of the particle increasing with time. 34 •• Sketch a v-versus-t curve for each of the following conditions: (a) Acceleration is zero and constant while velocity is not zero. (b) Acceleration is constant but not zero. (c) Velocity and acceleration are both positive. (d) Velocity and acceleration are both negative. (e) Velocity is positive and acceleration is negative. (f) Velocity is negative and acceleration is positive. (g) Velocity is momentarily zero but the acceleration is not zero. Determine the Concept Acceleration is the slope of a velocity-versus-time curve. (a) Acceleration is zero and constant while velocity is not zero. t v (b) Acceleration is constant but not zero. t v (c) Velocity and acceleration are both positive. t v (d) Velocity and acceleration are both negative. t v
Motion in One Dimension 73 (e) Velocity is positive and acceleration is negative. t v (f) Velocity is negative and acceleration is positive. t v (g) Velocity is momentarily zero but the acceleration is not zero. t v 35 •• Figure 2-37 shows nine graphs of position, velocity, and acceleration for objects in motion along a straight line. Indicate the graphs that meet the following conditions: (a) Velocity is constant, (b) velocity reverses its direction, (c) acceleration is constant, and (d) acceleration is not constant. (e) Which graphs of position, velocity, and acceleration are mutually consistent? Determine the Concept Velocity is the slope and acceleration is the slope of the slope of a position-versus-time curve. Acceleration is the slope of a velocity- versus-time curve. (a) Graphs ( ) ( ) ( )ifa and,, describe motion at constant velocity. For constant velocity, x versus t must be a straight line; v-versus-t must be a horizontal straight line; and a versus t must be a straight horizontal line at a = 0. (b) Graphs ( ) ( )dc and describe motion in which the velocity reverses its direction. For velocity to reverse its direction x-versus-t must have a slope that changes sign and v versus t must cross the time axis. The acceleration cannot remain zero at all times. (c) Graphs ( ) ( ) ( ) ( ) ( ) ( )ihfeda and,,,,, describe motion with constant acceleration. For constant acceleration, x versus t must be a straight horizontal line or a parabola, v versus t must be a straight line, and a versus t must be a horizontal straight line.
Chapter 274 (d) Graphs ( ) ( ) ( )gcb and,, describe motion with non-constant acceleration. For non-constant acceleration, x versus t must not be a straight line or a parabola; v versus t must not be a straight line, or a versus t must not be a horizontal straight line. (e) The following pairs of graphs are mutually consistent: ),(and)( ia ),(and)( hd and ).(and)( if For two graphs to be mutually consistent, the curves must be consistent with the definitions of velocity and acceleration. Estimation and Approximation 36 • While engrossed in thought about the scintillating lecture just delivered by your physics professor you mistakenly walk directly into the wall (rather than through the open lecture hall door). Estimate the magnitude of your average acceleration as you rapidly come to a halt. Picture the Problem The speed of one’s walk varies from person to person, but 1.0 m/s is reasonable. We also need to estimate a distance within which you would stop in such a case. We’ll assume a fairly short stopping distance of 1.5 cm. We’ll also assume (unrealistically) that you experience constant acceleration and choose a coordinate system in which the direction you are walking is the +x direction. Using a constant-acceleration equation, relate your final speed to your initial speed, acceleration, and displacement while stopping: xavv xΔ22 i 2 f += ⇒ x vv ax Δ2 2 i 2 f − = Substitute numerical values and evaluate the magnitude of your acceleration: ( ) ( ) 2 2 2 2 m/s33 m105.12 s m 0.10 = × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = −xa 37 • [SSM] Occasionally, people can survive falling large distances if the surface they land on is soft enough. During a traverse of the Eiger’s infamous Nordvand, mountaineer Carlos Ragone’s rock anchor gave way and he plummeted 500 feet to land in snow. Amazingly, he suffered only a few bruises and a wrenched shoulder. Assuming that his impact left a hole in the snow 4.0 ft deep, estimate his average acceleration as he slowed to a stop (that is, while he was impacting the snow).
Motion in One Dimension 75 Picture the Problem In the absence of air resistance, Carlos’ acceleration is constant. Because all the motion is downward, let’s use a coordinate system in which downward is the positive direction and the origin is at the point at which the fall began. Using a constant-acceleration equation, relate Carlos’ final velocity v2 to his velocity v1 just before his impact, his stopping acceleration as upon impact, and his stopping distance Δy: yavv Δ2 s 2 1 2 2 += ⇒ y vv a Δ2 2 1 2 2 s − = or, because v2 = 0, y v as Δ2 2 1 −= (1) Using a constant-acceleration equation, relate Carlos’ speed just before impact to his acceleration during free-fall and the distance he fell h: havv fall-free 2 0 2 1 2+= or, because v0 = 0 and ga =fall-free , ghv 22 1 = Substituting for 2 1v in equation (1) yields: y gh as Δ2 2 −= Substitute numerical values and evaluate as: ( )( ) ( ) 23 2 m/s102.1 ft0.42 ft500m/s81.92 ×−= −=a Remarks: The magnitude of this acceleration is about 125g! 38 •• When we solve free-fall problems near Earth, it’s important to remember that air resistance may play a significant role. If its effects are significant, we may get answers that are wrong by orders of magnitude if we ignore it. How can we tell when it is valid to ignore the affects of air resistance? One way is to realize that air resistance increases with increasing speed. Thus, as an object falls and its speed increases, its downward acceleration decreases. Under these circumstances, the object's speed will approach, as a limit, a value called its terminal speed. This terminal speed depends upon such things as the mass and cross-sectional area of the body. Upon reaching its terminal speed, its acceleration is zero. For a ″typical″ skydiver falling through the air, a typical the terminal speed is about 50 m/s (roughly 120 mph). At half its terminal speed, the skydiver’s acceleration will be about g4 3 . Let’s take half the terminal speed as a reasonable ″upper bound″ beyond which we shouldn’t use our constant acceleration free-fall relationships. Assuming the skydiver started from rest, (a) estimate how far, and for how long, the skydiver falls before we can no longer neglect air resistance. (b) Repeat the analysis for a ping-pong ball, which has a terminal speed of about 5.0 m/s. (c) What can you conclude by comparing your answers for Parts (a) and (b)?
