1. INSTRUCTOR'S SOLUTIONS MANUAL FOR SERWAY AND JEWETT'S PHYSICS FOR SCIENTISTS AND ENGINEERS SIXTH EDITION Ralph V. McGrew Broome Community College James A. Currie Weston High School Australia Canada Mexico Singapore Spain United Kingdom United States

2. 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 d yr 1 74 109 109 yr d d s d s s . . ~ FH G IK J + = FH G IK J = . 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

3. 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, Ldiag = L2 + L2 . Thus, since the atoms are separated by a distance L = 0.200 nm, the diagonal planes are separated by 1 2 L2 + L2 = 0.141 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 3 6 37 106 1 08 10 3 21 3 r = e . mj = . m . Its density is then = = 5 98 10 1 08 10 m = V 5 52 10 24 21 3 . kg . 3 3 . 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 kg m3 . The average density of the Earth is significantly higher, so higher-density material must be down below the surface. P1.3 With V = abase areafbheightg V = e r 2 jh and = m V , we have = = FH G 1 m r 2h 2 a fa f IK J = 9 19 5 39 0 4 3 10 1 2 15 10 kg mm mm mm m kg m 3 3 . . . . *P1.4 Let V represent the volume of the model, the same in = m V for both. Then iron = 9.35 kg V and gold m V gold = . Next, gold iron gold kg = m 9.35 and mgold 3 3 19.3 10 kg / m = kg 23 0 3 . 3 kg kg / m FH G IK J 9 35 = 7 86 10 . . . P1.5 V = Vo Vi = r r 4 e 3 3 j 3 2 1 = m V e j e j 4 3 3 , so m V r r r r = = FH G IK J 3 2 = 4 2 1 3 3 3 1 .

4. Chapter 1 3 P1.6 For either sphere the volume is V = r 4 3 3 and the mass is m = V = 4 r 3 3. We divide this equation for the larger sphere by the same equation for the smaller: m m r r r 4 3 4 3 A = A = A = s s rs 5 3 3 3 3 . Then rA = rs 3 5 = 4.50 cma1.71f = 7.69 cm . P1.7 Use 1 u = 1.66 1024 g . (a) For He, m0 FH G IK J 1.66 10 g . u = . = 4 00 6 64 10 24 1 u g -24 . (b) For Fe, m0 FH G IK J 1.66 10 g . u = . = 55 9 9 29 10 23 1 u g -24 . (c) For Pb, m0 24 FH G 1 66 10 IK J g 207 22 = = 3 44 10 u 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 = Nm0 . The first assertion is that the mass of one aluminum atom is m0 = 27.0 u = 27.0 u 1.66 1027 kg 1 u = 4.48 1026 kg . Then the mass of 6.02 1023 atoms is m = Nm = = = 0 6.02 1023 4.48 10 26 kg 0.027 0 kg 27.0 g . Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be written m = Nm0 . 0 . 027 = 26 6 . 02 10 0 027 0 6 02 1023 . kg = . m0 , so m0 23 = 4 48 10 . kg kg , in agreement with the first assertion. (b) The general equation m = Nm0 applied to one mole of any substance gives M g = NM u, where M is the numerical value of the atomic mass. It divides out exactly for all substances, giving 1.000 000 0 103 kg = N1.660 540 2 1027 kg . With eight-digit data, we can be quite sure of the result to seven digits. For one mole the number of atoms is N = FH G IK J 1 + = 1 660 540 2 10 3 27 6 022 137 1023 . . . (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 H2O is 2b1.008 0g+ 15.999 u = 18.0 u. Then the molar mass is 18.0 g . (d) For CO2 we have 12.011 g + 2b15.999 gg = 44.0 g as the mass of one mole.

5. 4 Physics and Measurement P1.9 Mass of gold abraded: m = = = FH G IK J 1 kg 10 g kg 3 b g . 3 80 3 35 0 45 0 45 = . g . g . g . g 4.5 10 4 Each atom has mass m0 27 1 66 10 kg 197 197 25 = = 3 27 10 1 FH G IK J = u u u kg . . . Now, m = N m0 , and the number of atoms missing is N m m = = = 0 4 25 4 . 5 10 kg 1 . 38 10 21 3 . 27 10 kg atoms . The rate of loss is N t N t = FH G IK J FH G IK J FH G IK J FH G IK J 1 38 10 50 = 1 1 1 1 21 atoms yr 8 72 10 11 . . . yr 365.25 d d 24 h h 60 min min 60 s atoms s P1.10 (a) m = L3 = 3 6 = = 3 16 19 e7.86 g cm je5.00 10 cmj 9.83 10 g 9.83 10 kg (b) N m m = = = 0 19 27 9 83 10 7 atoms e j 55 9 1 66 10 1 06 10 . . . . kg u kg 1 u P1.11 (a) The cross-sectional area is a fa f a fa f A= + = 2 0 . 150 0 . 010 0 . 340 0 . 010 6 . 40 10 3 . m m m m m2 . The volume of the beam is V = AL = e6.40 103 m2 ja1.50 mf = 9.60 103 m3 . Thus, its mass is m = V = e7.56 103 kg / m3 je9.60 103 m3 j = 72.6 kg . FIG. P1.11 (b) The mass of one typical atom is m0 27 a f 1 . 66 10 kg 55 9 26 kg . Now = 9 28 10 1 FH G IK J = . u . u m = Nm0 and the number of atoms is N m m = = 72 6 26 = 0 26 9 28 10 7 82 10 . . . kg kg atoms .

6. Chapter 1 5 P1.12 (a) The mass of one molecule is m0 27 FH G 1 66 10 IK J . kg 18 0 26 = = 2 99 10 . u . 1 u kg . The number of molecules in the pail is N m 1 20 25 10 = = molecules pail m kg = 2.99 kg 0 26 4 02 10 . . . (b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere. N N m both pail M pail total (4.02 1025 molecules) kg kg = FH G IK J = . FH G IK J 1 20 1 . 32 1021 , or Nboth = 3.65 104 molecules . 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 = kamtn has dimensions of L = eLT2 j aTf m n or L T L T 1 0 2 = m n m. The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m = 1 . Likewise, equating terms in T, we see that n 2m 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 L2 L e j1 2 2 / = , so this must be area (c). Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension LeL2 j = L3 , so it is (b). Thus, (a) = ii; (b) = iii, (c) = i .

7. 6 Physics and Measurement P1.15 (a) This is incorrect since the units of ax are m2 s2 , while the units of v are m s . (b) This is correct since the units of y are m, and cosakxf is dimensionless if k is in m1 . *P1.16 (a) a or a k F m = represents the proportionality of acceleration to resultant force and F m the inverse proportionality of acceleration to mass. If k has no dimensions, we have a k F m = , L T 1 F 2 M = , F M L T2 = . (b) In units, M L T kg m = 2 s2 , so 1 newton = 1 kg m s2 . P1.17 Inserting the proper units for everything except G, kg m 2 kg m s LN M OQ P = G 2 2 . Multiply both sides by m 2 and divide by kg 2 ; the units of G are 3 2 m kg s . Section 1.5 Conversion of Units *P1.18 Each of the four walls has area a8.00 ftfa12.0 ftf = 96.0 ft2 . Together, they have area m2 e jFH G 4 96 0 1 3 28 2 m . ft IK J 2 35 7 = . . ft . P1.19 Apply the following conversion factors: 1 in = 2.54 cm, 1 d = 86 400 s , 100 cm= 1 m, and 109 nm= 1 m 1 2 54 cm in 10 2 m cm 10 9 nm m in day 9 19 nm s 32 86 400 s day G FH IK J = . . b ge je j . This means the proteins are assembled at a rate of many layers of atoms each second! 3 = 3 m3 FH G IK J *P1.20 8 50 8 50 0 025 4 1 39 10 3 . m . in . in . 4 1 in =

8. 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 a30 mfa50 mf = 1 500 m2 . 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 = FH G IK J FH G IK J a f a f = 1 390 m2 m2 1 3 281 m ft 100 150 = 1 3 281 ft ft 1 39 103 m ft . . . . 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 = a40.0 mfa20.0 mfa12.0 mf = 9.60 103 m3 b g . V = 9.60 103 m3 3.28 ft 1 m 3 = 3 39 105 ft3 (b) The mass of the air is e1.20 kg m3 je9.60 103 m3 j 1.15 104 kg. m = airV = = The student must look up weight in the index to find Fg = mg = e1.15 104 kgje9.80 m s2 j = 1.13 105 N. Converting to pounds, Fg = e1.13 105 Njb1 lb 4.45 Ng = 2.54 104 lb . P1.23 (a) Seven minutes is 420 seconds, so the rate is r = = 30 0 7 14 10 2 . 420 . gal s gal s . (b) Converting gallons first to liters, then to m3 , r r = FH G IK J FH G IK J e j = 7 14 10 3 786 10 2 70 10 2 3 4 . . gal s . . L 1 gal m 1 L m s 3 3 (c) At that rate, to fill a 1-m3 tank would take t = FH G IK J FH G IK J m = 1 2 70 10 1 1 03 4 m s h 3 600 h 3 . 3 . .

