Instructor’s Solutions for Serway and Jewett’s Physics for Scientists and Engineers Sixth Edition, Volume One Ralph V. McGrew James A. CurrieINSTRUCTOR'S SOLUTIONS MANUAL FoR SERWAY AND JEWETT’S PHYSICS FOR SCIENTISTS AND ENGINEERS SIXTH EDITION, VOLUME ONE Ralph V. McGrew Broome Community College James A. Currie Weston High Schoo! THOMSON BROOKS/COLE ‘Australia + Canada + Mexico « Singapore + Spein + United Kingdom + United StatesPreface Chapter ‘Answers to questions Solutions to problems ‘Answers to even-numbered problems Chapter? Answers to questions Solutions to problems ‘Answers to even-numbered problems Chapter’ Answers to questions Solutions to problems ‘Answers to even-numbered problems Chapter Answers to questions Solutions to problems ‘Answers to even-numbered problems Chapters “Answers to questions Solutions to problems ‘Answers to even-numbered problems Chapter 6 Answers to questions Solutions to problems “Answers to even-numbered problems Chapter? Answers to questions Solutions to problems Answers to even-numbered problems Chapters ‘Answers to questions Solutions to problems “Answers to even-numbered problems Chapter9 “Answers to questions ‘Solutions to problems Answers to even-numbered problems CONTENTS a 2B 53 55 36 Res u BES 187 160 189 wm 193 24 215 Bae Chapter 10 “Answers to questions Solutions to problems “Answers to even-numbered problems Chapter 11 Answers to questions Solutions to problems ‘Answers to even-numbered problems ‘Chapter 12 Answers to questions Solutions to problems ‘Answers to ever-numbered problems Chapter 13, Answers to questions Solutions to problems “Answers to even-numbered problems ‘Chapter 14 Answers to questions Solutions to problems Answers to even-numbered problems Chapter 15 Answers to questions Solutions to problems Answers to even-numbered problems Chapter 16 Answers to questions Solutions to problems Answers to even-numbered problems Chapter 17 Answers to questions Solutions to problems ‘Answers to even-numbered problems Chapter 18, ‘Answers to questions Solutions to problems ‘Answers to even-numbered problems 285 287 323 325 337 9 351 381 383 10 a 28 89 a1 a a3 474 Bes geeeas 388 ‘suaiqoad parequunu-tase 0} siomstry suraqqosd 0} suonn}og suonsonb 01 siamsuy ezandey suiajqoud paraquinuszan2 0) sraMsuy suiajgoud 03 Sonn|OS suoysanb oj siamsuy qesdeqy sus us 19g, suajqoud paroquinu-uasa 0 siomsuy ‘swayqoud 01 suonnjog suonsanb 0} siemsuy orzadey3 suajgord pazaquinu-uasa 0} siamsuy ‘sulajgord oy suonnjog, suopsonb a siaasuy erradey>1 Physics and Measurement CHAPTER OUTLINE ANSWERS TO QUESTIONS 11 Standards of Leng Mas. sndioe QLA Atomic docks are based on electromagmetic waves which atoms Mater 12 Mair and Moet sting ‘mst Also, pulsars are highly regular astronomical clocks. 12 Olney 18 Comrsion of Uris Q12 Density vanes with temperature and pressure, It would be 18 Sean en Orel necessary to measure both mass and volume very accurately in ty am cau order to use the density of water as a standard. Qu Qi9 qu oun 13 People have different size hands. Defining the unit precisely would be cumbersome. QL (a) 03 millimeters (b) 50 microseconds (c) 72-kilograms QLS (bj and (@). Youcannot add or subtract quantities of different dimension. QL A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. I'an ‘equation is not dimensionally correct, it cannot be correct. If were a runner, [might walk or run 10" miles per day. Since Iam a college professor, I walk about 10° miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on vacation. (On February 7, 2001, 1am 35 years and 39 days old. 365.25 4 . 86400 s sof Ste ann") “Many college students are just approaching 1 Gs. 17410" ¢~10° 5. Zero digits, An ordet-of- magnitude calculation is accurate only within a factor of 10, “The mass ofthe forty-six chapter textbook is on the order of 10° kg With one datum known to one significant digit, we have 80 million yr + 24 yt = 80 million yr.2 Physics and Measurement OLDS SOS a As a SESE SES Section 1.1 Standards of Length, Mass, and Time No problems in this section Section 1.2 Matter and Modet-Bullding FLI_ From the figure, we may sce that the spacing between diagonal planes is half the distance between iagonally adjacent atoms on a fat plane. This diagonal distance may be obtained from the Pythagorean theorem, Lag = VE +17 . Thus, since the atoms are separated by a distance 1+ 0280 mth gl liner ae paid y MET = [ST Section 1.3 Density and Atomic Mass srg Nodeing erase sphre wetnd elas Ser al 65700 108108 wy density is then on ee ERI? kg/m” |. This valueis intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2.000 to 3000 kg/m, The average density ofthe Earth is significanly higher, to higher-density material smust be down below the surface. PL3 With V=(base area)(height) V=(rr?)h and p=, we have v 7 kg 10° an) ork” Zissmme oem Tm) i510" kg/m *PIA Let Vrepresent the volume ofthe model, the saime in = for both. Then gq 2 ATE Next, 208 TE and gus 995 kel 21 BL , Pet = New To 535 ig ne Mad 995 6 Tae kg im? | LOS eA] 3PLE Foreithersphere the volume is V=4:rr® and the massis Chapters 3 =pV= phar. We divide this equation for the larger sphere by the same equation for the smaller: PL Uelu=1esx10 g. @) ©) © “PLE (@) ®) © @ me pares im, parr33 Toon so ret my anise