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2. 1 Physics and Measurement CHAPTER OUTLINE 1.1ANSWERS TO QUESTIONSStandards of Length, Mass, and Time Matter and Model-Building Density and Atomic Mass Dimensional Analysis Conversion of Units Estimates and Order-ofMagnitude Calculations Significant FiguresAtomic clocks are based on electromagnetic waves which atoms emit. Also, pulsars are highly regular astronomical clocks.Q1.2Density 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. People have different size hands. Defining the unit precisely would be cumbersome.Q1.4(a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilogramsQ1.5(b) and (d). You cannot add or subtract quantities of different dimension.Q1.61.7Q1.1Q1.31.2 1.3 1.4 1.5 1.6A 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.Q1.7If I were a runner, I might walk or run 10 1 miles per day. Since I am a college professor, I walk about 10 0 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on vacation.Q1.8On February 7, 2001, I am 55 years and 39 days old. 55 yrF 365.25 d I + 39 d = 20 128 dFG 86 400 s IJ = 1.74 10 GH 1 yr JK H 1d K9s ~ 10 9 s .Many college students are just approaching 1 Gs. Q1.9Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10.Q1.10The mass of the forty-six chapter textbook is on the order of 10 0 kg .Q1.11With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.1 3. 2Physics and MeasurementSOLUTIONS TO PROBLEMS Section 1.1Standards of Length, Mass, and TimeNo problems in this sectionSection 1.2 P1.1Matter and Model-BuildingFrom 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 distanceL = 0.200 nm , the diagonal planes are separated bySection 1.3 *P1.21 2 L + L2 = 0.141 nm . 2Density and Atomic MassModeling the Earth as a sphere, we find its volume as4 3 4 r = 6.37 10 6 m 3 3ej3= 1.08 10 21 m 3 . Itsm 5.98 10 24 kg = = 5.52 10 3 kg m3 . This value is intermediate between the V 1.08 10 21 m 3 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. density is then =P1.3afbge jWith V = base area height V = r 2 h and ==afa = 2.15 10 kg m3393.m for both. Then iron = 9.35 kg V and V 19.3 10 3 kg / m3 = 23.0 kg . = 9.35 kg 7.86 10 3 kg / m3Let V represent the volume of the model, the same in = gold =P1.5F 10 mm I f GH 1 m JK1 kg m = 2 r h 19.5 mm 2 39.0 mm 4*P1.4m , we have Vm gold VV = Vo Vi ==. Next, gold iron4 3 r2 r13 3e=m gold 9.35 kgand m goldF GHjFG IJ e H Ke3 4 r2 r13 m 4 3 , so m = V = r2 r13 = V 3 3jj.I JK 4. Chapter 1P1.634 4 3 r and the mass is m = V = r 3 . We divide this equation 3 3 for the larger sphere by the same equation for the smaller:For either sphere the volume is V = 4 r 3 3 r 3 m = = = 5. m s 4 rs3 3 rs3a fThen r = rs 3 5 = 4.50 cm 1.71 = 7.69 cm . P1.7Use 1 u = 1.66 10 24 g .F 1.66 10 GH 1 u F 1.66 10 = 55.9 uG H 1u F 1.66 10 = 207 uG H 1u-24I = 6.64 10 JK gI JK = 9.29 10 gI JK = 3.44 10 g24g .23g .22g .(a)(b)For Fe, m 0(c) *P1.8For He, m 0 = 4.00 uFor Pb, m 0(a)The mass of any sample is the number of atoms in the sample times the mass m 0 of one atom: m = Nm 0 . The first assertion is that the mass of one aluminum atom is-2424m 0 = 27.0 u = 27.0 u 1.66 10 27 kg 1 u = 4.48 10 26 kg . Then the mass of 6.02 10 23 atoms is m = Nm 0 = 6.02 10 23 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 = Nm 0 . 0.027 0 kg = 6.02 10 23 m 0 , so m 0 =0.027 kg 6.02 10 23= 4.48 10 26 kg ,in agreement with the first assertion. (b)The general equation m = Nm 0 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 10 3 kg = N 1.660 540 2 10 27 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=F 1 I 10 GH 1.660 540 2 JK3 + 27= 6.022 137 10 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 2 O is 2 1.008 0 + 15.999 u = 18.0 u. Then the molar mass is 18.0 g .(d)For CO 2 we have 12.011 g + 2 15.999 g = 44.0 g as the mass of one mole.bgbg 5. 4Physics and MeasurementP1.91 b gFGH 10kgg IJK = 4.5 10 kg I JK = 3.27 10 kg .Mass of gold abraded: m = 3.80 g 3.35 g = 0.45 g = 0.45 gF 1.66 10 GH 1 u27Each atom has mass m 0 = 197 u = 197 u34kg .25Now, m = N m 0 , and the number of atoms missing is N =m=m04.5 10 4 kg 3.27 10 25 kg= 1.38 10 21 atoms .The rate of loss is N t N t=FG H1 yr 1.38 10 21 atoms 365.25 d 50 yr= 8.72 10 11 atoms s .ejejP1.11(a)m = L3 = 7.86 g cm 3 5.00 10 6 cm(b)P1.10IJ FG 1 d IJ FG 1 h IJ FG 1 min IJ K H 24 h K H 60 min K H 60 s KN=(a)3= 9.83 10 16 g = 9.83 10 19 kgThe cross-sectional area is9.83 10 19 kg m = = 1.06 10 7 atoms m 0 55.9 u 1.66 10 27 kg 1 uejaf afafaA = 2 0.150 m 0.010 m + 0.340 m 0.010 m = 6.40 1032m .f.The volume of the beam isjaefV = AL = 6.40 10 3 m 2 1.50 m = 9.60 10 3 m3 . Thus, its mass iseFIG. P1.11je9.60 10 m j = 72.6 kg . F 1.66 10 kg I = 9.28 10 The mass of one typical atom is m = a55.9 ufG H 1 u JK 3m = V = 7.56 10 kg / m(b)333270m = Nm 0 and the number of atoms is N =26kg . Now72.6 kg m = = 7.82 10 26 atoms . 26 m 0 9.28 10 kg 6. Chapter 1P1.12(a)F 1.66 10 GH 1 u27The mass of one molecule is m 0 = 18.0 ukgI = 2.99 10 JK265kg . The number ofmolecules in the pail is N pail = (b)1.20 kg m = = 4.02 10 25 molecules . m 0 2.99 10 26 kgSuppose that enough time has elapsed for thorough mixing of the hydrosphere. N both = N pailF m I = (4.02 10 GH M JK pail25F 1.20 kg I , GH 1.32 10 kg JKmolecules)total21orN both = 3.65 10 4 molecules .Section 1.4 P1.13Dimensional AnalysisThe term x has dimensions of L, a has dimensions of LT 2 , and t has dimensions of T. Therefore, the equation x = ka m t n has dimensions ofeL = LT 2j aTf mnor L1 T 0 = Lm T n 2 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 .e jExpression (i) has dimension L L21/2= L2 , so this must be area (c).Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L L2 = L3 , so it is (b). Thus, (a) = ii; (b) = iii, (c) = i .e j 7. 6Physics and Measurement*P1.16(a)This is incorrect since the units of ax are m 2 s 2 , while the units of v are m s .(b)P1.15This is correct since the units of y are m, and cos kx is dimensionless if k is in m 1 .(a)a faFor a = kFrepresents the proportionality of acceleration to resultant force and m m the inverse proportionality of acceleration to mass. If k has no dimensions, we have a = k(b) P1.17In units,M L T2=kg m s2F F L M L , F = , 2 =1 . m T M T2, so 1 newton = 1 kg m s 2 .Inserting the proper units for everything except G,LM kg m OP = G kg Ns Q m22Multiply both sides by mSection 1.5 *P1.18.m32and divide by kg ; the units of G arekg s 2.Conversion of UnitsafafEach of the four walls has area 8.00 ft 12.0 ft = 96.0 ft 2 . Together, they have areae4 96.0 ft 2 P1.1922m jFGH 3.128 ft IJK2= 35.7 m 2 .Apply the following conversion factors: 1 in = 2.54 cm , 1 d = 86 400 s , 100 cm = 1 m , and 10 9 nm = 1 mFG 1 H 32IJ b2.54 cm inge10 m cmje10 K 86 400 s day 2in day9j=nm m9.19 nm s .This means the proteins are assembled at a rate of many layers of atoms each second! *P1.208.50 in 3 = 8.50 in 3FG 0.025 4 m IJ H 1 in K3= 1.39 10 4 m 3 8. Chapter 1P1.21Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A 30 m 50 m = 1 500 m 2 .afafCategorize: We model the lot as a perfect rectangle to use Area = Length Width. Use the conversion: 1 m = 3.281 ft .aAnalyze: A = LW = 100 ft1m 1m f FGH 3.281 ft IJK a150 ftfFGH 3.281 ft IJK = 1 390 m2= 1.39 10 3 m 2 .Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m 2 . Unit conversion is a common technique that is applied to many problems. P1.22(a)afafafV = 40.0 m 20.0 m 12.0 m = 9.60 10 3 m 3 33bgV = 9.60 10 m 3.28 ft 1 m (b)3= 3.39 10 5 ft 3The mass of the air isejejm = air V = 1.20 kg m3 9.60 10 3 m3 = 1.15 10 4 kg . The student must look up weight in the index to findejejFg = mg = 1.15 10 4 kg 9.80 m s 2 = 1.13 10 5 N . Converting to pounds,jbegFg = 1.13 10 5 N 1 lb 4.45 N = 2.54 10 4 lb . P1.23(a)Seven minutes is 420 seconds, so the rate is r=(b)30.0 gal = 7.14 10 2 gal s . 420 sConverting gallons first to liters, then to m3 ,er = 7.14 10 2 gal sjFGH 3.1786 L IJK FGH 10 1 Lm IJK gal 33r = 2.70 10 4 m3 s . (c)At that rate, to fill a 1-m3 tank would take t=F 1m GH 2.70 1034m3IF 1 h I = s J H 3 600 K KG J1.03 h .7 9. 8Physics and Measurement*P1.24(a)(b)(c)(d) P1.25FG 1.609 km IJ = 560 km = 5.60 10 m = 5.60 10 cm . H 1 mi K F 0.304 8 m IJ = 491 m = 0.491 km = 4.91 10 cm . Height of Ribbon Falls = 1 612 ftG H 1 ft K F 0.304 8 m IJ = 6.19 km = 6.19 10 m = 6.19 10 cm . Height of Denali = 20 320 ftG H 1 ft K F 0.304 8 m IJ = 2.50 km = 2.50 10 m = 2.50 10 cm . Depth of Kings Canyon = 8 200 ftG H 1 ft K 5Length of Mammoth Cave = 348 mi743535From Table 1.5, the density of lead is 1.13 10 4 kg m 3 , 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 ==23.94 g 2.10 cm 3F 1 kg I FG 100 cm IJ GH 1 000 g JK H 1 m K3m . We must convert to SI units in the calculation. V= 1.14 10 4 kg m3At 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 cm 3 , and objects that float must be less dense than water. P1.26It 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 a1 acrefFGH 640mi IJK FGH 1 609 m IJK acres mi 2*P1.27 P1.28The weight flow rate is 1 200FG Hton 2 000 lb h ton2= 4.05 10 3 m 2 .IJ FG 1 h IJ FG 1 min IJ = K H 60 min K H 60 s K667 lb s .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 19(a)F 6 10 $ I F 1 h I FG 1 day IJ F 1 yr I = GH 1 000 $ s JK GH 3 600 s JK H 24 h K GH 365 days JK(b)P1.29The circumference of the Earth at the equator is 2 6.378 10 3 m = 4.01 10 7 m . The length12190 yearsejof one dollar bill is 0.155 m so that the length of 6 trillion bills is 9.30 10 11 m. Thus, the 6 trillion dollars would encircle the Earth 9.30 10 11 m = 2.32 10 4 times . 