9. 8 Physics and Measurement *P1.24 (a) Length of Mammoth Cave = FH G IK J km mi 1 609 1 mi 560 5 60 105 5 60 107 348 = = = km m cm . . . . (b) Height of Ribbon Falls = FH G IK J 0 .304 8 m ft 491 m 0 491 4 91 104 1 612 = = = 1 ft . km . cm . (c) Height of Denali = FH G IK J 0 .304 8 m ft 6 6 19 103 6 19 105 20 320 = = = 1 ft .19 km . m . cm . (d) Depth of Kings Canyon = FH G IK J 0 .304 8 m ft 2 2 50 103 2 50 105 8 200 = = = 1 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. = FH G IK J FH G IK J . 4 . . kgm3 = g cm 23 94 2 10 kg g 1 1 000 100 1 3 cm m 1 14 10 3 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 a f1 mi 2 1 609 m 2 1 acre 4 05 G FH 10 3 m 2 640 acres mi IK J FH G IK J = . . *P1.27 The weight flow rate is 1 200 2 000 1 1 667 lb s G FH ton h lb ton h 60 min min 60 s IK JFH GIK J FH G IK J = . P1.28 1 mi = 1 609 m= 1.609 km; thus, to go from mph to km h, multiply by 1.609. (a) 1 mi h = 1.609 km h (b) 55 mi h = 88.5 km h (c) 65 mi h = 104.6 km h. Thus, v = 16.1 km h .

10. Chapter 1 9 P1.29 (a) 6 10 1 1 1 190 12 FH G IK J FH G IK J FH G IK J FH G IK J $ = 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 e6.378 103 mj = 4.01 107 m. 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 times . 7 11 . 4 . m = 4.01 0 m P1.30 N m 1 99 10 = = atoms atoms m Sun atom kg = 1.67 10 kg 119 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 1 3 P1.32 V = Bh = = a 13 0 acres fe 43 560 ft acre ja f 3 481 9 08 107 . . , ft ft 2 3 or e j B V= FH G IK J = 9 08 10 2 83 10 1 7 2 57 10 2 6 . . . ft m ft 3 m 3 3 3 h FIG. P1.32 P1.33 Fg = b2.50 tons blockge2.00 106 blocksjb2 000 lb tong = 1.00 1010 lbs *P1.34 The area covered by water is a fe j a fa fe 6 j2 2 . Aw = 0.70AEarth = 0.70 4 RE2arth = 0.70 4 6.37 10 m = 3.6 1014 m The average depth of the water is d = a2.3 milesfb1 609 m l mileg = 3.7 103 m. The volume of the water is V = Awd = e3.6 1014 m2 je3.7 103 mj = 1.3 1018 m3 and the mass is m = V = e1 000 kg m3 je1.3 1018 m3 j = 1.3 1021 kg .

11. 10 Physics and Measurement e 2 . 40 10 j . 3 , or = ft P1.35 (a) d d d atom, scale atom, real nucleus, scale nucleus, real d m ft 1.06 10 m FH G IK J = FH G IK J = 300 15 6 79 10 10 dnucleus, scale = e6.79 103 ftjb304.8 mm 1 ftg = 2.07 mm (b) V V r r d d r r atom nucleus atom nucleus atom nucleus atom nucleus m m times as large = =FH G IK J = FH G IK J = FH G IK J = 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 scale = e . km j . 3 m km distance factor . 9 FH G m IK J FH G IK J = FH G IK J = 4 0 10 7 0 10 1 4 10 13 200 P1.37 The scale factor used in the dinner plate model is S = 0 25 = 1 0 10 . 6 2 10 5 . m lightyears .5 m lightyears . The distance to Andromeda in the scale model will be = = 2.0 10 lightyears 2.5 10 m lightyears = m e je j 5.0 . Dscale DactualS 6 6 P1.38 (a) A A r r r r Earth Moon Earth 2 Moon Earth Moon m cmm cm = =FH G IK J = 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 cmm cm = =FH G IK J = 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, mFe = mAl or FeVFe = AlVAl Fe Fe Al Al Al Fe Fe Al a cm f . cm . 4 3 4 3 2 00 7 86 2 70 2 86 3 3 1 3 1 3 FH G IK J = FH G IK J = FH G IK J = FH G IK J = r r r r / / . . .

12. Chapter 1 11 P1.40 The mass of each sphere is = = Al Al m V r Al Al Al 4 3 3 and = = Fe Fe m V r Fe Fe Fe 4 3 3 . Setting these masses equal, 3 3 AlrAl FerFe 4 3 4 = and rAl rFe 3 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 3 2 87 10 m3 FH G . 5 . m 2 IK J = . Therefore, one can fit about 48 ping-pong balls in the room. 2 87 10 10 5 6 . ~ 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 1.67 106 0.740~106 . 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 b50 000 migb5 280 ft migb1 rev 8 ftg = 3 107 rev~ 107 rev . P1.43 In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least 1 16 in2 = 43 105 ft2 . 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= = 7 . = total area area per blade acre ft acre ft blade 2.5 10 blades blades 2 2 7 0 25 43 560 43 10 ~ 10 5 a fe j .

13. 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 in3 . Since 1 acre = 43 560 ft2 , the volume of water required to cover it to a approximately 4 103 depth of 1 inch is 2 2 a fa f a f ft 1 acre 1 inch = 1 acre in 6.3 106 in3 1 acre in ft FH G IK J FH G IK J 43 560 144 1 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 = a0.5fa1.3 mfa0.5 mfa0.3 mf = 0.10 m3 . The mass of this volume of water is = = e1 000 kg m3 je0.10 m3 j = 100 kg ~102 kg . mwater waterV Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is = = e8 920 kg m3 je0.10 m3 j = 892 kg ~103 kg . mcopper copperV 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 e250 106 cans dayjb365 days yearg 1011 cans are thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, we estimate this represents e1011 cansjb0.1 oz cangb1 lb 16 ozgb1 ton 2 000 lbg 3.1 105 tons year . ~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 FH G IK J FH G IK J (107 ) = 100 .

14. Chapter 1 13 Section 1.7 Significant Figures *P1.48 METHOD ONE We treat the best value with its uncertainty as a binomial a21.3 0.2f cm a9.8 0.1f cm, A = 21.3a9.8f 21.3a0.1f 0.2a9.8f a0.2fa0.1f 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 cm2 4 cm2 . METHOD TWO We add the fractional uncertainties in the data. A= + FH G IK J a cmfa cmf cm cm cm 0 2 21 3 0 1 9 8 . . . . . . 209 2 2% 209 2 4 2 21 3 9 8 = = a f ( . ) ( . )( . ) ( . ) r 2 2 = = + = P1.49 (a) 10 5 0 2 10 5 2 10 5 0 2 0 2 346 13 . m . m 2 2 m m m m m 2 m 2 (b) 2 r = 2 a10.5 m 0.2 mf = 66.0 m 1.3 m P1.50 (a) 3 (b) 4 (c) 3 (d) 2 a f a f a f c h P1.51 r m 6 50 0 20 6 50 0 20 10 1 85 0 02 = = = = . . cm . . m . . kg m r 2 4 3 3 also, = + m m r r 3 . In other words, the percentages of uncertainty are cumulative. Therefore, . a f , 3 0 20 6 50 = 0 02 + = 1 85 0 103 . . . . c h . e mj = . = 3 3 . 1 85 6 5 10 1 61 10 4 3 2 3 kg m and = a1.61 0.17f 103 kg m3 = a1.6 0.2f103 kg m3 .