Chapter 276 Picture the Problem Because we’re assuming that the accelerations of the skydiver and a ping-pong ball are constant to one-half their terminal velocities, we can use constant-acceleration equations to find the times required for them to reach their ″upper-bound″ velocities and their distances of fall. Let’s use a coordinate system in which downward is the +y direction. (a) Using a constant-acceleration equation, relate the upper-bound velocity to the free-fall acceleration and the time required to reach this velocity: tgvv Δ+= 0boundupper or, because v0 = 0, tgv Δ=boundupper ⇒ g v t boundupper =Δ Substitute numerical values and evaluate Δt: s6.2s55.2 m/s9.81 m/s52 Δ 2 ≈==t Using a constant-acceleration equation, relate the skydiver’s terminal speed to his/her acceleration and distance of fall: ( ) ygvv Δ+= 22 0 2 t2 1 or, because v0 = 0, ( ) ygv Δ= 2 2 t2 1 ⇒ ( ) g v y 2 2 t2 1 =Δ Substitute numerical values and evaluate Δy: ( )[ ] ( ) m32 m/s81.92 m/s50 2 2 2 1 ≈=Δy (b) Proceed as in (a) with m/s0.5boundupper =v to obtain: ( ) s26.0s255.0 m/s9.81 m/s0.5 Δ 2 2 1 ≈==t and ( )[ ] ( ) cm32 m/s81.92 m/s5 2 2 2 1 ≈=Δy (c) The analysis of the motion of a ping-pong ball requires the inclusion of air resistance for almost any situation, whereas the analysis of the motion of the sky diver doesn’t require it until the fall distances and times are considerably longer. 39 •• On June 14, 2005 Asafa Powell of the Jamaica set a world’s record for the 100-m dash with a time t = 9.77 s. Assuming he reached his maximum speed in 3.00 s, and then maintained that speed until the finish, estimate his acceleration during the first 3.00 s. Picture the Problem This is a constant-acceleration problem. Choose a coordinate system in which the direction Powell is running is the +x direction. During the first 3 s of the race his acceleration is positive and during the rest of the race it is zero. The pictorial representation summarizes what we know about Powell’s race.
Motion in One Dimension 77 00 =t s77.92 =t 00 =x 00 =v 1x max1 vv = m1002 =x max2 vv = s00.31 =t Express the total distance covered by Powell in terms of the distances covered in the two phases of his race: 1201m100 xx Δ+Δ= (1) Express the distance he runs getting to his maximum velocity: ( ) ( )2 2 12 01012 1 01001 s3atatvx =Δ+Δ=Δ The distance covered during the rest of the race at the constant maximum velocity is given by: ( ) ( ) ( )( )s77.6s00.3 ΔΔ ΔΔΔ 1201 2 12122 1 12max12 a tta tatvx = = += Substitute for these displacements in equation (1) to obtain: ( ) ( )( )s77.6s00.3s00.3m100 2 2 1 aa += Solving for a yields: ( ) ( )( ) 2 2 2 1 m/s03.4 s77.6s00.3s00.3 m100 = + =a 40 •• The photograph in Figure 2-38 is a short-time exposure (1/30 s) of a juggler with two tennis balls in the air. (a) The tennis ball near the top of its trajectory is less blurred than the lower one. Why is that? (b) Estimate the speed of the ball that he is just releasing from his right hand. (c) Determine how high the ball should have gone above the launch point and compare it to an estimate from the picture. (Hint: You have a built-in distance scale if you assume some reasonable value for the height of the juggler.) Determine the Concept This is a constant-acceleration problem with a = −g if we take upward to be the positive direction. At the maximum height the ball will reach, its speed will be near zero and when the ball has just been tossed in the air its speed is near its maximum value.
Chapter 278 (a) Because the ball is moving slowly its blur is relatively short (i.e., there is less blurring). (b) The average speed of the ball is given by: timeelapsed traveleddistance av =v Estimating how far the ball has traveled in 1/30 s yields: s 30 1 diametersball2 av =v The diameter of a tennis ball is 6.5 cm: ( ) m/s9.3 s 30 1 cm6.52 av ≈=v (c) Use a constant-acceleration equation to relate the initial and final speeds of the ball to its maximum height h: havv y22 0 2 += ⇒ ya vv h 2 2 0 2 − = or, because v = 0 and ay = g, g v h 2 2 0 −= Substitute numerical values and evaluate h: ( ) ( ) cm78 m/s81.92 m/s9.3 2 2 ≈ − −=h If we assume that the juggler is approximately 6 ft (1.8 m) tall, then our calculated value for h seems to be a good approximation to the height shown in the photograph. 41 •• A rough rule of thumb for determining the distance between you and a lightning strike is to start counting the seconds that elapse ("one-Mississippi, two-Mississippi, …") until you hear the thunder (sound emitted by the lightning as it rapidly heats the air around it). Assuming the speed of sound is about 750 mi/h, (a) estimate how far away is a lightning strike if you counted about 5 s until you heard the thunder. (b) Estimate the uncertainty in the distance to the strike in Part (a). Be sure to explain your assumptions and reasoning. (Hint: The speed of sound depends on the air temperature and your counting is far from exact!) Picture the Problem We can use the relationship between distance, speed, and time to estimate the distance to the lightning strike. (a) Relate the distance Δd to the lightning strike to the speed of sound in air v and the elapsed time Δt: tvd ΔΔ =
Motion in One Dimension 79 Substitute numerical values and evaluate Δd: ( ) mi1km7.1 s5 h mi 0.6818 s m 3048.0 h mi 750Δ ≈≈ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =d (b) You are probably lucky if the uncertainty in your time estimate is less than 1 s (±20%), so the uncertainty in the distance estimate is about 20% of 1.7 km or approximately 300 m. This is probably much greater than the error made by assuming v is constant. Speed, Displacement, and Velocity 42 • (a) An electron in a television tube travels the 16-cm distance from the grid to the screen at an average speed of 4.0 × 107 m/s. How long does the trip take? (b) An electron in a current-carrying wire travels at an average speed of 4.0 × 10–5 m/s. How long does it take to travel 16 cm? Picture the Problem Think of the electron as traveling in a straight line at constant speed and use the definition of average speed. (a) Using its definition, express the average speed of the electron: t s Δ Δ = = flightoftime traveleddistance speedAverage Solve for and evaluate the time of flight: ns0.4s100.4 sm100.4 m16.0 speedAverage Δ Δ 9 7 =×= × == − s t (b) Calculate the time of flight for an electron in a 16-cm long current carrying wire similarly. min67 s60 min1 s100.4 sm100.4 m16.0 speedAverage Δ Δ 3 5 =××= × == − s t 43 • [SSM] A runner runs 2.5 km, in a straight line, in 9.0 min and then takes 30 min to walk back to the starting point. (a) What is the runner’s average velocity for the first 9.0 min? (b) What is the average velocity for the time spent walking? (c) What is the average velocity for the whole trip? (d) What is the average speed for the whole trip?