4.01 0 7 m 1.99 10 30 kg mSun = = 1.19 10 57 atoms m atom 1.67 10 27 kgP1.30N atoms =P1.31V = At so t =V 3.78 10 3 m 3 = = 1.51 10 4 m or 151 m 2 A 25.0 mbafe13.0 acres 43 560 ft 2 acre 1 V = Bh = 3 3 7 3 = 9.08 10 ft ,P1.32gj a481 ftf horeV = 9.08 10 7 ft 3jFGH 2.83 110 ft2m33I JKBFIG. P1.32= 2.57 10 6 m3 P1.33 *P1.34bgejbgFg = 2.50 tons block 2.00 10 6 blocks 2 000 lb ton = 1.00 10 10 lbs The area covered by water isa fej a fa fej2 A w = 0.70 AEarth = 0.70 4 REarth = 0.70 4 6.37 10 6 m2= 3.6 10 14 m 2 .The average depth of the water isafbgd = 2.3 miles 1 609 m l mile = 3.7 10 3 m . The volume of the water isejejV = A w d = 3.6 10 14 m 2 3.7 10 3 m = 1.3 10 18 m 3 and the mass isejejm = V = 1 000 kg m3 1.3 10 18 m3 = 1.3 10 21 kg . 11. 10 P1.35Physics and Measurement(a)d nucleus, scale = d nucleus, realeFd I = 2.40 10 m FG 300 ft IJ = 6.79 10 jH 1.06 10 m K GH d JK e ft jb304.8 mm 1 ft g = 2.07 mm atom, scale1510atom, reald nucleus, scale = 6.79 10 3(b)Vatom = Vnucleus3 4 ratom 3 3 4 rnucleus 3=FG r HrIJ = FG d K Hd 3atomnucleusIJ = F 1.06 10 K GH 2.40 10 3atomnucleus10 15m mI JK3= 8.62 10 13 times as large *P1.36scale distance betweenP1.37=FG real IJ FG scale IJ = e4.0 10 H distanceK H factorK13m jFGH 71..0410 m IJK = 10 3km200 km9The scale factor used in the dinner plate model is 0.25 mS=51.0 10 lightyears= 2.5 10 6 m lightyears .The distance to Andromeda in the scale model will beejejDscale = Dactual S = 2.0 10 6 lightyears 2.5 10 6 m lightyears = 5.0 m .FG HF e6.37 10 mjb100 cm mg I =G GH 1.74 10 cm JJK = 13.4 IJ = FG e6.37 10 mjb100 cm mg IJ = 49.1 K GH 1.74 10 cm JKP1.39(a)(b)P1.382 AEarth 4 rEarth r = = Earth 2 A Moon 4 rMoon rMoonVEarth = VMoon3 4 rEarth 3 3 4 rMoon 3Fr =G HrIJ K8Moon363Earth2628To balance, m Fe = m Al or FeVFe = Al VAl FeFG 4IJ r H 3KFe3FG 4 IJ r H 3K FG IJ = a2.00 cmfFG 7.86 IJ H 2.70 K H K= AlAl1/3rAl = rFeFe Al31/3= 2.86 cm .3ft , or 12. Chapter 1P1.4011The mass of each sphere is m Al = Al VAl =4 Al rAl 3 3m Fe = FeVFe =4 Fe rFe 3 . 3andSetting these masses equal, 4 Al rAl 3 4 Fe rFe 3 = and rAl = rFe 3 Fe . 3 3 AlSection 1.6 P1.41Estimates and Order-of-Magnitude CalculationsModel 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 isFG H4 0.038 m 3 2IJ K3= 2.87 10 5 m3 .48 ~ 10 6 ping-pong balls in the room. 2.87 10 5 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 1 packing fraction is 2 = 0.74 so that at least 26% of the space will be empty. Therefore, the 6 above estimate reduces to 1.67 10 6 0.740 ~ 10 6 .Therefore, one can fit aboutP1.42A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus,bgbgbgthe tire would make 50 000 mi 5 280 ft mi 1 rev 8 ft = 3 10 7 rev ~ 10 7 rev . P1.43In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least 1 in 2 = 43 10 5 ft 2 . Since 1 acre = 43 560 ft 2 , the number of blades of grass to be expected on a 16 quarter-acre plot of land is about n=afej0.25 acre 43 560 ft 2 acre total area = = 2.5 10 7 blades ~ 10 7 blades . area per blade 43 10 5 ft 2 blade 13. 12 P1.44Physics and MeasurementA typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then approximately 4 10 3 in 3 . Since 1 acre = 43 560 ft 2 , the volume of water required to cover it to a depth of 1 inch isa1 acrefa1 inchf = a1 acre infFGH 431560 ft acre2I F 144 in I 6.3 10 JK GH 1 ft JK 226in 3 .The number of raindrops required is n= *P1.45volume of water required 6.3 10 6 in 3 = = 1.6 10 9 ~ 10 9 . volume of a single drop 4 10 3 in 3Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is thena fafafafV = 0.5 1.3 m 0.5 m 0.3 m = 0.10 m3 . The mass of this volume of water isejejm water = water V = 1 000 kg m3 0.10 m3 = 100 kg ~ 10 2 kg . Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required isejejmcopper = copper V = 8 920 kg m3 0.10 m3 = 892 kg ~ 10 3 kg . P1.46The 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, soe250 106jbgcans day 365 days year 10 11 cansare thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, we estimate this representse10 P1.4711jbgbgbgcans 0.1 oz can 1 lb 16 oz 1 ton 2 000 lb 3.1 10 5 tons year . ~ 10 5 tonsAssume: Total population = 10 7 ; 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 ~F 1 tuner I F 1 piano I (10 GH 1 000 pianos JK GH 100 people JK7people) = 100 . 14. Chapter 1Section 1.7 *P1.4813Significant FiguresMETHOD ONE We treat the best value with its uncertainty as a binomial 21.3 0.2 cm 9.8 0.1 cm ,a f a A = 21.3a9.8f 21.3a0.1f 0.2a9.8 f a0.2 fa0.1f cmf2.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 cm 2 4 cm 2 . METHOD TWO We add the fractional uncertainties in the data.a0 0 f FGH 21..23 + 9..1 IJK = 209 cm 8faA = 21.3 cm 9.8 cm P1.49af r 2 = 10.5 m 0.2 m(a)2 2% = 209 cm 2 4 cm 22= (10.5 m) 2 2(10.5 m)(0.2 m) + ( 0.2 m) 2 = 346 m 2 13 m 2af2 r = 2 10.5 m 0.2 m = 66.0 m 1.3 m(b)3P1.50(a)P1.514a f a m = a1.85 0.02f kg(b)3(c)(d)fr = 6.50 0. 20 cm = 6.50 0.20 10 2 m=mc h r 4 33also, m 3 r = + . m rIn other words, the percentages of uncertainty are cumulative. Therefore,a f 0.02 3 0.20 = + = 0.103 , 6.50 1.85 =anda1.85c h e6.5 10 4 3f2jm3= 1.61 10 3 kg m 3af = 1.61 0.17 10 3 kg m3 = 1.6 0.2 10 3 kg m3 .2 15. 14 P1.52Physics and Measurement756.?? 37.2? 0.83 + 2.5? 796.53 = 797 //(b)0.003 2 2 s.f. 356.3 4 s.f. = 1.140 16 = 2 s.f.(c) *P1.53(a)5.620 4 s.f. > 4 s.f. = 17.656 = 4 s.f.afaaf aP1.55afaff1.117.66We work to nine significant digits: 1 yr = 1 yrP1.54fF 365.242 199 d I FG 24 h IJ FG 60 min IJ FG 60 s IJ = GH 1 yr JK H 1 d K H 1 h K H 1 min K31 556 926.0 s .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 mbV = 2V1 + 2V2 = 2 V1 + V2ga fa fa f V = a10.0 mfa1.0 mfa0.090 mf = 0.900 m V = 2e1.70 m + 0.900 m j = 5.2 m U 0.12 m = = 0.0063 | 19.0 m | V w 0.01 m | = 0.006 + 0.010 + 0.011 = 0.027 = = = 0.010 V w 1.0 m |V t 0.1 cm | = = 0.011 | t 9.0 cm W V1 = 17.0 m + 1.0 m + 1.0 m 1.0 m 0.09 m = 1.70 m 3 32333FIG. P1.551113%111Additional Problems P1.56It is desired to find the distance x such that 1 000 m x = 100 m x (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 thatafbgx 2 = 100 m 1 000 m = 1.00 10 5 m 2 and therefore x = 1.00 10 5 m 2 = 316 m . 16. Chapter 1*P1.57Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has massam 0 = 197 ufFGH 1.66 110u27kgI = 3.27 10 JK25kg .So, the number of atoms in the cube is N=19 300 kg 3.27 10 25 kg= 5.90 10 28 .The imagined cubical volume of each atom is d3 =1 m3= 1.69 10 29 m 3 .5.90 10 28So d = 2.57 10 10 m .P1.58A total P1.59F I V a fe j FG V IJ e A j = GG V JJ e4 r j H K H K F 3V IJ = 3FG 30.0 10 m IJ = 4.50 m =G H r K H 2.00 10 m KAtotal = N A drop =total6total2total 4 r 3 3dropdrop325One month isbgbgbg1 mo = 30 day 24 h day 3 600 s h = 2.592 10 6 s . Applying units to the equation,ej ejV = 1.50 Mft 3 mo t + 0.008 00 Mft 3 mo 2 t 2 . Since 1 Mft 3 = 10 6 ft 3 ,ej ejV = 1.50 10 6 ft 3 mo t + 0.008 00 10 6 ft 3 mo 2 t 2 . Converting months to seconds, V=e1.50 10 6 ft 3 mo 2.592 10 6 s moj et+j0.008 00 10 6 ft 3 mo 2e2.592 10Thus, V [ft 3 ] = 0.579 ft 3 s t + 1.19 10 9 ft 3 s 2 t 2 .6s moj2t2.15 17. 16Physics and MeasurementafafP1.61 (deg) (rad)tan sin difference15.0 20.0 25.0 24.0 24.4 24.5 24.6 24.7P1.600.262 0.349 0.436 0.419 0.426 0.428 0.429 0.4310.268 0.364 0.466 0.445 0.454 0.456 0.458 0.4600.259 0.342 0.423 0.407 0.413 0.415 0.416 0.4183.47% 6.43% 10.2% 9.34% 9.81% 9.87% 9.98% 10.1%24.62 r = 15.0 m r = 2.39 m h = tan 55.0 r h = 2.39 m tan( 55.0 ) = 3.41 mafh55 rFIG. P1.61 *P1.62Let d represent the diameter of the coin and h its thickness. The mass of the gold is m = V = At = F 2 d GH 42I JK+ dh twhere t is the thickness of the plating.LM a2.41f NM 4m = 19.3 22a fa+ 2.41 0.178= 0.003 64 gramsfOPPe0.18 10 j Q 4cost = 0.003 64 grams $10 gram = $0.036 4 = 3.64 cents This is negligible compared to $4.98. P1.63The actual number of seconds in a year isb86 400 s daygb365.25 day yrg = 31 557 600 s yr . The percent error in the approximation ise 107j bs yr 31 557 600 s yr 31 557 600 s yrg 100% =0.449% . 18. Chapter 1P1.64V = L3 , A = L2 , h = L(a)V = A h L3 = L2 L = L3 . Thus, the equation is dimensionally correct. (b)eVrectangular object = P1.65(a)j wh = a w fh = Ah , whereVcylinder = R 2 h = R 2 h = Ah , where A = R 2The speed of rise may be found from v=(b)aVol rate of flowf = 16.5 cm (Area:(a)a D2 4 )3 6 .30 cm 4fs 2= 0.529 cm s .Likewise, at a 1.35 cm diameter, v=P1.66A= w16.5 cm 3 sa 1.35 cm 4f2= 11.5 cm s .1 cubic meter of water has a massejejejm = V = 1.00 10 3 kg cm3 1.00 m 3 10 2 cm m (b)3= 1 000 kgAs a rough calculation, we treat each item as if it were 100% water. cell:m = V = FG 4 R IJ = FG 1 D IJ = e1 000 kg m jFG 1 IJ e1.0 10 H3 K H6 K H6 K 3336jm3= 5.2 10 16 kg kidney: m = V = FG 4 R IJ = e1.00 10 H3 K 33kg cm34 jFGH 3 IJK ( 4.0 cm)3= 0.27 kg fly:m=FG D hIJ = e1 10 H4 K 23kg cm 3= 1.3 10 5 kgP1.67V20 mpg =(10 8 cars)(10 4 mi yr ) = 5.0 10 10 gal yr 20 mi galV25 mpg =(10 8 cars)(10 4 mi yr ) = 4.0 10 10 gal yr 25 mi galFuel saved = V25 mpg V20 mpg = 1.0 10 10 gal yrjFGH IJK a2.0 mmf a4.0 mmfe10 4 21jcm mm317 19. 18 P1.68P1.69Physics and MeasurementF GHIF JK GHIF JK GHIF JK GHI FG JK HIJ FG KHI JKfurlongs 220 yd 0.914 4 m 1 fortnight 1 day 1 hr = 8.32 10 4 m s fortnight 1 furlong 1 yd 14 days 24 hrs 3 600 s This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth. v = 5.00The volume of the galaxy isej e10 r 2 t = 10 21 m219jm ~ 10 61 m3 .If the distance between stars is 4 10 16 m , then there is one star in a volume on the order ofe4 10 The number of stars is aboutP1.7010503m starThe density of each material is =Al:Cu:Brass:Sn:Fe:P1.7110 61 m 3(a) (b)=ab4 51.5 ggf a3.75 cmf 4b56.3 g g = = a1.23 cmf a5.06 cmf 4b94.4 g g = = a1.54 cmf a5.69 cmf 4b69.1 g g = = a1.75 cmf a3.74 cmf 4b 216.1 g g = = a1.89 cmf a9.77 cmf 2.52 cmj29.3628.9127.682~ 10 11 stars .7.88g cm3g cm3FG g IJ is H cm K F g IJ is The tabulated value G 8.92 H cm K The tabulated value 2.70This would takee2% smaller.35% smaller.cm3 g cm3 g cm3FG HThe tabulated value 7.863.16 10 7 s yr3 4 3 4 r = 5.00 10 7 m = 5.24 10 19 m3 3 3 1 m3 = = 1.91 10 18 micrometeorites 5.24 10 19 m3Vmm =3gb3 600 s hrgb24 hr daygb365.25 days yrg = Vcube Vmm3m ~ 10 50 m 3 .4m m m . = = V r 2h D2h= 2.75216j1.91 10 18 micrometeorites 3.16 10 7 micrometeorites yr= 6.05 10 10 yr .g cm3IJ is K0.3% smaller. 