s) Bban10™ g fork my694{ EO 8) TG] fect m2 2078220") Co GG ‘The mass of any sample isthe number of atoms in the sample times the mass mp of one atom: m= Nig, The fits assertion is that the mass of one aluminum atom is amy © 27.0 = 27.0 x1.66%10™ kg/1 w= 4.48 x10 kg. ‘Then the mass of 6.0210 atoms is m= Nrig = 60210 + 44810 kg =0.027 0 kg =270 g ‘Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be written m= Ning OTE «44810 kg, 1027 0 kg = 6.10210 my, 90 my = in agreement with the first assertion, ‘The general equation m= Nor applied to one mole of any substance gives Mg =NM u, ‘where M is the numerical value of the atomic mass. It divides out exactly forall substances, giving 1.000 000010" kg = N1.660 540 210 kg. With eight-digit data, we can be quite sure ofthe result to seven digits. For one mole the number of atoms is 6022157 x10" ‘The atomic mass of hydrogen’ 1,008 0 u and that of oxygen is 15.999 u. The mass of one molecule of 1,0 is 2(1.008 0) + 15.995 u=180 u. Then the molar mass is [180 g For CO; we have 12.011 g +2(15.999 g)=[440g] as the mass of one mole.4 Physics and ifeasurement Plo PLO PLad 5B | 45x10" kg. iy | 154107 27 x10" kg. Mons ok sas || 80 52256085 g=(065¢ 278] Each atom has mass mip =197 u =197 Now, | Are|=[AN mtg, and the number of atoms missing is ‘The rate of loss is [an] 181g atom lye Vid Yih Lia) ar SO yr 35.254 Wh, 2A 60s La] [amasa0" wanes a (m= pL? =(2.86 g/em?)(5.00 «10° om) =[983 10 g ]=9.83 x10 kg 2 Nee PIS _ 160" atoms mio" Boefi65«10 tgs) (a) Thecross-sectional areas 215.0emn—) A=2{0.150 m}(0.010 m) + (0.340 mX{0.010 mm) 6.40107 m?, . The volume of the beam is V= AL = (6.4010 m?](1.50 m) = 9.60210 m?, ‘Thus, its mass is m= pV =(756%10? kg/m°\(9.60x10" m°)=[726 Kg]. (©) The mass of one typical atom is mg = (55.9 ul ES) oan as0™ kg- Now 72.6 kg, m= Nmg and the number of atoms is N ="4 «5S _ [7.8210 atoms mg and the number of 2 mp 9.28%10™ kgChapter? 5 Te 99 x10"™ kg. The number of a0? Pu @) Temas ofone malls m =1804{ 8 a0 ‘molecules in the pail is m 120 kg Npat Tao” 299x10 hg 40210 molecules (©) Suppose that enough time has elapsed for thorough mixing of the hydrosphere. Gres 210 my —_boks ti ate 2) Saou ‘molecules Section 1.4 Dimensional Analysis PL13 The term x has dimensions of L, a has dimensions of LT, and ¢ has dimensions of T. Therefore, the ‘equation 2= la" has dimensions of L=(Lr) (ry or LA =U", ‘The powers of L and T must be the same on each side of the equation. Therefore, Us" and [mat Likewise, equating terms in, we see that n~2m must equal 0. Thus, dimensionless constant, [cannot be oblained by dimensional analysis *PLI¢ (a) Circumference has dimensions of L. |. The value of ka (©) Volume has dimensions of L?. (©) Area has dimensions of Expression (9 has dimension L(L?)"” =12, so this must be area () Expression (i) has dimension L, so itis (a). Expression (ii) has dimension L(L?) =L’, soit is (o). Thus,8 Physics ana Measurement PIS (a) [This sincoizect] since the units off are m/s?, while the units of {p] are nys. (&) [ise correet] since the units of [y] are m, and cos(kx) is dimensionless if k] isin m*. criss (we Eorent 2 spe te proporioliy fae sant fen the inverse proportionality of acceleration to mass. If has no dimensions, we have ML Puy ‘Multiply both sides by [}* and divide by [ke}’; the units of G are Section 1.6 Conversion of Units *P1.18 Each of the four walls has area (8,00 ft)/120 ft) - 96.0 #7, Together, they have area atm jt dfos0 2 55) = [BF me F119 Apply the following conversion factors: Tin=254.em, 1 d= 86 400 5, 100 em=1 m,and 10? nm=1m 254 cxyin\10™ aycm)(10° norm 86400 siday . injday) 39 nas |. ‘This means the proteins are assembled at arate of many layers of atoms each second! *PL20 8.50 in? =850 inChapter 1 7 PLO Conceptuatize: We must calculate the area and convert units, Since a meter is about 3 feet, we should expect the area tobe about A »(50-m)(50 m)=1500 m?, Categorize: We model the lot as a perfect rectangle to use Area = Length x Width. Use the conversion: 1 m=3.281 ft. Analyse A= LW = 0049 (= a0 139 xI0 mt J. Finalize: Our calculated result agrees reesonably well with our inital estimate and has the proper units of m?, Unit conversion is a common technique that is applied to many problems. rim @ ) PLB fa) © © ‘V = (40.0 m)(20.0 m)(12.0 m) ~ 9.6010? m? ¥ =9.60%10? m*(3.28 ft mJ? =[539x10" A The mass of the aris m= pa¥=(120 kg/n*)(9.60x10" m?) =115% ‘The student must look up weight in the index to find Converting to pounds, 10° kg (1.15 10" bg)(2.80 as!) =1.13.107 W. F, =(113% 10° NY fb/4.45N) =[254% 108 ‘Seven minutes is 420 seconds, so the rate is, 300 gal 205 —— Converting gallons first to liters, then to m?, r=(714e107 galjs 3.786 10> m* iL Teal = [Ero a], At that rate, to illa 1-m? tank would take ) Tosh]. 