Chapter 280 Picture the Problem In this problem the runner is traveling in a straight line but not at constant speed - first she runs, then she walks. Let’s choose a coordinate system in which her initial direction of motion is taken as the +x direction. (a) Using the definition of average velocity, calculate the average velocity for the first 9 min: min/km28.0 min0.9 km5.2 Δ Δ av === t x v (b) Using the definition of average velocity, calculate her average velocity for the 30 min spent walking: min/m83 min30 km5.2 Δ Δ av −= − == t x v (c) Express her average velocity for the whole trip: 0 Δ 0 Δ Δ tripround av === tt x v (d) Finally, express her average speed for the whole trip: min/km13.0 min0.9min30 )km5.2(2 timeelapsed traveleddistance speedav = + = = 44 • A car travels in a straight line with an average velocity of 80 km/h for 2.5 h and then with an average velocity of 40 km/h for 1.5 h. (a) What is the total displacement for the 4.0-h trip? (b) What is the average velocity for the total trip? Picture the Problem The car is traveling in a straight line but not at constant speed. Let the direction of motion be the +x direction. (a) The total displacement of the car for the entire trip is the sum of the displacements for the two legs of the trip: 21total xxx Δ+Δ=Δ Find the displacement for each leg of the trip: ( )( ) km200 h5.2km/h8011,1 = =Δ=Δ tvx av and ( )( ) km0.60 h5.1km/h4022,2 = =Δ=Δ tvx av Add the individual displacements to get the total displacement: m106.2 km0.60km200ΔΔΔ 5 21total ×= +=+= xxx
Motion in One Dimension 81 (b) As long as the car continues to move in the same direction, the average velocity for the total trip is given by: total total av Δ Δ t x v = Substitute numerical values and evaluate vav: hkm65 h5.1h5.2 m106.2 5 av = + × =v 45 • One busy air route across the Atlantic Ocean is about 5500 km. The now-retired Concord, a supersonic jet capable of flying at twice the speed of sound was used to travel such routes. (a) Roughly how long did it take for a one- way flight? (Use 343 m/s for the speed of sound.) (b) Compare this time to the time taken by a subsonic jet flying at 0.90 times the speed of sound. Picture the Problem However unlikely it may seem, imagine that both jets are flying in a straight line at constant speed and use the definition of average speed to find the flight times. (a) The time of flight is the ratio of the distance traveled to the speed of the supersonic jet. ( )( ) h2.2h23.2 s/h3600m/s3432 km5500 supersonic Atlantic supersonic == = = v s t (b) Express the ratio of the time for the trip at supersonic speed to the time for the trip at subsonic speed and simplify to obtain: supersonic subsonic subsonic Atlantic supersonic Atlantic subsonic supersonic v v v s v s t t == Substitute numerical values and evaluate the ratio of the flight times: ( )( ) ( )( ) 45.0 m/s3432 m/s34390.0 subsonic supersonic == t t 46 •• The speed of light, designated by the universally recognized symbol c, has a value, to two significant figures, of 3.0 × 108 m/s. (a) How long does it take for light to travel from the Sun to Earth, a distance of 1.5 × 1011 m? (b) How long does it take light to travel from the Moon to Earth, a distance of 3.8 × 108 m? Picture the Problem In free space, light travels in a straight line at constant speed, c.