20. Chapter 1ANSWERS TO EVEN PROBLEMS 5.52 10 3 kg m3 , between the densities of aluminum and iron, and greater than the densities of surface rocks.P1.341.3 10 21 kgP1.36200 kmP1.423.0 kgP1.38(a) 13.4; (b) 49.1P1.67.69 cmP1.8(a) and (b) see the solution, N A = 6.022 137 10 23 ; (c) 18.0 g; (d) 44.0 gP1.2P1.40Al(a)P1.18 P1.20534(a) 3.39 10 ft ; (b) 2.54 10 lb 5115.9 m 316 m4.50 m 2P1.601.39 10 4 m3(a) 797; (b) 1.1; (c) 17.66P1.5835.7 m 2M L ; (b) 1 newton = 1 kg m s 2 T2(a) 3; (b) 4; (c) 3; (d) 2P1.56P1.16a209 4f cmP1.52(a) ii; (b) iii; (c) i~ 10 11 cans; ~ 10 5 tonsP1.54P1.14~ 10 9 raindropsP1.50(a) 4.02 10 25 molecules; (b) 3.65 10 4 moleculesP1.44P1.48P1.12~ 10 7 revP1.46(a) 9.83 10 16 g ; (b) 1.06 10 7 atomsP1.2413FeP1.42P1.10P1.22rAl = rFeFG IJ H Ksee the solution; 24.6P1.623.64 cents ; noP1.64see the solutionP1.66(a) 1 000 kg; (b) 5.2 10 16 kg ; 0. 27 kg ;27(a) 560 km = 5.60 10 m = 5.60 10 cm ; (b) 491 m = 0.491 km = 4.91 10 4 cm ; (c) 6.19 km = 6.19 10 3 m = 6.19 10 5 cm ; (d) 2.50 km = 2.50 10 3 m = 2.50 10 5 cmP1.264.05 10 3 m 2P1.28(a) 1 mi h = 1.609 km h ; (b) 88.5 km h ;1.3 10 5 kg(c) 16.1 km hP1.688.32 10 4 m s ; a snailP1.301.19 10 57 atomsP1.70see the solutionP1.322.57 10 6 m319 21. 2 Motion in One Dimension CHAPTER OUTLINE 2.1 2.2 2.3 2.4 2.5 2.6 2.7Position, Velocity, and Speed Instantaneous Velocity and Speed Acceleration Motion Diagrams One-Dimensional Motion with Constant Acceleration Freely Falling Objects Kinematic Equations Derived from CalculusANSWERS TO QUESTIONS Q2.1If I count 5.0 s between lightning and thunder, the sound has traveled 331 m s 5.0 s = 1.7 km . The transit time for the light is smaller bybga f3.00 10 8 m s = 9.06 10 5 times, 331 m s so it is negligible in comparison. Q2.2Yes. Yes, if the particle winds up in the +x region at the end.Q2.3Zero.Q2.4Yes. Yes.Q2.5No. 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.Q2.6We assume the object moves along a straight line. If its average x 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 t 0 in the graph, the slope of the curve is zero, and thus the instantaneous velocity at that time is also zero.t0tFIG. Q2.6 Q2.7Yes. 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. 22Motion in One DimensionafQ2.8Yes. If you drop a doughnut from rest v = 0 , 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.Q2.9No: 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.Q2.10Yes. Consider throwing a ball straight up. As the ball goes up, its v velocity is upward v > 0 , and its acceleration is directed down v0 a < 0 . 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.aa fft 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 WestQ2.12No. Constant acceleration only. Yes. Zero is a constant.Q2.13The 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.Q2.14Once the objects leave the hand, both are in free fall, and both experience the same downward acceleration equal to the free-fall acceleration, g.Q2.15They 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.Q2.16With h =1 2 gt , 2af1 2 g 0.707t . The time is later than 0.5t. 2(a)0.5 h =(b)The distance fallen is 0.25 h =a f1 2 g 0.5t . The elevation is 0.75h, greater than 0.5h. 2 23. Chapter 2Q2.17Above. Your ball has zero initial speed and smaller average speed during the time of flight to the passing point.SOLUTIONS TO PROBLEMS Section 2.1Position, Velocity, and SpeedP2.1v = 2.30 m s(a) (b)x 57.5 m 9.20 m = = 16.1 m s t 3.00 s(c) *P2.2v= v=x 57.5 m 0 m = = 11.5 m s t 5.00 s(a)v=x 20 ft 1m = t 1 yr 3.281 ftFG HIJ FG 1 yr IJ = 2 10 m s or in particularly windy times K H 3.156 10 s K x 100 ft F 1 m I F = v= GH 3.281 ft JK GH 3.1561 yr s IJK = 1 10 m s . t 1 yr 10 777(b)The time required must have been t =FG H3 000 mi 1 609 m x = v 10 mm yr 1 miv=x 10 m = = 5 ms t 2sv=5m = 1.2 m s 4s(c)v=x 2 x1 5 m 10 m = = 2.5 m s 4 s2 s t 2 t1(d)v=x 2 x 1 5 m 5 m = = 3.3 m s 7 s4 s t 2 t1(e)P2.4(a) (b)P2.36v=x 2 x1 0 0 = = 0 ms 80 t 2 t1x = 10t 2 : Foraf xamf ts= 2.02.13.0=44.19040(a)v=x 50 m = = 50.0 m s t 1.0 s(b)v=x 4.1 m = = 41.0 m s t 0.1 sIJ FG 10 mm IJ = KH 1 m K 35 10 8 yr .23 24. 24 P2.5Motion in One Dimension(a)Let d represent the distance between A and B. Let t1 be the time for which the walker has d the higher speed in 5.00 m s = . Let t 2 represent the longer time for the return trip in t1 d d d 3.00 m s = . Then the times are t1 = and t 2 = . The average speed t2 5.00 m s 3.00 m s is:bv=v= (b)Section 2.2 P2.6(a)e2 15.0 m 2 s 2 8.00 m sj=bbd+d = + 3 .00d m sd 5.00 m sg2db8.00 m sgd g e15.0 m s jg b23.75 m sInstantaneous Velocity and Speede Thus, at t = 3.00 s: x = e3.00 m s ja3.00 sf =jAt any time, t, the position is given by x = 3.00 m s 2 t 2 . 22ijae27.0 m .f2At t f = 3.00 s + t : x f = 3.00 m s 2 3.00 s + t , orbja fg ex f = 27.0 m + 18.0 m s t + 3.00 m s 2 t (c)t 0(a)F x x I = lim e18.0 m s + e3.00 m s jtj = GH t JK fi2t 0x f xi t f ti=a2.0 8.0f m = 6.0 m = a4 1.5f s 2.5 s18.0 m s .2.4 m sThe slope of the tangent line is found from points C and D. tC = 1.0 s, x C = 9.5 m and t D = 3.5 s, x D = 0 ,bgbgv 3.8 m s . (c).at ti = 1.5 s , x i = 8.0 m (Point A) at t f = 4.0 s , x f = 2.0 m (Point B) v=(b)2The instantaneous velocity at t = 3.00 s is: v = limP2.72She starts and finishes at the same point A. With total displacement = 0, average velocity = 0 .i(b)Total distance = Total timegThe velocity is zero when x is a minimum. This is at t 4 s .FIG. P2.7 25. Chapter 2P2.8(a)(b)58 m 2.5 s 54 m At t = 4.0 s, the slope is v 3s 49 m At t = 3.0 s, the slope is v 3.4 s 36 m At t = 2.0 s , the slope is v 4.0 sAt t = 5.0 s, the slope is v 23 m s . 18 m s . 14 m s . 9.0 m s .v 23 m s 4.6 m s 2 t 5.0 s(c) (d) (a)v=v=(c)v=(d) *P2.10Initial velocity of the car was zero .(b)P2.9a=v=(5 0 ) m (1 0) s= 5 ms(5 10) m (4 2) s= 2.5 m s(5 m 5 m) (5 s 4 s) 0 (5 m) (8 s 7 s )= 0= +5 m sFIG. P2.9Once it resumes the race, the hare will run for a time of t=x f xi vx=1 000 m 800 m = 25 s . 8 msIn this time, the tortoise can crawl a distanceafx f xi = 0.2 m s ( 25 s)= 5.00 m .25 26. 26Motion in One DimensionSection 2.3 P2.11AccelerationChoose the positive direction to be the outward direction, perpendicular to the wall. v f = vi + at : a =P2.12(a)afv 22.0 m s 25.0 m s = = 1.3410 4 m s 2 . t 3.50 103 sAcceleration is constant over the first ten seconds, so at the end,chv f = vi + at = 0 + 2.00 m s 2 (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 tochv f = vi + at = 20.0 m s + 3.00 m s 2 (5.00 s)= 5.00 m s . (b)In the first ten seconds, 1 2 1 2 at = 0 + 0 + 2.00 m s 2 (10.0 s) = 100 m . 2 2cx f = x i + vi t +hOver the next five seconds the position changes to x f = xi + vi t +af1 2 at = 100 m + 20.0 m s (5.00 s)+ 0 = 200 m . 2And at t = 20.0 s , x f = x i + vi t + *P2.13(a)af1 2 1 2 at = 200 m + 20.0 m s (5.00 s)+ 3.00 m s 2 (5.00 s) = 262 m . 2 2chdistance traveled . During the first t quarter mile segment, Secretariats average speed wasThe average speed during a time interval t is v =v1 =0.250 mi 1 320 ft = = 52.4 ft s 25.2 s 25.2 sb35.6 mi hg .During the second quarter mile segment, v2 =1 320 ft = 55.0 ft s 24.0 sb37.4 mi hg .For the third quarter mile of the race, v3 =1 320 ft = 55.5 ft s 23.8 sb37.7 mi hg ,and during the final quarter mile, v4 = continued on next page1 320 ft = 57.4 ft s 23.0 sb39.0 mi hg . 27. Chapter 2(b)27Assuming that v f = v 4 and recognizing that vi = 0 , the average acceleration during the race was v f via= P2.14(a)=total elapsed time57.4 ft s 0 = 0.598 ft s 2 . ( 25. 2 + 24.0 + 23.8 + 23.0) sAcceleration is the slope of the graph of v vs t.a (m/s2) 2.0For 0 < t < 5.00 s, a = 0 .1.6For 15.0 s < t < 20.0 s , a = 0 . For 5.0 s < t < 15.0 s , a =a=v f vi t f ti1.0.8.00 (8.00) 15.0 5.000.0= 1.60 m s 2t (s) 05a=v f vi t f ti(i)For 5.00 s < t < 15.0 s , ti = 5.00 s , vi = 8.00 m s , t f = 15.0 s v f = 8.00 m s a=(ii)x = 2.00 + 3.00t t 2 , v = At t = 3.00 s :v f vi t f ti=af8.00 8.00 = 1.60 m s 2 . 15.0 5.00ti = 0 , vi = 8.00 m s , t f = 20.0 s , v f = 8.00 m s a=P2.1515FIG. P2.14We can plot a(t ) as shown. (b)10v f vi t f ti=8.00 (8.00) 20.0 0dx dv = 3.00 2.00t , a = = 2.00 dt dt(a)x = ( 2.00 + 9.00 9.00) m = 2.00 m(b)v = (3.00 6.00) m s = 3.00 m s(c)a = 2.00 m s 2= 0.800 m s 220 28. 28 P2.16Motion in One Dimension(a)2At t = 2.00 s , x = 3.00( 2.00) 2.00( 2.00)+ 3.00 m = 11.0 m.a fAt t = 3.00 s , x = 3.00 9.002a f 2.00 3.00 + 3.00 m = 24.0 mso v= (b)x 24.0 m 11.0 m = 13.0 m s . = 3.00 s 2.00 s tAt all times the instantaneous velocity is v=d 3.00t 2 2.00t + 3.00 = (6.00t 2.00) m s dtchAt 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 . v 16.0 m s 10.0 m s = = 6.00 m s 2 t 3.00 s 2.00 s(c) (d)At all times a =(a)a=(b)Maximum positive acceleration is at t = 3 s, and is approximately 2 m s 2 .(c)a = 0 , at t = 6 s , and also for t > 10 s .(d)P2.17a=Maximum negative acceleration is at t = 8 s, and is approximately 1.5 m s 2 .Section 2.4 P2.18d (6.00 2.00)= 6.00 m s 2 . (This includes both t = 2.00 s and t = 3.00 s ). dtv 8.00 m s = = 1.3 m s 2 t 6.00 sMotion Diagrams(a) (b) (c) (d) (e) continued on next page 29. Chapter 2(f)Section 2.5 P2.1929One 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.One-Dimensional Motion with Constant AccelerationcFrom v 2 = vi2 + 2 ax , we have 10.97 10 3 m s fh2= 0 + 2 a( 220 m) , so that a = 2.7410 5 m s 2which is a = 2.79 10 4 times g . P2.20x f xi =(b) P2.21(a)a=v f vi taf1 1 vi + v f t becomes 40 m = vi + 2.80 m s (8.50 s) which yields vi = 6.61 m s . 