15. 14 Physics and Measurement P1.52 (a) 756.?? 37.2? 0.83 + 2.5? 796. /5/3 = 797 (b) 0.003 2a2 s.f.f 356.3a4 s.f.f = 1.140 16 = a2 s.f.f 1.1 (c) 5.620a4 s.f.f a>4 s.f.f = 17.656=a4 s.f.f 17.66 *P1.53 We work to nine significant digits: = s FH G 1 1 365 242 199 d 24 60 60 yr yr 31 556 926 0 1 yr h 1 d min 1 h s 1 min IK J FH G IK J FH G IK J FH G IK J = . . . 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 1 2 2 2 1 2 170 10 10 10 009 170 10 0 1 0 0 090 0 900 2 1 70 0 900 5 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 A A 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 1 000 x 100 m x 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 = a100 mfb1 000 mg = 1.00 105 m2 and therefore x = 1.00 105 m2 = 316 m .

16. 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 FH G IK J a f 1.66 10 kg . kg . = 197 = 3 27 10 25 u 1 u So, the number of atoms in the cube is N = 19 300 kg 28 = 5 90 10 25 3.27 10 kg . . The imagined cubical volume of each atom is d3 1 m = 29 5 90 10 = 1 69 10 28 m 3 3 . . . So d = 2.57 1010 m . a fe j e j e j P1.58 A NA V V A V 4 r 2 total drop total drop drop r total 4 = = FH G IK J = F H GG I K JJ 3 3 A V = total FH G m total r 3 m 2 m IK J = FH G IK J = 3 3 30 0 10 2 00 10 4 50 6 5 . . . P1.59 One month is 1 mo = b30 daygb24 h daygb3 600 s hg = 2.592 106 s . Applying units to the equation, V = e1.50 Mft3 mojt + e0.008 00 Mft3 mo2 jt 2 . Since 1 Mft3 = 106 ft3 , V = e1.50 106 ft3 mojt + e0.008 00 106 ft3 mo2 jt 2 . Converting months to seconds, 1 50 106 0 008 00 106 . ft mo . ft mo 2 2.592 10 s mo V = t + t 2 2.592 10 s mo 3 6 3 2 e 6 j . Thus, V [ft3 ] = e0.579 ft3 sjt + e1.19 109 ft3 s2 jt 2 .

17. 16 Physics and Measurement P1.60 (deg) (rad) tana f sina 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 2 150 2 39 2 39 55 0 3 41 r r h r h = = = = = . m . m tan 55.0 a . mftan( . ) . m 55 h r FIG. P1.61 *P1.62 Let d represent the diameter of the coin and h its thickness. The mass of the gold is IK J m V At d FH G 2 4 2 = = = + dh t where t is the thickness of the plating. L N MM a 2 . 41 f 2 a fa fe j 4 grams 19 3 2 . 4 m= + O Q PP . . . 2 41 0 178 0 18 10 = = = = . . $10 $0. . 0 003 64 0 003 64 036 4 3 64 cost grams gram cents This is negligible compared to $4.98. P1.63 The actual number of seconds in a year is b86 400 s daygb365.25 day yrg = 31 557 600 s yr . The percent error in the approximation is e jb g = 10 31 557 600 31 557 600 100% 0 449% 7 s yr s yr s yr . .

18. Chapter 1 17 P1.64 (a) V = L3 , A = L2 , h = L V = A h L3 = L2L = L3 . Thus, the equation is dimensionally correct. (b) Vcylinder = R2h = e R2 jh = Ah , where A = R2 Vrectangular object = Awh = aAwfh = Ah , where A = Aw P1.65 (a) The speed of rise may be found from v Vol rate of flow (Area: = = = D cm s cm s 3 cm a f 2 a 6 30 f2 4 4 16 5 0 529 ) . . . . (b) Likewise, at a 1.35 cm diameter, 16 5 v= = 11 5 1.35 4 2 . . cm s cm s 3 a cmf . P1.66 (a) 1 cubic meter of water has a mass e . kg cm je . m je cm mj kg m = V = 1 00 103 3 1 00 3 102 = 1 000 3 (b) As a rough calculation, we treat each item as if it were 100% water. = FH G e j e j 3 3 3 6 3 = = FH G IK J = FH G IK J IK J 4 3 1 6 1 6 1 000 1 0 10 cell: m V R D kg m m kg = e 3 j G FH = 5 2 10 = = FH G 4 3 4 3 m V R IK J IK J kidney: . kg cm cm kg 1 00 10 = FH G IK J 4 0 0 27 . e j a fa fe 3 2 = J = IK GFH m Dh fly: kg cm mm mm cm mm 4 1 10 4 2 0 4 0 10 16 3 3 3 2 3 1 . ( . ) . . . j3 = 1.3 105 kg P1.67 V20 mpg 10 8 4 cars mi yr (10 )(10 ) = = 5.0 10 gal yr mi gal 20 V25 mpg 10 8 4 cars mi yr (10 )(10 ) = = 4.0 10 gal yr mi gal 25 Fuel saved = V25 mpg V20 mpg = 1.0 1010 gal yr

19. 18 Physics and Measurement P1.68 v =FH G IK J FH G IK J FH G IK J FH G IK J FH G IK J FH G IK J 220 1 0 914 4 1 1 14 1 24 1 3 600 . 5 00 = . 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 2t 21 2 19 61 = e10 mj e10 mj~10 m3 . If the distance between stars is 4 1016 m, then there is one star in a volume on the order of 4 1016 10 3 50 e mj ~ m3 . The number of stars is about 10 10 10 61 50 m 11 m star stars 3 3 ~ . = = = m V P1.70 The density of each material is m r h 4 m 2 D2h . Al: 4 515 . FH G g = = 2 52 3 75 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 IK J = = FH G IK J = = = = = = 2 75 2 70 2% 4 56 3 1 5 9 36 8 92 5% 4 94.4 g 1 5 8 91 4 691 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 fa f b g a fa f b g a fa f The tabulated value is smaller. 3 7.86 0.3% FH G g cm IK J P1.71 (a) b3 600 s hrgb24 hr daygb365.25 days yrg = 3.16 107 s yr e j 3 7 3 19 3 = = = (b) V r V V mm cube mm . m . m 18 m m 1.91 10 micrometeorites = = 4 3 4 3 5 00 10 5 24 10 1 5 24 10 3 19 3 . This would take 1 91 10 3 16 10 6 05 10 18 7 . . 10 . = micrometeorites micrometeorites yr yr .

20. Chapter 1 19 ANSWERS TO EVEN PROBLEMS P1.2 5.52 103 kg m3 , 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 P1.4 23.0 kg P1.38 (a) 13.4; (b) 49.1 P1.6 7.69 cm 1 3 P1.40 rAl rFe Fe Al = FH G IK J P1.8 (a) and (b) see the solution, NA = 6.022 137 1023 ; (c) 18.0 g; P1.42 ~107 rev (d) 44.0 g P1.44 ~109 raindrops P1.10 (a) 9.83 1016 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 a209 4f cm2 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 newton = 1 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 1.39 104 m3 P1.58 4.50 m2 P1.22 (a) 3.39 105 ft3 ; (b) 2.54 104 lb P1.60 see the solution; 24.6 P1.24 (a) 560 km= 5.60 105 m= 5.60 107 cm; (b) 491 m= 0.491 km= 4.91 104 cm; P1.62 3.64 cents ; no (c) 6.19 km= 6.19 103 m= 6.19 105 cm; P1.64 see the solution (d) 2.50 km= 2.50 103 m= 2.50 105 cm P1.66 (a) 1 000 kg; (b) 5.2 1016 kg ; 0.27 kg ; 1.3 105 kg P1.26 4.05 103 m2 P1.28 (a) 1 mi h = 1.609 km h; (b) 88.5 km h ; (c) 16.1 km h P1.68 8.32 104 m s ; a snail P1.30 1.19 1057 atoms P1.70 see the solution P1.32 2.57 106 m3

