1. 1 CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model-Building 1.3 Density and Atomic Mass 1.4 Dimensional Analysis 1.5 Conversion of Units 1.6 Estimates and Order-of- Magnitude Calculations 1.7 Significant Figures Physics and Measurement ANSWERS TO QUESTIONS Q1.1 Atomic clocks are based on electromagnetic waves which atoms emit. Also, pulsars are highly regular astronomical clocks. Q1.2 Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard. Q1.3 People have different size hands. Defining the unit precisely would be cumbersome. Q1.4 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms Q1.5 (b) and (d). You cannot add or subtract quantities of different dimension. Q1.6 A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. If an equation is not dimensionally correct, it cannot be correct. Q1.7 If I were a runner, I might walk or run 101 miles per day. Since I am a college professor, I walk about 100 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on vacation. Q1.8 On February 7, 2001, I am 55 years and 39 days old. 55 365 25 1 39 20 128 86 400 1 1 74 10 109 9 yr d yr d d s d s s . . ~ F HG I KJ+ = F HG I KJ = . Many college students are just approaching 1 Gs. Q1.9 Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10. Q1.10 The mass of the forty-six chapter textbook is on the order of 100 kg . Q1.11 With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr. 1

2. 2 Physics and Measurement SOLUTIONS TO PROBLEMS Section 1.1 Standards of Length, Mass, and Time No problems in this section Section 1.2 Matter and Model-Building P1.1 From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, L L Ldiag = +2 2 . Thus, since the atoms are separated by a distance L = 0 200. nm , the diagonal planes are separated by 1 2 0 1412 2 L L+ = . nm . Section 1.3 Density and Atomic Mass *P1.2 Modeling the Earth as a sphere, we find its volume as 4 3 4 3 6 37 10 1 08 103 6 3 21 3 r = = . .m me j . Its density is then = = = m V 5 98 10 1 08 10 5 52 10 24 21 3 3 3. . . kg m kg m . This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 3 kg m . The average density of the Earth is significantly higher, so higher-density material must be down below the surface. P1.3 With V = base area heighta fb g V r h= 2 e j and = m V , we have = = F HG I KJ = m r h2 2 9 4 3 1 19 5 39 0 10 1 2 15 10 kg mm mm mm m kg m 3 3 . . . . a f a f *P1.4 Let V represent the volume of the model, the same in = m V for both. Then iron kg= 9 35. V and gold gold = m V . Next, gold iron gold kg = m 9 35. and mgold 3 3 3 kg 19.3 10 kg / m kg / m kg= F HG I KJ =9 35 7 86 10 23 03 . . . . P1.5 V V V r ro i= = 4 3 2 3 1 3 e j = m V , so m V r r r r = = F HG I KJ = 4 3 4 3 2 3 1 3 2 3 1 3 e j e j .

3. Chapter 1 3 P1.6 For either sphere the volume is V r= 4 3 3 and the mass is m V r= = 4 3 3 . We divide this equation for the larger sphere by the same equation for the smaller: m m r r r rs s s = = = 4 3 4 3 5 3 3 3 3 . Then r rs= = =5 4 50 1 71 7 693 . . .cm cma f . P1.7 Use 1 u . g= 1 66 10 24 . (a) For He, m0 24 4 00 6 64 10= F HG I KJ = . .u 1.66 10 g 1 u g -24 . (b) For Fe, m0 23 55 9 9 29 10= F HG I KJ = . .u 1.66 10 g 1 u g -24 . (c) For Pb, m0 24 22 207 1 66 10 3 44 10= F HG I KJ = u g 1 u g . . . *P1.8 (a) The mass of any sample is the number of atoms in the sample times the mass m0 of one atom: m Nm= 0. The first assertion is that the mass of one aluminum atom is m0 27 26 27 0 27 0 1 66 10 1 4 48 10= = = . . . .u u kg u kg . Then the mass of 6 02 1023 . atoms is m Nm= = = = 0 23 26 6 02 10 4 48 10 0 027 0 27 0. . . .kg kg g . Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be written m Nm= 0. 0 027 0 6 02 1023 0. .kg = m , so m0 23 260 027 6 02 10 4 48 10= = . . . kg kg , in agreement with the first assertion. (b) The general equation m Nm= 0 applied to one mole of any substance gives M NMg u= , where M is the numerical value of the atomic mass. It divides out exactly for all substances, giving 1 000 000 0 10 1 660 540 2 103 27 . . = kg kgN . With eight-digit data, we can be quite sure of the result to seven digits. For one mole the number of atoms is N = F HG I KJ = +1 1 660 540 2 10 6 022 137 103 27 23 . . . (c) The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one molecule of H O2 is 2 1 008 0 15 999 18 0. . .b g+ =u u. Then the molar mass is 18 0. g . (d) For CO2 we have 12 011 2 15 999 44 0. . .g g g+ =b g as the mass of one mole.

4. 4 Physics and Measurement P1.9 Mass of gold abraded: m = = = F HG I KJ = 3 80 3 35 0 45 0 45 1 4 5 10 4 . . . . .g g g g kg 10 g kg3b g . Each atom has mass m0 27 25 197 197 1 66 10 1 3 27 10= = F HG I KJ = u u kg u kg . . . Now, m N m= 0 , and the number of atoms missing is N m m = = = 0 4 25 214 5 10 3 27 10 1 38 10 . . . kg kg atoms. The rate of loss is N t N t = F HG I KJF HG I KJF HG I KJF HG I KJ = 1 38 10 50 1 1 1 1 8 72 10 21 11 . . . atoms yr yr 365.25 d d 24 h h 60 min min 60 s atoms s P1.10 (a) m L= = = = 3 3 6 3 16 19 7 86 5 00 10 9 83 10 9 83 10. g cm cm g kge je j. . . (b) N m m = = = 0 19 27 79 83 10 55 9 1 66 10 1 06 10 . . . . kg u kg 1 u atoms e j P1.11 (a) The cross-sectional area is A = + = 2 0 150 0 010 0 340 0 010 6 40 10 3 . . . . . . m m m m m2 a fa f a fa f. The volume of the beam is V AL= = = 6 40 10 1 50 9 60 103 3 . . .m m m2 3 e ja f . Thus, its mass is m V= = = 7 56 10 9 60 10 72 63 3 . . .kg / m m kg3 3 e je j . FIG. P1.11 (b) The mass of one typical atom is m0 27 26 55 9 1 66 10 1 9 28 10= F HG I KJ = . . .u kg u kga f . Now m Nm= 0 and the number of atoms is N m m = = = 0 26 2672 6 9 28 10 7 82 10 . . . kg kg atoms .

5. Chapter 1 5 P1.12 (a) The mass of one molecule is m0 27 26 18 0 1 66 10 2 99 10= F HG I KJ = . . .u kg 1 u kg . The number of molecules in the pail is N m m pail kg 2.99 kg molecules= = = 0 26 251 20 10 4 02 10 . . . (b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere. N N m M both pail pail total 25 (4.02 10 molecules) kg kg = F HG I KJ = F HG I KJ 1 20 1 32 1021 . . , or Nboth molecules= 3 65 104 . . Section 1.4 Dimensional Analysis P1.13 The term x has dimensions of L, a has dimensions of LT2 , and t has dimensions of T. Therefore, the equation x ka tm n = has dimensions of L LT T= 2 e j a fm n or L T L T1 0 2 = m n m . The powers of L and T must be the same on each side of the equation. Therefore, L L1 = m and m = 1 . Likewise, equating terms in T, we see that n m 2 must equal 0. Thus, n = 2 . The value of k, a dimensionless constant, cannot be obtained by dimensional analysis . *P1.14 (a) Circumference has dimensions of L. (b) Volume has dimensions of L3 . (c) Area has dimensions of L2 . Expression (i) has dimension L L L2 1 2 2 e j / = , so this must be area (c). Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L L L2 3 e j= , so it is (b). Thus, (a) ii; (b) iii, (c) i= = = .

6. 6 Physics and Measurement P1.15 (a) This is incorrect since the units of ax are m s2 2 , while the units of v are m s. (b) This is correct since the units of y are m, and cos kxa f is dimensionless if k is in m1 . *P1.16 (a) a F m or a k F m = represents the proportionality of acceleration to resultant force and the inverse proportionality of acceleration to mass. If k has no dimensions, we have a k F m = , L T 1 F M2 = , F M L T2 = . (b) In units, M L T kg m s2 2 = , so 1 1newton kg m s2 = . P1.17 Inserting the proper units for everything except G, kg m s kg m 2 L NM O QP= G 2 2 . Multiply both sides by m 2 and divide by kg 2 ; the units of G are m kg s 3 2 . Section 1.5 Conversion of Units *P1.18 Each of the four walls has area 8 00 12 0 96 0. . .ft ft ft2 a fa f= . Together, they have area 4 96 0 1 3 28 35 72 2 . .ft m . ft m2 e jF HG I KJ = . P1.19 Apply the following conversion factors: 1 2 54in cm= . , 1 86 400d s= , 100 1cm m= , and 10 19 nm m= 1 32 2 54 10 10 9 19 2 9 in day cm in m cm nm m 86 400 s day nm s F HG I KJ = . . b ge je j . This means the proteins are assembled at a rate of many layers of atoms each second! *P1.20 8 50 8 50 0 025 4 1 39 10 3 4 . . . .in in m 1 in m3 3 3 = F HG I KJ =

7. Chapter 1 7 P1.21 Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A m m m2 =30 50 1 500a fa f . Categorize: We model the lot as a perfect rectangle to use Area = Length Width. Use the conversion: 1 m 3.281 ft= . Analyze: A LW= = F HG I KJ F HG I KJ = 100 1 3 281 150 1 3 281 1 39 103 ft m ft ft m ft = 1 390 m m2 2 a f a f. . . . Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m2 . Unit conversion is a common technique that is applied to many problems. P1.22 (a) V = = 40.0 m 20.0 m 12.0 m . m3 a fa fa f 9 60 103 V = = 9 60 10 3 39 103 5 3 . m 3.28 ft 1 m ft3 3 b g . (b) The mass of the air is m V= = = air 3 3 kg m 9.60 10 m . kg1 20 1 15 103 4 .e je j . The student must look up weight in the index to find F mgg = = = 1.15 10 kg 9.80 m s 1.13 10 N4 2 5 e je j . Converting to pounds, Fg = = 1 13 10 2 54 105 4 . N 1 lb 4.45 N lbe jb g . . P1.23 (a) Seven minutes is 420 seconds, so the rate is r = = 30 0 420 7 14 10 2. . gal s gal s . (b) Converting gallons first to liters, then to m3 , r r = F HG I KJ F HG I KJ = 7 14 10 3 786 10 2 70 10 2 3 4 . . . . gal s L 1 gal m 1 L m s 3 3 e j (c) At that rate, to fill a 1-m3 tank would take t = F HG I KJ F HG I KJ = 1 2 70 10 1 1 034 m m s h 3 600 h 3 3 . . .

8. 8 Physics and Measurement *P1.24 (a) Length of Mammoth Cave = F HG I KJ = = = 348 1 609 1 560 5 60 10 5 60 105 7 mi km mi km m cm . . . . (b) Height of Ribbon Falls = F HG I KJ = = = 1 612 0 1 491 m 0 491 4 91 104 ft .304 8 m ft km cm. . . (c) Height of Denali = F HG I KJ = = = 20 320 0 1 6 6 19 10 6 19 103 5 ft .304 8 m ft .19 km m cm. . . (d) Depth of Kings Canyon = F HG I KJ = = = 8 200 0 1 2 2 50 10 2 50 103 5 ft .304 8 m ft .50 km m cm. . . P1.25 From Table 1.5, the density of lead is 1 13 104 . kg m3 , so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks. Density is defined as mass per volume, in = m V . We must convert to SI units in the calculation. = F HG I KJF HG I KJ = 23 94 2 10 1 1 000 100 1 1 14 103 3 4. . g cm kg g cm m . kg m3 At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important common-sense check on density values is that objects which sink in water must have a density greater than 1 g cm3 , and objects that float must be less dense than water. P1.26 It is often useful to remember that the 1 600-m race at track and field events is approximately 1 mile in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to 1 1 640 1 609 4 05 10 2 3 acre mi acres m mi m 2 2 a fF HG I KJF HG I KJ = . . *P1.27 The weight flow rate is 1 200 2 000 1 1 667 ton h lb ton h 60 min min 60 s lb s F HG I KJF HG I KJF HG I KJ = . P1.28 1 1 609 1 609mi m km= = . ; thus, to go from mph to km h, multiply by 1.609. (a) 1 1 609mi h km h= . (b) 55 88 5mi h km h= . (c) 65 104 6mi h km h= . . Thus, v = 16 1. km h .