1-(—1e 270x107 m/s8 Physics and Measurement “PLA (a) Lang fant cave atm 22) n= FROF TT SEGA) 1) Hag crntbonae cia 28) Bl m= 0491 k= 491010" cm (©) Height of Denali =20320 (een) TR 619 m= 619 x10 m= 61910" em 1) epheing Can ta of 889) RETO RT| From Table 1.5, the density oflead is 11910 kg/m? so we should expect our calculated value to Pas 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, Dena is fined 3 i yer vlune in p= Wem convert fo STunisin the clean, 23945 { kg men 210 cm? (10002, 1m 114410" kg/m’ Atone step in the calculation, we note that one milion cubic centimeters make one cubic meter, Our resultis 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 havea density greater than 1 g/cm®, and objects tha oat must be less dense than water. Itis offen useful to remember that the 1 600-m race at track and field events is app in length. To be precise there are 1 609 meters in a mile. Thus, 1 acre is equal in area to 1609.m > = [05107 aD acres @ et 1 ih min. *PL27 The weight flow rate is 1200 £08 h Ser (2ane 1 sin) ton 80s P128 1609 m = 609 km; thus, to go from mph to km/h, multiply by 1.609. 1.609 Kenji 85 km/h 65 mi/h~1046 kml, Thus, ao=| Tet key]. ately ¥ mile oeChapter 1 190 years 6x10" SY th 1000 8/s 3.6008 neo [ (@) The circumference of the Earth at the equator is 2n{6.378x10° ma}= 40110" m. The length of one dollar bill is 0.155 m so that the length of 6 tilion bills is 93010" m. Thus, the 6 tilion dollars would encircle the Earth 9.30%10" m_j =[232x10" times J. 401%07 m a ~ [119507 atoms Vi _ 378x107 en =[ 51x10 m (or 151 an) 1, _ |(03.0 acres){43560 ft PLS vat ah 5 Piso PIsL (481 ft) =9.08x10 ©, or «107 m3 Vena” (2 at = [Berto PL33__F, =(250 tons/block}{2.00 x 10° blocks|(2000 Th/ton}=[ 1.0010" Ibs. “P34 The area covered by wateris Ay =070A pay, =(0.70Y dr R3yng) =(0.70X'47)(637 x108 m)" =36x10" m?. ‘The average depth of the water is d= (23 miles{1 609 m/l mile) = 3.7 10° m. ‘The volume of the water is V2 bd =(86. 10! m2)(37 «20? mm) =13 x20! a? and the mass is m= pV =(1000 kg/m?Y1.3%10" mn) = [735107 Fg],10 F135 “P1368 Pis7 138 F139 Physies and Measurement 4, decay ete ~ act =(6.79%10% (t)(3048 mm/l ft}=[2.07 mm 3 > 4, 1.0610 m {Hao} J (iss) 240x108 Ts ei 8.6210 times as large 70x10 300 ft mace |= (2.40107 mj — sono ol oa } scale distance ( real \ea between “{asence) fcr) =(42%40° 4 14x10 mn Tre scale factor used in the “dinner plate” model is ‘The distance to Andromeda in the scale model will be Dang = Dunas S=(20%10° lightyears}]25%10° mylightyeas) = » (637 «10 m)(00 exn/mn))” at 6 ‘174x10" em ol oe . . Veaum se {resco (6.37%10° m)(100 cxvm Wan = (ss) (auc #4 Tobalance, mpe="a1 OF preVee = PavAs oS on Sp 79x10" ft, or 30m). vuern( 22) "000 {28)” [anaPiao ‘Section 4 PLat Piaz Chopter? 14 ‘The mass of each sphere is ad m= Pay = eeu and M50 = PpeVge = Setting these masses equal 3 2 Paral Pete and| 6 Estimates and Order-of- Magnitude Calcutations Model the room as a rectangular solid with dimensions 4 m by 4m by 3 m, and each ping-pong ball as a sphere of diameter 0.038 m, The volume of the room is 4x43 48 m®, while the volume of ‘one ballis ae(ons m 0 agra? a. [10°] ping-pong balls in the room. As an aside, the actual number is smaller than this because there will bea lot of space in the ‘oom that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best packing fraction” is pnt '=074 so that at least 26% of the space will be empty. Therefore, the above estimate reduces to 1.67 10° x0.740~10°. ‘A reasonable guess for the diameter of tire might be 25 ft, with a circumference of about Bt. Thus, the tire would make (50.000 mi)(5 280 ft/mi)(1 rev/8 ft) =3 107 rev~[10" rev]. In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least Lig ttf Since 29819501 the number of ade Fs oe expected ona quarter-acre plot of land is about otal area __ (0.25 acre 43 560 ft7/acre) om total area .5 x10" blades ~[10" bad ‘area per blade” 43x10 #/blade Le bi12 Physics and Measurement pias PLS Piss PLA? ‘Atypical raindrop is spherical and might have a radius of about 1 inch. Its vohume is then approximately 4520” in’, Since 1 acre = 43 560 f2, the volume of water required to cover itto.a depth of 1 inch is 43560 ft? ¥ 144 in? 5 ( (1 inch) = (1 acre-in} ————- 6.310% (1 acre} inch) =( of oad eral 08 in ‘The number of raindrops required is i in? n. -Yolume of water required _63%10 in? 4 5 aoe volume ofa single drop 4x10 in® a Assume the tub measures 1.3 m by 0.5 m by 03 m. One-half ofits volume is then V=(05X13 m\(O5 m}03 m)=010 m? ‘The mass ofthis volume of water is: 000 kg/m? 010 m®" ~107 kg. Maer = Prac Pennies are now mostly zine, but consider copper pennies filling SO%, of the volume of the tub. The ‘mass of copper required is reopps~ Pes =(8920 g/m n9 1} = 092 kg 10" gp]. ‘The typical person probably drinks 2to 3 soft drinks daily, Perhaps half ofthese were in aluminum cans. Thus, we will estimate 1 aluminum ean disposal per person per day. In the U.S. there are ~250 million people, and 365 days in a year, 0 (250% 10° cans/day)(365 days/year)=[10" cans are thrown away ot recycled each year. Guessing that each can weighs around 1/10 of an ounce, we estimate this represents {10 cans}(0.1 o2/can}(1 Th/16 02)(1 tor/2.000 Ib) ~31x10° tons/year. [~10° tons Assume: Total population = 107; one out of every 100 people has a piano; one tuner can serve about 11000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year. Therefore, {Atunes_Y_1 pian Jagr se ootey al # tuners (cats ao people 1” People) = [100]Chapter? 