21. 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 b331 m sga5.0 sf = 1.7 km. 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 t0 t 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

22. 22 Motion in One Dimension Q2.8 Yes. If you drop a doughnut from rest av = 0f, 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 av > 0f, and its acceleration is directed down aa < 0f. 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 , 0 707 2 . h = ga . tf . The time is later than 0.5t. (a) 0 5 1 2 0 5 2 . h = ga . tf . The elevation is 0.75h, greater than 0.5h. (b) The distance fallen is 0 25 1 2

23. 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 = = FH G x t IK J ft 7 1 yr FH G IK J 20 1 1 = 3 156 10 2 10 7 m 3.281 ft yr s m s . or in particularly windy times v = = FH G x t IK J ft 6 FH G IK J 100 1 1 = 3 156 10 1 10 7 1 yr m 3.281 ft yr s m s . . (b) The time required must have been = = FH G t x v IK J FH G IK J = 3 000 1 609 10 5 10 3 mi 8 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 = = 5 10 2 1 = 2 1 2 2 5 m m 4 s s . m s (d) v x x t t = 5 5 2 1 = = 2 1 4 3 3 m m 7 s s . m s (e) v x x t t = = 0 0 8 0 2 1 = 2 1 0 m s P2.4 x = 10t2 : For a f a f t x s m = = 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

24. 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 . ms= d the higher speed in 5 00 1 t . Let t2 represent the longer time for the return trip in 3 00 = 2 . ms d t . Then the times are t d = b . m sg and t 1 5 00 d = b . m sg. The average speed 2 3 00 is: v d d d v 2 b g b g b g d d d Total distance Total time = = + + = m s m s = = . . m s m s m s m s m s 2 2 8 00 15 0 5 00 3 00 2 2 2 15 0 8 00 3 75 . . . . . 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 e j. P2.6 (a) At any time, t, the position is given by x t = 3 00 2 . m s2 Thus, at ti = 3.00 s: xi = 3 00 3 00 = 27 0 2 e . m s2 ja . sf . m . (b) At t f = 3.00 s + t: x f = 3 00 3 00 + t 2 e . m s2 ja . s f , or b g e ja f . x t t f = + + 27 0 18 0 3 00 2 . . . m m s m s2 (c) The instantaneous velocity at t = 3.00 s is: v x x lim lim . . . t t t f i t = FH G IK J = + = 0 0 e18 0 m s e3 00 m s2 j j 18 0 m s . P2.7 (a) at ti = 1 5 . s, xi = 8 0 . m (Point A) at t f = 4 0 . s, x f = 2 0 . m (Point B) v x x t t f i f i = = a 2 0 8 0 f = 6 0 a f = . 4 15 2 4 . . . . m s m 2.5 s m s (b) The slope of the tangent line is found from points C and D. btC = 1.0 s, xC = 9.5 mg and btD = 3.5 s, xD = 0g, v 3 8 . m s . FIG. P2.7 (c) The velocity is zero when x is a minimum. This is at t 4 s .

25. 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 f xi =a0.2 m sf(25 s)= 5.00 m .

26. 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 f = vi +at : a v t = = m s2 a f . . 4 = 22 0 25 0 3 50 10 1 34 10 3 . . m s m s s . P2.12 (a) Acceleration is constant over the first ten seconds, so at the end, v f = vi +at=0+c2.00 m s2 h(10.0 s)= 20.0 m s . 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 f = vi +at= 20.0 m s+c3.00 m s2 h(5.00 s)= 5.00 m s . (b) In the first ten seconds, x f = xi +vit+ at = + + ( ) = 1 2 0 0 1 2 2 2 00 10 0 2 100 c . m s2 h . s m. Over the next five seconds the position changes to x f = xi +vit+ at = + ( )+ = 1 2 2 100 m a20.0 m sf 5.00 s 0 200 m. And at t= 20 0 . s, 2 3 00 5 00 2 262 m a . m sf . s c . m s2 h . s m . x f = xi +vit+ at = + ( )+ ( ) = 1 2 200 20 0 5 00 1 2 *P2.13 (a) The average speed during a time interval t is v = distance traveled t . During the first quarter mile segment, Secretariats average speed was v1 = 0 . 250 mi 1 320 = = 52 . 4 35 . 6 25.2 s ft 25.2 s ft s b mi hg. During the second quarter mile segment, v2 1 320 ft = = 55 0 37 4 24.0 s . ft s b . mi hg. For the third quarter mile of the race, v3 1 320 ft = = 55 5 37 7 23.8 s . ft s b . mi hg, and during the final quarter mile, v4 1 320 ft = = 57 4 39 0 23.0 s . ft s b . mi hg. continued on next page

27. Chapter 2 27 (b) Assuming that v f = v4 and recognizing that vi =0 , the average acceleration during the race was a v f vi = = ft s2 57 4 0 ( + + + ) = total elapsed time ft s s 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

28. 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 = 3.00a9.00f2 2.00a3.00f+ 3.00 m= 24.0 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 = c3.00t 2 2.00t+3.00h=(6.00t2.00) m s At t= 2.00 s , v= 6.00(2.00)2.00 m s= 10.0 m s . At t=3.00 s , v= 6.00(3.00)2.00 m s= 16.0 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 8 00 = = t 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

29. 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 f vi ax c . m sh = + a( m), so that a=2.74105 m s2 2 = 2 +2 , we have 10 97 103 2 0 2 220 which is a= 2.79104 times g . P2.20 (a) x f xi = vi +v f t 1 2c h becomes 40 m= avi +2.80 m sf(8.50 s) which yields vi = 6.61 m s . 1 2 (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 xi =3.00 cm(t=0), and at t= 2.00 s , x f =5.00 cm, x f xi = vit+ at 1 2 1 2 2 00 2 . . . . a . 2 : 5 003 00=12 0(2 00)+ ( ) 8.00= 24.0+2a 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 2 2 v v a x x xf xi x f i a a x x 2 2 0 60 2 121 0 1 3 600 242 5 280 1 21 8 21 8 1 609 1 9 75 = + = + FH G IK J = FH G IK J FH G IK J = = FH G IK J FH G IK J = d i b mi hg a ft f 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, b g a f 0 80 2 a 211 0 mi h ft b g b g 6 400 5 280 422 3 600 mi h s 22 2 mi h s 9 94 m s2 2 = + = = = a x x . . . (c) Let i be moving at 80 mi h and f be moving at 60 mi h. 2 2 d i v v a x x xf xi x f i b mi h g b mi h g a a ft ft f a x 2 2 x 2 60 80 2 211 121 b g a fb g 2 800 5 280 2 90 3600 mi h s 22 . 8 mi h s 10 . 2 m s2 . = + = + = = =

30. 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 t f i xi x = + + 1 2 2 : m= +a m sft+ c3.5 m s2 ht 2 c1.75 m s2 ht2 a30 m sft+100 m=0 . 100 0 30 1 2 We use the quadratic formula: t = 2 4 b b ac a 2 t= ( ) = = . 30 900 4 1 75 100 2 175 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) vxf = vxi + axt = 30 m s e3.5 m s2 j4.53 s = 14.1 m s P2.24 (a) Total displacement = area under the av, tf curve from t=0 to 50 s. x x b ga f b ga f b ga f = + + = 1 2 50 15 50 40 15 1 2 m s s m s s m s s m 50 10 1 875 (b) From t=10 s to t= 40 s , displacement is x = 1 + + = 2 b50 m s 33 m sga5 sf b50 m sga25 sf 1 457 m . FIG. P2.24 (c) 0t15 s : a = v t 1 ( ) 50 0 15 0 = = 3 3 m s s . m s2 15 s