9. Chapter 1 9 P1.29 (a) 6 10 1 1 1 190 12 F HG I KJ F HG I KJF HG I KJF HG I KJ = $ 1 000 $ s h 3 600 s day 24 h yr 365 days years (b) The circumference of the Earth at the equator is 2 6 378 10 4 01 103 7 . . = m me j . The length of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9 30 1011 . m. Thus, the 6 trillion dollars would encircle the Earth 9 30 10 2 32 10 11 4. . = m 4.01 0 m times7 . P1.30 N m m atoms Sun atom kg 1.67 kg atoms= = = 1 99 10 10 1 19 10 30 27 57. . P1.31 V At= so t V A = = = 3 78 10 25 0 1 51 10 151 3 4. . . m m m or m 3 2 b g P1.32 V Bh= = = 1 3 13 0 43 560 3 481 9 08 107 . . , acres ft acre ft ft 2 3 a fe j a f or V = F HG I KJ = 9 08 10 2 83 10 1 2 57 10 7 2 6 . . . ft m ft m 3 3 3 3 e j B h B h FIG. P1.32 P1.33 Fg = = 2 50 2 00 10 2 000 1 00 106 10 . . .tons block blocks lb ton lbsb ge jb g *P1.34 The area covered by water is A A Rw = = = = 0 70 3 6 102 14 . 0.70 4 0.70 4 6.37 10 m . mEarth Earth 6 2 2 a fe j a fa fe j . The average depth of the water is d = = 2.3 miles 1 609 m l mile . ma fb g 3 7 103 . The volume of the water is V A dw= = = 3 6 10 3 7 10 1 3 1014 2 3 18 3 . m . m . me je j and the mass is m V= = = 1 000 1 3 10 1 3 103 18 3 21 kg m . m kge je j . .

10. 10 Physics and Measurement P1.35 (a) d d d d nucleus, scale nucleus, real atom, scale atom, real m ft 1.06 10 m ft= F HG I KJ = F HG I KJ = 2 40 10 300 6 79 1015 10 3 . .e j , or dnucleus, scale ft mm 1 ft mm= = 6 79 10 304 8 2 073 . . .e jb g (b) V V r r d d r r atom nucleus atom nucleus atom nucleus atom nucleus m m times as large = = F HG I KJ = F HG I KJ = F HG I KJ = 4 3 4 3 3 3 10 15 3 13 3 3 1 06 10 2 40 10 8 62 10 . . . *P1.36 scale distance between real distance scale factor km m m km= F HG I KJF HG I KJ = F HG I KJ = 4 0 10 7 0 10 1 4 10 20013 3 9 . . . e j P1.37 The scale factor used in the dinner plate model is S = = 0 25 1 0 10 2 105 6. . m lightyears .5 m lightyears . The distance to Andromeda in the scale model will be D D Sscale actual 6 6 2.0 10 lightyears 2.5 10 m lightyears m= = = e je j 5 0. . P1.38 (a) A A r r r r Earth Moon Earth Moon 2 Earth Moon m cm m cm = = F HG I KJ = F H GG I K JJ = 4 4 6 37 10 100 1 74 10 13 4 2 2 6 8 2 . . . e jb g (b) V V r r r r Earth Moon Earth Moon 3 3 Earth Moon m cm m cm = = F HG I KJ = F H GG I K JJ = 4 3 4 3 6 8 3 3 6 37 10 100 1 74 10 49 1 . . . e jb g P1.39 To balance, m mFe Al= or Fe Fe Al AlV V= Fe Fe Al Al Al Fe Fe Al cm cm 4 3 4 3 2 00 7 86 2 70 2 86 3 3 1 3 1 3 F HG I KJ = F HG I KJ = F HG I KJ = F HG I KJ = r r r r / / . . . . .a f

11. Chapter 1 11 P1.40 The mass of each sphere is m V r Al Al Al Al Al = = 4 3 3 and m V r Fe Fe Fe Fe Fe = = 4 3 3 . Setting these masses equal, 4 3 4 3 3 3 Al Al Fe Fer r = and r rAl Fe Fe Al = 3 . Section 1.6 Estimates and Order-of-Magnitude Calculations P1.41 Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball as a sphere of diameter 0.038 m. The volume of the room is 4 4 3 48 = m3 , while the volume of one ball is 4 3 0 038 2 87 10 3 5 . . m 2 m3F HG I KJ = . Therefore, one can fit about 48 2 87 10 105 6 . ~ ping-pong balls in the room. As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called best packing fraction is 1 6 2 0 74 = . so that at least 26% of the space will be empty. Therefore, the above estimate reduces to 1 67 10 0 740 106 6 . . ~ . P1.42 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus, the tire would make 50 000 5 280 1 3 107 mi ft mi rev 8 ft rev~ 10 rev7 b gb gb g= . P1.43 In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least 1 16 in 43 10 ft2 5 2 = . Since 1 acre 43 560 ft2 = , the number of blades of grass to be expected on a quarter-acre plot of land is about n = = = total area area per blade acre ft acre ft blade 2.5 10 blades blades 2 2 7 0 25 43 560 43 10 105 7 . ~ a fe j .

12. 12 Physics and Measurement P1.44 A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then approximately 4 10 3 in3 . Since 1 acre 43 560 ft2 = , the volume of water required to cover it to a depth of 1 inch is 1 acre 1 inch 1 acre in ft 1 acre in ft 6.3 10 in6 3 a fa f a f= F HG I KJ F HG I KJ 43 560 144 1 2 2 2 . The number of raindrops required is n = = = volume of water required volume of a single drop in in . 6 3 10 4 10 1 6 10 10 6 3 3 3 9 9. ~ . *P1.45 Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then V = =0 5 1 3 0 5 0 3 0 10. . . . .a fa fa fa fm m m m3 . The mass of this volume of water is m Vwater water 3 3 kg m m kg kg= = = 1 000 0 10 100 102 e je j. ~ . Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is m Vcopper copper 3 3 kg m m kg kg= = = 8 920 0 10 892 103 e je j. ~ . P1.46 The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are ~250 million people, and 365 days in a year, so 250 10 365 106 11 cans day days year canse jb g are thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, we estimate this represents 10 0 1 1 1 3 1 1011 5 cans oz can lb 16 oz ton 2 000 lb tons yeare jb gb gb g. . . ~105 tons P1.47 Assume: Total population = 107 ; one out of every 100 people has a piano; one tuner can serve about 1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore, # tuners ~ 1 tuner 1 000 pianos 1 piano 100 people people F HG I KJ F HG I KJ =( )10 1007 .

13. Chapter 1 13 Section 1.7 Significant Figures *P1.48 METHOD ONE We treat the best value with its uncertainty as a binomial 21 3 0 2 9 8 0 1. . . . a f a fcm cm , A = 21 3 9 8 21 3 0 1 0 2 9 8 0 2 0 1. . . . . . . .a f a f a f a fa f cm2 . The first term gives the best value of the area. The cross terms add together to give the uncertainty and the fourth term is negligible. A = 209 42 2 cm cm . METHOD TWO We add the fractional uncertainties in the data. A = + F HG I KJ = = 21 3 9 8 0 2 21 3 0 1 9 8 209 2% 209 42 2 2 . . . . . . cm cm cm cm cma fa f P1.49 (a) r2 2 2 2 2 2 10 5 0 2 10 5 2 10 5 0 2 0 2 346 13 = = + = . m . m m m m m m m a f ( . ) ( . )( . ) ( . ) (b) 2 2 10 5 0 2 66 0 1 3 r = = . m . m m ma f . . P1.50 (a) 3 (b) 4 (c) 3 (d) 2 P1.51 r m m r = = = = 6 50 0 20 6 50 0 20 10 1 85 0 02 2 4 3 3 . . cm . . m . . kg a f a f a f c h also, = + m m r r 3 . In other words, the percentages of uncertainty are cumulative. Therefore, = + = 0 02 1 85 3 0 20 6 50 0 103 . . . . . a f , = = 1 85 6 5 10 1 61 10 4 3 2 3 3 3. . . c h e jm kg m and = = 1 61 0 17 10 1 6 0 2 103 3 . . . .a f a fkg m kg m3 3 .

14. 14 Physics and Measurement P1.52 (a) 756.?? 37.2? 0.83 + 2.5? 796./5/3 = 797 (b) 0 003 2 356 3 1 140 16 1 1. 2 s.f. . 4 s.f. . 2 s.f.a f a f a f = = . (c) 5.620 4 s.f. >4 s.f. 17.656= 4 s.f. 17.66a f a f a f = *P1.53 We work to nine significant digits: 1 1 365 242 199 24 60 60 31 556 926 0yr yr d 1 yr h 1 d min 1 h s 1 min s= F HG I KJF HG I KJF HG I KJF HG I KJ = . . . P1.54 The distance around is 38.44 m 19.5 m 38.44 m 19.5 m 115.88 m+ + + = , but this answer must be rounded to 115.9 m because the distance 19.5 m carries information to only one place past the decimal. 115 9. m P1.55 V V V V V V V V = + = + = + + = = = = + = 2 2 2 17 0 1 0 1 0 1 0 0 09 1 70 10 0 1 0 0 090 0 900 2 1 70 0 900 5 2 1 2 1 2 1 2 3 b g a fa fa f a fa fa f e j . . . . . . . . . . . . . m m m m m m m m m m m m m 3 3 3 3 1 1 1 1 1 1 0 12 0 0063 0 01 0 010 0 1 0 011 0 006 0 010 0 011 0 027 3% = = = = = = U V ||| W ||| = + + = = . . . . . . . . . . m 19.0 m m 1.0 m cm 9.0 cm w w t t V V FIG. P1.55 Additional Problems P1.56 It is desired to find the distance x such that x x100 1 000 m m = (i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that x2 5 100 1 000 1 00 10= = m m m2 a fb g . and therefore x = =1 00 10 3165 . m m2 .

15. Chapter 1 15 *P1.57 Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has mass m0 27 25 197 3 27 10= F HG I KJ = u 1.66 10 kg 1 u kga f . . So, the number of atoms in the cube is N = = 19 300 5 90 1025 28kg 3.27 10 kg . . The imagined cubical volume of each atom is d3 28 291 5 90 10 1 69 10= = m m 3 3 . . . So d = 2 57 10 10 . m . P1.58 A N A V V A V r rtotal drop total drop drop total 4 = = F HG I KJ = F H GG I K JJa fe j e j e j 3 3 2 4 A V r total total 3 2m m m= F HG I KJ = F HG I KJ = 3 3 30 0 10 2 00 10 4 50 6 5 . . . P1.59 One month is 1 30 24 3 600 2 592 106 mo day h day s h s= = b gb gb g . . Applying units to the equation, V t t= +1 50 0 008 00 2 . .Mft mo Mft mo3 3 2 e j e j . Since 1 106 Mft ft3 3 = , V t t= + 1 50 10 0 008 00 106 6 2 . .ft mo ft mo3 3 2 e j e j . Converting months to seconds, V t t= + 1 50 10 0 008 00 106 6 2 2. .ft mo 2.592 10 s mo ft mo 2.592 10 s mo 3 6 3 2 6 e j . Thus, V t tft ft s ft s3 3 3 2 [ ] . .= + 0 579 1 19 10 9 2 e j e j .