43 Section 1.7 Significant Figures PLs3 Piss PISI METHOD ONE ‘We treat the best value with its uncertainty a8 a binomial (213:+0.2) em (98:0.1) em, A= [213(0.8) 21.3(04) £0.2(98) (0.2X01)] om? ‘The first term gives the best value of the area. The ctoss terms add together to give the uncertainty and the fourth term is negligible. A= [209 om Ede METHOD TWO. We add the fractional uncertainties in the data, A 208 cm? + 27% = 209 em? 3:4 em? 2 a1 2.3 cayosena( 22-2) 3 = x(105 m£02 m)? ~r[a05 m? +2005 moa m +(02m7] =| 346m? £13 m? @ oar Bar =2a(105 m + 02 m)=[G50mETS ® B o ££ o B ® = (6.50020) cm=(6.5020.20) x10 m m= (1852002) kg Oe also, Se _ om 3br pom Inother words the percentages of uncertainty are cumulative, Therefore, 02 , 30.20) 5 650 0103, pe (He(65107 my Tars1” kg/m and p26) [CaEORT RIO Kym” ]~(1.6402)x10? kg/m.14 Physics and Measurement PIS2 © (@) 73627 37.22 083 2252 796.58 ca ©) 0008 2(2.s.£)3563(48.6)=114016= (2s) [ET © 5620(4 8.) x (>a s.f)=17.656= (48.6) [17 *PLS3. We work to nine significant digits: lyr-1 31556 92605 Th Ada seers ‘60 min 608 PLS4 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 195 m carries information to anly one place past the decimal. [T359 m PISS -Vn2¥,+2¥, <2, 44) V_ = (17.0 m1. m+1.0 mX{L.0 009 my Vo = (10.0 my(.0 my(0.090 m) = 0.900 mn? ¥=2{170 1m? +0900 m?)=[52 m0 010 AZ 0006+ 0010+ 0011 =| 00m = 011 ‘Additional Problems P156 —_Itis desired to find the distance x such that x _ 1000 m ome (Ge, such that zis the same multiple of 140 m asthe multiple that 1 000 mis of x). Thus, itis seen that (200 #52000 m) = 1,00.10° m? and therefore x=¥100010 ao? - [Siem]*PLs7 PLS PLs9 Chapter 15 Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg, One atom of gold has mass (7 of 2 srr a, So, the number of atoms in the cube is 27x10 kg ‘The imagined cubical volume of each atom is =169x10" m? 6 mn? tae Si) a | (One month is 1 mo = (30 day}(24 h/day)(3 600 5/h) = 2.592 x 108 s Applying units to the equation, ¥ =(150 Mtt?/mo}#+(000800 M3 /mo?)¥* since Ute =108 #8, Vv =(150%20° 4°/mo}t (0008 00.10 &2/mo?) 2 Converting months to seconds, 22502108 £2/mo | 0.008.00%108 48/0? p> 2E9210 ai" (asm 10" we) ‘thus [VIIA [OSR MALMO? WTYE],16 Physics and Measurement Pie PLst “Pian a'(deg) a(rad) tan(a) sin(a) | difference 15.0 0.262 0.268, 0.259 ‘347% 20.0 0349 0.364 0.342 6.43% Bo | tas | ome | tan | wae xo | tas | om | tar | “33m aa | tae | ose | ako | sane ms | tus | ose | aus ] sane we | tas | oss | tae ) joe | Caz ar_| cir | oso | ois | sie aere50m vans m an s80" = (239 m) tan(55.0") FIG. PL6L Let d represent the diameter of the coin and h its thickness. The mass of the gold is and? + mdi} esa} mop attnef ‘where tis the thickness of the plating, : natn ae namosr9sec0 4 1003 64 grams cost = 0.003 64 grams x$10/gram =$0.036 4~ [364 cents “This is negligible compared to $4.98. ‘The actual number of seconds in a year is (86-400 s/day}365.25 day yr) = 31 557 600 s/yr. The percent error in the approximation is (= 10" s/yr)-(31557 600 s/yr)| 31557600 oye 100% = [OHPL Pues: PL66 PLS7 @ @ ) @ (b) Yoo, Vesa Fuel saved = Vis npg ~ Varmpy =[L0<10™ gal/yr M=U. [l= Bb (™1=[4]01 L = L=L°. Thus, the equation is dimensionally correct. Veyanir = R24=(aR*)t= Al, where [AzaR™ Vrecanguarcect ~ tht (Ez) Ak, where [A= Gu The speed of rise may be found from ise, ata 1.35 cm diameter, 165 om'/s sank sar TS os 1 cubic meter of water has a mass = (1.00% 10 kg/em?)(.00 m?)(10% myn)" ‘As.a rough calculation, we treat each item as if it were 100% water. cet: me pv =,{4an)= 2 e0°)=(100 tg/m'(L-]0s10° a) Bax10 kg stnep meta Sas? 0.30? halen? (Se) em? fy: m=. AE0% = (110 kg/aw?( Z)(20 mm*ca0 smm(20" ryan)” =[13x10 kg _ 108 carsy10% mye 20 migal = 208 cars)10" mai’yr) _ 49, 4910 os avg sally 5.010" gal/yr48 Physics and Measurement furlongs Y_200 yd_Y09144m/1 fortnight Y 1 day = >= {soo foas ad osteoma aera) Raan107 ays ms { ‘ornight )7 furiong Lyd 1ddays Ea} 7 This speed is almost I mav's so we might guess the Sessa ana, or perhaps a sloth. PL69 The volume of the galaxy is {10 x)? (00” mm) ~10"" n? If the distance between stars is 4x10" m, then there is one star in a volume on the order of (410% an) ~10% on? wo! mgr ‘he numberof stars is about <2 _. [30 sa sais oe gta EO S| F170 The density of each material is ptt tai F] te wastes vine (2m-£5) [3] smut. WR calzone am en 3%] smaller. cu pena M8698) —[og6_B5] the abused vale ao2 85) (1.23 em}(5.06 em) |_cm* } em oi4s) | (1.54 cm)?(5.69 crm) sn: p= — M18) _ “(175 em)?(B74 Gm) = A261 8) __[7 5g 8 The tblaedvalue 78685 i 5H] smaller. wan o7en) | em om Brass: p= PLTL — (@) (3600 sftui24 hr/day (365.25 days/ye)= [BA6x10" syn ® 1.91x10™ micrometeorites ‘This would take LL*100 micrometeores F555 Foye], {ake 576x107 micrometcortey yr TIO FEPL Pia Fis Ps PLlo Pia Plas PLas PLis F120 Piz Pia P12 P30 P32 5.52% 10° kg/m? , between the densities ‘of aluminum and iron, and greater than the densities of surface rocks. 20kg 769m {@) and (b) see the solution, Nq = 6002213710; (¢) 180 g (440g (@) 98310 (0) 106107 atoms (@) 40210 molecules; (b) 365% 10' molecules G)iis @)iie (Vi () SEE: newton =1 kg y's? 357 m? 1.39%104 m? (a) 339% 108 f°; (b) 254% 108 Ib (2) 560 km=5.60 x10" m=8.60x10 em; (©) 491 m=0491 km=491 10" cm; (© 619m =619x10 m=6.19%10° can; (d) 2.50 km = 250 x 10° m=2.50x10° cm. 405% 10? m? (@)1 mijin=1.609 kmh; (b) 88.5 Jem; (©) 361 kmyh 119% 10 atoms. 