31. Chapter 2 31 1 2 1 2 c h or x1 t 0 2 2 3 3 = + = . m s2 (d) (i) x1 a1t t 2 1 67 = . m s2 c h = (15 s) 50 m s0 +a50 m sf( 15 s) or x2 =a50 m sft375 m 1 2 (ii) x2 t (iii) For 40 st50 s, x v t 3 a3 t t t 2 0 1 2 = 40 50 40 = FH G IK J + ( ) + ( ) area under vs from to 40 s s a m sf s or = m+ m+ e5.0 m s2 ja 40 sf + b50 m sga 40 sf 1 2 2 375 1 250 x3 t t which reduces to = b250 m sg e2.5 m s2 j 2 4 375 m . x3 t t total displacement total elapsed time m s (e) v= = = m s 1 875 50 37 5 . P2.25 (a) Compare the position equation x= 2.00+3.00t4.00t 2 to the general form x x v t at f 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 f = vi +at , is then v f =3.00 m sc8.00 m s2 ht . The particle changes direction when v f =0 , which occurs at t= 3 8 s . The position at this time is: x= + FH G IK J FH G IK J 3 8 2 00 3 00 4 00 = 3 8 2 56 2 . m a . m sf s c . m s2 h s . m . (b) From x x v t at f i i = + + 1 2 2 , observe that when x x f = i , the time is given by t v a = i 2 . Thus, when the particle returns to its initial position, the time is t= = a f 2 3 00 8 00 3 4 . . m s m s s 2 and the velocity is v f = FH G IK J 3 4 . m s c . m s2 h s 3.00 m s . 3 00 8 00 =

32. 32 Motion in One Dimension *P2.26 The time for the Ford to slow down we find from = + + x x v v t f i xi xf t x v v xi xf = + = + = 1 2 2 2 250 71 5 0 6 99 d i a mf m s s . . . Its time to speed up is similarly t= ( ) + = 2 350 0 715 9 79 m m s s . . . The whole time it is moving at less than maximum speed is 6.99 s+5.00 s+9.79 s= 21.8 s . The Mercedes travels x f = xi + vxi + vxf t = + + = 1 2 0 1 2 71 5 71 5 21 8 1 558 d i a . . fbm sga . sf m while the Ford travels 250 + 350 m= 600 m, to fall behind by 1 558 m 600 m= 958 m . P2.27 (a) vi =100 m s , a=5 00 . m s2 2 = 2 +2 c h so , v f = vi +at so 0=1005t , v f vi a x f xi 0 100 2500 0 =( )2 ( . ) x f c 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 at f i i = + + 1 2 2 to find a . m sf c 2.00 m s2 h 2 1 2 x f =0+ 30 0t + t when t is in seconds x f =c30.0tt 2 h m . To find an equation for the velocity, use v f = vi + at = 30.0 m s + e2.00 m s2 jt , v f =(30.02.00t) 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 f =(3.002.00t) m s , v f =0 when t=15.0 s. Then xmax =c30.0tt2 h m=(30.0)(15.0)(15.0)2 = 225 m .

33. Chapter 2 33 P2.29 In the simultaneous equations: = + v v at xf xi x = + x x v v t f i xi xf RS | T| UV | W| 1 2c h we have c 2 h c h . = ( ) v v . . m s s = + ( ) . . m v v s xf xi xi xf RS | T | UV | W | 5 60 4 20 62 4 1 2 4 20 So substituting for vxi gives 62 4 1 2 . m= vxf + 56.0 m s2 (4.20 s)+vxf (4.20 s) c h 14 9 1 2 . m s= vxf + 5.60 m s2 (4.20 s) c h . Thus vxf = 3.10 m s . P2.30 Take any two of the standard four equations, such as = + v v at xf xi x = + x x v v t f i xi xf RS | T| UV | W| 1 2c h. Solve one for vxi , and substitute into the other: vxi = vxf axt x f xi = vxf axt+vxf t 1 2c h. Thus x f xi = vxf t axt 1 2 2 . Back in problem 29, 62 4 4 20 1 2 5 60 4 20 2 . m= vxf ( . s) . m s2 ( . s) 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 3 600 e j . ft s2 m s2 (b) x f = vit + at = FH G IK J 2 a f a . f a662fa1.40f2 649 ft 198 m 1 632 = = 2 5 280 3 600 1 40 1 2

34. 34 Motion in One Dimension P2.32 (a) The time it takes the truck to reach 20.0 m s is found from v f = vi +at . Solving for t yields t v v a f i = = = 20 0 0 2 00 10 0 . . . m s m s m s s 2 . The total time is thus 10.0 s+20.0 s+5.00 s= 35.0 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 10 0 100 = = + FH G x1 vt 0 200 2 IK J ( )= . . m. With a being 0 for this interval, the distance traveled during the next 20.0 s is 1 2 2 = + =(20.0)(20.0)+0= 400 m. x2 vit at The distance traveled in the last 5.00 s is 5 00 50 0 = = + FH G x3 vt 20 0 0 2 IK J ( )= . . . m. The total distance x= x1 +x2 +x3 =100+400+50=550 m, and the average velocity is given by v = = = 550 x t 35 0 15 7 . . m s . P2.33 We have vi = 2.00104 m s, v f =6.00106 m s , x f xi =1.50102 m. (a) x f xi = vi +v f t 1 2c h: t 9 c h c . h 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 . . . m m s m s s 2 = 2 + 2 d i: (b) v f vi ax x f xi a 2 e je j 2 2 6 2 4 2 v v f i f i = x x x = = 2 15 2 6 00 10 2 00 10 2 1 50 10 1 20 10 ( ) . . ( . ) . m s m s m m s

35. Chapter 2 35 2 = 2 +2 c h: 0 01 3 108 0 2 40 *P2.34 (a) vxf vxi ax x f xi 2 . c m sh = + ax( m) ax = c h m s2 = 3 10 80 1 12 10 6 2 11 m s m . (b) We must find separately the time t1 for speeding up and the time t2 for coasting: d i : m e m s j f i = xf + xi = + x x v v t t t = 1 2 40 1 2 3 10 0 2 67 10 1 6 1 1 5 . s d i : e j f i = xf + xi = + x x v v t t m m s m s t = 1 2 60 1 2 3 10 3 10 2 00 10 2 6 6 2 2 5 . s total time = 4.67105 s . *P2.35 (a) Along the time axis of the graph shown, let i=0 and f =tm . Then vxf = vxi +axt gives vc =0+amtm a v m t c m = . (b) The displacement between 0 and tm is = + = + m = c m 1 x x v t at v t c m f i xi x t v t 2 0 1 2 1 2 2 2 . The displacement between tm and t0 is 0 a f . x f xi = vxit+ axt = vc t tm + 1 2 2 0 The total displacement is x = vctm + vct vctm = vc t tm FH G IK J 1 2 1 0 0 2 . (c) For constant vc and t0 , x is minimized by maximizing tm to tm =t0 . Then 1 0 2 2 x v t t v t c c min = FH GIK J 0 0 = . (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= vc at0 0f= vct0 . (e) This cannot be attained because the acceleration must be finite.

36. 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, = + + 1 2 x x v t at f i xi x = + + = = + v t a t v t i d d d d 0 v v a t d i d 1 2 1 2 2 A 2 (a) The speed halfway through the photogate in space is given by 2 2 2 2 vhs vi a vi avd td 2 = + FH G IK J A = + . vhs = vi2 +avdtd and this is not equal to vd unless a=0 . = + d FH G (b) The speed halfway through the photogate in time is given by v v a t ht i IK J 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 f xi =9.00 m. Then 2 2 2 02 2 0 500 9 00 = + = + = v v ax x v f i f i f 3 00 d i e . ja . f . . m s m m s 2 (b) x f xi = vit+ at 1 2 2 e 2 jt . . 2 = + = 9 00 0 1 2 0 500 6 . 00 m s s 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 f xi =15.00 m. 2 = 2 +2 c h gives v f vi a x f xi a v f v i x x a . f c h 0 300 f i = = ( ) = 2 2 2 2 0 300 2 15 0 . . 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 x f i =8 00 . m, a=0 300 . m s2 2 2 2 2 3 00 2 0 300 8 00 4 20 = + = + = = v v ax x v f i f i f 2 05 d i b . g e . ja . f . . . m s m s m m s m s 2 22