16. 16 Physics and Measurement P1.60 (deg) (rad) tan a f sin a f difference 15.0 0.262 0.268 0.259 3.47% 20.0 0.349 0.364 0.342 6.43% 25.0 0.436 0.466 0.423 10.2% 24.0 0.419 0.445 0.407 9.34% 24.4 0.426 0.454 0.413 9.81% 24.5 0.428 0.456 0.415 9.87% 24.6 0.429 0.458 0.416 9.98% 24.6 24.7 0.431 0.460 0.418 10.1% P1.61 2 15 0 2 39 2 39 55 0 3 41 r r h r h = = = = = . m . m tan 55.0 . m ( . ) ma ftan . 5555 h rr h FIG. P1.61 *P1.62 Let d represent the diameter of the coin and h its thickness. The mass of the gold is m V At d dh t= = = + F HG I KJ 2 4 2 where t is the thickness of the plating. m = + L N MM O Q PP = = = = 19 3 2 2 41 4 2 41 0 178 0 18 10 0 003 64 0 003 64 036 4 3 64 2 4 . . . . . . . $10 $0. . a f a fa f e j grams cost grams gram cents This is negligible compared to $4.98. P1.63 The actual number of seconds in a year is 86 400 s day 365.25 day yr 31 557 600 s yrb gb g= . The percent error in the approximation is = 10 31 557 600 31 557 600 100% 0 449% 7 s yr s yr s yr e j b g . .

17. Chapter 1 17 P1.64 (a) V = L3 , A = L2 , h = L V A h= L L L L3 2 3 = = . Thus, the equation is dimensionally correct. (b) V R h R h Ahcylinder = = = 2 2 e j , where A R= 2 V wh w h Ahrectangular object = = =a f , where A w= P1.65 (a) The speed of rise may be found from v D = = = Vol rate of flow (Area: cm s cm s 3 cm a f a f 2 2 4 6 30 4 16 5 0 529 ) . . . . (b) Likewise, at a 1.35 cm diameter, v = = 16 5 11 5 1.35 4 2 . . cm s cm s 3 cma f . P1.66 (a) 1 cubic meter of water has a mass m V= = = 1 00 10 1 00 10 1 0003 3 3 2 3 . kg cm . m cm m kge je je j (b) As a rough calculation, we treat each item as if it were 100% water. cell: kg m m kg kidney: . kg cm cm kg fly: kg cm mm mm cm mm 3 3 3 2 m V R D m V R m D h = = F HG I KJ = F HG I KJ = F HG I KJ = = = F HG I KJ = F HG I KJ = = F HG I KJ = F HG I KJ 4 3 1 6 1 000 1 6 1 0 10 5 2 10 4 3 1 00 10 4 3 4 0 0 27 4 1 10 4 2 0 4 0 10 3 3 6 3 16 3 3 3 2 3 1 e j e j e j e j a f a fe . . ( . ) . . . j3 5 1 3 10= . kg P1.67 V20 mpg 10cars mi yr mi gal 5.0 10 gal yr= = ( )( )10 10 20 8 4 V25 mpg 10cars mi yr mi gal 4.0 10 gal yr= = ( )( )10 10 25 8 4 Fuel saved gal yr25 mpg 20 mpg= = V V 1 0 1010 .

18. 18 Physics and Measurement P1.68 v = F HG I KJ F HG I KJ F HG I KJ F HG I KJF HG I KJF HG I KJ = 5 00 220 1 0 914 4 1 1 14 1 24 1 3 600 8 32 10 4 . . . furlongs fortnight yd furlong m yd fortnight days day hrs hr s m s This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth. P1.69 The volume of the galaxy is r t2 21 2 19 61 10 10 10= m m m3 e j e j~ . If the distance between stars is 4 1016 m, then there is one star in a volume on the order of 4 10 1016 3 50 m m3 e j ~ . The number of stars is about 10 10 10 61 50 11m m star stars 3 3 ~ . P1.70 The density of each material is = = = m V m r h m D h2 2 4 . Al: g cm cm g cm The tabulated value g cm is smaller. Cu: g .23 cm .06 cm g cm The tabulated value g cm is smaller. Brass: .54 cm .69 cm g cm Sn: g .75 cm .74 cm g cm Fe: .89 cm .77 cm g cm 3 3 3 3 3 3 3 = = F HG I KJ = = F HG I KJ = = = = = = 4 51 5 2 52 3 75 2 75 2 70 2% 4 56 3 1 5 9 36 8 92 5% 4 94.4 g 1 5 8 91 4 69 1 1 3 7 68 4 216.1 g 1 9 7 88 2 2 2 2 2 . . . . . . . . . . . . b g a f a f b g a f a f b g a f a f b g a f a f b g a f a f The tabulated value g cm is smaller.3 7 86 0 3%. . F HG I KJ P1.71 (a) 3 600 s hr 24 hr day 365.25 days yr s yrb gb gb g= 3 16 107 . (b) V r V V mm cube mm 18 . m . m m m 1.91 10 micrometeorites = = = = = 4 3 4 3 5 00 10 5 24 10 1 5 24 10 3 7 3 19 3 3 19 3 e j . This would take 1 91 10 3 16 10 6 05 10 18 7 10. . . = micrometeorites micrometeorites yr yr .

19. Chapter 1 19 ANSWERS TO EVEN PROBLEMS P1.2 5 52 103 3 . kg m , between the densities of aluminum and iron, and greater than the densities of surface rocks. P1.34 1 3 1021 . kg P1.36 200 km P1.38 (a) 13.4; (b) 49.1P1.4 23.0 kg P1.40 r rAl Fe Fe Al = F HG I KJ 1 3 P1.6 7.69 cm P1.8 (a) and (b) see the solution, NA = 6 022 137 1023 . ; (c) 18.0 g; P1.42 ~10 rev7 (d) 44.0 g P1.44 ~109 raindrops P1.10 (a) 9 83 10 16 . g ; (b) 1 06 107 . atoms P1.46 ~1011 cans; ~105 tons P1.12 (a) 4 02 1025 . molecules; (b) 3 65 104 . molecules P1.48 209 4 2 a fcm P1.14 (a) ii; (b) iii; (c) i P1.50 (a) 3; (b) 4; (c) 3; (d) 2 P1.16 (a) M L T2 ; (b) 1 1newton kg m s2 = P1.52 (a) 797; (b) 1.1; (c) 17.66 P1.54 115.9 m P1.18 35 7. m2 P1.56 316 m P1.20 1 39 10 4 . m3 P1.58 4 50. m2 P1.22 (a) 3 39 105 3 . ft ; (b) 2 54 104 . lb P1.60 see the solution; 24.6 P1.24 (a) 560 5 60 10 5 60 105 7 km m cm= = . . ; P1.62 3 64. cents ; no(b) 491 m 0 491 4 91 104 = = . .km cm ; (c) 6 6 19 10 6 19 103 5 .19 km m cm= = . . ; P1.64 see the solution (d) 2 2 50 10 2 50 103 5 .50 km m cm= = . . P1.66 (a) 1 000 kg; (b) 5 2 10 16 . kg ; 0 27. kg ; 1 3 10 5 . kg P1.26 4 05 103 . m2 P1.28 (a) 1 1 609mi h km h= . ; (b) 88 5. km h ; P1.68 8 32 10 4 . m s ; a snail(c) 16 1. km h P1.70 see the solution P1.30 1 19 1057 . atoms P1.32 2 57 106 3 . m

20. 2 CHAPTER OUTLINE 2.1 Position, Velocity, and Speed 2.2 Instantaneous Velocity and Speed 2.3 Acceleration 2.4 Motion Diagrams 2.5 One-Dimensional Motion with Constant Acceleration 2.6 Freely Falling Objects 2.7 Kinematic Equations Derived from Calculus Motion in One Dimension ANSWERS TO QUESTIONS Q2.1 If I count 5.0 s between lightning and thunder, the sound has traveled 331 5 0 1 7m s s kmb ga f. .= . The transit time for the light is smaller by 3 00 10 331 9 06 10 8 5. . = m s m s times, so it is negligible in comparison. Q2.2 Yes. Yes, if the particle winds up in the +x region at the end. Q2.3 Zero. Q2.4 Yes. Yes. Q2.5 No. Consider a sprinter running a straight-line race. His average velocity would simply be the length of the race divided by the time it took for him to complete the race. If he stops along the way to tie his shoe, then his instantaneous velocity at that point would be zero. Q2.6 We assume the object moves along a straight line. If its average velocity is zero, then the displacement must be zero over the time interval, according to Equation 2.2. The object might be stationary throughout the interval. If it is moving to the right at first, it must later move to the left to return to its starting point. Its velocity must be zero as it turns around. The graph of the motion shown to the right represents such motion, as the initial and final positions are the same. In an x vs. t graph, the instantaneous velocity at any time t is the slope of the curve at that point. At t0 in the graph, the slope of the curve is zero, and thus the instantaneous velocity at that time is also zero. x tt0 FIG. Q2.6 Q2.7 Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, or is a constant. 21

21. 22 Motion in One Dimension Q2.8 Yes. If you drop a doughnut from rest v = 0a f, then its acceleration is not zero. A common misconception is that immediately after the doughnut is released, both the velocity and acceleration are zero. If the acceleration were zero, then the velocity would not change, leaving the doughnut floating at rest in mid-air. Q2.9 No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past. Q2.10 Yes. Consider throwing a ball straight up. As the ball goes up, its velocity is upward v > 0a f, and its acceleration is directed down a < 0a f. A graph of v vs. t for this situation would look like the figure to the right. The acceleration is the slope of a v vs. t graph, and is always negative in this case, even when the velocity is positive. v t v0 FIG. Q2.10 Q2.11 (a) Accelerating East (b) Braking East (c) Cruising East (d) Braking West (e) Accelerating West (f) Cruising West (g) Stopped but starting to move East (h) Stopped but starting to move West Q2.12 No. Constant acceleration only. Yes. Zero is a constant. Q2.13 The position does depend on the origin of the coordinate system. Assume that the cliff is 20 m tall, and that the stone reaches a maximum height of 10 m above the top of the cliff. If the origin is taken as the top of the cliff, then the maximum height reached by the stone would be 10 m. If the origin is taken as the bottom of the cliff, then the maximum height would be 30 m. The velocity is independent of the origin. Since the change in position is used to calculate the instantaneous velocity in Equation 2.5, the choice of origin is arbitrary. Q2.14 Once the objects leave the hand, both are in free fall, and both experience the same downward acceleration equal to the free-fall acceleration, g. Q2.15 They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity equal to vi. This velocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds will also be the same. Q2.16 With h gt= 1 2 2 , (a) 0 5 1 2 0 707 2 . .h g t= a f . The time is later than 0.5t. (b) The distance fallen is 0 25 1 2 0 5 2 . .h g t= a f . The elevation is 0.75h, greater than 0.5h.

22. Chapter 2 23 Q2.17 Above. Your ball has zero initial speed and smaller average speed during the time of flight to the passing point. SOLUTIONS TO PROBLEMS Section 2.1 Position, Velocity, and Speed P2.1 (a) v = 2 30. m s (b) v x t = = m m s = 16.1 m s 57 5 9 20 3 00 . . . (c) v x t = = = 57 5 0 11 5 . . m m 5.00 s m s *P2.2 (a) v x t = = F HG I KJ F HG I KJ = 20 1 1 3 156 10 2 107 7ft 1 yr m 3.281 ft yr s m s . or in particularly windy times v x t = = F HG I KJ F HG I KJ = 100 1 1 3 156 10 1 107 6ft 1 yr m 3.281 ft yr s m s . . (b) The time required must have been t x v = = F HG I KJF HG I KJ = 3 000 1 609 10 5 10 3 8mi 10 mm yr m 1 mi mm 1 m yr . P2.3 (a) v x t = = = 10 5 m 2 s m s (b) v = = 5 1 2 m 4 s m s. (c) v x x t t = = = 2 1 2 1 5 10 2 2 5 m m 4 s s m s. (d) v x x t t = = = 2 1 2 1 5 5 4 3 3 m m 7 s s m s. (e) v x x t t = = =2 1 2 1 0 0 8 0 0 m s P2.4 x t= 10 2 : For t x s m a f a f = = 2 0 2 1 3 0 40 44 1 90 . . . . (a) v x t = = = 50 50 0 m 1.0 s m s. (b) v x t = = = 4 1 41 0 . . m 0.1 s m s

23. 24 Motion in One Dimension P2.5 (a) Let d represent the distance between A and B. Let t1 be the time for which the walker has the higher speed in 5 00 1 . m s = d t . Let t2 represent the longer time for the return trip in = 3 00 2 . m s d t . Then the times are t d 1 5 00 = . m sb g and t d 2 3 00 = . m sb g. The average speed is: v d d d v d d d = = + + = = = Total distance Total time m s m s m s m s m s m s m s 2 2 2 25 00 3 00 8 00 15 0 2 2 15 0 8 00 3 75 . . . . . . . b g b g b g e j e j (b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0 . Section 2.2 Instantaneous Velocity and Speed P2.6 (a) At any time, t, the position is given by x t= 3 00 2 . m s2 e j . Thus, at ti = 3 00. s: xi = =3 00 3 00 27 0 2 . . .m s s m2 e ja f . (b) At t tf = +3 00. s : x tf = +3 00 3 00 2 . .m s s2 e ja f , or x t tf = + +27 0 18 0 3 00 2 . . .m m s m s2 b g e ja f . (c) The instantaneous velocity at t = 3 00. s is: v x x t t t f i t = F HG I KJ = + = lim lim . . . 0 0 18 0 3 00 18 0m s m s m s2 e je j . P2.7 (a) at ti = 1 5. s, xi = 8 0. m (Point A) at tf = 4 0. s , x f = 2 0. m (Point B) v x x t t f i f i = = = = 2 0 8 0 4 1 5 6 0 2 4 . . . . . a f a f m s m 2.5 s m s (b) The slope of the tangent line is found from points C and D. t xC C= =1 0 9 5. .s, mb g and t xD D= =3 5 0. s,b g, v 3 8. m s . FIG. P2.7 (c) The velocity is zero when x is a minimum. This is at t 4 s .