237x108 m* Piso Pisa pias Piss Pag F150 Pis2 Pisa PLs6 PLS Pie PLen PL64 P1466 PLS Puzo 1.3% 10 kg, 200 ken (a) 134; (6) 49.1 2 mar 2 won (2a) ~107 rev ~10° raindrops ~10" cans; ~10* tons (2094) em? 350)45 3:2 (a) 797; (0) 1.1; (6) 17.66 159m 36m 450m? see the solution; 24.6" 3.64 cents; no see the solution (2) 1000 kgs (0) 52.197 igs 13x10 kg 832410 mys; snail see the solutionMotion in One Dimension (CHAPTER OUTLINE ANSWERS TO QUESTIONS 24 22 23 24 25 28 27 Qs 6 ‘Speed Q21 If count 50s between lightning and thunder, the sound has fees traveled (331 msX5.0 s)=17 km. The tranuit time forthe light ‘Aceterason issmaller by ‘tr Constant Acceleration 30010" 8 _ 9.05 4108 times, Frey Faling Objects 351 ays Derved om Cals so itis negligible in comparison. Q22 Yes. Yes, ifthe particle winds up in the +2 region at the end. 23 Zero. na Yes. Yes. 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. We assume the object moves along straight line. fits average velocity is zero, then the displacement must be zero over the time interval, according to Equation 22. The object might be stationary throughout the interval. Ifitis moving to the right at first, it must later move tothe left to eum to its starting point. Is 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 xvs.t graph, the instantaneous velocity at any time Fs the slope of the curve at that point. At fp in the graph, the slope of the curve is zero, and thus the instantaneous velocity at that time is also zero. x 5 r FIG. Q2.6 Yes. 1f the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, oris a constant. a22 Motion in One Dimension ms Q2.10 Qu on nad Qaas Qni6 Yes. 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 zeso. ifthe acceleration wete zero, then the velocity would not change, leaving the doughnut Aoating at restin mid-air. No: Car A might have greater acyeleration than B, but they might both have zero acceleration, or otherwise equal accelerations; othe driver of B might have tramped hard on the gas pedal in the regent past. Yes. Consider throwing a bal straight up. As the ball goes up, its velocity is upward (o> 0), and its acceleration is directed down {<0}. A graph of o vs. ¢ for ths situation would look like the figure tothe right. The acceleration is the slope of av vs. t graph, ands always negative inthis case, even when the velocity is positive, FIG.Q2.10 (@) Accelerating East ©) Braking East (©) Cruising East (a) Braking West (©) Accelerating West (f)_—Cruising West (® Stopped but stating to move East (t) Stopped but starting to move West No. Constant acceleration only. Yes. Zero is a constant. ‘The position does depend on the origin of the coordinate system. Assume that the cliffis 20 m tall, and that the stone reaches a maximum height of 10 m above the top ofthe cliff. If the origin is taken {as the top of the cliff then the maximum height reached by the stone would be 10 m. Ifthe origin is taken as the bottom of the cif, then the maximum height would be 30m. ‘The velocity is independent ofthe origin. Since the change in position is used to calculate the instantaneous velocity in Equation 25, the choice of origin is arbitrary. ‘Once te objects leave the hand, both are in fee fall, and both experience the same downward acceleration equal tothe free-fall acceleration, -3. ‘They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downwatd velocity equal to o,. This velocity isthe same as the velocity of the second ball s0 after they fall through equal heights their impact speeds will also be the same. 1 With hod gi? th heat, 1 (2) @5h=5 g{07071)*. The time is later than OSE. ‘The distance fallen is 025h= 7 (0517 ‘The elevation is 0.75h, greater than 05k.Q2a7 passing point. Pisa Section 2.1 Position, Velocity, and Speed P21 @) 230 mys ® & 2S m-900 9 ep a pet. 575m-0 @ saat HSmom isa] ma) Tr Sara ora) ae pede _100ft(_im i __) [pan ae age (same) ages) me (© Thetime required must have been _ Ax 3000 mi (1609 m/10? mm) _ 7 oe Teal Tm) LSxt ye}: m3 @) b) pemice Sm-10m OF en asa @ @. =33 mys © pop 0) = 20 21 30 ra re @ SO mys » Ts ‘Above. Your ball has zero initial speed and smaller average speed during the time of flight to the or in particularly windy times24 Motion in One Dimension Let d represent the distance between A and B. Let, be the time for which the walker has 4 P25 (a) ©) Section 2.2 P26 fa) % © &) © the higher speed in 5.00 m/s ~300 mys: Let ty represent the longer time for the return trip in Then the times are f, = The average speed a a G0 mys 4 G0 may pa Total distance dtd “Toaline alan aan BE, 2(150 m/s!) 8.00 mys 375 as ‘She starts and finishes atthe same point A. With total displacement = 0, average velocity, -[0 Instantaneous Velocity and Speed Atay time, the postion s given by x=(3.00 m/s"). Thus, at f= 30s: x, = (300 m/s2}c.00 5)? = [37m]. Atty =3:008+at: x, =(3.00 nys*}3.0054a4)?, or sy [270m 80 mys)an [200 wl" f aH ‘The instantaneous velocity at # = 3.00 s is: 2S )-tno80 mys(000 of} = RT] (om) at ty =405, x, =20 m (Point BY 2 1 ae 20-A0)m 60m pe a XV (G-15)s 28s 6 ‘ z ‘The slope of the tangent line is found from pointsCand > dD D. (le =105, x¢ =95 m) and (fp =358, xp =0), ° #68) os [38 ays ‘The velocity is zero when xis a minimum. Thisis at t= [4sChapter? 25 P28 (a) x(m) (m/s) 20. +68) 168) Oy At £2505 the slopes 72 3 Bi mys]. 54 35 ae AtE=405, the slope is v2 3 At1= 305, the slope is ne 2 Teme Teas]. 