Chapter 282 (a) Using the definition of average speed, express the time Δt required for light to travel from the Sun to Earth: speedaverage Δ Δ s t = where Δs is the distance from the Sun to Earth. Substitute numerical values and evaluate Δt: min3.8 s105.0 m/s103.0 m101.5 Δ 2 8 11 ≈ ×= × × =t (b) Proceed as in (a) this time using the Moon-Earth distance: s3.1 m/s100.3 m108.3 Δ 8 8 ≈ × × =t 47 • [SSM] Proxima Centauri, the closest star to us besides our own sun, is 4.1 × 1013 km from Earth. From Zorg, a planet orbiting this star, a Gregor places an order at Tony’s Pizza in Hoboken, New Jersey, communicating via light signals. Tony’s fastest delivery craft travels at 1.00 × 10–4 c (see Problem 46). (a) How long does it take Gregor’s order to reach Tony’s Pizza? (b) How long does Gregor wait between sending the signal and receiving the pizza? If Tony’s has a "1000-years-or-it’s-free" delivery policy, does Gregor have to pay for the pizza? Picture the Problem In free space, light travels in a straight line at constant speed, c. We can use the definition of average speed to find the elapsed times called for in this problem. (a) Using the definition of average speed (equal here to the assumed constant speed of light), solve for the time Δt required to travel the distance to Proxima Centauri: lightofspeed traveleddistance Δ =t Substitute numerical values and evaluate Δt: y4.3 s1037.1 sm100.3 m101.4 Δ 8 8 16 = ×= × × =t
Motion in One Dimension 83 (b) The delivery time (Δttotal) is the sum of the time for the order to reach Hoboken and the travel time for the delivery craft to travel to Proxima Centauri: ( )( ) y103.4y103.4y3.4 s103.156 y1 sm100.31000.1 km101.4 33.4 ΔΔΔ 44 784 13 deliveredbe order to Hobokensent to beorder tototal ×≈×+= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ××× × += += − y ttt Because 4.3 × 104 y >> 1000 y, Gregor does not have to pay. 48 • A car making a 100-km journey travels 40 km/h for the first 50 km. How fast must it go during the second 50 km to average 50 km/h? Picture the Problem The time for the second 50 km is equal to the time for the entire journey less the time for the first 50 km. We can use this time to determine the average speed for the second 50 km interval from the definition of average speed. Using the definition of average speed, find the time required for the total journey: h0.2 hkm50 km100Δ av total ===Δ v x t Find the time required for the first 50 km: h25.1 hkm40 km50 km501st ==Δt Find the time remaining to travel the last 50 km: h0.75 h1.25h0.2km50st1totalkm502nd = −=−=Δ ttt Finally, use the time remaining to travel the last 50 km to determine the average speed over this distance: hkm67 h75.0 km50 Δ Δ km502nd km502nd km502ndav, = == t x v 49 •• Late in ice hockey games, the team that is losing sometimes "pulls" their goalkeeper off the ice to add an additional offensive player and increase their chances of scoring. In such cases, the goalie on the opposing team might have an opportunity to score into the unguarded net that is 55.0 m away. Suppose you are the goaltender for your university team and are in just such a situation. You launch a shot (in hopes of getting your first career goal) on the frictionless ice. You hear a disappointing ″clang″ as the puck strikes a goalpost (instead of going in!) exactly 2.50 s later. In this case, how fast did the puck travel? You should assume 343 m/s for the speed of sound.
Chapter 284 Picture the Problem The distance over which both the puck and the sound from the puck hitting the goalpost must travel is 55.0 m. The time between the shot being released and the sound reaching the goalie’s ear can be found by expressing the total elapsed time as the sum of the elapsed times for the shot to travel 55.0 m and for the sound to travel back to you. The total elapsed time as the sum of the elapsed times for the shot and for the sound to travel back to you : soundshottotal ΔΔΔ ttt += Express the time for the shot to travel to the other net a distance Δx away: shot shot Δ Δ v x t = Express the time for the sound to travel a distance Δx back to you: sound sound Δ Δ v x t = Substitute in the expression for totalΔt to obtain: soundshot total ΔΔ Δ v x v x t += Solving this equation for shotv yields: xtv xv v ΔΔ Δ totalsound sound shot − = Substitute numerical values and evaluate shotv : ( )( ) ( )( ) m/s5.23 m0.55s50.2m/s343 m0.55m/s343 shot = − =v 50 •• Cosmonaut Andrei, your co-worker at the International Space Station, tosses a banana at you with a speed of 15 m/s. At exactly the same instant, you fling a scoop of ice cream at Andrei along exactly the same path. The collision between banana and ice cream produces a banana split 7.2 m from your location 1.2 s after the banana and ice cream were launched. (a) How fast did you toss the ice cream? (b) How far were you from Andrei when you tossed the ice cream? (Neglect any effects due to gravity.) Picture the Problem Let the subscript b refer to the banana and the subscript ic refer to the ice cream. Then the distance covered by the ice cream before collision is given by Δxic = vicΔt and the distance covered by the banana is Δxb = vbΔt. The distance between you and Andrei is then the sum of these distances: Δxtot = Δxic + Δxb. (a) The speed of the ice cream is given by: t x v Δ Δ ic ic = where Δt is the time-to-collision.