24. Chapter 2 25 P2.8 (a) (b) At t = 5 0. s, the slope is v 58 23 m 2.5 s m s . At t = 4 0. s, the slope is v 54 18 m 3 s m s . At t = 3 0. s, the slope is v 49 m 14 3.4 s m s . At t = 2 0. s , the slope is v 36 m 9 4.0 s .0 m s . (c) a v t = 23 5 0 4 6 m s s m s2 . . (d) Initial velocity of the car was zero . P2.9 (a) v = ( ) ( ) = 5 0 1 0 5 m s m s (b) v = ( ) ( ) = 5 10 4 2 2 5 m s m s. (c) v = ( ) ( ) = 5 5 5 4 0 m m s s (d) v = ( ) ( ) = + 0 5 8 7 5 m s s m s FIG. P2.9 *P2.10 Once it resumes the race, the hare will run for a time of t x x v f i x = = = 1 000 25 m 800 m 8 m s s . In this time, the tortoise can crawl a distance x xf i = ( )=0 2 25 5 00. .m s s ma f .

25. 26 Motion in One Dimension Section 2.3 Acceleration P2.11 Choose the positive direction to be the outward direction, perpendicular to the wall. v v atf i= + : a v t = = = 22 0 25 0 3 50 10 1 34 103 4. . . . m s m s s m s2a f . P2.12 (a) Acceleration is constant over the first ten seconds, so at the end, v v atf i= + = + ( )=0 2 00 10 0 20 0. . .m s s m s2 c h . Then a = 0 so v is constant from t =10 0. s to t =15 0. s. And over the last five seconds the velocity changes to v v atf i= + = + ( )=20 0 3 00 5 00 5 00. . . .m s m s s m s2 c h . (b) In the first ten seconds, x x v t atf i i= + + = + + ( ) = 1 2 0 0 1 2 2 00 10 0 1002 2 . .m s s m2 c h . Over the next five seconds the position changes to x x v t atf i i= + + = + ( )+ = 1 2 100 20 0 5 00 0 2002 m m s s m. .a f . And at t = 20 0. s, x x v t atf i i= + + = + ( )+ ( ) = 1 2 200 20 0 5 00 1 2 3 00 5 00 2622 2 m m s s m s s m2 . . . .a f c h . *P2.13 (a) The average speed during a time interval t is v t = distance traveled . During the first quarter mile segment, Secretariats average speed was v1 0 250 1 320 52 4 35 6= = = . . . mi 25.2 s ft 25.2 s ft s mi hb g. During the second quarter mile segment, v2 1 320 55 0 37 4= = ft 24.0 s ft s mi h. .b g. For the third quarter mile of the race, v3 1 320 55 5 37 7= = ft 23.8 s ft s mi h. .b g, and during the final quarter mile, v4 1 320 57 4 39 0= = ft 23.0 s ft s mi h. .b g. continued on next page

26. Chapter 2 27 (b) Assuming that v vf = 4 and recognizing that vi = 0 , the average acceleration during the race was a v vf i = = + + +( ) = total elapsed time ft s s ft s257 4 0 25 2 24 0 23 8 23 0 0 598 . . . . . . . P2.14 (a) Acceleration is the slope of the graph of v vs t. For 0 5 00<

27. 28 Motion in One Dimension P2.16 (a) At t = 2 00. s, x = ( ) ( )+ =3 00 2 00 2 00 2 00 3 00 11 0 2 . . . . . .m m. At t = 3 00. s , x = + =3 00 9 00 2 00 3 00 3 00 24 0 2 . . . . . .a f a f m m so v x t = = = 24 0 11 0 2 00 13 0 . . . . m m 3.00 s s m s . (b) At all times the instantaneous velocity is v d dt t t t= + = ( )3 00 2 00 3 00 6 00 2 002 . . . . .c h m s At t = 2 00. s, v = ( ) =6 00 2 00 2 00 10 0. . . .m s m s . At t = 3 00. s , v = ( ) =6 00 3 00 2 00 16 0. . . .m s m s . (c) a v t = = = 16 0 10 0 3 00 2 00 6 00 . . . . . m s m s s s m s2 (d) At all times a d dt = ( )=6 00 2 00 6 00. . . m s2 . (This includes both t = 2 00. s and t = 3 00. s ). P2.17 (a) a v t = = = 8 00 6 00 1 3 . . . m s s m s2 (b) Maximum positive acceleration is at t = 3 s, and is approximately 2 m s2 . (c) a = 0 , at t = 6 s , and also for t >10 s . (d) Maximum negative acceleration is at t = 8 s, and is approximately 1 5. m s2 . Section 2.4 Motion Diagrams P2.18 (a) (b) (c) (d) (e) continued on next page

28. Chapter 2 29 (f) One way of phrasing the answer: The spacing of the successive positions would change with less regularity. Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude and direction. Section 2.5 One-Dimensional Motion with Constant Acceleration P2.19 From v v axf i 2 2 2= + , we have 10 97 10 0 2 2203 2 . = + ( )m s mc h a , so that a = 2 74 105 . m s2 which is a g= 2 79 104 . times . P2.20 (a) x x v v tf i i f = + 1 2 c h becomes 40 1 2 2 80 8 50m m s s= + ( )vi . .a f which yields vi = 6 61. m s . (b) a v v t f i = = = 2 80 6 61 8 50 0 448 . . . . m s m s s m s2 P2.21 Given vi =12 0. cm s when x ti = =( )3 00 0. cm , and at t = 2 00. s, x f =5 00. cm, x x v t atf i i = + 1 2 2 : = ( )+ ( )5 00 3 00 12 0 2 00 1 2 2 00 2 . . . . .a = +8 00 24 0 2. . a a = = 32 0 2 16 0 . . cm s2 . *P2.22 (a) Let i be the state of moving at 60 mi h and f be at rest v v a x x a a xf xi x f i x x 2 2 2 2 0 60 2 121 0 1 3 600 242 5 280 1 21 8 21 8 1 609 1 9 75 = + = + F HG I KJ = F HG I KJF HG I KJ = = F HG I KJF HG I KJ = d i b g a fmi h ft mi 5 280 ft mi h ft 1 mi h 3 600 s mi h s mi h s m 1 mi h 3 600 s m s 2 2 . . . . (b) Similarly, 0 80 2 211 0 6 400 5 280 422 3 600 22 2 9 94 2 = + = = = mi h ft mi h s mi h s m s2 b g a f b g b g a a x x . . . (c) Let i be moving at 80 mi h and f be moving at 60 mi h. v v a x x a a xf xi x f i x x 2 2 2 2 2 60 80 2 211 121 2 800 5 280 2 90 3 600 22 8 10 2 = + = + = = = d i b g b g a f b g a fb g mi h mi h ft ft mi h s mi h s m s2 . . .

29. 30 Motion in One Dimension *P2.23 (a) Choose the initial point where the pilot reduces the throttle and the final point where the boat passes the buoy: xi = 0 , x f =100 m, vxi = 30 m s, vxf =?, ax =3 5. m s2 , t =? x x v t a tf i xi x= + + 1 2 2 : 100 0 30 1 2 3 5 2 m m s m s2 = + + a f c ht t. 1 75 30 100 02 . m s m s m2 c h a ft t + = . We use the quadratic formula: t b b ac a = 2 4 2 t = ( ) = = 30 900 4 1 75 100 2 1 75 30 14 1 3 5 12 6 m s m s m s m m s m s m s m s s 2 2 2 2 2 . . . . . c h c h or 4 53. s . The smaller value is the physical answer. If the boat kept moving with the same acceleration, it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s. (b) v v a txf xi x= + = =30 3 5 4 53 14 1m s m s s m s2 . . .e j P2.24 (a) Total displacement = area under the v t,a f curve from t = 0 to 50 s. x x = + + = 1 2 50 15 50 40 15 1 2 50 10 1 875 m s s m s s m s s m b ga f b ga f b ga f (b) From t =10 s to t = 40 s , displacement is x = + + = 1 2 50 33 5 50 25 1 457m s m s s m s s mb ga f b ga f . FIG. P2.24 (c) 0 15 t s: a v t 1 50 0 15 0 3 3= = ( ) = m s s m s2 . 15 40s s<

30. Chapter 2 31 (d) (i) x a t t1 1 2 2 0 1 2 1 2 3 3= + = . m s2 c h or x t1 2 1 67= . m s2 c h (ii) x t2 1 2 15 50 0 50 15= ( ) + ( )s m s m s sa f or x t2 50 375= m s ma f (iii) For 40 50s s t , x v t t a t t3 3 2 0 1 2 40 50 40= = F HG I KJ+ ( ) + ( ) area under vs from to 40 s s m s sa f or x t t3 2 375 1 250 1 2 5 0 40 50 40= + + + m m m s s m s s2 .e ja f b ga f which reduces to x t t3 2 250 2 5 4 375= m s m s m2 b g e j. . (e) v = = = total displacement total elapsed time m s m s 1 875 50 37 5. P2.25 (a) Compare the position equation x t t= + 2 00 3 00 4 00 2 . . . to the general form x x v t atf i i= + + 1 2 2 to recognize that xi = 2 00. m, vi = 3 00. m s, and a =8 00. m s2 . The velocity equation, v v atf i= + , is then v tf = 3 00 8 00. .m s m s2 c h . The particle changes direction when v f = 0, which occurs at t = 3 8 s . The position at this time is: x = + F HG I KJ F HG I KJ =2 00 3 00 3 8 4 00 3 8 2 56 2 . . . .m m s s m s s m2 a f c h . (b) From x x v t atf i i= + + 1 2 2 , observe that when x xf i= , the time is given by t v a i = 2 . Thus, when the particle returns to its initial position, the time is t = = 2 3 00 8 00 3 4 . . m s m s s2 a f and the velocity is v f = F HG I KJ= 3 00 8 00 3 4 3 00. . .m s m s s m s2 c h .

31. 32 Motion in One Dimension *P2.26 The time for the Ford to slow down we find from x x v v t t x v v f i xi xf xi xf = + + = + = + = 1 2 2 2 250 71 5 0 6 99 d i a f m m s s . . . Its time to speed up is similarly t = ( ) + = 2 350 0 71 5 9 79 m m s s . . . The whole time it is moving at less than maximum speed is 6 99 5 00 9 79 21 8. . . .s s s s+ + = . The Mercedes travels x x v v tf i xi xf= + + = + + = 1 2 0 1 2 71 5 71 5 21 8 1 558 d i a fb ga f. . .m s s m while the Ford travels 250 350 600+ =m m, to fall behind by 1 558 600 958m m m = . P2.27 (a) vi =100 m s, a =5 00. m s2 , v v atf i= + so 0 100 5= t, v v a x xf i f i 2 2 2= + c h so 0 100 2 5 00 0 2 =( ) ( ) . x fc h. Thus x f = 1 000 m and t = 20 0. s . (b) At this acceleration the plane would overshoot the runway: No . P2.28 (a) Take ti = 0 at the bottom of the hill where xi = 0 , vi = 30 0. m s, a =2 00. m s2 . Use these values in the general equation x x v t atf i i= + + 1 2 2 to find x t tf = + + 0 30 0 1 2 2 00 2 . .m s m s2 a f c h when t is in seconds x t tf = 30 0 2 .c hm . To find an equation for the velocity, use v v at tf i= + = + 30 0 2 00. .m s m s2 e j , v tf = ( )30 0 2 00. . m s . (b) The distance of travel x f becomes a maximum, xmax , when v f = 0 (turning point in the motion). Use the expressions found in part (a) for v f to find the value of t when x f has its maximum value: From v tf = ( )3 00 2 00. . m s, v f = 0 when t =15 0. s. Then x t tmax . . . .= =( )( )( ) =30 0 30 0 15 0 15 0 2252 2 c hm m .