36m Att=20s, the slopeis v2 = a[50 nye]. © eB eae @ P29 (a) ) 45 nye FIG. P29 "7240 Once it resumes the race, the hare will run for a time of 1.000 m~800:m @ ays In this time, the tortoise can crawl a distance y—% = (02 nys\25 6)= [500m28 Metion in One Dimension Section 2.3 Acceleration PLI1 Choose the positive direction tobe the outward direction, perpendicular tothe wal. v2) cc) 213 @) Do _ 220 mys—(-25.0 mys) | ar 350x105 bpm ota Taso" ay Acceleration is constant over the first ten seconds, so at the end, vy =0, +at=0-4(200 m/s?)(100s)=[200 mys ‘Then a=0 sovis constant from t= 10.0 s to #=15.0 s, And over the last five seconds the velocity changes 10 =O +at= 200 mys-+(300 m/s*\5.00 s) =[500 mys In the first ten seconds, spasetote batt 04044 (200 m/s" {1003)* 200m. (Over the next five seconds the position changes to x sottdet = 100 m+(20.0 mofs}{5.00 8)-+0= 200 m. Andat i= 2005, ayaa tet Satt= 200 m+ (200 mysXS.00 s)+4{-300 m/s*)500 5)? = [Bem ‘The average speed during a time interval At is Sisence wavelet During the first Quarter mile segment, Secretariat’ average speed was p< 0.250-mi _ 1320 ft I Sa ps DEERE] (656 mit) During the second quarter mile segment, = 21m B= pe SORE] (74 min). For the third quarter mile ofthe race, 5, = BBM 3338s BSS Rs] (977 mifh). and during the final quarter mile, 374 ffs] (390 mi/h). continued on next pagePag Pas ©) @ eo @ b) © Chaptor 2 27 Assuming that 0, =0, and recognizing that v, =0, the average acceleration during the race 7 7 Tlalelapsed Gime ~ (SI4240 SxS ~ OS ‘Acceleration i the slope ofthe graph of» vs coms) 20 For 010 5 |. “Maximum negative acceleration sat 18s, andi approximately [i ays Motion Diagrams Oe Oe oe oe [= treading order mee veliy SES | ecntraton continued on next pageChapter 2 28 () 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 @) through (e). Within one drawing, the accelerations vectors would vary in magnitude and direction. Section 2.5 One-Dimensional Motion with Constant Accoleration P219 From v} =v}-+2ar, we have (109710 mys)" =0+2a(220 m), so that [a=2.74x10° m/s which is [@= 2.7910" times g |. 1 220 (a) x3, (vj -40,)# becomes 40 make; +280 my/s)(8.50 8) which yields v, =[EAT mys 2 go 2L2% _ 280 mys-661 mys 7 8505 ® 0.448 m/s? P221 Given 0, =120 ems when x, = 3.0 em(t=0),and at f= 2.00 5, x, =-5.00.em, sycacoott Tat: 80-300 1am) «22007 = Ee ant "7222 (a) __Lot be the state of moving at 69 mifh and fbe atrost 8.00 = 240424 0-0) 420,(x;-n) 2 imi 0=(60 mph zea -of at) 3600 mi/5280 st ah 7 “Sen (im seas) 218 mys =-218 mijh (ea. Baal Ti) 3e0ps )"L 22 v2" (©) Similarly, (0=(80 mi/h)” +22,(211 ft-0) 16. 400(5 280) . “mea (© Let#be moving at 80 mijh and fbe moving at 60 mi/h. Baa mips 998 ms? 0% =o), +24,(44-m) (60 mi/h)? = (90 mish)” +20, (211 ft 121 £8) -2.800(5 280) 2(903(3600) mi/h.30° Motion in One Dimerision "e223 (a) ) raza (a) ) © ‘Choose the intial point where the pilot reduces the throttle and the final point where the boat passes the buoy: + j= 100 m, 945 =30 YS, dy =?, a, =-35 mys®, f=? ayaa togtt pat 100 m=0+(30 mypsisia5 mys?) (1.75 my/s*)}t? -(20 mys)t +100 m=0. We use the quadratic formula: pach Tae ~ ta 30 m/s-+/900 m?/s? — 42.75 rey's*}100 m) pa Boman ae? 475 eV QOO) 39 mls18 me yo og GT], ‘21.75 mys?) 35 mys? ‘The smaller value is the physical answer. Ifthe boat kept moving with the same acceleration, it would stop and move backward, then gain speed, and pass the buoy again at 1265. by =0y +2,t=30 mjs(35 ayfs?}453 s°[TaT mya] Total displacement = area under the (o, £) curve from ¢=0 10.505. (50 9/505 5} +(50 my/sj40~15) +t00 ys\{10 8) ax- [1m From ¢=10 s to t= 40 s, displacement is Ax= (50 m/s+33 nys)(3 s)+(50 nys\(25 s)=[1457 m eiseq Da tO38 Motion in One Dimension Section 2.6 Freely Falling Objects 2.40 Poat ‘Choose the origin (y=0, 10) at the starting point ofthe bal and take upward as positive. Then ¥,=0, 0,50, and a=—g=~930 m/s? The position and the velocity at time f become: 1 1 ~ymotstatty,e-Sgr Yea edtoayp 8H and @ at t= 1.005: yy =F 0.0 ys*).00 5)? ono n/s?y2005)*=|=B5R| (080 s/s?(3.00 5) (6) at #=1008: 0, = (9.80 m/s*)(1.00 5) =| 00 5: 2 = {9.80 my/s*)(2.00 5) 0 5: vp = {9.80 m/s? X50 — as ms ‘Assume that air resistance may be neglected, Then, the acceleration a ll times during the Rights, that due to gravity, a=—g=—9.80 m/s”. During the flight, Goff went 1 mile (1 609m) up and then. 1 mile back down. Determine his speed just after launch by considering his upward Night: 2 2alyy-v) =f -2(0.80 mys2)(1 609 m) 1y=178 mys. 4 His time in the air may be found by considering his motion from just ater launch to just before impact: we matador = (178 mjs}t —hes80 mys?) ‘The root f= 0 describes launch; the other root, t= 36.2, deseribes his flight time. His rate of pay may then be found from $1.00 3628 pay rate = SEO. (a0276 8/5360 s/h)=[S93TR]. ‘We have assumed that the workman's Aight time, “a mile", and “a dollar”, were measured to three- digit precision. We have interpreted “up in the sky" as referring tothe freefall time, not to the launch and landing times. Both the takeoff and landing times must be several seconds away fom the job, in order for Goff to survive to resume work.Chapter? 