Motion in One Dimension 85 Substitute numerical values and evaluate vic: m/s0.6 s2.1 m2.7 ic ==v (b) Express the distance between yourself and Andrei as the sum of the distances the ice cream and the banana travel: bictotal ΔΔΔ xxx += Because tvx Δ=Δ bb : tvxx ΔΔΔ bictotal += Substitute numerical values and evaluate totalΔx : ( )( ) m25s2.1m/s15m2.7Δ total =+=x 51 •• Figure 2-39 shows the position of a particle as a function of time. Find the average velocities for the time intervals a, b, c, and d indicated in the figure. Picture the Problem The average velocity in a time interval is defined as the displacement divided by the time elapsed; that is txv ΔΔ= /av . (a) Δxa = 0 0av =v (b) Δxb = 1 m and Δtb = 3 s m/s3.0av =v (c) Δxc = –6 m and Δtc = 3 s m/s2av −=v (d) Δxd = 3 m and Δtd = 3 s m/s1av =v 52 •• It has been found that, on average, galaxies are moving away from Earth at a speed that is proportional to their distance from Earth. This discovery is known as Hubble’s law, named for its discoverer, astrophysicist Sir Edwin Hubble. He found that the recessional speed v of a galaxy a distance r from Earth is given by v = Hr, where H = 1.58 × 10–18 s–1 is called the Hubble constant. What are the expected recessional speeds of galaxies (a) 5.00 × 1022 m from Earth and (b) 2.00 × 1025 m from Earth? (c) If the galaxies at each of these distances had traveled at their expected recessional speeds, how long ago would they have been at our location? Picture the Problem In free space, light travels in a straight line at constant speed c. We can use Hubble’s law to find the speed of the two planets. (a) Using Hubble’s law, calculate the speed of the first galaxy: ( )( ) m/s1090.7 s1058.1m1000.5 4 11822 a ×= ××= −− v
Chapter 286 (b) Using Hubble’s law, calculate the speed of the second galaxy: ( )( ) m/s1016.3 s1058.1m1000.2 7 11825 b ×= ××= −− v (c) Using the relationship between distance, speed, and time for both galaxies, express how long ago Δt they were both located at the same place as Earth: HrH r v r t 1 Δ === Substitute numerical values and evaluate Δt: y1020s1033.6Δ 917 ×≈×=t 53 •• [SSM] The cheetah can run as fast as 113 km/h, the falcon can fly as fast as 161 km/h, and the sailfish can swim as fast as 105 km/h. The three of them run a relay with each covering a distance L at maximum speed. What is the average speed of this relay team for the entire relay? Compare this average speed with the numerical average of the three individual speeds. Explain carefully why the average speed of the relay team is not equal to the numerical average of the three individual speeds. Picture the Problem We can find the average speed of the relay team from the definition of average speed. Using its definition, relate the average speed to the total distance traveled and the elapsed time: timeelapsed traveleddistance av =v Express the time required for each animal to travel a distance L: cheetah cheetah v L t = , falcon falcon v L t = and sailfish sailfish v L t = Express the total time Δt: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ++=Δ sailfishfalconcheetah 111 vvv Lt Use the total distance traveled by the relay team and the elapsed time to calculate the average speed: km/h122km/h03.122 km/h105 1 km/h161 1 km/h113 1 3 av == ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ++ = L L v
Motion in One Dimension 87 Calculating the average of the three speeds yields: av speedsthree 04.1 km/h126km/h33.126 3 km/h105km/h161km/h113 Average v= == ++ = The average speed of the relay team is not equal to the numerical average of the three individual speeds because the ″runners″ did not run for the same interval of time. The average speed would be equal to one-third the sum of the three speeds if the three speeds were each maintained for the same length of time instead of for the same distance. 54 •• Two cars are traveling along a straight road. Car A maintains a constant speed of 80 km/h and car B maintains a constant speed of 110 km/h. At t = 0, car B is 45 km behind car A. (a) How much farther will car A travel before car B overtakes it? (b) How much ahead of A will B be 30 s after it overtakes A? Picture the Problem Let the position of car A at t = 0 be the origin of our coordinate system. Then we can use a constant-acceleration equation to express the positions of both cars as functions of time and equate these expressions to determine the time at which car A is overtaken by car B. (a) Car B overtakes car A when their x coordinates are the same: ( ) ( )txtx BA = (1) Using a constant-acceleration equation with a = 0, express the position of car A as a function of time: ( ) tvxtx A0AA += where x0A is the position of car A at t = 0. Because we’ve let car A be at the origin at t = 0: ( ) tvtx AA = (2) Using a constant-acceleration equation, express the position of car B as a function of time: ( ) tvxtx BB0B += where x0B is the position of car B at t = 0. Substitute for xA(t) and xB(t) in equation (1) to obtain: tvxtv BB0A += ⇒ BA B0 vv x t − = Substitute numerical values and evaluate the time t at which car B overtakes car A: h50.1 km/h110km/h80 km45 = − − =t
Chapter 288 Now we can evaluate equation (2) at t = 1.50 h to obtain: ( ) ( ) m102.1 km120h50.1 h km 80h50.1 5 A ×= =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =x (b) The separation of the cars as a function of time is given by: ( ) ( ) ( ) tvtvxtxtxtx AB0BABΔ −−=−= Substitute numerical values and evaluate ( ) ( )hh50.1s30h50.1 120 1 +Δ=+Δ xx to obtain: ( ) ( ) km0.25hh50.1 h km 80 h km 110km45hh50.1 120 1 120 1 =+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+−=+Δx Remarks: One can use a graphing calculator or a spreadsheet program to solve this problem. A spreadsheet program was used to plot the following graph: 0 50 100 150 200 250 300 0 0.5 1 1.5 2 2.5 3 t, h x,km Car A Car B Note that this graph confirms our result that the cars are at the same location at t = 1.5 h. 55 •• [SSM] A car traveling at a constant speed of 20 m/s passes an intersection at time t = 0. A second car traveling at a constant speed of 30 m/s in the same direction passes the same intersection 5.0 s later. (a) Sketch the position functions x1(t) and x2(t) for the two cars for the interval 0 ≤ t ≤ 20 s. (b) Determine when the second car will overtake the first. (c) How far from the intersection will the two cars be when they pull even? (d) Where is the first car when the second car passes the intersection? Picture the Problem One way to solve this problem is by using a graphing calculator to plot the positions of each car as a function of time. Plotting these positions as functions of time allows us to visualize the motion of the two cars