32. Chapter 2 33 P2.29 In the simultaneous equations: v v a t x x v v t xf xi x f i xi xf = + = + R S| T| U V| W|1 2 c h we have v v v v xf xi xi xf = ( ) = + ( ) R S| T| U V| W| 5 60 4 20 62 4 1 2 4 20 . . . . m s s m s 2 c h c h . So substituting for vxi gives 62 4 1 2 56 0 4 20 4 20. . . .m m s s s2 = + ( )+ ( )v vxf xfc h 14 9 1 2 5 60 4 20. . .m s m s s2 = + ( )vxf c h . Thus vxf = 3 10. m s . P2.30 Take any two of the standard four equations, such as v v a t x x v v t xf xi x f i xi xf = + = + R S| T| U V| W|1 2 c h . Solve one for vxi , and substitute into the other: v v a txi xf x= x x v a t v tf i xf x xf = + 1 2 c h . Thus x x v t a tf i xf x = 1 2 2 . Back in problem 29, 62 4 4 20 1 2 5 60 4 20 2 . . . .m s m s s2 = ( ) ( )vxf c h vxf = = 62 4 49 4 3 10 . . . m m 4.20 s m s . P2.31 (a) a v v t f i = = = = 632 1 40 662 202 5 280 3 600e j . ft s m s2 2 (b) x v t atf i= + = F HG I KJ = = 1 2 632 5 280 3 600 1 40 1 2 662 1 40 649 1982 2 a f a f a fa f. . ft m

33. 34 Motion in One Dimension P2.32 (a) The time it takes the truck to reach 20 0. m s is found from v v atf i= + . Solving for t yields t v v a f i = = = 20 0 0 2 00 10 0 . . . m s m s m s s2 . The total time is thus 10 0 20 0 5 00 35 0. . . .s s s s+ + = . (b) The average velocity is the total distance traveled divided by the total time taken. The distance traveled during the first 10.0 s is x vt1 0 20 0 2 10 0 100= = +F HG I KJ( )= . . m. With a being 0 for this interval, the distance traveled during the next 20.0 s is x v t ati2 21 2 20 0 20 0 0 400= + =( )( )+ =. . m. The distance traveled in the last 5.00 s is x vt3 20 0 0 2 5 00 50 0= = +F HG I KJ( )= . . . m. The total distance x x x x= + + = + + =1 2 3 100 400 50 550 m, and the average velocity is given by v x t = = = 550 35 0 15 7 . . m s . P2.33 We have vi = 2 00 104 . m s, v f = 6 00 106 . m s, x xf i = 1 50 10 2 . m. (a) x x v v tf i i f = + 1 2 c h : t x x v v f i i f = + = + = 2 2 1 50 10 2 00 10 6 00 10 4 98 10 2 4 6 9 c h c h. . . . m m s m s s (b) v v a x xf i x f i 2 2 2= + d i: a v v x x x f i f i = = = 2 2 6 2 4 2 2 15 2 6 00 10 2 00 10 2 1 50 10 1 20 10 ( ) . . ( . ) . m s m s m m s2e j e j

34. Chapter 2 35 *P2.34 (a) v v a x xxf xi x f i 2 2 2= + c h: 0 01 3 10 0 2 408 2 . = + ( )m s mc h ax ax = = 3 10 80 1 12 10 6 2 11 m s m m s2 c h . (b) We must find separately the time t1 for speeding up and the time t2 for coasting: x x v v t t t f i xf xi = + = + = 1 2 40 1 2 3 10 0 2 67 10 1 6 1 1 5 d i e j: m m s s. x x v v t t t f i xf xi = + = + = 1 2 60 1 2 3 10 3 10 2 00 10 2 6 6 2 2 5 d i e j: . m m s m s s total time = 4 67 10 5 . s . *P2.35 (a) Along the time axis of the graph shown, let i = 0 and f tm= . Then v v a txf xi x= + gives v a tc m m= +0 a v t m c m = . (b) The displacement between 0 and tm is x x v t a t v t t v tf i xi x c m m c m = + = + = 1 2 0 1 2 1 2 2 2 . The displacement between tm and t0 is x x v t a t v t tf i xi x c m = + = + 1 2 02 0a f . The total displacement is x v t v t v t v t tc m c c m c m= + = F HG I KJ1 2 1 2 0 0 . (c) For constant vc and t0 , x is minimized by maximizing tm to t tm = 0 . Then x v t t v t c c min = F HG I KJ=0 0 01 2 2 . (e) This is realized by having the servo motor on all the time. (d) We maximize x by letting tm approach zero. In the limit x v t v tc c= =0 00a f . (e) This cannot be attained because the acceleration must be finite.

35. 36 Motion in One Dimension *P2.36 Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit, x x v t a t v t a t v t v v a t f i xi x i d d d d d i d = + + = + + = = + 1 2 0 1 2 1 2 2 2 (a) The speed halfway through the photogate in space is given by v v a v av ths i i d d 2 2 2 2 2 = + F HG I KJ= + . v v av ths i d d= +2 and this is not equal to vd unless a = 0 . (b) The speed halfway through the photogate in time is given by v v a t ht i d = + F HG I KJ 2 and this is equal to vd as determined above. P2.37 (a) Take initial and final points at top and bottom of the incline. If the ball starts from rest, vi = 0 , a = 0 500. m s2 , x xf i = 9 00. m. Then v v a x x v f i f i f 2 2 2 2 0 2 0 500 9 00 3 00 = + = + = d i e ja f. . . . m s m m s 2 (b) x x v t atf i i = + 1 2 2 9 00 0 1 2 0 500 6 00 2 . . . = + = m s s 2 e jt t (c) Take initial and final points at the bottom of the planes and the top of the second plane, respectively: vi = 3 00. m s, v f = 0, x xf i =15 00. m. v v a x xf i f i 2 2 2= + c h gives a v v x x f i f i = = ( ) = 2 2 2 2 0 3 00 2 15 0 0 300 c h a f. . . m s m m s2 . (d) Take the initial point at the bottom of the planes and the final point 8.00 m along the second: vi = 3 00. m s, x xf i = 8 00. m, a =0 300. m s2 v v a x x v f i f i f 2 2 2 2 3 00 2 0 300 8 00 4 20 2 05 = + = + = = d i b g e ja f. . . . . . m s m s m m s m s 2 2 2

36. Chapter 2 37 P2.38 Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sues car. For her we have xis = 0, vis = 30 0. m s, as =2 00. m s2 so her position is given by x t x v t a t t ts is is s( )= + + = + 1 2 30 0 1 2 2 002 2 . .m s m s2 a f c h . For the van, xiv =155 m, viv = 5 00. m s , av = 0 and x t x v t a t tv iv iv v( )= + + = + + 1 2 155 5 00 02 . m sa f . To test for a collision, we look for an instant tc when both are at the same place: 30 0 155 5 00 0 25 0 155 2 2 . . . . t t t t t c c c c c = + = + From the quadratic formula tc = ( ) ( ) = 25 0 25 0 4 155 2 13 6 2 . . . s or 11 4. s . The smaller value is the collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position 155 5 00 11 4 212m m s s m+ ( )=. .a f . *P2.39 As in the algebraic solution to Example 2.8, we let t represent the time the trooper has been moving. We graph x tcar = +45 45 and x ttrooper = 1 5 2 . . They intersect at t = 31 s . x (km) t (s) 10 20 30 40 0.5 1 1.5 car police officer FIG. P2.39

37. 38 Motion in One Dimension Section 2.6 Freely Falling Objects P2.40 Choose the origin y t= =0 0,a f at the starting point of the ball and take upward as positive. Then yi = 0 , vi = 0 , and a g= =9 80. m s2 . The position and the velocity at time t become: y y v t atf i i = + 1 2 2 : y gt tf = = 1 2 1 2 9 802 2 . m s2 e j and v v atf i= + : v gt tf = = 9 80. m s2 c h . (a) at t =1 00. s: y f = ( ) = 1 2 9 80 1 00 4 90 2 . . .m s s m2 c h at t = 2 00. s: y f = ( ) = 1 2 9 80 2 00 19 6 2 . . .m s s m2 c h at t = 3 00. s : y f = ( ) = 1 2 9 80 3 00 44 1 2 . . .m s s m2 c h (b) at t =1 00. s: v f = ( )= 9 80 1 00 9 80. . .m s s m s2 c h at t = 2 00. s: v f = ( )= 9 80 2 00 19 6. . .m s s m s2 c h at t = 3 00. s : v f = ( )= 9 80 3 00 29 4. . .m s s m s2 c h P2.41 Assume that air resistance may be neglected. Then, the acceleration at all times during the flight is that due to gravity, a g= =9 80. m s2 . During the flight, Goff went 1 mile (1 609 m) up and then 1 mile back down. Determine his speed just after launch by considering his upward flight: v v a y y v v f i f i i i 2 2 2 2 0 2 9 80 1 609 178 = + = = d i e jb g: . . m s m m s 2 His time in the air may be found by considering his motion from just after launch to just before impact: y y v t atf i i = + 1 2 2 : 0 178 1 2 9 80 2 = m s m s2 a f c ht t. . The root t = 0 describes launch; the other root, t = 36 2. s, describes his flight time. His rate of pay may then be found from pay rate = = = $1. . . $99. 00 36 2 0 027 6 3 600 3 s $ s s h hb gb g . We have assumed that the workmans flight time, a mile, and a dollar, were measured to three- digit precision. We have interpreted up in the sky as referring to the free fall time, not to the launch and landing times. Both the takeoff and landing times must be several seconds away from the job, in order for Goff to survive to resume work.

38. Chapter 2 39 P2.42 We have y gt v t yf i i= + + 1 2 2 0 4 90 8 00 30 02 = +. . .m s m s m2 c h a ft t . Solving for t, t = + 8 00 64 0 588 9 80 . . . . Using only the positive value for t, we find that t = 1 79. s . P2.43 (a) y y v t atf i i = + 1 2 2 : 4 00 1 50 4 90 1 50 2 . . . .=( ) ( )( )vi and vi = 10 0. m s upward . (b) v v atf i= + = ( )( )=10 0 9 80 1 50 4 68. . . . m s v f = 4 68. m s downward P2.44 The bill starts from rest vi = 0 and falls with a downward acceleration of 9 80. m s2 (due to gravity). Thus, in 0.20 s it will fall a distance of y v t gti= = ( ) = 1 2 0 4 90 0 20 0 202 2 . . .m s s m2 c h . This distance is about twice the distance between the center of the bill and its top edge 8 cma f. Thus, David will be unsuccessful . *P2.45 (a) From y v t ati= + 1 2 2 with vi = 0 , we have t y a = = ( ) = 2 2 23 9 80 2 17 a f m m s s2 . . . (b) The final velocity is v f = + ( )= 0 9 80 2 17 21 2. . .m s s m s2 c h . (c) The time take for the sound of the impact to reach the spectator is t y v sound sound m 340 m s s= = = 23 6 76 10 2 . , so the total elapsed time is ttotal s s s= + 2 17 6 76 10 2 232 . . . .