2 2c=h2.80 m s 6.61 m s = 0.448 m s 2 8.50 sGiven vi = 12.0 cm s when x i = 3.00 cm(t = 0) , and at t = 2.00 s , x f = 5.00 cm , 1 1 2 2 at : 5.00 3.00 = 12.0( 2.00)+ a( 2.00) 2 2 32.0 a = = 16.0 cm s 2 . 8.00 = 24.0 + 2 a 2x f x i = vi t +*P2.22(a)Let i be the state of moving at 60 mi h and f be at restd 0 = b60 mi hg2 2 v xf = v xi + 2 a x x f xiax(b)Similarly,mi fFGH 5 1280 ft IJK 3 600 mi F 5 280 ft I F 1 h I = G JG J = 21.8 mi h s 242 h H 1 mi K H 3 600 s K F 1 609 m IJ FG 1 h IJ = 9.75 m s = 21.8 mi h s G H 1 mi K H 3 600 s K 2a+ 2 a x 121 ft 02b0 = 80 mi h ax = (c)ibg26 400 5 280ba+ 2 a x 211 ft 0422 3 600gg2.fmi h s = 22.2 mi h s = 9.94 m s 2 .Let i be moving at 80 mi h and f be moving at 60 mi h .d i b60 mi hg = b80 mi hg + 2a a211 ft 121 ftf 2 800b5 280 g a = mi h s = 22.8 mi h s = 10.2 m s 2a90fb3 600g 2 2 v xf = v xi + 2 a x x f x i 22xx2. 30. 30 *P2.23Motion in One Dimension(a)Choose the initial point where the pilot reduces the throttle and the final point where the boat passes the buoy: x i = 0 , x f = 100 m , v xi = 30 m s , v xf = ?, a x = 3.5 m s 2 , t = ? x f = xi + v xi t +1 axt 2 : 2af100 m = 0 + 30 m s t +1 3.5 m s 2 t 2 2chc1.75 m s ht a30 m sft + 100 m = 0 . 22We use the quadratic formula: t=t=cb b 2 4ac 2ah30 m s 900 m 2 s 2 4 1.75 m s 2 (100 m)c2 1.75 m s2h=30 m s 14.1 m s 3.5 m s 2= 12.6 s 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.ej(b) P2.24v xf = v xi + a x t = 30 m s 3.5 m s 2 4.53 s = 14.1 m s(a)Total displacement = area under the v , t curve from t = 0 to 50 s.a fb bga f b ga fgaf1 50 m s 15 s + 50 m s 40 15 s 2 1 + 50 m s 10 s 2 x = 1 875 m x =(b)From t = 10 s to t = 40 s , displacement is(c)bga f bga f1 50 m s + 33 m s 5 s + 50 m s 25 s = 1 457 m . 2 v (50 0) m s 0 t 15 s : a1 = = = 3.3 m s 2 t 15 s 0 15 s < t < 40 s : a 2 = 0 x =40 s t 50 s : a 3 = continued on next pagev (0 50) m s = = 5.0 m s 2 t 50 s 40 sFIG. P2.24 31. Chapter 2(d)1 1 a1 t 2 = 3.3 m s 2 t 2 or x1 = 1.67 m s 2 t 2 2 2chch(i)x1 = 0 +(ii)1 x 2 = (15 s) 50 m s 0 + 50 m s (t 15 s) or x 2 = 50 m s t 375 m 2(iii)For 40 s t 50 s ,ax3 =fafFG area under v vs t IJ + 1 a (t 40 s) + a50 m sf(t 40 s) H from t = 0 to 40 sK 2 23or x 3 = 375 m + 1 250 m +ja1 5.0 m s 2 t 40 s 2ef + b50 m sgat 40 sf 2which reduces tobg ejx 3 = 250 m s t 2.5 m s 2 t 2 4 375 m . total displacement 1 875 m = = 37.5 m s total elapsed time 50 s(e) P2.25v=(a)Compare the position equation x = 2.00 + 3.00t 4.00t 2 to the general form x f = xi + vi t +1 2 at 2to recognize that x i = 2.00 m, vi = 3.00 m s, and a = 8.00 m s 2 . The velocity equation, v f = vi + at , is thenchv f = 3.00 m s 8.00 m s 2 t . The particle changes direction when v f = 0 , which occurs at t = time is:ax = 2.00 m + 3.00 m s(b)fFGH 3 sIJK c4.00 m s hFGH 3 sIJK 8 8 23 s . The position at this 82= 2.56 m .2v 1 2 at , observe that when x f = xi , the time is given by t = i . Thus, a 2 when the particle returns to its initial position, the time isFrom x f = xi + vi t +t=ca2 3.00 m s 8.00 m sand the velocity is v f = 3.00 m s 8.00 m s 22f=3 shFGH 3 sIJK = 443.00 m s .31 32. 32 *P2.26Motion in One DimensionThe time for the Ford to slow down we find from 1 v xi + v xf t 2 2 250 m 2 x t= = = 6.99 s . v xi + v xf 71.5 m s + 0x f = xi +daifIts time to speed up is similarly t=2(350 m) 0 + 71.5 m s= 9.79 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 travelsafb ga1 1 v xi + v xf t = 0 + 71.5 + 71.5 m s 21.8 s 2 2 = 1 558 mdx f = xi +ifwhile the Ford travels 250 + 350 m = 600 m, to fall behind by 1 558 m 600 m = 958 m . P2.27(a)chvi = 100 m s , a = 5.00 m s 2 , v f = vi + at so 0 = 100 5t , v 2 = vi2 + 2 a x f xi so f 2ch0 = (100 ) 2(5.00) x f 0 . Thus x f = 1 000 m and t = 20.0 s . (b) P2.28At this acceleration the plane would overshoot the runway: No .(a)Take ti = 0 at the bottom of the hill where x i = 0 , vi = 30.0 m s, a = 2.00 m s 2 . Use these values in the general equation x f = xi + vi t + to finda1 2 at 2f 1 c2.00 m s ht 2 2x f = 0 + 30.0t m s + when t is in secondsc2hx f = 30.0t t 2 m .ejTo find an equation for the velocity, use v f = vi + at = 30.0 m s + 2.00 m s 2 t , v f = (30.0 2.00t ) m s . (b)The distance of travel x f becomes a maximum, x max , 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 = (3.00 2.00t ) m s , v f = 0 when t = 15.0 s. Thench2x max = 30.0t t 2 m = (30.0)(15.0)(15.0) = 225 m . 33. Chapter 2P2.2933In the simultaneous equations:R | Sx | Tv xf = v xi + a x t f xi =U | | we have Rv = v c5.60 m s h(4.20 s)U . S 62.4 m = 1 v + v (4.20 s) | V V t| hW | | c h 2 T W xf1 v xi + v xf 2cSo substituting for v xi gives 62.4 m =2xixixf1 v xf + 56.0 m s 2 ( 4.20 s)+ v xf ( 4.20 s) 2ch14.9 m s = v xf +1 5.60 m s 2 ( 4.20 s). 2chThus v xf = 3.10 m s .P2.30Take any two of the standard four equations, such as substitute into the other: v xi = v xf a x t x f xi =R | Sx | Tv xf = v xi + a x t f xi =1 v xi + v xf 2cU |. Solve one for v ht V | W1 v xf a x t + v xf t . 2chThus 1 x f xi = v xf t a x t 2 . 2 Back in problem 29, 62.4 m = v xf ( 4.20 s) v xf =P2.31v f vi(a)a=(b)x f = vi t +t=632e j= 5 280 3 6001.40afFGH1 2 5.60 m s 2 ( 4. 20 s) 2ch62.4 m 49.4 m = 3.10 m s . 4.20 s662 ft s 2 = 202 m s 2I a f a fa f JK5 280 1 2 1 1.40 662 1.40 at = 632 2 3 600 22= 649 ft = 198 mxi ,and 34. 34 P2.32Motion in One Dimension(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 f vi a=20.0 m s 0 m s 2.00 m s 2= 10.0 s .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 x 1 = vt =FG 0 + 20.0 IJ(10.0)= 100 m . H 2 KWith a being 0 for this interval, the distance traveled during the next 20.0 s is x 2 = vi t +1 2 at = ( 20.0)( 20.0)+ 0 = 400 m. 2The distance traveled in the last 5.00 s is x 3 = vt =FG 20.0 + 0 IJ(5.00)= 50.0 m. H 2 KThe total distance x = x1 + x 2 + x 3 = 100 + 400 + 50 = 550 m , and the average velocity is x 550 = 15.7 m s . given by v = = t 35.0 P2.33We have vi = 2.00 10 4 m s, v f = 6.00 10 6 m s , x f xi = 1.50 102 m .cchh2 1.50 102 m 2 x f xi 1 = = 4.98 109 s vi + v f t : t = 4 6 2 vi + v f 2.00 10 m s + 6.00 10 m sch(a)x f xi =(b)v 2 = vi2 + 2 a x x f xi : fdax =iv 2 vi2 f 2( x f xi )e6.00 10 =6msj e2.00 10 22(1.50 10 2 m)4msj2= 1.20 10 15 m s 2 35. Chapter 2*P2.34(a)chc2 2 v xf = v xi + 2 a x x f x i : 0.01 3 10 8 m sax (b)c310 =6h2ms80 m= 0 + 2 a x ( 40 m)h2= 1.12 10 11 m s 2We must find separately the time t1 for speeding up and the time t 2 for coasting: x f xi =x f xi =1 1 v xf + v xi t1 : 40 m = 3 10 6 m s + 0 t1 2 2 t1 = 2.67 10 5 sdiej1 1 v xf + v xi t 2 : 60 m = 3 10 6 m s + 3 10 6 m s t 2 2 2 t 2 = 2.00 10 5 sdiejtotal time = 4.67 105 s . *P2.35(a)Along the time axis of the graph shown, let i = 0 and f = t m . Then v xf = v xi + a x t gives v c = 0 + am tm am =(b)vc . tmThe displacement between 0 and t m is x f xi = v xi t +1 vc 2 1 1 axt 2 = 0 + t m = v c tm . 2 tm 2 2The displacement between t m and t 0 is x f xi = v xi t +af1 a x t 2 = v c t0 tm + 0 . 2The total displacement is x =FG H1 1 v c t m + v c t 0 v c t m = v c t 0 tm 2 2IJ K.(c)For constant v c and t 0 , x is minimized by maximizing t m to t m = t 0 . Then v t 1 x min = v c t 0 t 0 = c 0 . 2 2(e)This is realized by having the servo motor on all the time.(d)We maximize x by letting t m approach zero. In the limit x = v c t 0 0 = v c t 0 .(e)This cannot be attained because the acceleration must be finite.FG HIJ Kaf35 36. 36 *P2.36Motion in One DimensionLet the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit, 1 axt 2 2 1 2 = 0 + vi t d + at d = v d t d 2 1 v d = vi + at d 2x f = xi + v xi t +(a)The speed halfway through the photogate in space is given by 2 v hs = vi2 + 2 aFG IJ = v H 2K2 i+ av d t d .v hs = vi2 + av d t d and this is not equal to v d unless a = 0 . (b)The speed halfway through the photogate in time is given by v ht = vi + aFG t IJ and this is H 2K dequal to v d 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 s 2 , x f xi = 9.00 m . Thendijaefv 2 = vi2 + 2 a x f xi = 0 2 + 2 0.500 m s 2 9.00 m f v f = 3.00 m s . (b)x f x i = vi t +1 2 at 21 0.500 m s 2 t 2 2 t = 6.00 se9.00 = 0 +(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.chv 2 = vi2 + 2 a x f xi gives f a= (d)jv 2 vi2 fc2 x f xih=a0 3.00 m s 2(15.0 m)f2= 0.300 m s 2 .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 f xi = 8.00 m , a = 0.300 m s 2di bv 2 = vi2 + 2 a x f xi = 3.00 m s f v f = 2.05 m s .g + 2e0.300 m s ja8.00 mf = 4.20 m 222s2 37. Chapter 2P2.3837Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sues car. For her we have x is = 0 , vis = 30.0 m s , a s = 2.00 m s 2 so her position is given by x s (t )= x is + vis t +af1 1 a s t 2 = 30.0 m s t + 2.00 m s 2 t 2 . 2 2chFor the van, x iv = 155 m, viv = 5.00 m s , a v = 0 and x v (t )= xiv + viv t +af1 a v t 2 = 155 + 5.00 m s t + 0 . 2To test for a collision, we look for an instant t c when both are at the same place: 2 30.0t c t c = 155 + 5.00t c 2 0 = t c 25.0t c + 155 .From the quadratic formula 2tc =25.0 ( 25.0) 4(155) 2= 13.6 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 positionaf155 m + 5.00 m s (11.4 s)= 212 m . *P2.39As in the algebraic solution to Example 2.8, we let t represent the time the trooper has been moving. We graph x car = 45 + 45tx (km) 1.5 car 1and 2x trooper = 1.5t . They intersect atpolice officer0.510 t = 31 s .2030FIG. P2.3940t (s) 38. 