39 We have yy Lee copy 0= {490 m/s?) ~ (800 mys}t-300 m. Solving fort 1 800s VOT 9.80 Using only the positive value for t we find that t=[77973]. ara QO ¥- tthe: L0= (0m, (4909259? an ©) vp =v, bat =10.0 -(9.60)01.50)=-468 mys Tee nafs downward % The bill stars from rest v,=0 and falls with a downward acceleration of 80 mm/s? (due to gravity) Thus, in 0.20 it wil fala distance of aynnt-is? =0-(490 m/s*X0.20 5)? = —0.20 m. ‘This distance is about twice the distance between the center ofthe bill and its op edge (#8 cm). "Thus, David will be unsuccessful 6) Frm ay-ntt a with 0, neha aay) _ {2-23 m) yD TY r980 mys 27s]. ©) Thefinal velocity se vy =0+(-980 mys?) 217 5) =[—BE2 mye}. (6) Thetime take for the sound of the impact to reach the spectator is Ay | 3m sto oem ap ape 86H" 8, foun 0 the total elapsed time Is fyaat = 217 8 +6.76% 10" s e[ 2338].40. Motion in One Dimension F246 At any time , the position of the ball released from rest is given by ys Jet? attimet, the position of the ball thrown vertically upward is desccbed by y2 = s-$s0) The time at which the 4 i—L?, which yields first bail has a position of yy = is found fom the first equation as. 2 £ Toreque tha the scond balhave a poston of y= at thi ive, he eon h equation to obtain 4 uation to obtain > batts [=f gt: vy =0 when !=3.00s, g=980 m/s*, Therefore, fe ~4s(e) gives the required initial upward velocity of the second P2a7 (a) = (680 m/s*)300 s) [25a mys], 4 youre zley tae (294 mfs)(3.00s) = [44 m Yer Hay *P248 (a) Consider the upward flight of the arrow. ene ayy —w) 0 (100 mys)? +2{-9.8 mey/s?)ay 10.000 m/s? _ 4 06 wet 310m ©) Consider the whole fight ofthe arow. wrneogiebeg o=0+(100 ayy 98 ay?) ‘The root t= 0 refers to the starting point. The time of flight is given by. mas]. F249 ‘Time fo fall 3.00 mis found from Eq. 2.12 with 7; =0, 3.0 m= 10.0 ays?}?, t= 0.7828. 2) With the horse galloping at 100 ns, the horizontal distance is vt [782m]. b) t= [0787sPst P2s2 Ghopter2 44 ‘Take downward as the pasitive y direction, (@) While the woman was in free fall, Ay=144 ft, v, =0,anda=g = 320 ffs? ‘Thus, aymntt tar +144 f= 0+ (160 ft/s?)? giving tay =3.00 5, Her velocity just before impacts: + gt =0+(320 fe/s*)(300 s) =[980 HS 96.0 Hs, vy =0, and Ay =180 in.=1.80 ft. Therefore, (©) While crushing the box, _ef-sP _ 0-060 ty? 2Gy) ALOR) toch p= SE = AY - H © Tmerosanton at StS MISO 307410" ts, or [a= 37 RIO? B/S? upward) Al=313x107 § y= 3.008: At f= 2.00, y= 3.00(2.00)’ = 24.0 m and Mes0ot =360 ms If he helicopter releases a small mailbag at this time, the equation of motion ofthe mailbag s penne tos Ze? 2404 50 Lennp Setting y, =0, 0=240+36.0F--49007, Solving for‘, (only positive values of count), [7=7368]. Consider the last 30 m of fall, We find its speed 30 m above the ground: 1 ypnsiteyte ya 1 2 das mey(15 )42(-98 mys%)a5 9 30 m+1.0m 15s Now consider the portion ofits fall above the 30 m point. We assume it starts from rest 2 = 85+ 2ayly) —s) (-126 mys) =0+2{-9.8 my/s*)ay 160 mis? “96 aye Tis original height was then 30 m-+[-816 m| = [382 =+126 nys. ay: = 816m,42 Motion in One Dimension Section 2.7 P253@) tb) Kinematic Equations Derived from Calculus. ry 0. constant a a= Jat an Jfat— +e, Fa ant a do=adt on fait fea) Li? sare, when £=0, $0 ¢2 = Tit atey, dx = vdt ae fodt=[($1 vats) 1 1 Ws Sat +0405 $0 ¢5 = Therefore, [x= 29? 4 Lat? 4 op tx; Ita)? = P70? ab + 2)ait + (7 +25ad) oat 42)(J 1 rat) Recall the expression for v: » FIP tate, Sooo) FIP +a. Therefore,P2s4 @ ) © (a) «) ) ) © @ See the graphs at the right. Choose x=0 at #=0. Ate 5,2=4(6 my/s\9s)—12m. Att=5s,.£=12m-+(B mys)(2s)=28 m. ALERT 5, x= 28 mL mysi28)—36m For 0<1<35,a~5.208 267 m/s? For3 100x107 Soom" s ‘New v= (-5:00x107 )3.00%10-*)* +-(3.00%10°(3.00%10->) 450 m/s+900 mys =[450 ays = 0.450 m4}1.38 m={0.500 mM ee ee +(1.30%210°)(3.00%10-Additional Problems “Pas7 ‘The distance the car travels at constant velocity, vo, during the reaction time is (Ax), = vpAt,. The time for the car to come to res, from initial velocity vg, after the brakes are applied is and the distance traveled during this braking period is woven Thus, the total distance traveled before coming to a stop is. Sap = (0), +402), *} Yee 3 (0) acarisa distance sug = rt SE (ee the outon te obi 257 rom the intersection of lengths, when the light tus yellow, the distance the car must travel before the fight tans red is ot Bx Soup = ty Bhs, ‘Assume the driver does not accelerate in an attempt to “beat the light” (an extremely dangerous prectice!). The time the light should remain yellow is then the time required for the car to travel distance ax at constant velocity ny. This is L242 ©) With) =16m, 960 kamjh, a=-20 mys? and At, 116, enya Abigye <1 s~ 2 VR “6m Tan) ein oat a@ ©) © ‘As we see from the graph, from about 50 s to 50s ‘Acela is cruising at a constant positive velocity in 200 the +x direction. From 50's to 200s, Acela acceleratesin the + direction reaching a top speed 99 of about 17D :ai/h. Around 2005, the engineer applies the brakes, and the train, stil travetingin the +x direction, slows down and then stops at 350s, Just after 350°, Acela reverses direction (> becomes negative) and steadily gains speed inthe ~100 =x direction. FIG. P259(a) ‘The peak acceleration between 45 and 170 mi/h is given by the slope of the steepest tangent to the v versus ¢ curve in this interval. From the tangent line shown, we find ap _ 055=45) mh seep a 00-50) = [22 (mifh)s ]=0.98 mys’. Let