39. 40 Motion in One Dimension P2.46 At any time t, the position of the ball released from rest is given by y h gt1 21 2 = . At time t, the position of the ball thrown vertically upward is described by y v t gti2 21 2 = . The time at which the first ball has a position of y h 1 2 = is found from the first equation as h h gt 2 1 2 2 = , which yields t h g = . To require that the second ball have a position of y h 2 2 = at this time, use the second equation to obtain h v h g g h g i 2 1 2 = F HG I KJ. This gives the required initial upward velocity of the second ball as v ghi = . P2.47 (a) v v gtf i= : v f = 0 when t = 3 00. s , g = 9 80. m s2 . Therefore, v gti = = ( )=9 80 3 00 29 4. . .m s s m s2 c h . (b) y y v v tf i f i = + 1 2 c h y yf i = = 1 2 29 4 3 00 44 1. . .m s s mb ga f *P2.48 (a) Consider the upward flight of the arrow. v v a y y y y yf yi y f i 2 2 2 2 0 100 2 9 8 10 000 19 6 510 = + = + = = d i b g e jm s m s m s m s m 2 2 2 2 . . (b) Consider the whole flight of the arrow. y y v t a t t t f i yi y= + + = + + 1 2 0 0 100 1 2 9 8 2 2 m s m s2 b g e j. The root t = 0 refers to the starting point. The time of flight is given by t = = 100 4 9 20 4 m s m s s2 . . . P2.49 Time to fall 3.00 m is found from Eq. 2.12 with vi = 0 , 3 00 1 2 9 80 2 . .m m s2 = c ht , t = 0 782. s. (a) With the horse galloping at 10 0. m s, the horizontal distance is vt = 7 82. m . (b) t = 0 782. s

40. Chapter 2 41 P2.50 Take downward as the positive y direction. (a) While the woman was in free fall, y =144 ft , vi = 0 , and a g= = 32 0. ft s2 . Thus, y v t at ti= + = + 1 2 144 0 16 02 2 ft ft s2 .c h giving tfall s= 3 00. . Her velocity just before impact is: v v gtf i= + = + ( )=0 32 0 3 00 96 0. . .ft s s ft s2 c h . (b) While crushing the box, vi = 96 0. ft s, v f = 0, and y = =18 0 1 50. .in. ft. Therefore, a v v y f i = = ( ) = 2 2 2 3 2 0 96 0 2 1 50 3 07 10 a f a f. . . ft s ft ft s2 , or a = 3 07 103 . ft s upward2 . (c) Time to crush box: t y v y v vf i = = = ( ) ++ 2 2 1 50 0 96 0 . . ft ft s or t = 3 13 10 2 . s . P2.51 y t= 3 00 3 . : At t = 2 00. s, y = =3 00 2 00 24 0 3 . . .a f m and v dy dt ty = = = A9 00 36 02 . . m s . If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is y y v t gt t tb bi i= + = + ( ) 1 2 24 0 36 0 1 2 9 802 2 . . . . Setting yb = 0, 0 24 0 36 0 4 90 2 = + . . .t t . Solving for t, (only positive values of t count), t = 7 96. s . *P2.52 Consider the last 30 m of fall. We find its speed 30 m above the ground: y y v t a t v v f i yi y yi yi = + + = + + = + = 1 2 0 30 1 5 1 2 9 8 1 5 30 11 0 12 6 2 2 m s m s s m m 1.5 s m s 2 . . . . . . a f e ja f Now consider the portion of its fall above the 30 m point. We assume it starts from rest v v a y y y y yf yi y f i 2 2 2 2 12 6 0 2 9 8 160 19 6 8 16 = + = + = = d i b g e j. . . . . m s m s m s m s m 2 2 2 2 Its original height was then 30 8 16 38 2m m m+ =. . .

41. 42 Motion in One Dimension Section 2.7 Kinematic Equations Derived from Calculus P2.53 (a) J da dt = = constant da Jdt= a J dt Jt c= = +z 1 but a ai= when t = 0 so c ai1 = . Therefore, a Jt ai= + a dv dt dv adt v adt Jt a dt Jt a t ci i = = = = + = + +z zb g 1 2 2 2 but v vi= when t = 0, so c vi2 = and v Jt a t vi i= + + 1 2 2 v dx dt dx vdt x vdt Jt a t v dt x Jt a t v t c x x i i i i i = = = = + + F HG I KJ = + + + = z z 1 2 1 6 1 2 2 3 2 3 when t = 0, so c xi3 = . Therefore, x Jt a t v t xi i i= + + + 1 6 1 2 3 2 . (b) a Jt a J t a Ja ti i i 2 2 2 2 2 2= + = + +a f a a J t Ja ti i 2 2 2 2 2= + +c h a a J Jt a ti i 2 2 2 2 1 2 = + + F HG I KJ Recall the expression for v: v Jt a t vi i= + + 1 2 2 . So v v Jt a ti i = +a f 1 2 2 . Therefore, a a J v vi i 2 2 2= + a f .

42. Chapter 2 43 P2.54 (a) See the graphs at the right. Choose x = 0 at t = 0. At t = 3 s, x = ( )= 1 2 8 3 12m s s ma f . At t = 5 s, x = + ( )=12 8 2 28m m s s ma f . At t = 7 s, x = + ( )=28 1 2 8 2 36m m s s ma f . (b) For 0 3<

43. 44 Motion in One Dimension P2.56 a dv dt v= =3 00 2 . , vi =1 50. m s Solving for v, dv dt v=3 00 2 . v dv dt v v t t v v v v v t t i i i = = z z= + = = 2 0 3 00 1 1 3 00 3 00 1 1 . . . .or When v vi = 2 , t vi = = 1 3 00 0 222 . . s . Additional Problems *P2.57 The distance the car travels at constant velocity, v0 , during the reaction time is x v tra f1 0= . The time for the car to come to rest, from initial velocity v0 , after the brakes are applied is t v v a v a v a f i 2 0 00 = = = and the distance traveled during this braking period is x vt v v t v v a v a f i a f2 2 2 0 0 0 2 2 0 2 2 = = +F HG I KJ = +F HG I KJ F HG I KJ = . Thus, the total distance traveled before coming to a stop is s x x v t v a rstop = + = a f a f1 2 0 0 2 2 . *P2.58 (a) If a car is a distance s v t v a rstop = 0 0 2 2 (See the solution to Problem 2.57) from the intersection of length si when the light turns yellow, the distance the car must travel before the light turns red is x s s v t v a si r i= + = +stop 0 0 2 2 . Assume the driver does not accelerate in an attempt to beat the light (an extremely dangerous practice!). The time the light should remain yellow is then the time required for the car to travel distance x at constant velocity v0 . This is t x v v t s v t v a s v r v a i r i light = = + = + 0 0 2 0 0 0 0 2 2 . (b) With si =16 m, v = 60 km h, a = 2 0. m s2 , and tr = 1 1. s, tlight 2 s km h m s m s km h m 60 km h km h m s s= F HG I KJ+ F HG I KJ =1 1 60 2 2 0 0 278 1 16 1 0 278 6 23. . . . . e j .

44. Chapter 2 45 *P2.59 (a) As we see from the graph, from about 50 s to 50 s Acela is cruising at a constant positive velocity in the +x direction. From 50 s to 200 s, Acela accelerates in the +x direction reaching a top speed of about 170 mi/h. Around 200 s, the engineer applies the brakes, and the train, still traveling in the +x direction, slows down and then stops at 350 s. Just after 350 s, Acela reverses direction (v becomes negative) and steadily gains speed in the x direction. t(s) 100 200 300 100 100 200 400 v t 50 0 0 FIG. P2.59(a) (b) The peak acceleration between 45 and 170 mi/h is given by the slope of the steepest tangent to the v versus t curve in this interval. From the tangent line shown, we find a v t = = = ( ) ( ) = =slope mi h s mi h s m s2 155 45 100 50 2 2 0 98. .a f . (c) Let us use the fact that the area under the v versus t curve equals the displacement. The trains displacement between 0 and 200 s is equal to the area of the gray shaded region, which we have approximated with a series of triangles and rectangles. x0 200 50 50 50 50 160 100 1 2 100 1 2 100 170 160 24 000 = + + + + + + + + = s 1 2 3 4 5area area area area area mi h s mi h s mi h s 50 s mi h s mi h mi h mi h s b ga f b ga f b ga f a fb g a fb g b ga f t(s) 100 200 300 100 200 400 1 2 4 3 5 0 0 FIG. P2.59(c) Now, at the end of our calculation, we can find the displacement in miles by converting hours to seconds. As 1 3 600h s= , x0 200 24 000 6 7 F HG I KJ =s mi 3 600 s s mia f . .

45. 46 Motion in One Dimension *P2.60 Average speed of every point on the train as the first car passes Liz: x t = = 8 60 5 73 . . m 1.50 s m s. The train has this as its instantaneous speed halfway through the 1.50 s time. Similarly, halfway through the next 1.10 s, the speed of the train is 8 60 7 82 . . m 1.10 s m s= . The time required for the speed to change from 5.73 m/s to 7.82 m/s is 1 2 1 50 1 2 1 10 1 30. . .s s s( )+ ( )= so the acceleration is: a v t x x = = = 7 82 5 73 1 30 1 60 . . . . m s m s s m s2 . P2.61 The rate of hair growth is a velocity and the rate of its increase is an acceleration. Then vxi =1 04. mm d and ax = F HG I KJ0 132. mm d w . The increase in the length of the hair (i.e., displacement) during a time of t = =5 00 35 0. .w d is x v t a t x xi x= + = + 1 2 1 04 35 0 1 2 0 132 35 0 5 00 2 . . . . .mm d d mm d w d wb ga f b ga fa f or x = 48 0. mm . P2.62 Let point 0 be at ground level and point 1 be at the end of the engine burn. Let point 2 be the highest point the rocket reaches and point 3 be just before impact. The data in the table are found for each phase of the rockets motion. (0 to 1) v f 2 2 80 0 2 4 00 1 000 =. .a f a fb g so v f =120 m s 120 80 0 4 00= +( ). . t giving t =10 0. s (1 to 2) 0 120 2 9 80 2 ( ) = ( ) . x xf ic h giving x xf i = 735 m 0 120 9 80 = . t giving t =12 2. s This is the time of maximum height of the rocket. (2 to 3) v f 2 0 2 9 80 1 735 = .a fb g v tf = = ( )184 9 80. giving t =18 8. s FIG. P2.62 (a) ttotal s= + + =10 12 2 18 8 41 0. . . (b) x xf i =c htotal km1 73. continued on next page

46. Chapter 2 47 (c) vfinal m s= 184 t x v a 0 Launch 0.0 0 80 +4.00 #1 End Thrust 10.0 1 000 120 +4.00 #2 Rise Upwards 22.2 1 735 0 9.80 #3 Fall to Earth 41.0 0 184 9.80 P2.63 Distance traveled by motorist = 15 0. m sa ft Distance traveled by policeman = 1 2 2 00 2 . m s2 c ht (a) intercept occurs when 15 0 2 . t t= , or t = 15 0. s (b) v tofficer m s m s2 ( )= =2 00 30 0. .c h (c) x tofficer m s m2 ( )= = 1 2 2 00 2252 .c h P2.64 Area A1 is a rectangle. Thus, A hw v txi1 = = . Area A2 is triangular. Therefore A bh t v vx xi2 1 2 1 2 = = b g. The total area under the curve is A A A v t v v t xi x xi = + = + 1 2 2 b g and since v v a tx xi x = A v t a txi x= + 1 2 2 . The displacement given by the equation is: x v t a txi x= + 1 2 2 , the same result as above for the total area. vx vx vxi 0 t t A2 A1 FIG. P2.64

47. 48 Motion in One Dimension P2.65 (a) Let x be the distance traveled at acceleration a until maximum speed v is reached. If this is achieved in time t1 we can use the following three equations: x v v ti= + 1 2 1a f , 100 10 2 1 = x v t.a fand v v ati= + 1 . The first two give 100 10 2 1 2 10 2 1 2 200 20 4 1 1 1 1 1 = F HG I KJ = F HG I KJ = . . . . t v t at a t tb g For Maggie: m s For Judy: m s 2 2 a a = = = = 200 18 4 2 00 5 43 200 17 4 3 00 3 83 . . . . . . a fa f a fa f (b) v a t= 1 Maggie: m s Judy: m s v v = = = = 5 43 2 00 10 9 3 83 3 00 11 5 . . . . . . a fa f a fa f (c) At the six-second mark x at v t= + 1 2 6 001 2 1.a f Maggie: m Judy: m x x = + = = + = 1 2 5 43 2 00 10 9 4 00 54 3 1 2 3 83 3 00 11 5 3 00 51 7 2 2 . . . . . . . . . . a fa f a fa f a fa f a fa f Maggie is ahead by 2 62. m . P2.66 a1 0 100= . m s2 a2 0 500= . m s2 x a t v t a t= = + +1 000 1 2 1 2 1 1 2 1 2 2 2 2 m t t t= +1 2 and v a t a t1 1 1 2 2= = 1 000 1 2 1 2 1 1 2 1 1 1 1 2 2 1 1 2 2 = + F HG I KJ+ F HG I KJa t a t a t a a a t a 1 000 1 2 11 1 2 1 2 = F HG I KJa a a t t1 20 000 1 20 129= = . s t a t a 2 1 1 2 12 9 0 500 26= = . . s Total time = =t 155 s