38Motion in One DimensionSection 2.6 P2.40Freely Falling ObjectsafChoose the origin y = 0 , t = 0 at the starting point of the ball and take upward as positive. Then yi = 0 , vi = 0 , and a = g = 9.80 m s 2 . The position and the velocity at time t become: y f yi = vi t +1 1 1 2 at : y f = gt 2 = 9.80 m s 2 t 2 2 2 2ejandchv f = vi + at : v f = gt = 9.80 m s 2 t . 1 2 9.80 m s 2 (1.00 s) = 4.90 m 2 1 2 at t = 2.00 s : y f = 9.80 m s 2 ( 2.00 s) = 19.6 m 2 1 2 at t = 3.00 s : y f = 9.80 m s 2 (3.00 s) = 44.1 m 2c c ch h h(a)at t = 1.00 s : y f = (b)at t = 1.00 s : v f = 9.80 m s 2 (1.00 s)= 9.80 m s at t = 2.00 s : v f at t = 3.00 s : v fP2.41c h = c9.80 m s h( 2.00 s)= 19.6 m s = c9.80 m s h(3.00 s)= 29.4 m s 2 2Assume 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 s 2 . 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:div 2 = vi2 + 2 a y f yi : fjbe0 = vi2 2 9.80 m s 2 1 609 m vi = 178 m s .gHis time in the air may be found by considering his motion from just after launch to just before impact: y f yi = vi t +af1 1 2 at : 0 = 178 m s t 9.80 m s 2 t 2 . 2 2chThe 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 =bgbg$1.00 = 0.027 6 $ s 3 600 s h = $99.3 h . 36.2 sWe have assumed that the workmans flight time, a mile, and a dollar, were measured to threedigit 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 2P2.42391 We have y f = gt 2 + vi t + yi 2h acf0 = 4.90 m s 2 t 2 8.00 m s t + 30.0 m . Solving for t, t=8.00 64.0 + 588 . 9.80Using only the positive value for t, we find that t = 1.79 s . P2.431 2 2 at : 4.00 = (1.50)vi (4.90)(1.50) and vi = 10.0 m s upward . 2(a)y f yi = vi t +(b)v f = vi + at = 10.0 (9.80)(1.50) = 4.68 m s v f = 4.68 m s downwardP2.44The bill starts from rest vi = 0 and falls with a downward acceleration of 9.80 m s 2 (due to gravity). Thus, in 0.20 s it will fall a distance of y = vi t 1 2 2 gt = 0 4.90 m s 2 (0. 20 s) = 0.20 m . 2chafThis distance is about twice the distance between the center of the bill and its top edge 8 cm . Thus, David will be unsuccessful . *P2.45(a)From y = vi t +1 2 at with vi = 0 , we have 2 t=a f=2 y2(23 m)a9.80 m s 2c= 2.17 s .h(b)The final velocity is v f = 0 + 9.80 m s 2 ( 2.17 s)= 21.2 m s .(c)The time take for the sound of the impact to reach the spectator is t sound =y v sound=23 m = 6.76 102 s , 340 m sso the total elapsed time is t total = 2.17 s + 6.76 10 2 s 2.23 s . 40. 40 P2.46Motion in One DimensionAt any time t, the position of the ball released from rest is given by y1 = h 1 2 gt . At time t, the 21 2 gt . The time at which the 2 h 1 h first ball has a position of y1 = is found from the first equation as = h gt 2 , which yields 2 2 2 h h . To require that the second ball have a position of y 2 = at this time, use the second t= g 2 position of the ball thrown vertically upward is described by y 2 = vi t equation to obtainF I GH JKh h 1 h = vi g . This gives the required initial upward velocity of the second 2 g 2 gball as vi = gh . P2.47(a)v f = vi gt : v f = 0 when t = 3.00 s , g = 9.80 m s 2 . Therefore,chvi = gt = 9.80 m s 2 (3.00 s)= 29.4 m s . (b)y f yi =1 v f + vi t 2chy f yi = *P2.48(a)bfga1 29.4 m s 3.00 s = 44.1 m 2Consider the upward flight of the arrow.d i 0 = b100 m sg + 2e 9.8 m s jy2 2 v yf = v yi + 2 a y y f yi 2y = (b)210 000 m 2 s 2 19.6 m s 2= 510 mConsider the whole flight of the arrow. y f = yi + v yi t +b1 ayt 2 2g0 = 0 + 100 m s t +1 9.8 m s 2 t 2 2ejThe root t = 0 refers to the starting point. The time of flight is given by t=P2.49100 m s 4.9 m s 2= 20.4 s .Time to fall 3.00 m is found from Eq. 2.12 with vi = 0 , 3.00 m =1 9.80 m s 2 t 2 , t = 0.782 s. 2ch(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 2P2.50Take 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 s 2 . Thus, y = vi t + before impact is:1 2 at 144 ft = 0 + 16.0 ft s 2 t 2 giving t fall = 3.00 s . Her velocity just 2chchv f = vi + gt = 0 + 32.0 ft s 2 (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=(c)v 2 vi2 fa f2 y=a0 96.0 ft sf22(1.50 ft )Time to crush box: t == 3.07 10 3 ft s 2 , or a = 3.07 10 3 ft s 2 upward .2(1.50 ft) y y = v +v = or t = 3.13 102 s . f i v 0 + 96.0 ft s 2P2.51a fy = 3.00t 3 : At t = 2.00 s , y = 3.00 2.003= 24.0 m andvy =Ady = 9.00t 2 = 36.0 m s . dtIf the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is y b = y bi + vi t 1 1 2 gt = 24.0 + 36.0t (9.80)t 2 . 2 2Setting y b = 0 , 0 = 24.0 + 36.0t 4.90t 2 . Solving for t, (only positive values of t count), t = 7.96 s . *P2.52Consider the last 30 m of fall. We find its speed 30 m above the ground: y f = yi + v yi t +1 ayt 2 2a f 1 e9.8 m s ja1.5 sf 2 20 = 30 m + v yi 1.5 s + v yi =230 m + 11.0 m = 12.6 m s . 1.5 sNow consider the portion of its fall above the 30 m point. We assume it starts from restdi b12.6 m sg = 0 + 2e9.8 m s jy 2 2 v yf = v yi + 2 a y y f yi 2y =2160 m 2 s 2 19.6 m s 2Its original height was then 30 m + 8.16 m = 38.2 m .= 8.16 m .41 42. 42Motion in One DimensionSection 2.7 P2.53(a)Kinematic Equations Derived from Calculus J=da = constant dt da = Jdtza = J dt = Jt + c 1 but a = ai when t = 0 so c 1 = ai . Therefore, a = Jt + ai dv dt dv = adt a=z zbv = adt =gJt + ai dt =but v = vi when t = 0, so c 2 = vi and v =1 2 Jt + ai t + c 2 21 2 Jt + ai t + vi 2dx dt dx = vdt v=z z FGHx = vdt =IJ K1 2 Jt + ai t + vi dt 21 3 1 2 Jt + ai t + vi t + c 3 6 2 x = xi x=when t = 0, so c 3 = xi . Therefore, x = (b)aa 2 = Jt + aicf2= J 2 t 2 + ai2 + 2 Jai ta 2 = ai2 + J 2 t 2 + 2 Jai tFG H1 3 1 2 Jt + ai t + vi t + xi . 6 2h1 2 a 2 = ai2 + 2 J Jt + ai t 2IJ KRecall the expression for v: v =af1 2 1 Jt + ai t + vi . So v vi = Jt 2 + ai t . Therefore, 2 2aa 2 = ai2 + 2 J v vif. 43. Chapter 2P2.54(a)See the graphs at the right. Choose x = 0 at t = 0. At t = 3 s, x =af1 8 m s (3 s)= 12 m . 2afAt t = 5 s, x = 12 m + 8 m s ( 2 s)= 28 m . At t = 7 s, x = 28 m +af1 8 m s ( 2 s)= 36 m . 28 ms = 2.67 m s 2 . 3s For 3 < t < 5 s, a = 0 .(b)(c)For 5 s < t < 9 s , a = (d)At t = 6 s, x = 28 m + 6 m s (1 s)= 34 m .(e)P2.55For 0 < t < 3 s, a =At t = 9 s, x = 36 m +(a)16 m s = 4 m s 2 . 4saa=faf1 8 m s ( 2 s)= 28 m . 2FIG. P2.54dv d = 5.00 10 7 t 2 + 3.00 10 5 t dt dtcha = 10.0 10 7 m s 3 t + 3.00 10 5 m s 2 Take x i = 0 at t = 0. Then v =dx dtz ze tt00x 0 = vdt =j5.00 10 7 t 2 + 3.00 10 5 t dtt3 t2 + 3.00 10 5 3 2 3 3 7 x = 1.67 10 m s t + 1.50 10 5 m s 2 t 2 .x = 5.00 10 7e(b)j echThe bullet escapes when a = 0 , at 10.0 10 7 m s 3 t + 3.00 10 5 m s 2 = 0 t=(c)jchcNew v = 5.00 10 7 3.00 1033.00 10 5 s = 3.00 103 s . 10.0 10 7h + c3.0010 hc3.0010 h 253v = 450 m s + 900 m s = 450 m s . (d)chcx = 1.67 10 7 3.00 103h + c1.5010 hc3.0010 h 3x = 0.450 m + 1.35 m = 0.900 m53 243 44. 44 P2.56Motion in One Dimensiona=dv = 3.00 v 2 , vi = 1.50 m s dtSolving for v,dv = 3.00 v 2 dtz vz tv 2 dv = 3.00 dtv = vi When v =t =01 1 1 1 + = 3.00t or 3.00t = . v vi v vivi 1 , t= = 0.222 s . 2 3.00 viAdditional Problems *P2.57a fThe distance the car travels at constant velocity, v 0 , during the reaction time is x 1 = v 0 t r . The time for the car to come to rest, from initial velocity v 0 , after the brakes are applied is t2 =v f vi a=0 v0 v = 0 a aand the distance traveled during this braking period isaxf2= vt 2 =Fv GH+ vif2I t = FG 0 + v IJ FG v IJ = v . JK H 2 K H a K 2 a 0202 0Thus, the total distance traveled before coming to a stop isa f + a x fsstop = x*P2.58(a)12= v 0 t r 2 v0 . 2a2 v0 (See the solution to Problem 2.57) from the 2a intersection of length s i when the light turns yellow, the distance the car must travel before the light turns red is v2 x = sstop + si = v 0 t r 0 + si . 2aIf a car is a distance sstop = v 0 t r 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 v 0 . This is v2t light (b)v s x v 0 t r 2 0 + si a = = = t r 0 + i . v0 v0 2 a v0With si = 16 m, v = 60 km h , a = 2.0 m s 2 , and t r = 1.1 s , t light = 1.1 s F 0.278 m s I + 16 m F 1 km h I = G J G J 2e 2.0 m s j H 1 km h K 60 km h H 0.278 m s K 60 km h26. 23 s . 45. Chapter 2*P2.59(a)(b)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.200 v100t 100 200 300 4000 50 0t (s)100FIG. P2.59(a)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 = slope =(c)45v (155 45) mi h = = 2. 2 mi h s = 0.98 m s 2 . (100 50) s taLet 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.f200x 0 200 s = area 1 + area 2 + area 3 + area 4 + area 55100 04 1 2 0b ga f b ga f + b160 mi hga100 sf 1 + a50 sfb100 mi hg 2 1 + a100 sfb170 mi h 160 mi hg 2 = 24 000bmi hgasf3100 200 300 400 50 mi h 50 s + 50 mi h 50 sFIG. 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 , x 0 200 s F 24 000 mi I asf = GH 3 600 s JK6.7 mi .t (s) 46. 46 *P2.60Motion in One DimensionAverage speed of every point on the train as the first car passes Liz: x 8.60 m = = 5.73 m s. 1.50 s t The train has this as its instantaneous speed halfway through the 1.50 s time. Similarly, halfway 8.60 m through the next 1.10 s, the speed of the train is = 7.82 m s . The time required for the speed 1.10 s to change from 5.73 m/s to 7.82 m/s is 1 1 (1.50 s)+ (1.10 s)= 1.30 s 2 2 so the acceleration is: a x =P2.61v x 7.82 m s 5.73 m s = = 1.60 m s 2 . t 1.30 sThe rate of hair growth is a velocity and the rate of its increase is an acceleration. Then mm d v xi = 1.04 mm d and a x = 0.132 . The increase in the length of the hair (i.e., displacement) w during a time of t = 5.00 w = 35.0 d isFG Hx = v xi t +bIJ K1 axt 2 2f 1 b0.132 mm d wga35.0 dfa5.00 wf 2gax = 1.04 mm d 35.0 d + or x = 48.0 mm . P2.62Let 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.a fv 2 80.0 f(0 to 1)2a fb= 2 4.00 1 000gso giving120 = 80.0 +( 4.00)tc2(1 to 2)0 (120) = 2(9.80) x f xihv f = 120 m s t = 10.0 sgivingx f xi = 735 m0 120 = 9.80t giving This is the time of maximum height of the rocket.afbv 2 0 = 2 9.80 1 735 f(2 to 3)gv f = 184 = (9.80)t (a)t total = 10 + 12.2 + 18.8 = 41.0 s(b)cxf xihtotal= 1.73 kmcontinued on next paget = 12.2 sgivingt = 18.8 sFIG. P2.62 47. Chapter 2(c)v final = 184 m s0 #1 #2 #3 P2.63t 0.0 10.0 22.2 41.0Launch End Thrust Rise Upwards Fall to Earthax 0 1 000 1 735 0fa +4.00 +4.00 9.80 9.80Distance traveled by motorist = 15.0 m s t 1 Distance traveled by policeman = 2.00 m s 2 t 2 2ch(a)intercept occurs when 15.0t = t 2 , or t = 15.0 s(b)v(officer)= 2.00 m s 2 t = 30.0 m s(c) P2.64v 80 120 0 184x(officer )=ch1 2.00 m s 2 t 2 = 225 m 2chArea A1 is a rectangle. Thus, A1 = hw = v xi t . 