Motion in One Dimension 89 relative to the (fixed) ground. More importantly, it allows us to see the motion of the two cars relative to each other. We can, for example, tell how far apart the cars are at any given time by determining the length of a vertical line segment from one curve to the other. (a) Letting the origin of our coordinate system be at the intersection, the position of the slower car, x1(t), is given by: x1(t) = 20t where x1 is in meters if t is in seconds. Because the faster car is also moving at a constant speed, we know that the position of this car is given by a function of the form: x2(t) = 30t + b We know that when t = 5.0 s, this second car is at the intersection (that is, x2(5.0 s) = 0). Using this information, you can convince yourself that: b = −150 m Thus, the position of the faster car is given by: ( ) 150302 −= ttx One can use a graphing calculator, graphing paper, or a spreadsheet to obtain the following graphs of x1(t) (the solid line) and x2(t) (the dashed line): 0 50 100 150 200 250 300 350 400 450 0 5 10 15 20 t, s x,m (b) Use the time coordinate of the intersection of the two lines to determine the time at which the second car overtakes the first: From the intersection of the two lines, one can see that the second car will "overtake" (catch up to) the first car at s.15=t
Chapter 290 (c) Use the position coordinate of the intersection of the two lines to determine the distance from the intersection at which the second car catches up to the first car: From the intersection of the two lines, one can see that the distance from the intersection is m.300 (d) Draw a vertical line from t = 5 s to the solid line and then read the position coordinate of the intersection of the vertical line and the solid line to determine the position of the first car when the second car went through the intersection. From the graph, when the second car passes the intersection, the first car was ahead.m100 56 •• Bats use echolocation to determine their distance from objects they cannot easily see in the dark. The time between the emission of high-frequency sound pulse (a click) and the detection of its echo is used to determine such distances. A bat, flying at a constant speed of 19.5 m/s in a straight line toward a vertical cave wall, makes a single clicking noise and hears the echo 0.15 s later. Assuming that she continued flying at her original speed, how close was she to the wall when she received the echo? Assume a speed of 343 m/s for the speed of sound. Picture the Problem The sound emitted by the bat travels at soundv and during the time interval Δt during which the sound travels to the wall (a distance L from the bat’s initial position) and back to the bat, the bat travels a distance of tvx Δ=Δ batbat , where batv is the bat’s flying speed. The distance of the bat from the wall when she received the echo of her click is tvLx Δ−=Δ bataway where batto backwallto ΔΔΔ ttt += . tv Δbat tvL Δ− bat L Express the distance of the bat from the wall of the cave when it hears the echo of its click: tvLx Δ−=Δ bataway (1) The elapsed time between the bat clicking and hearing the sound is: batto backwallto ΔΔΔ ttt +=
Motion in One Dimension 91 Substituting for walltoΔt and batto backΔt gives: sound bat sound Δ v tvL v L t Δ− += Solving for L yields: ( ) tvvL Δ+= batsound2 1 Substitute for L in equation (1) and simplify to obtain: ( ) ( ) tvv tvtvvx Δ−= Δ−Δ+=Δ batsound2 1 batbatsound2 1 away Substitute numerical values and evaluate awayΔx : ( )( ) m24 s15.0m/s5.19m/s343Δ 2 1 away = −=x 57 ••• A submarine can use sonar (sound traveling through water) to determine its distance from other objects. The time between the emission of a sound pulse (a ″ping″) and the detection of its echo can be used to determine such distances. Alternatively, by measuring the time between successive echo receptions of a regularly timed set of pings, the submarine's speed may be determined by comparing the time between echoes to the time between pings. Assume you are the sonar operator in a submarine traveling at a constant velocity underwater. Your boat is in the eastern Mediterranean Sea, where the speed of sound is known to be 1522 m/s. If you send out pings every 2.000 s, and your apparatus receives echoes reflected from an undersea cliff every 1.980 s, how fast is your submarine traveling? Picture the Problem Both the pulses sent out by the submarine and the pulses returning from the sea-wall are traveling at 1522 m/s. Consequently, we can determine the distance in water between two successive echo (or emitted) pulses of sound which were emitted with a time interval Δtemitted between them. The actual distance in the seawater between the echoed pulses is given by Δx = vsoundΔtemitted. We need to find the time Δtreceived between successive pulses received by the submarine. We start our ″clock″, as it were, when the submarine passes one of two successive pulses that approach it, separated by the distance Δx. After passing the first pulse, the next sound pulse moves toward the submarine at vsound and the submarine moves toward the pulse at speed vsub. The distance between successive pulses Δx may be divided into Δxsub and Δxsound, which are equal to vsubΔtreceived and vsubΔtreceived, respectively. The distance between successive pulses is given by: receivedsub ΔΔΔ xxx += Substituting for all three terms in this equation yields: receivedsoundreceivedsubemittedsound ΔΔΔ tvtvtv +=
Chapter 292 Solve for subv to obtain: ( ) received receivedemittedsound sub Δ ΔΔ t ttv v − = Substitute numerical values and evaluate subv : ( )( ) m/s15 s.0002 s1.980s000.2m/s1522 sub = − =v Acceleration 58 • A sports car accelerates in third gear from 48.3 km/h (about 30 mi/h) to 80.5 km/h (about 50 mi/h) in 3.70 s. (a) What is the average acceleration of this car in m/s2 ? (b) If the car maintained this acceleration, how fast would it be moving one second later? Picture the Problem In Part (a), we can apply the definition of average acceleration to find aav. In Part (b), we can find the change in the car’s velocity in one second and add this change to its velocity at the beginning of the interval to find its speed one second later. (a) The definition of average acceleration is: t v a Δ Δ av = Substitute numerical values and evaluate aav: sh km 8.70 s3.70 km/h48.3km/h80.5 av ⋅ = − =a Convert aav to m/s2 : 2 3 av m/s42.2 s3600 h1 sh m 108.70 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ×=a (b) Express the speed of the car at the end of 4.7 s: ( ) ( ) s1 s1 Δkm/h5.80 Δs70.3s7.4 v vvv += += Find the change in the speed of the car in 1.00 s: ( ) km/h8.70 s1.00 sh km 8.70ΔΔ av = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ == tav Substitute and evaluate v(4.7 s): ( ) km/h2.89 km/h7.8km/h5.80s7.4 = +=v