37. 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 2 a . m sf c 2.00 m s2 h 2 . xs(t)= xis +vist+ ast = t+ t 1 2 30 0 1 2 For the van, xiv =155 m, viv = 5.00 m s , av =0 and xv(t)= xiv +vivt+ avt = + t+ 1 2 2 155 a5.00 m sf 0 . To test for a collision, we look for an instant tc when both are at the same place: 2 = + 30 0 155 5 00 . . t t t c c c 2 = + 0 t 250 t 155 . . 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 m+a5.00 m sf(11.4 s)= 212 m . *P2.39 As in the algebraic solution to Example 2.8, we let t represent the time the trooper has been moving. We graph xcar = 45+45t and xtrooper = 1.5t2 . They intersect at t = 31 s . x (km) t (s) 10 20 30 40 1.5 1 0.5 car police officer FIG. P2.39

38. 38 Motion in One Dimension Section 2.6 Freely Falling Objects P2.40 Choose the origin ay=0, t=0f 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 f yi = vit+ at 1 2 2 : y gt t f = = 1 2 1 2 e j 9 80 2 2 . m s2 and c h. v v at f i = + : v gt t f = = 9 80 . m s2 (a) at t=1.00 s : y f = ( ) = 1 2 9 80 1 00 4 90 2 c . m s2 h . s . m at t= 2.00 s : y f = ( ) = 1 2 9 80 2 00 19 6 2 c . m s2 h . s . m at t=3.00 s : y f = ( ) = 1 2 9 80 3 00 44 1 2 c . m s2 h . s . m (b) at t=1.00 s : v f =c9.80 m s2 h(1.00 s)= 9.80 m s at t= 2.00 s : v f =c9.80 m s2 h(2.00 s)= 19.6 m s at t=3.00 s : v f =c9.80 m s2 h(3.00 s)= 29.4 m s 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: d i: e . jb g 2 = 2 + 2 0 = 2 2 9 80 1 609 v v ay y v f i f i i v i 178 = . 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 f yi = vit+ at 1 2 =a m sft c9.80 m s2 ht2 . 2 : 0 178 1 2 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 b $ sgb s hg h . 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.

39. Chapter 2 39 P2.42 We have y f = gt +vit+yi 1 2 2 0=c4.90 m s2 ht2 a8.00 m sft+30.0 m. 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 f yi = vit+ at 1 2 2 : 4 00 1 50 4 90 1 50 2 . =( . )vi ( . )( . ) and vi = 10.0 m s upward . (b) v f = vi +at=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= vit gt = ( ) = 1 2 2 0 4 90 0 20 2 0 20 c . m s2 h . s . m. This distance is about twice the distance between the center of the bill and its top edge a 8 cmf. Thus, David will be unsuccessful . *P2.45 (a) From y= vit+ at 1 2 2 with vi =0 , we have t y a = = ( ) = 2 2 23 s . 2 9 80 2 17 a f m m s . . (b) The final velocity is v f =0+c9.80 m s2 h(2.17 s)= 21.2 m s . (c) The time take for the sound of the impact to reach the spectator is t y = = 23 = s v sound sound m 340 m s 6.76 10 2 , so the total elapsed time is ttotal = 2.17 s + 6.76 102 s 2.23 s .

40. 40 Motion in One Dimension P2.46 At any time t, the position of the ball released from rest is given by y1 h gt 1 2 2 = . At time t, the 1 2 2 = . The time at which the position of the ball thrown vertically upward is described by y2 vit gt first ball has a position of y h = is found from the first equation as 1 2 h h gt 2 1 2 = 2 , which yields t h g = . To require that the second ball have a position of y h = at this time, use the second 2 2 equation to obtain = FH G h h h v g 2 i g g 1 2 IK J . This gives the required initial upward velocity of the second ball as vi = gh . P2.47 (a) v v gt f i = : v f =0 when t=3 00 . s , g =9 80 . m s2 . Therefore, vi = gt=c9.80 m s2 h(3.00 s)= 29.4 m s . (b) y f yi = v f +vi t 1 2c h y f yi = = 1 2 b29.4 m sga3.00 sf 44.1 m *P2.48 (a) Consider the upward flight of the arrow. 2 2 d i v v a y y y yf yi y f i y b m sg 2 e m s j 2 0 100 2 98 10 000 19 6 510 = + = + m s m s = = m 2 2 2 2 . . (b) Consider the whole flight of the arrow. f = i + yi + y = + + y y v t at t t 1 2 0 0 100 1 2 9 8 2 b m sg e . m s2 j 2 The root t=0 refers to the starting point. The time of flight is given by 100 4 9 t= = . . 20 4 m s m s s . 2 P2.49 Time to fall 3.00 m is found from Eq. 2.12 with vi =0 , 3 00 1 2 . m= c9.80 m s2 ht2 , 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

41. 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= vit+ at = + t 1 2 2 144 ft 0 c16.0 ft s2 h 2 giving tfall =3.00 s. Her velocity just before impact is: v f = vi + gt=0+c32.0 ft s2 h(3.00 s)= 96.0 ft s . (b) While crushing the box, vi =96.0 ft s , v f =0 , and y = 18.0 in.= 1.50 ft . Therefore, a v v f i = y = ( ) = 2 2 2 3 2 0 960 2 150 3 07 10 a f a . f . . ft s ft ft s2 , or a=3.07103 ft s2 upward . (c) Time to crush box: t y v y = = = v f vi ( ) 2 150 0 960 + + 2 . . ft ft s or t=3.13102 s . P2.51 y=3.00t3 : At t= 2.00 s , y = 3 00 2 00 = 24 0 3 . a . f . m and v y = =9.00t2 =36.0 m s A. dy dt If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is yb = ybi +vit gt = + t ( )t 1 2 24 0 36 0 1 2 2 . . 9.80 2 . Setting yb =0, 0= 24.0+36.0t4.90t 2 . 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 at v v f i yi y yi yi = + + a f e ja f . . 2 . . = + + = + = 1 2 0 30 15 1 2 9 8 1 5 30 11 0 12 6 2 m s ms s 2 m m . m s . 1.5 s Now consider the portion of its fall above the 30 m point. We assume it starts from rest 2 2 d i v v a y y y yf yi y f i b . g 2 e . j y 2 12 6 0 2 9 8 160 19 6 8 16 = + = + = = . . . m s m s m s m s m 2 2 2 2 Its original height was then 30 m+ 8.16 m = 38.2 m .

42. 42 Motion in One Dimension Section 2.7 Kinematic Equations Derived from Calculus P2.53 (a) J da dt = = constant da= Jdt a= Jzdt= Jt+c1 i abut a = ai when t=0 so c 1 = ai . Therefore, a = Jt + a dv dt = = = z = zb + g = 1 + + dv adt v adt Jt ai dt Jt ait c 2 2 2 but v= vi when t=0, so c2 = vi and v= Jt +ait+vi 1 2 2 v dx dt = = = = + + FH G dx vdt x vdt Jt at v dt i i z z 1 2 1 6 1 2 2 3 2 x Jt at i vt i c x x i IK J = + + + = 3 when t=0, so c3 = xi . Therefore, x= Jt + ait +vit+xi 1 6 1 2 3 2 . 2 =a + f2 = 2 2 + 2 +2 a ai J t Jait (b) a Jt ai J t ai Jait 2 2 222 = +c + h a ai J Jt ait = + 1 + 2 G FH 2 2 2 2 IK J Recall the expression for v: v= Jt +ait+vi 1 2 2 . So v v Jt a t a i f= + i 1 2 2 . Therefore, 2 = 2 +2 a f . a ai J v vi

43. 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 a8 m sf 3 s 12 m. At t=5 s, x=12 m+a8 m sf(2 s)= 28 m. At t=7 s, x= 28 + ( )= m a8 m sf 2 s 36 m. 1 2 8 3 (b) For 0

44. 44 Motion in One Dimension P2.56 a dv dt = =3.00v2 , vi =1.50 m s Solving for v, dv dt =3.00v2 z 2 = z v dv dt 1 1 v v 0 t t v v v v v t t 3 00 i i i = = + = = 3 00 3 00 1 1 . . or . . When v = vi 2 , t = = 1 3 00 vi 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 0 0 = and the distance traveled during this braking period is x vt v v t v v a v a a f f i 2 2 2 0 0 0 2 2 0 2 2 = = + FH G IK J = + FH G IK J FH G IK J = . Thus, the total distance traveled before coming to a stop is 2 2 t s = v a x f + a x f = v 0 stop 1 2 0 r a . 2 2 t *P2.58 (a) If a car is a distance s v 0 v (See the solution to Problem 2.57) from the stop r a = 0 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 2 2 v 0 a = stop + i = 0 r + si . 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 x v 2 v t s t v t v a s v r v a i r i light = = + = + 0 0 2 0 0 0 0 2 . (b) With si =16 m, v=60 km h, a = 2 0 . m s2 , and tr = 1 1 . s, 60 2 20 km h m s tlight 2 s m s 0 278 1 16 1 = . 6 23 s e j . . km h m 60 km h km h m s FH G IK J + FH G IK J 1 1 = 0 278 . . .