48. Chapter 2 49 P2.67 Let the ball fall 1.50 m. It strikes at speed given by v v a x xxf xi f i 2 2 2= + c h: vxf 2 0 2 9 80 1 50= + ( ). .m s m2 c h vxf =5 42. m s and its stopping is described by v v a x x a a xf xi x f i x x 2 2 2 2 2 3 2 0 5 42 2 10 29 4 2 00 10 1 47 10 = + = + = = + d i b g e j. . . . . m s m m s m m s 2 2 2 Its maximum acceleration will be larger than the average acceleration we estimate by imagining constant acceleration, but will still be of order of magnitude ~103 m s2 . *P2.68 (a) x x v t a tf i xi x= + + 1 2 2 . We assume the package starts from rest. = + + 145 0 0 1 2 9 80 2 m m s2 .c ht t = ( ) = 2 145 9 80 5 44 m m s s2 . . (b) x x v t a tf i xi x= + + = + + ( ) = 1 2 0 0 1 2 9 80 5 18 1312 2 . .m s s m2 c h distance fallen = =x f 131 m (c) speed = = + = + =v v a txf xi x 0 9 8 5 18 50 8. . .m s s m s2 e j (d) The remaining distance is 145 131 5 13 5m m m =. . . During deceleration, vxi =50 8. m s, vxf = 0, x xf i =13 5. m v v a x xxf xi x f i 2 2 2= + c h: 0 50 8 2 13 5 2 = + ( ). .m s ma f ax ax = = + = 2 580 2 13 5 95 3 95 3 m s m m s m s upward 2 2 2 2 . . . a f .

49. 50 Motion in One Dimension P2.69 (a) y v t at t tf i= + = = + ( )1 2 21 2 50 0 2 00 1 2 9 80. . . , 4 90 2 00 50 0 02 . . .t t+ = t = + ( ) ( ) ( ) 2 00 2 00 4 4 90 50 0 2 4 90 2 . . . . . Only the positive root is physically meaningful: t = 3 00. s after the first stone is thrown. (b) y v t atf i= +2 21 2 and t = =3 00 1 00 2 00. . . s substitute 50 0 2 00 1 2 9 80 2 002 2 . . . .= ( )+ ( )( )vi : vi2 15 3= . m s downward (c) v v atf i1 1 2 00 9 80 3 00 31 4= + = +( )( )=. . . . m s downward v v atf i2 2 15 3 9 80 2 00 34 8= + = +( )( )=. . . . m s downward P2.70 (a) d t= ( ) 1 2 9 80 1 2 . d t= 336 2 t t1 2 2 40+ = . 336 4 90 2 402 2 2 t t= . .a f 4 90 359 5 28 22 02 2 2. . .t t + = t2 2 359 5 359 5 4 4 90 28 22 9 80 = ( )( ). . . . . t2 359 5 358 75 9 80 0 076 5= = . . . . s so d t= =336 26 42 . m (b) Ignoring the sound travel time, d = ( )( ) = 1 2 9 80 2 40 28 2 2 . . . m, an error of 6 82%. . P2.71 (a) In walking a distance x, in a time t, the length of rope is only increased by xsin . The pack lifts at a rate x t sin . v x t v x v x x h = = = + sin boy boy 2 2 (b) a dv dt v dx dt v x d dt = = + F HG I KJboy boy 1 a v v v x d dt = boy boy boy 2 , but d dt v v x = = boy = F HG I KJ= = + a v x v h h v x h boy 2 boy 2 boy 2 1 2 2 2 2 2 2 2 3 2 c h (c) v h boy 2 , 0 (d) vboy , 0 FIG. P2.71

50. Chapter 2 51 P2.72 h= 6 00. m, vboy m s= 2 00. v x t v x v x x h = = = + sin boy boy 2 2 1 2 c h . However, x v t= boy : = + = + v v t v t h t t boy 2 boy 2 2 2 1 2 2 1 2 4 4 36c h c h . (a) t vs m s 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.32 0.63 0.89 1.11 1.28 1.41 1.52 1.60 1.66 1.71 a f b g FIG. P2.72(a) (b) From problem 2.71 above, a h v x h h v v t h t = + = + = + 2 2 2 3 2 2 2 2 3 2 2 3 2 144 4 36 boy 2 boy 2 boy 2 c h c h c h . t as m s 0 0.5 1 1.5 2 2.5 3. 3.5 4. 4.5 5 0.67 0.64 0.57 0.48 0.38 0.30 0.24 0.18 0.14 0.11 0.09 2 a f e j FIG. P2.72(b) P2.73 (a) We require x xs k= when t ts k= + 1 00. x t t x t t t s k k k k k k = + = = + = = 1 2 3 50 1 00 1 2 4 90 1 00 1 183 5 46 2 2 . . . . . . . m s m s s 2 2 e jb g e jb g (b) xk = = 1 2 4 90 5 46 73 0 2 . . .m s s m2 e ja f (c) vk = =4 90 5 46 26 7. . .m s s m s2 e ja f vs = =3 50 6 46 22 6. . .m s s m s2 e ja f

51. 52 Motion in One Dimension P2.74 Time t (s) Height h (m) h (m) t (s) v (m/s) midpt time t (s) 0.00 5.00 0.75 0.25 3.00 0.13 FIG. P2.74 0.25 5.75 0.65 0.25 2.60 0.38 0.50 6.40 0.54 0.25 2.16 0.63 0.75 6.94 0.44 0.25 1.76 0.88 1.00 7.38 0.34 0.25 1.36 1.13 1.25 7.72 0.24 0.25 0.96 1.38 1.50 7.96 0.14 0.25 0.56 1.63 1.75 8.10 0.03 0.25 0.12 1.88 2.00 8.13 0.06 0.25 0.24 2.13 2.25 8.07 0.17 0.25 0.68 2.38 2.50 7.90 0.28 0.25 1.12 2.63 2.75 7.62 0.37 0.25 1.48 2.88 3.00 7.25 0.48 0.25 1.92 3.13 3.25 6.77 0.57 0.25 2.28 3.38 3.50 6.20 0.68 0.25 2.72 3.63 3.75 5.52 0.79 0.25 3.16 3.88 4.00 4.73 0.88 0.25 3.52 4.13 4.25 3.85 0.99 0.25 3.96 4.38 4.50 2.86 1.09 0.25 4.36 4.63 4.75 1.77 1.19 0.25 4.76 4.88 5.00 0.58 TABLE P2.74 acceleration = slope of line is constant. a = =1 63 1 63. .m s m s downward2 2

52. Chapter 2 53 P2.75 The distance x and y are always related by x y L2 2 2 + = . Differentiating this equation with respect to time, we have 2 2 0x dx dt y dy dt + = Now dy dt is vB , the unknown velocity of B; and dx dt v= . From the equation resulting from differentiation, we have dy dt x y dx dt x y v= F HG I KJ= ( ). B O y A x L v x y FIG. P2.75 But y x = tan so v vB = F HG I KJ1 tan . When = 60 0. , v v v vB = = = tan . . 60 0 3 3 0 577 . ANSWERS TO EVEN PROBLEMS P2.2 (a) 2 10 7 m s; 1 10 6 m s; P2.24 (a) 1.88 km; (b) 1.46 km; (c) see the solution;(b) 5 108 yr (d) (i) x t1 2 1 67= . m s2 e j ; P2.4 (a) 50 0. m s ; (b) 41 0. m s (ii) x t2 50 375= m s mb g ; (iii) x t t3 2 250 2 5 4 375= m s m s m2 b g e j. ;P2.6 (a) 27 0. m; (e) 37 5. m s(b) 27 0 18 0 3 00 2 . . .m m s m s2 + +b g e ja f t t ; (c) 18 0. m s P2.26 958 m P2.8 (a), (b), (c) see the solution; 4 6. m s2 ; (d) 0 P2.28 (a) x t tf = 30 0 2 .e jm; v tf = 30 0 2.a f m s ; (b) 225 mP2.10 5.00 m P2.12 (a) 20 0. m s; 5 00. m s ; (b) 262 m P2.30 x x v t a tf i xf x = 1 2 2 ; 3 10. m s P2.14 (a) see the solution; P2.32 (a) 35.0 s; (b) 15 7. m s(b) 1 60. m s2 ; 0 800. m s2 P2.34 (a) 1 12 1011 . m s2 ; (b) 4 67 10 5 . sP2.16 (a) 13 0. m s; (b) 10 0. m s; 16 0. m s; (c) 6 00. m s2 ; (d) 6 00. m s2 P2.36 (a) False unless the acceleration is zero; see the solution; (b) TrueP2.18 see the solution P2.38 Yes; 212 m; 11.4 sP2.20 (a) 6 61. m s; (b) 0 448. m s2 P2.40 (a) 4 90. m; 19 6. m; 44 1. m; P2.22 (a) = 21 8 9 75. .mi h s m s2 ; (b) 9 80. m s; 19 6. m s; 29 4. m s (b) = 22 2 9 94. .mi h s m s2 ; (c) = 22 8 10 2. .mi h s m s2 P2.42 1.79 s

53. 54 Motion in One Dimension P2.44 No; see the solution P2.60 1 60. m s2 P2.46 The second ball is thrown at speed v ghi = P2.62 (a) 41.0 s; (b) 1.73 km; (c) 184 m s P2.64 v t a txi x+ 1 2 2 ; displacements agree P2.48 (a) 510 m; (b) 20.4 s P2.66 155 s; 129 sP2.50 (a) 96 0. ft s; (b) a = 3 07 103 . ft s upward2 ; P2.68 (a) 5.44 s; (b) 131 m; (c) 50 8. m s ; (c) t = 3 13 10 2 . s (d) 95 3. m s upward2 P2.52 38.2 m P2.70 (a) 26.4 m; (b) 6.82% P2.54 (a) and (b) see the solution; (c) 4 m s2 ; (d) 34 m; (e) 28 m P2.72 see the solution P2.74 see the solution; ax = 1 63. m s2 P2.56 0.222 s P2.58 (a) see the solution; (b) 6.23 s

54. 3 CHAPTER OUTLINE 3.1 Coordinate Systems 3.2 Vector and Scalar Quantities 3.3 Some Properties of Vectors 3.4 Components of a Vector and Unit Vectors Vectors ANSWERS TO QUESTIONS Q3.1 No. The sum of two vectors can only be zero if they are in opposite directions and have the same magnitude. If you walk 10 meters north and then 6 meters south, you wont end up where you started. Q3.2 No, the magnitude of the displacement is always less than or equal to the distance traveled. If two displacements in the same direction are added, then the magnitude of their sum will be equal to the distance traveled. Two vectors in any other orientation will give a displacement less than the distance traveled. If you first walk 3 meters east, and then 4 meters south, you will have walked a total distance of 7 meters, but you will only be 5 meters from your starting point. Q3.3 The largest possible magnitude of R A B= + is 7 units, found when A and B point in the same direction. The smallest magnitude of R A B= + is 3 units, found when A and B have opposite directions. Q3.4 Only force and velocity are vectors. None of the other quantities requires a direction to be described. Q3.5 If the direction-angle of A is between 180 degrees and 270 degrees, its components are both negative. If a vector is in the second quadrant or the fourth quadrant, its components have opposite signs. Q3.6 The books displacement is zero, as it ends up at the point from which it started. The distance traveled is 6.0 meters. Q3.7 85 miles. The magnitude of the displacement is the distance from the starting point, the 260-mile mark, to the ending point, the 175-mile mark. Q3.8 Vectors A and B are perpendicular to each other. Q3.9 No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude. 55