1 1 Area A 2 is triangular. Therefore A 2 = bh = t v x v xi . 2 2 The total area under the curve isbA = A1 + A 2 = v xi t +bvxgvx vx A2vxigA1 v xi t 20and since v x v xi = a x tt FIG. P2.64A = v xi t +1 axt 2 . 2The displacement given by the equation is: x = v xi t + same result as above for the total area.1 a x t 2 , the 2t47 48. 48 P2.65Motion in One Dimension(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=aaff1 v + vi t1 , 100 x = v 10.2 t1 and v = vi + at1 . 2The first two giveFG 1 t IJ v = FG10.2 1 t IJ at H 2 K H 2 K 200 a= b20.4 t gt .100 = 10.2 11For Maggie: a =11200a18.4fa2.00f = 200 For Judy: a = a17.4fa3.00f =(b)15.43 m s 2 3.83 m s 2v = a1 ta fa f Judy: v = a3.83fa3.00f =Maggie: v = 5.43 2.00 = 10.9 m s(c)11.5 m sAt the six-second mark x=fa fa f + a10.9fa4.00f = 54.3 m a fa f + a11.5fa3.00f = 51.7 m1 5.43 2.00 2 1 Judy: x = 3.83 3.00 2Maggie: x =a1 2 at1 + v 6.00 t1 2 22Maggie is ahead by 2.62 m . P2.66a1 = 0.100 m s 2 1 1 2 2 x = 1 000 m = a1 t1 + v1 t 2 + a 2 t 2 2 2 at at 1 1 2 1 000 = a1 t1 + a1 t1 1 1 + a 2 1 1 2 a2 2 a2FG HIJ KFG IJ H Ka 2 = 0.500 m s 2 t = t1 + t 2 and v1 = a1 t1 = a 2 t 2 21 000 = t1 =t2 =a1 t1 12.9 = 26 s a 2 0.500FG HIJ Ka 1 2 a1 1 1 t1 2 a220 000 = 129 s 1.20Total time = t = 155 s 49. Chapter 2P2.67Let the ball fall 1.50 m. It strikes at speed given bych2 2 v xf = v xi + 2 a x f x i :ch2 v xf = 0 + 2 9.80 m s 2 (1.50 m)v xf = 5.42 m s and its stopping is described byd2 2 v xf = v xi + 2 a x x f x ib0 = 5.42 m s ax =g2e2.00 10j+ 2 a x 10 2 m29.4 m 2 s 2 2im= +1.47 10 3 m s 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 ~ 10 3 m s 2 . *P2.68(a)x f = xi + v xi t +1 a x t 2 . We assume the package starts from rest. 2 145 m = 0 + 0 +t=(b)x f = xi + v xi t +1 9.80 m s 2 t 2 2c2(145 m) 9.80 m s 2h= 5. 44 s1 1 2 a x t 2 = 0 + 0 + 9.80 m s 2 (5.18 s) = 131 m 2 2chdistance fallen = x f = 131 mej(c)speed = v xf = v xi + a x t = 0 + 9.8 m s 2 5.18 s = 50.8 m s(d)The remaining distance is 145 m 131.5 m = 13.5 m . During deceleration, v xi = 50.8 m s, v xf = 0, x f xi = 13.5 mch2 2 v xf = v xi + 2 a x x f x i :a0 = 50.8 m s ax =f2+ 2 a x (13.5 m)2 580 m 2 s 2 = +95.3 m s 2 = 95.3 m s 2 upward . 2 13.5 maf49 50. 50 P2.69Motion in One Dimension(a)1 1 2 at = 50.0 = 2.00t + (9.80)t 2 , 2 2 4.90t 2 + 2.00t 50.0 = 0 y f = v i1 t +t=2.00 + 2.00 2 4( 4.90)(50.0) 2( 4.90)Only the positive root is physically meaningful: t = 3.00 s after the first stone is thrown. (b)1 2 at and t = 3.00 1.00 = 2.00 s 2 1 2 substitute 50.0 = vi 2 ( 2.00)+ (9.80)( 2.00) : 2 y f = vi 2 t +vi2 = 15.3 m s downward (c)v1 f = vi1 + at = 2.00 +(9.80)(3.00)= 31.4 m s downward v 2 f = vi 2 + at = 15.3 +(9.80)( 2.00)= 34.8 m s downwardP2.70(a)1 2 d = (9.80)t1 2 t1 + t 2 = 2.40 2 4.90t 2d = 336 t 2a336t 2 = 4.90 2.40 t 2 359.5t 2 + 28.22 = 0t2 =359.5 358.75 = 0.076 5 s t2 = 9.80 (b) P2.71(a)(d)359.5 359.5 2 4( 4.90)( 28.22) 9.80d = 336 t 2 = 26.4 mIn walking a distance x , in a time t , the length of rope is only increased by x sin . x sin . The pack lifts at a rate t x x sin = v boy = v boy tx 2x + h2FG IJ HKd 1 dv v boy dx = + v boy x dt dt dt v boy v boy x d x d = v = v boy a = v boy 2 , but dt dt 2 2 2 v boy v boy h 2 h 2 v boy x2 a= = 1 2 = 2 3 2 x 2 + h2 a=F GH(c)21 2 Ignoring the sound travel time, d = (9.80)( 2.40) = 28.2 m , an error of 6.82% . 2v=(b)sof2 v boyh,0v boy , 0I JKchFIG. P2.71 51. Chapter 2P2.72h = 6.00 m, v boy = 2.00 m s v = However, x = v boy t : v = (a)c2 v boy tc2 v boy t 2+hc 4t4t 2+ 36h12.0.32 0.631.50.8921.112.51.2831.413.51.5241.604.51.6651.71FIG. P2.72(a)From problem 2.71 above, a =a f aem s j2 h 2 v boycx2+hh2 3 2=2 h 2 v boyc2 v boy t 2+hh2 3 2=c4t144 2+ 36h32.2ts 00.670.50.6410.571.50.4820.382.50.303.0.243.50.184.0.144.50.115 (a)=00.5P2.73h2 12ha f vb m s gts 0 1(b)v boy x x x sin = v boy = . 12 t x 2 + h20.09FIG. P2.72(b)We require x s = x k when t s = t k + 1.00jb1 3.50 m s 2 t k + 1.00 2 t k + 1.00 = 1.183t k xs =et k = 5.46 s .ja1 4.90 m s 2 5.46 s 2ef2(b)xk =(c)v k = 4.90 m s 2 5.46 s = 26.7 m s vse ja f = e3.50 m s ja6.46 sf = 2= 73.0 m22.6 m sg2=jb g1 4.90 m s 2 t k 2e2= xk51 52. 52 P2.74Motion in One Dimension3.000.130.252.600.380.252.160.630.440.251.760.880.340.251.361.130.240.250.961.380.140.250.561.630.030.250.121.880.060.250.242.130.170.250.682.380.280.251.122.630.370.251.482.880.480.251.923.130.570.252.283.380.680.252.723.630.790.253.163.880.880.253.524.130.990.253.964.381.090.254.364.631.190.254.764.886.941.000.256.400.75midpt time t (s)5.750.50v (m/s)0.540.25t (s)0.65Height h (m) 5.00h (m) 0.75Time t (s) 0.007.381.257.721.507.961.758.102.008.132.258.072.507.902.757.623.007.253.256.773.506.203.755.524.004.734.253.854.502.864.751.775.000.58 TABLE P2.74acceleration = slope of line is constant.a =1.63 m s 2 = 1.63 m s 2 downwardFIG. P2.74 53. 53Chapter 2P2.75The distance x and y are always related by x 2 + y 2 = L2 . Differentiating this equation with respect to time, we havey B xdy dx 2x + 2y =0 dt dtLydy dx = v . is v B , the unknown velocity of B; and dt dt From the equation resulting from differentiation, we haveNowOFG IJ H KFG HAxdy x dx x = (v). = dt y dt y ButvFIG. P2.75IJ Ky v v 3 1 = = 0.577 v . = tan so v B = v . When = 60.0 , v B = tan 60.0 3 x tan ANSWERS TO EVEN PROBLEMS (a) 2 10 7 m s ; 1 10 6 m s ;P2.24(b) 5 10 yr(a) 1.88 km; (b) 1.46 km; (c) see the solution; (d) (i) x 1 = 1.67 m s 2 t 2 ;P2.4(a) 50.0 m s ; (b) 41.0 m s(ii) x 2 = 50 m s t 375 m ;P2.6(a) 27.0 m ; 2 (b) 27.0 m + 18.0 m s t + 3.00 m s 2 t ;(iii) x 3P2.28bg eeja f(c) 18.0 m s P2.8(a), (b), (c) see the solution; 4.6 m s 2 ; (d) 0P2.10(a) 20.0 m s ; 5.00 m s ; (b) 262 mP2.14P2.16b g = b 250 m sgt e 2.5 m s jt 22 4 375 m ;(e) 37.5 m s P2.26958 mP2.28(a) x f = 30.0t t 2 m; v f = 30.0 2t m s ;5.00 mP2.12jejaf(b) 225 m 1 a x t 2 ; 3.10 m s 2P2.30x f xi = v xf t (a) see the solution; (b) 1.60 m s 2 ; 0.800 m s 2P2.32(a) 35.0 s; (b) 15.7 m s(a) 13.0 m s; (b) 10.0 m s; 16.0 m s;P2.34(a) 1.12 10 11 m s 2 ; (b) 4.67 10 5 sP2.36(a) False unless the acceleration is zero; see the solution; (b) True(c) 6.00 m s 2 ; (d) 6.00 m s 2 P2.18see the solutionP2.20(a) 6.61 m s; (b) 0. 448 m s 2P2.38Yes; 212 m; 11.4 sP2.22(a) 21.8 mi h s = 9.75 m s 2 ;P2.40(a) 4.90 m ; 19.6 m; 44.1 m; (b) 9.80 m s; 19.6 m s; 29.4 m sP2.421.79 s(b) 22.2 mi h s = 9.94 m s 2 ; (c) 22.8 mi h s = 10.2 m s 2 54. 54Motion in One DimensionP2.44No; see the solutionP2.601.60 m s 2P2.46The second ball is thrown at speed vi = ghP2.62(a) 41.0 s; (b) 1.73 km; (c) 184 m sP2.48(a) 510 m; (b) 20.4 sP2.64v xi t +P2.50(a) 96.0 ft s ;P2.66155 s; 129 sP2.68(a) 5.44 s; (b) 131 m; (c) 50.8 m s ;32(b) a = 3.07 10 ft s upward ; (c) t = 3.13 10 2 s P2.521 a x t 2 ; displacements agree 2(d) 95.3 m s 2 upward38.2 mP2.70(a) 26.4 m; (b) 6.82%2P2.54(a) and (b) see the solution; (c) 4 m s ; (d) 34 m; (e) 28 mP2.72see the solutionP2.560.222 sP2.74see the solution; a x = 1.63 m s 2P2.58(a) see the solution; (b) 6.23 s 55. 3 Vectors CHAPTER OUTLINE 3.1 3.2Coordinate Systems Vector and Scalar Quantities Some Properties of Vectors Components of a Vector and Unit VectorsANSWERS TO QUESTIONS 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.Q3.23.3 3.4Q3.1No, 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.Q3.3The 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.Q3.4Only force and velocity are vectors. None of the other quantities requires a direction to be described.Q3.5If 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.Q3.6The books displacement is zero, as it ends up at the point from which it started. The distance traveled is 6.0 meters.Q3.785 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.Q3.8Vectors A and B are perpendicular to each other.Q3.9No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude.55 56. 56VectorsQ3.10Any vector that points along a line at 45 to the x and y axes has components equal in magnitude.Q3.11A x = B x and A y = B y .Q3.12Addition of a vector to a scalar is not defined. Think of apples and oranges.Q3.13One difficulty arises in determining the individual components. The relationships between a vector and its components such as A x = A 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, 2 2 A = A x + A y only holds true if the two component vectors, A x and A y , are perpendicular.Q3.14If 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 P3.1P3.2Coordinate Systemsa f a fa f y = r sin = a5.50 mf sin 240 = a5.50 mfa 0.866f =x = r cos = 5.50 m cos 240 = 5.50 m 0.5 = 2.75 m(a)4.76 mx = r cos and y = r sin , therefore x1 = 2.50 m cos 30.0 , y1 = 2.50 m sin 30.0 , andafafbx , y g = a2.17 , 1.25f m x = a3.80 mf cos 120 , y = a3.80 mf sin 120 , and bx , y g = a1.90, 3.29f m . 