45. 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) 100 200 300 200 100 0 100 400 v t 50 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 155 45 100 50 2.2 a f 0.98 . v mi h s m s2 t = = = ( ) ( ) mi h s slope = = (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 = + + + + s area1 area2 area3 area4 area5 b ga fb ga f b ga f a fb g a fb g + + + 50 50 50 50 160 100 1 2 mi h s mi h s mi h s 50 s 100 mi h 1 2 + = 100 170 160 24 000 s mi h mi h mi h s b ga f t (s) 4 3 100 200 300 200 100 400 1 2 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 h = 3 600 s , 6 7 FH G x0 200 24 000 IK J mi s = 3 600 s asf . mi .

46. 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 8 60 through the next 1.10 s, the speed of the train is 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 ( . s)+ (1.10 s)=1.30 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 = FH G IK J 0.132 mm d w . The increase in the length of the hair (i.e., displacement) during a time of t=5.00 w=35.0 d is 1 2 x v t a t x = xi + x = + 1 04 35 0 1 2 0 132 35 0 5 00 2 b . mm dga . df b . mm d wga . dfa . wf 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. 2 a80.0f = 2a4.00fb1 000g so v f =120 m s 120=80.0+(4.00)t giving t=10.0 s (0 to 1) vf 2 (1 to 2) 0 120 2 9 80 ( )2 = ( . ) x x f i c h giving x f xi =735 m 0120=9.80t giving t=12.2 s This is the time of maximum height of the rocket. (2 to 3) vf 2 0 = 2a9.80fb1 735g v f =184=(9.80)t giving t=18.8 s FIG. P2.62 (a) ttotal =10+12.2+18.8= 41.0 s (b) cx f xi h = total 1.73 km continued on next page

47. Chapter 2 47 (c) vfinal = 184 m s 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 =a15.0 m sft Distance traveled by policeman = 1 2 2 00 2 . m s2 c ht (a) intercept occurs when 15.0t=t2 , or t= 15.0 s (b) v(officer)=c2.00 m s2 ht= 30.0 m s (c) x(officer)= m s2 t = m 1 2 c2.00 h 2 225 P2.64 Area A1 is a rectangle. Thus, A1 = hw = vxit . = = b g. 1 2 1 2 Area A2 is triangular. Therefore A2 bh t vx vxi The total area under the curve is b g = + = + x xi A A A v t v v t xi 1 2 2 and since vx vxi = axt A = vxit + axt 1 2 2 . The displacement given by the equation is: x = vxit + axt 1 2 2 , the same result as above for the total area. vx vx vxi A2 A1 0 t t FIG. P2.64

48. 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: 2 1 a f , 100x= va10.2t1 f and v= vi +at1 . x= v+vi t 1 The first two give . . 100 10 2 1 2 t v t at 10 2 1 2 200 20 4 1 1 1 1 1 = FH G IK J = FH G IK J = . . a b t gt 200 For Maggie: m s 18 4 2 00 For Judy: m s 2 2 a a = = = = 5 43 200 17 4 3 00 3 83 . . . . . . a fa f a fa f (b) v= a1t a fa f a fa f Maggie: v m s Judy: v m s = = = = 5 . 43 2 . 00 10 . 9 3 . 83 3 . 00 11 . 5 (c) At the six-second mark 2 a . 1 f x= at +v t 1 2 1 6 00 1 2 a fa f 2 a fa f a fa f a fa f Maggie: x m 5 . 43 2 . 00 10 . 9 4 . 00 54 . 3 1 2 2 Judy: x m = + = = + = . . . . . 3 83 3 00 11 5 3 00 51 7 Maggie is ahead by 2.62 m . P2.66 a1 0 100 = . m s2 a2 0 500 = . m s2 1 2 1 2 1 2 2 2 2 m t=t1 +t2 and v1 = a1t1 =a2t2 x = 1 000 = a t + v t + a t 1 1 2 1 000 1 2 1 2 1 1 1 1 2 1 1 2 2 1 1 2 2 = + FH G IK J + FH G IK J a t a t a t a a a t a 1 000 = 1 1 1 2 2 G 1 1 2 FH IK J a 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

49. Chapter 2 49 P2.67 Let the ball fall 1.50 m. It strikes at speed given by 2 = 2 +2 c h: vxf vxi a x f xi 2 =0+2c9.80 m s2 h(1.50 m) vxf vxf =5.42 m s and its stopping is described by 2 2 d i = + = + v v a x x b . g a e j xf xi x f i a m s x m m s x 2 2 2 3 2 0 542 2 10 29 4 2 00 10 1 47 10 = = + . . . . 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 t f i xi x = + + 1 2 2. We assume the package starts from rest. 1 2 m c 9.80 m s2 ht2 145 =0+0+ t= ( ) = 2 145 9 80 . 5 44 m m s s . 2 (b) x f = xi +vxit+ axt = + + ( ) = 1 2 0 0 1 2 2 9 80 5 18 2 131 c . m s2 h . s m distance fallen = x f = 131 m (c) speed = vxf = vxi + axt = 0 + e9.8 m s2 j5.18 s = 50.8 m s (d) The remaining distance is 145 m131.5 m=13.5 m. During deceleration, vxi =50.8 m s, vxf =0, x f xi =13.5 m 2 = 2 +2 c h: vxf vxi ax x f xi 0 50 8 2 13 5 2 =a . m sf + ax( . m) ax = = + = 2 580 2 135 95 3 95 3 m s m m s m s upward 2 2 2 2 . a f . . .

50. 50 Motion in One Dimension P2.69 (a) y f = vi1t+ at = = t+ ( )t 1 2 50 0 2 00 2 2 1 2 . . 9.80 , 4.90t 2 +2.00t50.0=0 t= . . 2 . . 2 00 + 2 00 4 ( 4 90 )( 50 0 ) ( ) 2 490 . Only the positive root is physically meaningful: t= 3.00 s after the first stone is thrown. 1 2 2 (b) y f = vi2t+ at and t=3.001.00= 2.00 s 2 . = vi ( . )+ ( . )( . ) : substitute 50 0 2 00 1 2 2 9 80 2 00 vi2 = 15.3 m s downward (c) v1 f = vi1 +at= 2.00+(9.80)(3.00)= 31.4 m s downward v2 f = vi2 +at=15.3+(9.80)(2.00)= 34.8 m s downward P2.70 (a) d = ( )t 1 2 . 2 d =336t2 9 80 1 2 t = . a . t f t1 +t2 = 2.40 336 2 4 90 2 40 2 2 . t . t2 + . = t2 4 90 2 359 5 28 22 0 359 5 359 52 4 4 90 28 22 9 80 = . . ( . )( . ) . t2 . . = . 359 5 358 75 = 0 076 5 9 80 . s so d =336t2 = 26.4 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 A is only increased by x sin . The pack lifts at a rate x t sin . v x t v x v x x h = = = + sin boy boy A 2 2 (b) a G FH v = dv = dx + d v x dt dt dt IK J boy 1 boy A A a v v v xd dt boy boy A A = boy A 2 , but A x d dt v v A = = boy FH G 2 = IK J 2 = = + a v x v h h v 2 x h boy boy boy 1 2 2 A A A A 2 2 2 2 23 2 c h (c) 2 v h boy , 0 (d) vboy , 0 FIG. P2.71