Motion in One Dimension 93 59 • [SSM] An object is moving along the x axis. At t = 5.0 s, the object is at x = +3.0 m and has a velocity of +5.0 m/s. At t = 8.0 s, it is at x = +9.0 m and its velocity is –1.0 m/s. Find its average acceleration during the time interval 5.0 s ≤ t ≤ 8.0 s. Picture the Problem We can find the change in velocity and the elapsed time from the given information and then use the definition of average acceleration. The average acceleration is defined as the change in velocity divided by the change in time: t v a Δ Δ av = Substitute numerical values and evaluate aav: ( ) ( ) ( ) ( ) 2 av m/s0.2 s0.5s0.8 m/s0.5m/s0.1 −= − −− =a 60 •• A particle moves along the x axis with velocity vx = (8.0 m/s2 ) t – 7.0 m/s. (a) Find the average acceleration for two different one-second intervals, one beginning at t = 3.0 s and the other beginning at t = 4.0 s. (b) Sketch vx versus t over the interval 0 < t < 10 s. (c) How do the instantaneous accelerations at the middle of each of the two time intervals specified in Part (a) compare to the average accelerations found in Part (a)? Explain. Picture the Problem The important concept here is the difference between average acceleration and instantaneous acceleration. (a) The average acceleration is defined as the change in velocity divided by the change in time: t v a Δ Δ av = Determine vx at t = 3.0 s, t = 4.0 s, and t = 5.0 s: ( ) ( )( ) m/s17 m/s0.7s0.3m/s0.8s0.3 2 = −=xv ( ) ( )( ) m/s25 m/s0.7s0.4m/s0.8s0.4 2 = −=xv ( ) ( )( ) m/s33 m/s0.7s0.5m/s0.8s0.5 2 = −=xv
Chapter 294 Find aav for the two 1-s intervals: 2 s4.0tos3.0av, m/s0.8 s1.0 m/s17m/s25 = − =a and 2 s5.0tos4.0av, m/s0.8 s1.0 m/s25m/s33 = − =a The instantaneous acceleration is defined as the time derivative of the velocity or the slope of the velocity- versus-time curve: ( )[ ] 2 2 m/s0.8 m/s0.7m/s0.8 = −== t dt d dt dv a x x (b) The given function and a spreadsheet program were used to plot the following graph of v-versus-t: -10 -5 0 5 10 15 20 25 30 35 0 1 2 3 4 5 t, s v,m/s (c) Because the particle’s speed varies linearly with time, these accelerations are the same. 61 •• [SSM] The position of a certain particle depends on time according to the equation x(t) = t2 – 5.0t + 1.0, where x is in meters if t is in seconds. (a) Find the displacement and average velocity for the interval 3.0 s ≤ t ≤ 4.0 s. (b) Find the general formula for the displacement for the time interval from t to t + Δt. (c) Use the limiting process to obtain the instantaneous velocity for any time t. Picture the Problem We can closely approximate the instantaneous velocity by the average velocity in the limit as the time interval of the average becomes small. This is important because all we can ever obtain from any measurement is the average velocity, vav, which we use to approximate the instantaneous velocity v.
Motion in One Dimension 95 (a) The displacement of the particle during the interval 3.0 s ≤ t ≤ 4.0 s is given by: ( ) ( )s0.3s0.4Δ xxx −= (1) The average velocity is given by: t x v Δ Δ av = (2) Find x(4.0 s) and x(3.0 s): x(4.0 s) = (4.0)2 – 5(4.0) + 1 = –3.0 m and x(3.0 s) = (3.0)2 – 5(3.0) + 1 = −5.0 m Substitute numerical values in equation (1) and evaluate Δx: ( ) ( ) m0.2m0.5m0.3Δ =−−−=x Substitute numerical values in equation (2) and evaluate vav: m/s0.2 s1.0 m0.2 av ==v (b) Find x(t + Δt): x(t + Δt) = (t + Δt)2 − 5(t + Δt) + 1 = (t2 + 2tΔt + (Δt)2 ) – 5(t + Δt) + 1 Express x(t + Δt) – x(t) = Δx: ( ) ( )2 52 tttx Δ+Δ−=Δ where Δx is in meters if t is in seconds. (c) From (b) find Δx/Δt as Δt → 0: ( ) ( ) tt t ttt t x Δ+−= Δ Δ+Δ− = Δ Δ 52 52 2 and ( ) 52Δ/Δlim 0Δ −== → ttxv t where v is in m/s if t is in seconds. Alternatively, we can take the derivative of x(t) with respect to time to obtain the instantaneous velocity. ( ) ( ) ( ) 52 2 12 −= += ++== t bat btat dt d dt tdx tv 62 •• The position of an object as a function of time is given by x = At2 – Bt + C, where A = 8.0 m/s2 , B = 6.0 m/s, and C = 4.0 m. Find the instantaneous velocity and acceleration as functions of time. Picture the Problem The instantaneous velocity is dtdx and the acceleration is dtdv .
Chapter 296 Using the definitions of instantaneous velocity and acceleration, determine v and a: [ ] BAtCBtAt dt d dt dx v −=+−== 22 and [ ] ABAt dt d dt dv a 22 =−== Substitute numerical values for A and B and evaluate v and a: ( ) ( ) m/s0.6m/s16 m/s0.6m/s0.82 2 2 −= −= t tv and ( ) 22 m/s16m/s0.82 ==a 63 ••• The one-dimensional motion of a particle is plotted in Figure 2-40. (a) What is the average acceleration in each of the intervals AB, BC, and CE? (b) How far is the particle from its starting point after 10 s? (c) Sketch the displacement of the particle as a function of time; label the instants A, B, C, D, and E on your graph. (d) At what time is the particle traveling most slowly? Picture the Problem We can use the definition of average acceleration (aav = Δv/Δt) to find aav for the three intervals of constant acceleration shown on the graph. (a) Using the definition of average acceleration, find aav for the interval AB: 2 ABav, m/s3.3 s0.3 m/s0.5m/s0.15 = − =a Find aav for the interval BC: 0 s3.0 m/s15.0m/s15.0 BCav, = − =a Find aav for the interval CE: 2 CEav, m/s5.7 s4.0 m/s15.0m/s15.0 −= −− =a (b) Use the formulas for the areas of trapezoids and triangles to find the area under the graph of v as a function of t. ( ) ( ) ( ) ( ) ( )( ) m75 s)m/s)(2.015.0( )sm/s)(2.00.15(s)m/s)(3.0(15.0s3.0m/s15.0m/s5.0 ΔΔΔΔΔ 2 1 2 1 2 1 EDDCCBBA = −+ +++= +++= →→→→ xxxxx