1122(b) P3.322d = ( x) 2 + ( y) 2 = 16.6 + 4.16 = 4.55 mThe 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 = x 2 + y 2 =(b) = tan 1FG 1 IJ = 26.6 ; r = H 2Ka2.00 mf + a1.00 mf2.24 m, 26.622= 5.00 m 2 = 2.24 m 57. Chapter 3P3.4(a)d=bx2 x1g + by 22 y1g2=c2.00 3.00 h + a4.00 3.00f 22d = 25.0 + 49.0 = 8.60 m (b)a2.00f + a4.00f = 20.0 = F 4.00 IJ = 63.4 = tan G H 2.00 K 2r1 =1r2 =24.47 m1a3.00f + a3.00f 22= 18.0 = 4.24 m 2 = 135 measured from the +x axis. P3.5We have 2.00 = r cos 30.0 r=2.00 = 2.31 cos 30.0and y = r sin 30.0 = 2.31 sin 30.0 = 1.15 . P3.6We have r = x 2 + y 2 and = tan 1 (a)FG y IJ . H xKThe radius for this new point isa x f2+ y2 = x2 + y2 = rand its angle is tan 1(b)FG y IJ = H x K180 .b g( 2 x) 2 + ( 2 y) 2 = 2r . This point is in the third quadrant if x , y is in the first quadrantb gor in the fourth quadrant if x , y is in the second quadrant. It is at an angle of 180+ . (c)b g( 3 x) 2 + ( 3 y) 2 = 3r . This point is in the fourth quadrant if x , y is in the first quadrantb gor in the third quadrant if x , y is in the second quadrant. It is at an angle of .57 58. 58VectorsSection 3.2Vector and Scalar QuantitiesSection 3.3Some Properties of VectorsP3.7x 100 m x = 100 m tan 35.0 = 70.0 mtan 35.0 =afFIG. P3.7 P3.8R = 14 km = 65 N of E R13 km 6 km1 kmFIG. 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.yA A+B(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.BBAB x OFIG. P3.10 59. Chapter 3P3.11(a)d = 10.0 i = 10.0 m since the displacement is in aCstraight 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=(c) P3.12595.00 m dBAFIG. P3.11b g1 2 r = 5 = 15.7 m 2If the circle is complete, d begins and ends at point A. Hence, d = 0 .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 . yF1 + F2F2F1x0 1 2 3 N FIG. P3.12 P3.13(a)The large majority of people are standing or sitting at this hour. Their instantaneous foot-tohead vectors have upward vertical components on the order of 1 m and randomly oriented horizontal components. The citywide sum will be ~ 10 5 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 ~ 10 5 0.03 m + 10 2 1 ma3~ 10 m upward .fa f 60. 60 P3.14Vectors NYour 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 the1m W15.0 metersRsketch can be converted according to the scale to be 7.9 m .3.50 meters30.08.20 metersESFIG. P3.14 P3.15To 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 *P3.16The three diagrams shown below represent the graphical solutions for the three vector sums: R 1 = A + B + C , R 2 = B + C + A , and R 3 = C + B + A . You should observe that R 1 = R 2 = R 3 , illustrating that the sum of a set of vectors is not affected by the order in which the vectors are added. 100 m CB AAB R1AR2C BFIG. P3.16R3C 61. Chapter 3P3.17The 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(Scale: 1 unit = 20 ft )d = 420 ft and = 3FIG. P3.17Section 3.4 P3.18Components of a Vector and Unit VectorsCoordinates of the super-hero are:a f a f y = a100 mf sina 30.0f =x = 100 m cos 30.0 = 86.6 m 50.0 m FIG. P3.18 P3.19A x = 25.0 A y = 40.0 2 2 A = Ax + Ay =a25.0f + a40.0f 22= 47.2 unitsWe observe that Aytan =.AxFIG. P3.19So = tan 1F A I = tan 40.0 = tan a1.60f = 58.0 . GH A JK 25.0 y1xThe diagram shows that the angle from the +x axis can be found by subtracting from 180: = 180 58 = 122 . P3.20afThe person would have to walk 3.10 sin 25.0 = 1.31 km north , andaf3.10 cos 25.0 = 2.81 km east .61 62. 62 P3.21Vectorsx = r cos and y = r sin , therefore:b g ejb g ejb g ej(a) (b)x = 3.30 cos 60.0 , y = 3.30 sin 60.0 , and x , y = 1.65 i + 2.86 j cm(c) P3.22x = 12.8 cos 150 , y = 12.8 sin 150 , and x , y = 11.1i + 6.40 j mx = 22.0 cos 215 , y = 22.0 sin 215 , and x , y = 18.0 i 12.6 j ina ad= *P3.23f a f f a f a25.0 mfi + a43.3 mf jx = d cos = 50.0 m cos 120 = 25.0 m y = d sin = 50.0 m sin 120 = 43.3 m(a)Her net x (east-west) displacement is 3.00 + 0 + 6.00 = +3.00 blocks, while her net y (northsouth) displacement is 0 + 4.00 + 0 = +4.00 blocks. The magnitude of the resultant displacement is R=b x g + by g net2net2=a3.00f + a4.00f 22= 5.00 blocksand the angle the resultant makes with the x-axis (eastward direction) is = tan 1FG 4.00 IJ = tan a1.33f = 53.1 . H 3.00 K 1The resultant displacement is then 5.00 blocks at 53.1 N of E . (b) *P3.24The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks . NLet i = east and j = north. The unicyclists displacement is, in meters 280 j + 220 i + 360 j 300 i 120 j + 60 i 40 j 90 i + 70 j . RR = 110 i + 550 j =a110 mf + a550 mf 22at tan 1= 561 m at 11.3 west of north .110 m west of north 550 mThe crows velocity is v= x 561 m at 11.3 W of N = t 40 s= 14.0 m s at 11.3 west of north .E FIG. P3.24 63. Chapter 3P3.2563+x East, +y North x = 250 +125 cos 30 = 358 m y = 75 +125 sin 30150 = 12.5 mc xh + c yh = a358f + a12.5f c yh = 12.5 = 0.0349 tan = c xh 358 2d=222= 358 m = 2.00 d = 358 m at 2.00 S of E P3.26The 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 DC east = d DA east + d AC east = 730 cos 5.00560 sin 21.0 = 527 miles. d DC north = d DA north + d AC north = 730 sin 5.00+560 cos 21.0 = 586 miles. By the Pythagorean theorem, d = ( d DC east ) 2 + ( d DC north ) 2 = 788 mi . Then tan =d DC north = 1.11 and = 48.0 . d DC eastThus, Chicago is 788 miles at 48.0 northeast of Dallas . P3.27(a)See figure to the right.(b)C = A + B = 2.00 i + 6.00 j + 3.00 i 2.00 j = 5.00 i + 4.00 j C = 25.0 + 16.0 at tan 1FG 4 IJ = H 5K6.40 at 38.7D = A B = 2.00 i + 6.00 j 3.00 i + 2.00 j = 1.00 i + 8.00 ja1.00f + a8.00f at tan FGH 81.00 IJK .00 D = 8.06 at b180 82.9g = 8.06 at 97.2 2D=P3.2821bx + x + x g + by + y + y g = a3.00 5.00 + 6.00f + a 2.00 + 3.00 + 1.00 f F 6.00 IJ = 56.3 = tan G H 4.00 K d=12322121232= 52.0 = 7.21 mFIG. P3.27 64. 64 P3.29VectorsWe have B = R A : A x = 150 cos120 = 75.0 cm A y = 150 sin 120 = 130 cm R x = 140 cos 35.0 = 115 cm R y = 140 sin 35.0 = 80.3 cm FIG. P3.29Therefore,a fejB = 115 75 i + 80.3 130 j = 190 i 49.7 j cm B = 190 2 + 49.7 2 = 196 cmFG H = tan 1 P3.30IJ K49.7 = 14.7 . 190A = 8.70 i + 15.0 j and B = 13. 2 i 6.60 j A B + 3C = 0 : 3C = B A = 21.9 i 21.6 j C = 7.30 i 7.20 j or C x = 7.30 cm ; C y = 7.20 cm (a)aA + Bf = e3 i 2 jj + e i 4 jj =2i 6 j(b)aA Bf = e3i 2 jj e i 4 jj =4i + 2 j(c)A + B = 2 2 + 6 2 = 6.32(d)A B = 4 2 + 2 2 = 4.47(e)P3.31 A+B = tan1 AB = tan1P3.32(a)FG 6 IJ = 71.6= H 2K FG 2 IJ = 26.6 H 4K288D = A + B + C = 2i + 4 j D = 2 2 + 4 2 = 4. 47 m at = 63.4(b)E = A B + C = 6 i + 6 j E = 6 2 + 6 2 = 8.49 m at = 135 65. Chapter 3P3.33ejd1 = 3.50 j mejd 2 = 8.20 cos 45.0 i + 8.20 sin 45.0 j = 5.80 i + 5.80 j mejd 3 = 15.0 i maf af e9.20i + 2.30 jj mR = d1 + d 2 + d 3 = 15.0 + 5.80 i + 5.80 3.50 j = (or 9.20 m west and 2.30 m north)The magnitude of the resultant displacement is 2 2 R = Rx + R y =FG 2.30 IJ = H 9.20 KThe direction is = arctan P3.3422= 9.48 m .166 . A = 10.0Refer to the sketch R = A + B + C = 10.0 i 15.0 j + 50.0 i = 40.0 i 15.0 ja f + a15.0fR = 40.0a9.20f + a2.30f22 12RB = 15.0 C = 50.0= 42.7 yardsFIG. P3.34 P3.35F = F1 + F2(a)aaa f F = 60.0 i + 104 j 20.7 i + 77.3 j = e39.3 i + 181 jj N ffafF = 120 cos 60.0 i + 120 sin 60.0 j 80.0 cos 75.0 i + 80.0 sin 75.0 jF = 39.3 2 + 181 2 = 185 N = tan 1 (b) P3.36F3 = F =East x 0m 1.41 0.500 +0.914 R=FG 181 IJ = H 39.3 K77.8e39.3 i 181 jj NWest y 4.00 m 1.41 0.866 4.55 22x + y = 4.64 m at 78.6 N of E65 66. 66VectorsA = 3.00 m, A = 30.0B = 3.00 m , B = 90.0A x = A cos A = 3.00 cos 30.0 = 2.60 mP3.37A y = A sin A = 3.00 sin 30.0= 1.50 mejA = A x i + A y j = 2.60 i + 1.50 j m Bx = 0 , By = 3.00 mejA + B = 2.60 i + 1.50 j + 3.00 j = P3.38B = 3.00 j msoe2.60i + 4.50 jj mLet the positive x-direction be eastward, the positive y-direction be vertically upward, and the positive z-direction be southward. The total displacement is thenejejejd = 4.80 i + 4.80 j cm + 3.70 j 3.70k cm = 4.80 i + 8.50 j 3.70k cm . 222(a) (b) P3.39The magnitude is d = ( 4.80) +(8.50) + (3.70) cm = 10.4 cm . Its angle with the y-axis follows from cos =8.50 , giving = 35.5 . 10. 4B = Bx i + By j + Bz k = 4.00 i + 6.00 j + 3.00k B = 4.00 2 + 6.00 2 + 3.00 2 = 7.81 = cos 1 = cos 1 = cos 1 P3.40FG 4.00 IJ = H 7.81 K FG 6.00 IJ = H 7.81 K FG 3.00 IJ = H 7.81 K59.2 39.8 67.4The y coordinate of the airplane is constant and equal to 7.60 10 3 m whereas the x coordinate is given by x = vi t where vi is the constant speed in the horizontal direction. isAt t = 30.0 s we have x = 8.0410 3 , so vi = 268 m s. The position vector as a function of timebg ejP = 268 m s t i + 7.60 10 3 m j . At t = 45.0 s , P = 1. 21 10 4 i + 7.60 10 3 j m. The magnitude isP=c1.2110 h + c7.6010 h 4 23 2m = 1.43 10 4 mand the direction is = arctanF 7.6010 I = GH 1.2110 JK 3432.2 above the horizontal . 67. Chapter 3A = 8.00 i + 12.0 j 4.00k B=(c) P3.42(a) (b)P3.41C = 3A = 24.0 i 36.0 j + 12.0kA = 2.00 i + 3.00 j 1.00k 4R = 75.0 cos 240 i + 75.0 sin 240 j + 125 cos 135 i + 125 sin 135 j + 100 cos 160 i + 100 sin 160 j R = 37.5 i 65.0 j 88.4i + 88.4 j 94.0 i + 34.2 j R = 220 i + 57.6 j 2R = (220 ) + 57.6 2 at arctanFG 57.6 IJ above the x-axis H 220 KR = 227 paces at 165 P3.43(a)e5.00 i 1.00 j 3.00kj mC=A+B= 222C = (5.00) +(1.00) +(3.00) m = 5.92 m (b)D = 2A B =e4.00i 11.0 j + 15.0kj m222D = ( 4.00) +(11.0) +(15.0) m = 19.0 m P3.44The position vector from radar station to ship isejejS = 17.3 sin 136 i + 17.3 cos 136 j km = 12.0 i 12.4 j km. From station to plane, the position vector isejP = 19.6 sin 153 i + 19.6 cos 153 j + 2.20k km, orejP = 8.90 i 17.5 j + 2.20k km. (a)To fly to the ship, the plane must undergo displacement D = S P =(b)e3.12 i + 5.02 j 2.20kj km .The distance the plane must travel is 222D = D = (3.12) +(5.02) +( 2.20) km = 6.31 km .67 68. 