55. 56 Vectors Q3.10 Any vector that points along a line at 45 to the x and y axes has components equal in magnitude. Q3.11 A Bx x= and A By y= . Q3.12 Addition of a vector to a scalar is not defined. Think of apples and oranges. Q3.13 One difficulty arises in determining the individual components. The relationships between a vector and its components such as A Ax = cos , are based on right-triangle trigonometry. Another problem would be in determining the magnitude or the direction of a vector from its components. Again, A A Ax y= +2 2 only holds true if the two component vectors, Ax and Ay , are perpendicular. Q3.14 If the direction of a vector is specified by giving the angle of the vector measured clockwise from the positive y-axis, then the x-component of the vector is equal to the sine of the angle multiplied by the magnitude of the vector. SOLUTIONS TO PROBLEMS Section 3.1 Coordinate Systems P3.1 x r= = = = cos cos . 5 50 240 5 50 0 5 2 75. m . m . ma f a fa f y r= = = = sin sin . 5 50 240 5 50 0 866 4 76. m . m . ma f a fa f P3.2 (a) x r= cos and y r= sin , therefore x1 2 50 30 0= . m .a fcos , y1 2 50 30 0= . m .a fsin , and x y1 1 2 17 1 25, . , . mb g a f= x2 3 80 120= . cosma f , y2 3 80 120= . sinma f , and x y2 2 1 90 3 29, . , . mb g a f= . (b) d x y= + = + =( ) ( ) . . . 2 2 16 6 4 16 4 55 m P3.3 The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m. (a) We can use the Pythagorean theorem to find the distance from the origin to the fly. distance m m m m2 = + = + = =x y2 2 2 2 2 00 1 00 5 00 2 24. . . .a f a f (b) = F HG I KJ = tan .1 1 2 26 6 ; r = 2 24 26 6. , .m

56. Chapter 3 57 P3.4 (a) d x x y y= + = + 2 1 2 2 1 2 2 2 2 00 3 00 4 00 3 00b g b g c h a f. . . . d = + =25 0 49 0 8 60. . . m (b) r1 2 2 2 00 4 00 20 0 4 47= + = =. . . .a f a f m 1 1 4 00 2 00 63 4= F HG I KJ = tan . . . r2 2 2 3 00 3 00 18 0 4 24= + = =. . . .a f a f m 2 135= measured from the +x axis. P3.5 We have 2 00 30 0. .= r cos r = = 2 00 30 0 2 31 . cos . . and y r= = =sin sin .30 0 2 31 30 0 1 15. . . . P3.6 We have r x y= +2 2 and = F HG I KJ tan 1 y x . (a) The radius for this new point is + = + =x y x y ra f2 2 2 2 and its angle is tan F HG I KJ = 1 180 y x . (b) ( ) ( ) + =2 2 22 2 x y r . This point is in the third quadrant if x y,b gis in the first quadrant or in the fourth quadrant if x y,b gis in the second quadrant. It is at an angle of 180+ . (c) ( ) ( )3 3 32 2 x y r+ = . This point is in the fourth quadrant if x y,b gis in the first quadrant or in the third quadrant if x y,b gis in the second quadrant. It is at an angle of .

57. 58 Vectors Section 3.2 Vector and Scalar Quantities Section 3.3 Some Properties of Vectors P3.7 tan . tan . . 35 0 100 100 35 0 70 0 = = = x x m m ma f FIG. P3.7 P3.8 R = = 14 65 km N of E R 13 km 6 km1 km FIG. P3.8 P3.9 = R 310 km at 57 S of W (Scale: 1 20unit km= ) FIG. P3.9 P3.10 (a) Using graphical methods, place the tail of vector B at the head of vector A. The new vector A B+ has a magnitude of 6.1 at 112 from the x-axis. (b) The vector difference A B is found by placing the negative of vector B at the head of vector A. The resultant vector A B has magnitude 14 8. units at an angle of 22 from the + x-axis. y x A + B A A B B B O FIG. P3.10

58. Chapter 3 59 P3.11 (a) d i= =10 0 10 0. . m since the displacement is in a straight line from point A to point B. (b) The actual distance skated is not equal to the straight-line displacement. The distance follows the curved path of the semi-circle (ACB). s r= = = 1 2 2 5 15 7 b g . m C B A 5.00 m d FIG. P3.11 (c) If the circle is complete, d begins and ends at point A. Hence, d = 0 . P3.12 Find the resultant F F1 2+ graphically by placing the tail of F2 at the head of F1 . The resultant force vector F F1 2+ is of magnitude 9 5. N and at an angle of 57 above the -axisx . 0 1 2 3 N x y F2 F1 F1 F2+ FIG. P3.12 P3.13 (a) The large majority of people are standing or sitting at this hour. Their instantaneous foot-to- head vectors have upward vertical components on the order of 1 m and randomly oriented horizontal components. The citywide sum will be ~105 m upward . (b) Most people are lying in bed early Saturday morning. We suppose their beds are oriented north, south, east, west quite at random. Then the horizontal component of their total vector height is very nearly zero. If their compressed pillows give their height vectors vertical components averaging 3 cm, and if one-tenth of one percent of the population are on-duty nurses or police officers, we estimate the total vector height as ~ . m m10 0 03 10 15 2 a f a f+ ~103 m upward .

59. 60 Vectors P3.14 Your sketch should be drawn to scale, and should look somewhat like that pictured to the right. The angle from the westward direction, , can be measured to be 4 N of W , and the distance R from the sketch can be converted according to the scale to be 7 9. m . 15.0 meters N EW S 8.20 meters 3.50 meters 1 m 30.0 R FIG. P3.14 P3.15 To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (Scale: 1 0 5unit m= . ) (a) A + B = 5.2 m at 60 (b) A B = 3.0 m at 330 (c) B A = 3.0 m at 150 (d) A 2B = 5.2 m at 300. FIG. P3.15 *P3.16 The three diagrams shown below represent the graphical solutions for the three vector sums: R A B C1 = + + , R B C A2 = + + , and R C B A3 = + + . You should observe that R R R1 2 3= = , illustrating that the sum of a set of vectors is not affected by the order in which the vectors are added. 100 m C A B R2 B A R1 C B A R3 C FIG. P3.16

60. Chapter 3 61 P3.17 The scale drawing for the graphical solution should be similar to the figure to the right. The magnitude and direction of the final displacement from the starting point are obtained by measuring d and on the drawing and applying the scale factor used in making the drawing. The results should be d = = 420 3ft and (Scale: 1 20unit ft= ) FIG. P3.17 Section 3.4 Components of a Vector and Unit Vectors P3.18 Coordinates of the super-hero are: x y = = = = 100 30 0 86 6 100 30 0 50 0 m m m m a f a f a f a f cos . . sin . . FIG. P3.18 P3.19 A A A A A x y x y = = = + = + = 25 0 40 0 25 0 40 0 47 22 2 2 2 . . . . .a f a f units We observe that tan = A A y x . FIG. P3.19 So = F HG I KJ = = = tan tan . . tan . .1 140 0 25 0 1 60 58 0 A A y x a f . The diagram shows that the angle from the +x axis can be found by subtracting from 180: = = 180 58 122 . P3.20 The person would have to walk 3 10 1 31. 25.0 km northsin . =a f , and 3 10 25 0 2 81. . km eastcos . =a f .

61. 62 Vectors P3.21 x r= cos and y r= sin , therefore: (a) x = 12 8 150. cos , y = 12 8 150. sin , and x y, . . mb g e j= +11 1 6 40i j (b) x = 3 30 60 0. cos . , y = 3 30 60 0. sin . , and x y, cmb g e j= +1 65 2 86. .i j (c) x = 22 0 215. cos , y = 22 0 215. sin , and x y, inb g e j= 18 0 12 6. .i j P3.22 x d= = = cos cos 50 0 120 25 0. m . ma f a f y d= = = = + sin sin . . . 50 0 120 43 3 25 0 43 3 . m m m m a f a f a f a fd i j *P3.23 (a) Her net x (east-west) displacement is + + = +3 00 0 6 00 3 00. . . blocks, while her net y (north- south) displacement is 0 4 00 0 4 00+ + = +. . blocks. The magnitude of the resultant displacement is R x y= + = + =net netb g b g a f a f2 2 2 2 3 00 4 00 5 00. . . blocks and the angle the resultant makes with the x-axis (eastward direction) is = F HG I KJ = = tan . . tan . .1 14 00 3 00 1 33 53 1a f . The resultant displacement is then 5 00 53 1. .blocks at N of E . (b) The total distance traveled is 3 00 4 00 6 00 13 0. . . .+ + = blocks . *P3.24 Let i = east and j = north. The unicyclists displacement is, in meters 280 220 360 300 120 60 40 90 70j i j i j i j i j+ + + + . R i j= + = + = 110 550 110 550 110 550 561 11 3 2 2 1 tan . . m m at m m west of north m at west of north a f a f The crows velocity is v x = = = t 561 11 3 40 14 0 11 3 m at W of N s m s at west of north . . . . R N E FIG. P3.24

62. Chapter 3 63 P3.25 +x East, +y North x y d x y y x = = = = = + = + = = = = = = 250 125 30 358 75 125 30 150 12 5 358 12 5 358 12 5 358 0 0349 2 00 358 2 00 2 2 2 2 + m + . m m . . m at S of E cos sin . tan . . c h c h a f a f c h c h d P3.26 The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to Chicago. In equation form: d d d d d d DC east DA east AC east DC north DA north AC north + . . miles. + . +560 . miles. = = = = = = 730 5 00 560 21 0 527 730 5 00 21 0 586 cos sin sin cos By the Pythagorean theorem, d d d= + =( ) ( )DC east DC north mi2 2 788 . Then tan = = d d DC north DC east .1 11 and = 48 0. . Thus, Chicago is 788 48 0miles at northeast of Dallas. . P3.27 (a) See figure to the right. (b) C A B i j i j i j= + = + + = +2 00 6 00 3 00 2 00 5 00 4 00. . . . . . C = + F HG I KJ = 25 0 16 0 6 401 . . tan .at 4 5 at 38.7 D A B i j i j i j= = + + = +2 00 6 00 3 00 2 00 1 00 8 00. . . . . . D = + F HG I KJ 1 00 8 00 8 00 1 00 2 2 1 . . tan . . a f a f at D = = 8 06 180 82 9 8 06 97 2. . . .at atb g FIG. P3.27 P3.28 d x x x y y y= + + + + + = + + + + = = = F HG I KJ = 1 2 3 2 1 2 3 2 2 2 1 3 00 5 00 6 00 2 00 3 00 1 00 52 0 7 21 6 00 4 00 56 3 b g b g a f a f m. . . . . . . . tan . . .

63. 64 Vectors P3.29 We have B R A= : A A R R x y x y = = = = = = = = 150 75 0 150 120 130 140 35 0 115 140 35 0 80 3 cos120 cm cm cm cm . sin cos . sin . . Therefore, FIG. P3.29 B i j i j B = + = = + = = F HG I KJ = 115 75 80 3 130 190 49 7 190 49 7 196 49 7 190 14 7 2 2 1 a f e j. . . tan . . . cm cm P3.30 A i j= +8 70 15 0. . and B i j= 13 2 6 60. . A B C + =3 0 : 3 21 9 21 6 7 30 7 20 C B A i j C i j = = = . . . . or Cx = 7 30. cm ; Cy = 7 20. cm P3.31 (a) A B i j i j i j+ = + = a f e j e j3 2 4 2 6 (b) A B i j i j i j = = +a f e j e j3 2 4 4 2 (c) A B+ = + =2 6 6 322 2 . (d) A B = + =4 2 4 472 2 . (e) A B+ = F HG I KJ= = tan .1 6 2 71 6 288 A B = F HG I KJ= tan .1 2 4 26 6 P3.32 (a) D A B C i j= + + = +2 4 D = + = = 2 4 4 47 63 42 2 . .m at (b) E A B C i j= + = +6 6 E = + = = 6 6 8 49 1352 2 . m at

64. Chapter 3 65 P3.33 d1 3 50= . je jm d2 8 20 45 0 8 20 45 0 5 80 5 80= + = +. cos . . sin . . .i j i je jm d3 15 0= . ie jm R i j i j= + + = + + = +d d d1 2 3 15 0 5 80 5 80 3 50 9 20 2 30. . . . . .a f a f e jm (or 9.20 m west and