Motion in One Dimension 97 (c) The graph of displacement, x, as a function of time, t, is shown in the following figure. In the region from B to C the velocity is constant so the x- versus-t curve is a straight line. 0 20 40 60 80 100 0 2 4 6 8 10 t, s x,m (d) Reading directly from the figure, we can find the time when the particle is moving the slowest. At point D, t = 8 s, the graph crosses the time axis; therefore v = 0. Constant Acceleration and Free-Fall 64 • An object projected vertically upward with initial speed v0 attains a maximum height h above its launch point. Another object projected up with initial speed 2v0 from the same height will attain a maximum height of (a) 4h, (b) 3h, (c) 2h, (d) h. (Air resistance is negligible.) Picture the Problem Because the acceleration is constant (–g) we can use a constant-acceleration equation to find the height of the projectile. Using a constant-acceleration equation, express the height of the object as a function of its initial speed, the acceleration due to gravity, and its displacement: ygvv Δ−= 22 0 2 or, because v(h) = 0, ghv 20 2 0 −= ⇒ 2 2 0 g v h = Express the ratio of the maximum height of the second object to that of the first object and simplify to obtain: ( ) ( ) 4 2 2 2 2 0 2 0 object1st object2nd == g v g v h h
Chapter 298 Solving for object2ndh yields: hh 4object2nd = ⇒( )a is correct. 65 •• A car traveling along the x axis starts from rest at x = 50 m and accelerates at a constant rate of 8.0 m/s2 . (a) How fast is it going after 10 s? (b) How far has it gone after 10 s? (c) What is its average velocity for the interval 0 ≤ t ≤ 10 s? Picture the Problem Because the acceleration of the car is constant we can use constant-acceleration equations to describe its motion. (a) Using a constant-acceleration equation, relate the velocity to the acceleration and the time: atvv += 0 Substitute numerical values and evaluate v: ( ) sm80s10 s m 0.80 2 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=v (b) Using a constant-acceleration equation, relate the displacement to the acceleration and the time: 2 00 2 t a tvxxx +=−=Δ Substitute numerical values and evaluate Δx: ( ) km40.0s10 s m 0.8 2 1 Δ 2 2 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =x (c) Use the definition of avv : m/s40 s10 m400 Δ Δ av === t x v Remarks: Because the area under a velocity-versus-time graph is the displacement of the object, we could solve this problem graphically. 66 • An object traveling along the x axis with an initial velocity of +5.0 m/s has a constant acceleration of +2.0 m/s2 . When its speed is 15 m/s, how far has it traveled? Picture the Problem Because the acceleration of the object is constant we can use constant-acceleration equations to describe its motion. Using a constant-acceleration equation, relate the speed of the object to its acceleration and displacement: xavv Δ+= 22 0 2 ⇒ a vv x 2 Δ 2 0 2 − = Substitute numerical values and evaluate Δx: ( ) ( ) m50 sm2.02 sm0.515 Δ 2 2222 = − =x
Motion in One Dimension 99 67 • [SSM] An object traveling along the x axis at constant acceleration has a velocity of +10 m/s when it is at x = 6.0 m and of +15 m/s when it is at x = 10 m. What is its acceleration? Picture the Problem Because the acceleration of the object is constant we can use constant-acceleration equations to describe its motion. Using a constant-acceleration equation, relate the velocity to the acceleration and the displacement: xavv Δ+= 22 0 2 ⇒ x vv a Δ − = 2 2 0 2 Substitute numerical values and evaluate a: ( ) ( ) 2 2222 sm16 m6.0m012 sm1015 = − − =a 68 • The speed of an object traveling along the x axis increases at the constant rate of +4.0 m/s each second. At t = 0.0 s, its velocity is +1.0 m/s and its position is +7.0 m. How fast is it moving when its position is +8.0 m, and how much time has elapsed from the start at t = 0.0 s? Picture the Problem Because the acceleration of the object is constant we can use constant-acceleration equations to describe its motion. Using a constant-acceleration equation, relate the velocity to the acceleration and the displacement: xavv Δ+= 22 0 2 ⇒ xavv Δ22 0 += Substitute numerical values and evaluate v to obtain: ( ) ( )( ) m/s0.3m7.0m0.8sm0.42sm0.1 22 =−+=v From the definition of average acceleration we have: av Δ Δ a v t = Substitute numerical values and evaluate Δt: s50.0 sm0.4 sm1.0sm0.3 Δ 2 = − =t 69 •• A ball is launched directly upward from ground level with an initial speed of 20 m/s. (Air resistance is negligible.) (a) How long is the ball in the air? (b) What is the greatest height reached by the ball? (c) How many seconds after launch is the ball 15 m above the release point? Picture the Problem In the absence of air resistance, the ball experiences constant acceleration. Choose a coordinate system with the origin at the point of release and the positive direction upward.
Chapter 2100 (a) Using a constant-acceleration equation, relate the displacement of the ball to the acceleration and the time: 2 2 1 0 attvy +=Δ Setting Δy = 0 (the displacement for a round trip), solve for the time required for the ball to return to its starting position: 2 trip round2 1 roundtrip00 attv += and g v t 0 trip round 2 = Substitute numerical values and evaluate trip roundt : ( ) s1.4 m/s9.81 m/s202 2 trip round ==t (b) Using a constant-acceleration equation, relate the final speed of the ball to its initial speed, the acceleration, and its displacement: yavv Δ+= 22 0 2 top or, because vtop = 0 and a = −g, ( )Hgv −+= 20 2 0 ⇒ g v H 2 2 0 = Substitute numerical values and evaluate H: ( ) ( ) m20 sm81.92 sm20 2 2 ==H (c) Using the same constant- acceleration equation with which we began part (a), express the displacement as a function of time: 2 2 1 0 attvy +=Δ Substitute numerical values to obtain: ( ) 2 2 2 m/s81.9 m/s20m15 tt ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= Use your graphing calculator or the quadratic formula to solve this equation for the times at which the displacement of the ball is 15 m: The solutions are s99.0=t (this corresponds to passing 15 m on the way up) and s1.3=t (this corresponds to passing 15 m on the way down). 70 •• In the Blackhawk landslide in California, a mass of rock and mud fell 460 m down a mountain and then traveled 8.00 km across a level plain. It has been theorized that the rock and mud moved on a cushion of water vapor. Assume that the mass dropped with the free-fall acceleration and then slid horizontally, losing speed at a constant rate. (a) How long did the mud take to drop the 460 m? (b) How fast was it traveling when it reached the bottom? (c) How long did the mud take to slide the 8.00 km horizontally?