68 P3.45VectorsThe hurricanes first displacement is isFG 41.0 km IJ(3.00 h) at 60.0 N of W, and its second displacement H h KFG 25.0 km IJ(1.50 h) due North. With i H h Krepresenting east and j representing north, its totaldisplacement is:FG 41.0 km cos 60.0IJ a3.00 hfe ij + FG 41.0 km sin 60.0IJ a3.00 hf j + FG 25.0 km IJ a1.50 hf j = 61.5 kme ij H h K H h K H hK +144 km j 22with magnitude (61.5 km) +(144 km) = 157 km . P3.46(a)aE= (b)ye15.1i + 7.72 jj cm a a f e7.72i + 15.1 jj cmfa a f e+7.72 i + 15.1 jj cm27.0 27.0fFFIG. P3.46A x = 3.00 , A y = 2.00 (a)A = A x i + A y j = 3.00 i + 2.00 j(b)2 2 A = A x + A y = (3.00) +( 2.00) = 3.612tan =Ay Ax=22.00 = 0.667 , tan1 (0.667)= 33.7 (3.00)af is in the 2 nd quadrant, so = 180+ 33.7 = 146 . (c)G EG = + 17.0 cm sin 27.0 i + 17.0 cm cos 27.0 j G=P3.47fF = 17.0 cm sin 27.0 i + 17.0 cm cos 27.0 j F=(c)afE = 17.0 cm cos 27.0 i + 17.0 cm sin 27.0 jR x = 0 , R y = 4.00 , R = A + B thus B = R A and Bx = R x A x = 0 (3.00)= 3.00 , By = R y A y = 4.00 2.00 = 6.00 . Therefore, B = 3.00 i 6.00 j .27.0 x 69. Chapter 3P3.48Let +x = East, +y = North, x 300 175 0 125y 0 303 150 453 y = 74.6 N of E x(a) (b) P3.49 = tan1R = x 2 + y 2 = 470 km(a)R x = 40.0 cos 45.0+30.0 cos 45.0 = 49.5yR y = 40.0 sin 45.030.0 sin 45.0+20.0 = 27.1AR = 49.5 i + 27.1 j (b)a49.5f + a27.1f = 56.4 F 27.1 IJ = 28.7 = tan G H 49.5 K 2R=2O1B45x45 CFIG. P3.49 P3.50Taking components along i and j , we get two equations:6.00 a 8.00b + 26.0 = 0 and8.00 a + 3.00b + 19.0 = 0 . Solving simultaneously, a = 5.00 , b = 7.00 . Therefore, 5.00A + 7.00B + C = 0 .69 70. 70VectorsAdditional Problems P3.51Let represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 180 , magnitude of R is then R = 2 A cosR /2 , and . The 2 2FG IJ . [Hint: apply the law of H 2Kcosines to the isosceles triangle and use the fact that B = A .] Again, A, B, and D = A B form an isosceles triangle with apex angle . Applying the law of cosines and the identityB AA DB FIG. P3.51a1 cos f = 2 sin FGH IJK 2 2FG IJ . H 2Kgives the magnitude of D as D = 2 A sin The problem requires that R = 100D . Thus, 2 A cosFG IJ = 200 A sinFG IJ . This gives tanFG IJ = 0.010 and H 2K H 2K H 2K = 1.15 . P3.52Let represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 180 , magnitude of R is then R = 2 A cos , and . The 2 2FG IJ . [Hint: apply the law of H 2Kcosines to the isosceles triangle and use the fact that B = A . ] Again, A, B, and D = A B form an isosceles triangle with apex angle . Applying the law of cosines and the identitya1 cos f = 2 sin FGH IJK 2 2FG IJ . H 2K F I F I The problem requires that R = nD or cosG J = n sinG J giving H 2K H 2K F 1I = 2 tan G J . H nKgives the magnitude of D as D = 2 A sin1FIG. P3.52 71. Chapter 3P3.53(a)R x = 2.00 , R y = 1.00 , R z = 3.00(b)2 2 2 R = R x + R y + R z = 4.00 + 1.00 + 9.00 = 14.0 = 3.74(c)cos x = cos y = cos z =*P3.54F I GH JK F R I = 74.5 from + y GH R JK F R I = 36.7 from + z GH R JKRx R x = cos1 x = 57.7 from + x R R Ry Ry y = cos1Rz z = cos1 RzTake the x-axis along the tail section of the snake. The displacement from tail to head isaaff240 m i + 420 240 m cos 180105 i 180 m sin 75 j = 287 m i 174 m j . 22Its magnitude is (287) +(174) m = 335 m . From v =distance , the time for each childs run is ta fa a fbfbgInge: t =distance 335 m h 1 km 3 600 s = = 101 s v 12 km 1 000 m 1 hOlaf: t =420 m s = 126 s . 3.33 mga fInge wins by 126 101 = 25.4 s . *P3.55The position vector from the ground under the controller of the first airplane isa fa f a = e17.4i + 8.11 j + 0.8k j km .faf afa fa f a = e16.5 i + 6.02 j + 1.1k j km .faf afr1 = 19.2 km cos 25 i + 19.2 km sin 25 j + 0.8 km kThe second is at r2 = 17.6 km cos 20 i + 17.6 km sin 20 j + 1.1 km kNow the displacement from the first plane to the second isejr2 r1 = 0.863 i 2.09 j + 0.3k km with magnitude 222(0.863 ) +( 2.09) +(0.3) = 2.29 km .71 72. 72 *P3.56Vectors3Let A represent the distance from island 2 to island 3. The displacement is A = A at 159 . Represent the displacement from 3 to 1 as B = B at 298 . We have 4.76 km at 37 +A + B = 0 .A 28 BFor x-componentsa4.76 kmf cos 37+ A cos 159+B cos 298 = 037169 C2 N E3.80 km 0.934 A + 0.469B = 0 FIG. P3.56B = 8.10 km + 1.99 A For y-components,a4.76 kmf sin 37+ A sin 159+B sin 298 = 0 2.86 km + 0.358 A 0.883B = 0 (a)We solve by eliminating B by substitution:af2.86 km + 0.358 A 0.883 8.10 km + 1.99 A = 0 2.86 km + 0.358 A + 7.15 km 1.76 A = 0 10.0 km = 1.40 A A = 7.17 km (b) *P3.57B = 8.10 km + 1.99(7.17 km)= 6.15 km(a)We first express the corners position vectors as sets of componentsa f a f B = a12 mf cos 30 i + a12 mf sin 30 j = 10.4 m i +6.00 mj .A = 10 m cos 50 i + 10 m sin 50 j = 6.43 m i +7.66 m jThe horizontal width of the rectangle is 10.4 m 6.43 m = 3.96 m . Its vertical height is 7.66 m 6.00 m = 1.66 m . Its perimeter is 2(3.96 + 1.66) m = 11.2 m . (b)The position vector of the distant corner is Bx i + A y j = 10.4 mi +7.66 mj = 10.4 2 + 7.66 2 m at 7.66 m = 12.9 m at 36.4 . tan1 10.4 m 73. Chapter 3P3.58Choose the +x-axis in the direction of the first force. The total force, y in newtons, is then 31 N 12.0 i + 31.0 j 8.40 i 24.0 j =e3.60 ij + e7.00 jj N8.4 NThe magnitude of the total force is 2x R.12 N 35.0 horizontal 24 N2(3.60) +(7.00) N = 7.87 N and the angle it makes with our +x-axis is given by tan =(7.00)(3.60) = 62.8 . Thus, its angle counterclockwise from the horizontal is 35.0+62.8 = 97.8 .P3.59FIG. P3.58 ,d 1 = 100 i d 2 = 300 ja f a f d = 200 cosa60.0fi + 200 sina60.0f j = 100 i + 173 j R = d + d + d + d = e 130 i 202 jj m R = a 130f + a202f = 240 m F 202 IJ = 57.2 = tan G H 130 Kd 3 = 150 cos 30.0 i 150 sin 30.0 j = 130 i 75.0 j 41234221FIG. P3.59 = 180 + = 237P3.60ejdr d 4 i + 3 j 2 t j = = 0 + 0 2 j = 2.00 m s j dt dtbgThe position vector at t = 0 is 4i + 3 j . At t = 1 s , the position is 4i + 1 j , and so on. The object is moving straight downward at 2 m/s, so dr represents its velocity vector . dt P3.61af afv = v x i + v y j = 300 + 100 cos 30.0 i + 100 sin 30.0 jejv = 387 i + 50.0 j mi h v = 390 mi h at 7.37 N of E73 74. 74 P3.62Vectors(a)ejYou start at point A: r1 = rA = 30.0 i 20.0 j m. The displacement to B is rB rA = 60.0 i + 80.0 j 30.0 i + 20.0 j = 30.0 i + 100 j .ejYou cover half of this, 15.0 i + 50.0 j to move to r2 = 30.0 i 20.0 j + 15.0 i + 50.0 j = 45.0 i + 30.0 j . Now the displacement from your current position to C is rC r2 = 10.0 i 10.0 j 45.0 i 30.0 j = 55.0 i 40.0 j . You cover one-third, moving to r3 = r2 + r23 = 45.0 i + 30.0 j +1 55.0 i 40.0 j = 26.7 i + 16.7 j . 3ejThe displacement from where you are to D is rD r3 = 40.0 i 30.0 j 26.7 i 16.7 j = 13.3 i 46.7 j . You traverse one-quarter of it, moving to r4 = r3 +bg1 1 rD r3 = 26.7 i + 16.7 j + 13.3 i 46.7 j = 30.0 i + 5.00 j . 4 4ejThe displacement from your new location to E is rE r4 = 70.0 i + 60.0 j 30.0 i 5.00 j = 100 i + 55.0 j of which you cover one-fifth the distance, 20.0 i + 11.0 j, moving to r4 + r45 = 30.0 i + 5.00 j 20.0 i + 11.0 j = 10.0 i + 16.0 j . The treasure is at (10.0 m, 16.0 m) . (b)Following the directions brings you to the average position of the trees. The steps we took numerically in part (a) bring you to rA +ar then to ar then toA+ rBf+r2 + rB + rC ACarA +rB23f+rf =araf FGHr + rB 1 rB rA = A 2 2IJ KrA + rB + rC 3A + rB + rCfr + rB + rC + rD = A 3 4 4 arA +rB +rC +rD f r + r + r + r + r rA + rB + rC + rD r B C D E 4 + E = A and last to . 4 5 5aD3fThis center of mass of the tree distribution is the same location whatever order we take the trees in. 75. Chapter 3*P3.63(a)75Let T represent the force exerted by each child. The xcomponent of the resultant force isaf a f a fT cos 0 + T cos 120+T cos 240 = T 1 + T 0.5 + T 0.5 = 0 . The y-component is T sin 0 + T sin 120 + T sin 240 = 0 + 0.866T 0.866T = 0 . FIG. P3.63Thus, F = 0. (b)P3.64If the total force is not zero, it must point in some direction. When each child moves one 360 space clockwise, the total must turn clockwise by that angle, . Since each child exerts N the same force, the new situation is identical to the old and the net force on the tire must still point in the original direction. The contradiction indicates that we were wrong in supposing that the total force is not zero. The total force must be zero.(a)From the picture, R 1 = a i + b j and R 1 = a 2 + b 2 .(b)R 2 = ai + b j + ck ; its magnitude is 2R1 + c 2 = a 2 + b 2 + c 2 .FIG. P3.64 76. 76 P3.65VectorsSince A + B = 6.00 j , we havebAxg ej+ Bx i + A y + B y j = 0 i + 6.00 j FIG. P3.65giving A x + B x = 0 or A x = Bx[1]A y + B y = 6.00 .[2]andSince both vectors have a magnitude of 5.00, we also have 2 2 2 2 A x + A y = Bx + By = 5.00 2 .From A x = Bx , it is seen that 2 2 A x = Bx . 2 2 2 2 Therefore, A x + A y = Bx + By gives 2 2 A y = By .Then, A y = By and Eq. [2] gives A y = By = 3.00 . Defining as the angle between either A or B and the y axis, it is seen that cos =Ay A=By B=3.00 = 0.600 and = 53.1 . 5.00The angle between A and B is then = 2 = 106 . 77. Chapter 3*P3.66Let represent the angle the x-axis makes with the horizontal. Since angles are equal if their sides are perpendicular right side to right side and left side to left side, is also the angle between the weight and our y axis. The x-components of the forces must add to zero:x y0.127 NTy 0.150 N sin + 0.127 N = 0 .0.150 N = 57.9(b) (a)FIG. P3.66The y-components for the forces must add to zero:af+Ty 0.150 N cos 57.9 = 0 , Ty = 0.079 8 N . (c) P3.67The angle between the y axis and the horizontal is 90.057.9= 32.1 .The displacement of point P is invariant under rotation of 2 the coordinates. Therefore, r = r and r 2 = r or,bgyb g + by g . Also, from the figure, = F y I F yI tan G J = tan G J H xK H x K y e j tan = x 1 + e j tan x 2 + y2 = x2Py21r1y xx tOy xFIG. P3.67Which we simplify by multiplying top and bott