51. Chapter 2 51 P2.72 h=6.00 m, vboy = 2.00 m s v x t = = =+ v x v x x h sin boy boy A 2 21 2 c h . However, x= vboyt : = 2 2 2 2 1 2 2 1 2 + = + v v t v t h t t boy boy 4 4 36 c h c h . a f b g (a) t s v 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 FIG. P2.72(a) (b) From problem 2.71 above, a 2 h v x h h v 2 c h c v 2 t h . h c t h = + = + = 144 + 2 2 2 3 2 2 2 23 2 2 3 2 4 36 boy boy boy a f e 2 j t s a m s 0 0.67 0.5 0.64 1 0.57 1.5 0.48 2 0.38 2.5 0.30 3. 0.24 3.5 0.18 4. 0.14 4.5 0.11 5 0.09 FIG. P2.72(b) P2.73 (a) We require xs = xk when ts = tk + 1.00 e 2 jb g e 2 jb g . m s . 2 . m s 2 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 1183 5 46 . s . (b) xk = = 1 2 4 90 5 46 73 0 2 e . m s2 ja . sf . m (c) vk = e4.90 m s2 ja5.46 sf = 26.7 m s vs = e3.50 m s2 ja6.46 sf = 22.6 m s

52. 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 m s2 = 1.63 m s2 downward

53. Chapter 2 53 P2.75 The distance x and y are always related by x2 +y2 =L2 . Differentiating this equation with respect to time, we have dx dt 2x 2 y 0 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 dx x v y FH G dt y IK J = ( ). B O y A x L v x y FIG. P2.75 But y x IK J =tan so vB = v FH G 1 tan . When =60.0, v v v B = v = = tan . . 60 0 3 3 0 577 . ANSWERS TO EVEN PROBLEMS P2.2 (a) 2 1 m s; 1 16 07 0 m s; P2.24 (a) 1.88 km; (b) 1.46 km; (b) 5 108 yr (c) see the solution; 2 1 67 = . m s2e j; (d) (i) x1 t P2.4 (a) 50.0 m s ; (b) 41.0 m s (ii) x t 2 = b50 m sg 375 m; P2.6 (a) 27 . 0 m; (iii) x3 = b250m sg t e2.5m s2 j t 2 4 375m; + b gt + e jatf ; (b) 27 0 18 0 3 00 (e) 37.5 m s 2 . . . m m s m s2 (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 f = e30.0t t 2 j m; v f = a30.0 2tf m s ; P2.10 5.00 m (b) 225 m P2.12 (a) 20.0 m s ; 5.00 m s ; (b) 262 m P2.30 x x v t a t f 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 P2.16 (a) 13.0 m s; (b) 10.0 m s; 16.0 m s; P2.34 (a) 1.12 1011 m s2 ; (b) 4.67 105 s (c) 6 00 . m s2 ; (d) 6 00 . m s2 P2.36 (a) False unless the acceleration is zero; P2.18 see the solution see the solution; (b) True P2.38 Yes; 212 m; 11.4 s P2.20 (a) 6 61 . m s; (b) 0 448 . m s2 P2.22 (a) 21.8 mi h s = 9.75 m s2 ; P2.40 (a) 4.90 m; 19.6 m; 44.1 m; (b) 9.80 m s; 19.6 m s; 29.4 m s (b) 22.2 mi h s = 9.94 m s2 ; (c) 22.8 mi h s = 10.2 m s2 P2.42 1.79 s

54. 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 vi = gh P2.62 (a) 41.0 s; (b) 1.73 km; (c) 184 m s P2.64 vxit + axt 1 2 2 ; displacements agree P2.48 (a) 510 m; (b) 20.4 s P2.50 (a) 96.0 ft s ; P2.66 155 s; 129 s (b) a = 3.07 103 ft s2 upward ; (c) t = 3.13 102 s P2.68 (a) 5.44 s; (b) 131 m; (c) 50.8 m s ; (d) 95.3 m s2 upward 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

55. 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

56. 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 Ax = Bx and Ay = By . 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 Ax = Acos , 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 = Ax2 + Ay2 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 = a5.50 mfcos 240= a5.50 mfa0.5f = 2.75 m y = r sin = a5.50 mfsin240= a5.50 mfa0.866f = 4.76 m P3.2 (a) x = r cos and y = r sin , therefore x1 = a2.50 mfcos30.0, y1 = a2.50 mfsin30.0, and bx1 , y1 g = a2.17, 1.25f m x2 = a3.80 mfcos120 , y2 = a3.80 mfsin120 , and bx2 , y2 g = a1.90, 3.29f m . (b) d = ( x)2 + ( y)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 = x2 + y2 = a2.00 mf2 + a1.00 mf2 = 5.00 m2 = 2.24 m (b) = FH G IK J tan1 1 = . 2 26 6 ; r = 2.24 m, 26.6

57. Chapter 3 57 b g 2 b g 2 c2.00 3.00 h 2 a 4.00 3.00f 2 P3.4 (a) d = x2 x1 + y 2 y 1 = + d = 25.0 + 49.0 = 8.60 m (b) r1 2 2 = a2.00f + a4.00f = 20.0 = 4.47 m 1 G FH = 1 4 . 00 63 4 2 . 00 IK J tan = . r2 2 2 = a3.00f + a3.00f = 18.0 = 4.24 m 2 = 135 measured from the +x axis. P3.5 We have 2.00 = r cos30.0 r = 2 00 = 30 0 2 31 . cos . . and y = r sin30.0= 2.31sin30.0= 1.15 . P3.6 We have r = x2 + y2 and = FH G IK J tan1 y x . (a) The radius for this new point is axf2 + y2 = x2 + y2 = r and its angle is tan y x FH G IK J 1 = 180 . (b) (2x)2 + (2y)2 = 2r . This point is in the third quadrant if bx, yg is in the first quadrant or in the fourth quadrant if bx, yg is in the second quadrant. It is at an angle of 180+ . (c) (3x)2 + (3y)2 = 3r . This point is in the fourth quadrant if bx, yg is in the first quadrant or in the third quadrant if bx, yg is in the second quadrant. It is at an angle of .

58. 58 Vectors Section 3.2 Vector and Scalar Quantities Section 3.3 Some Properties of Vectors P3.7 tan . x = = = tan . . 35 0 100 100 35 0 70 0 x m a mf m FIG. P3.7 P3.8 R = 14 65 km = N of E R 13 km 1 km 6 km FIG. P3.8 P3.9 R = 310 km at 57 S of W (Scale: 1 unit = 20 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 B B A A B O FIG. P3.10

59. Chapter 3 59 P3.11 (a) d = 10.0i = 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 b2 g 5 15.7 m C 5.00 m d B A FIG. P3.11 (c) If the circle is complete, d begins and ends at point A. Hence, d = 0 . P3.12 Find the resultant F1 + F2 graphically by placing the tail of F2 at the head of F1 . The resultant force vector F1 + F2 is of magnitude 9.5 N and at an angle of 57 above the x-axis . 0 1 2 3 N x y F 2 F 1 F 1 F + 2 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 ~105 a0.03 mf+ 102 a1 mf ~103 m upward .

60. 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 . N 15.0 meters W E 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 unit = 0.5 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: R1 =A+B+C, R2 =B+C+A, and R3 =C+B+A. You should observe that R1 =R2 =R3 , 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

61. 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 ft and = 3 (Scale: 1 unit = 20 ft ) FIG. P3.17 Section 3.4 Components of a Vector and Unit Vectors P3.18 Coordinates of the super-hero are: x y a f a f a f a f = = = = 100 30 0 86 6 100 30 0 50 0 m m m m cos . . sin . . FIG. P3.18 P3.19 A x y = 25 . 0 = 40 . 0 = + = + = A A A A 2 2 25 0 2 40 0 2 47 2 x y a . f a . f . units We observe that tan = A A y x . FIG. P3.19 So = FH GIK J 1 tan 1 . . 40 0 . . tan = tan = = 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 sina25.0f = 1.31 km north , and 3.10 cosa25.0f = 2.81 km east .