us use the fact that the area under the v versus curve equals the displacement. The train's displacement between 0 and 200 sis equal to the area of the gray shaded region, which we have approximated witha series of triangles and rectangles. Asosinoy atta; +area, + area areas afeay (60 mifhX.50 5) +(50 mi/h)G0s) (160 mifhy00 8) FIG. P255(6) + 415059100 miyh) + (000 5}470 miyh—160 min) =24.000(mi/hY(s) New, at the end of our calculation, we can find the displacement in miles by converting hours to seconds, As 1h=3 600s, ar mt wn) ‘3600s48 Motion in One Dimonsion =P2.60 Average speed of every point on the train as the frst car passes Liz: ~ Ax _ 860m a =573 mys. Br 150s 7? mi “The train has ths as its instantaneous speed halfway through the 1.50 time. Similarly, halfway through the next L105, the speed of the train’ Sem 79 ‘ays. The time required for the speed. to change from 573 mvs to 782 mvs 8 $as05)+2(108)=1305 72 ms 3.73 8 ay ayGt 0, the acceleration is: a, = 2 $0 tion is a, = er Tera a gow vey adie tis ince an aceon, Then v0 and, =055{ 8) reine etn a. Sipe durga tine of w=380 4 lig Arzogt+da,t vite ax=(104 mayd)(350. aye Z(oase mm/d-w)35.0 aX5.00 w) or [Beam] Let point 0 be at ground level and point 1 be at the end of the engine burn. Let re point 2 be the highest point the rocket reaches and point 3 be just before NX impact. The data in the table are found for each phase of the rocket’s motion. fy (tot) v7 ~(80.0)" = 2(4.00)(1 000) 80 pp =120 nys i 120=800+(400% giving t=1005 FIG. P2.62 (1t2) 0-(1207? 0~120 =—9.80t giving ‘This is the time of maximum height of the rocket: 29.80), siving @to3) vf -0=2(-9.80)(~1735) vy = -184=(-9.80)¢ giving 121885 @ tua =10+1224188 = [410 ©) ya ER conned net geP2463 © va = [Rae Cheptor2 47 = 2 a 0 [iaunch 00) o ‘0 [a0 #_| End Thrust 100, 7000) io | +400 #2 | Rise Upwards | 222 1735 0. 3.80) #3__[Pall to Barth 41.0 a se ~980 Distance traveled by motorist =(15.0 ny) Distance taveled by policeman = $200 m/s) (a) intercept occurs when 150t=19, or f= [1505 (©) elotticer)=(2.00 m/s*)=[300 mys (0 xloffcer)=4(200 nys?}?? = [25 a ‘Area A, isa rectangle, Thus, A; = hm=0,t. ‘Area A; is triangular. Therefore A; =o ho, - 4). ‘The total area under the curve is, Ay + Aye vgs eal and since », -9y =e,48 Motion in One Dimension P28 (@) Let be the distance traveled at acceleration a until maximum speed » is reached. If this is time fy we can use the following three equations: aeLeto he 100—2= o(10.2-1,) and v= 0, ah, “The fist two give so0-(102-41,}o- (102. Yh “ 5.43 js? 3.83 ys For Maggie: « = 200___| “Tiaay2.00) | 20 Forked «=F 5¥50)~ © Maggie: v= (6.43200) =[109 mys Judy: »=@.83\3.00) = [115 mys (©) Atthe sbosecomd mark Jat?-+0(6.00- xa atl + (60-1) Maggie: ¥= H(6A0)20) = (109K4.0) =543 m Judy: x= F250ya.007 +(115)(.00)=51.7 m Maggies ahead by [2m]. =0100 mys? 4, =-0500 ny/s? fy tty and 0 sat = apt 1 1 ra10t0 meLalt sob thea E 1a (y_4 area 1000-5) ( as 20000 Tabs Total time = 1 ah) row -2af ea(- odo nt 4Let the ball fall 1.50 m. Tt stikes at speed given by + 2alay 2 =0+2{-980 an/s*¥-150 m) yg =-5.42 mys and its stopping is described by 2} =0), +24,(2)-%) 0=(-5.42 m/s)? +22,(107 m) 29.4 m/s Pa I ar et0? myst =2.00%107 m ws {ts maximum acceleration wll be larger than the average acceteration we estimate by imagining constant acceleration, but will still be of order of magnitude [~10" m/s? (0) xpnactogtay?. Wesssume the pacage sans rome. 145 ma0+04(-960 mys?) as my 380 mys? Bas ©) epaa toate Fed =04042(-980 m/e25u85)* = 131 m distance fallen =e. +a,f=[04(98 nys?)5.18 ¢ (@) The remaining distance is. atm BOR ays © 145 m—I1315 m=135 m. During deceleration, vg =—508 ms, vy =0, xp~x;=—135 m vy muh taal ay —: O==(-508 m/s)” +2a,(—135 m) -[555 nm] upward.50. Motion in One Dimension 7269) ypangit Ze =500= 200+ 298084, 4908? +2.00¢—50. 2.00-+ 2.00 — 44 90,—50.0) 2450) Only the postive root is physically meaningful: BW S| after the first stone is thrown. ©) sponte taandest0-1tb=3005 ‘substitute 500: (200) +5 (9804200): 9%, = [153 mys] downward (©) ty 04 +at= 200+ (0803.00) = [31 ws] downward Pay 20g +at=153+(980)200) =[348 mys | downward P70 (a) d= 336t, 326, = 4.90(2.40~ ty)” [5595 —KASO(BED, 49083 ~35951, + 2822=0 1, = SESE YS595" = 8490/2824) ‘980 359.84 35875 ,- BEEIONTS «o07655 so 4=3364, (e) Ignoring the sound travel time, d = esoy240)* = 28.2 m, an error of [682%]. P271 (@)__Inwalkinga distance Az,ina time At, the length of rope ¢ is only increased by Arsing, “Tpke ataate snd ©) 2) see [FE ation © @Pu 600m, ty 200 me 9= Sindy However, x0, @ ) @ 2) © Chepter2 51 Prey! “ey 24m) hey a fagt tHE) (40 +96) 4) | mys) | os) 05 | 032 1/03 1s | 089 2 jadi 23 |1.28 3 faa 35 |152 4/160 45 |1.66 FIG. P2.734a) 5 fam ee ea from problem 271 above, a= t igs «ly. ae Geen)” (as a6) (8) | a{nr/'s? amis) ae oe 05 | 0.64 1 |os7 15 | 048 2 |os8 25 | 030 3. joa 35 | 018 4 fos 45/01 5 }o09 _ +100 = (350 m/s tf, +1.00 = 1.183t, [Bass]: ne lao m/s*)\5.46 5)? -[730m 1, = (490 m/e?) 46 «)= [BRT a] =(350 my/s?)6.46 5) = (3:80 m)sin120°, and (22, ¥2)= [EE 329m ) de (x yy? - M65 eai [EE ‘The x distance out tothe flys 200m and they distance up tothe fy is 1.00 m. (2) We can use the Pythagorean theorem to find the distance from the origin tothe fly. distance = yx? +y? = (2.00 my? + (40m)? = Dim, Be