CHAPTER 2 KINEMATICS IN ONE DIMENSION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS


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1 Physics h Ediion Cunell Johnson Young Sdler Soluions Mnul Soluions Mnul, Answer keys, Insrucor's Resource Mnul for ll chpers re included. Compleed downlod links: hps://esbnkre.com/downlod/physicshediionsoluionsmnulcunelljohnsonyoungsdler/ Tes Bnk for Physics h Ediion By John D. Cunell, Kenneh W. Johnson, Did Young, Shne Sdler Compleed downlod link: hps://esbnkre.com/downlod/physicshediionesbnkcunelljohnsonyoungsdler/ CHAPTER KINEMATICS IN ONE DIMENSION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS. (b) Displcemen, being ecor, coneys informion bou mgniude nd direcion. Disnce coneys no informion bou direcion nd, hence, is no ecor.. (c) Since ech runner srs he sme plce nd ends he sme plce, he hree displcemen ecors re equl. 3. (c) The erge speed is he disnce of 6. km diided by he elpsed ime of. h. The erge elociy is he displcemen of km diided by he elpsed ime. The displcemen is km, becuse he jogger begins nd ends he sme plce. 4. () Since he bicycle coers he sme number of meers per second eerywhere on he rck, is speed is consn. 5. (e) The erge elociy is he displcemen (. km due norh) diided by he elpsed ime (.5 h), nd he direcion of he elociy is he sme s he direcion of he displcemen. 6. (c) The erge ccelerion is he chnge in elociy (finl elociy minus iniil elociy) diided by he elpsed ime. The chnge in elociy hs mgniude of 5. km/h. Since he chnge in elociy poins due es, he direcion of he erge ccelerion is lso due es. 7. (d) This is lwys he siuion when n objec res begins o moe. 8. (b) If neiher he mgniude nor he direcion of he elociy chnges, hen he elociy is consn, nd he chnge in elociy is zero. Since he erge ccelerion is he chnge in elociy diided by he elpsed ime, he erge ccelerion is lso zero.
2 Chper Answers o Focus on Conceps Quesions () The runners re lwys moing fer he rce srs nd, herefore, he nonzero erge speed. The erge elociy is he displcemen diided by he elpsed ime, nd he displcemen is zero, since he rce srs nd finishes he sme plce. The erge ccelerion is he chnge in he elociy diided by he elpsed ime, nd he elociy chnges, since he conesns sr res nd finish while running.. (c) The equions of kinemics cn be used only when he ccelerion remins consn nd cnno be used when i chnges from momen o momen.. () Velociy, no speed, ppers s one of he ribles in he equions of kinemics. Velociy is ecor. The mgniude of he insnneous elociy is he speed.. (b) According o one of he equion of kinemics x, wih m/s displcemen is proporionl o he squre of he elociy., he 3. (d) According o one of he equion of kinemics x, wih m/s displcemen is proporionl o he ccelerion., he 4. (b) For single objec ech equion of kinemics conins four ribles, one of which is he unknown rible. 5. (e) An equion of kinemics elociy, he finl elociy, nd he ime re known. 6. (c) An equion of kinemics gies he nswer direcly, since he iniil x elociy, he finl elociy, nd he ime re known. 7. (e) An equion of kinemics x elociy, he finl elociy, nd he ccelerion re known. gies he nswer direcly, since he iniil gies he nswer direcly, since he iniil 8. (d) This semen is flse. Ner he erh s surfce he ccelerion due o griy hs he pproxime mgniude of 9.8 m/s nd lwys poins downwrd, owrd he cener of he erh. 9. (b) Freefll is he moion h occurs while he ccelerion is solely he ccelerion due o griy. While he rocke is picking up speed in he upwrd direcion, he ccelerion is no jus due o griy, bu is due o he combined effec of griy nd he engines. In fc, he effec of he engines is greer hn he effec of griy. Only when he engines shu down does he freefll moion begin.. (c) According o n equion of kinemics x, wih m/s speed is proporionl o he squre roo of he mximum heigh., he lunch
3 44 KINEMATICS IN ONE DIMENSION. () An equion of kinemics gies he nswer direcly.. (d) The ccelerion due o griy poins downwrd, in he sme direcion s he iniil elociy of he sone hrown from he op of he cliff. Therefore, his sone picks up speed s i pproches he nes. In conrs, he ccelerion due o griy poins opposie o he iniil elociy of he sone hrown from he ground, so h his sone loses speed s i pproches he nes. The resul is h, on erge, he sone hrown from he op of he cliff rels fser hn he sone hrown from he ground nd his he nes firs s 4. () The slope of he line in posiion ersus ime grph gies he elociy of he moion. The slope for pr A is posiie. For pr B he slope is negie. For pr C he slope is posiie. 5. (b) The slope of he line in posiion ersus ime grph gies he elociy of he moion. Secion A hs he smlles slope nd secion B he lrges slope. 6. (c) The slope of he line in posiion ersus ime grph gies he elociy of he moion. Here he slope is posiie ll imes, bu i decreses s ime increses from lef o righ in he grph. This mens h he posiie elociy is decresing s ime increses, which is condiion of decelerion.
4 Chper Problems 45 CHAPTER KINEMATICS IN ONE DIMENSION PROBLEMS. REASONING The disnce reled by he Spce Shule is equl o is speed muliplied by he ime. The number of foobll fields is equl o his disnce diided by he lengh L of one foobll field. SOLUTION The number of foobll fields is m / s 3 s x Number = 9. L L 9.4 m. REASONING The displcemen is ecor h poins from n objec s iniil posiion o is finl posiion. If he finl posiion is greer hn he iniil posiion, he displcemen is posiie. On he oher hnd, if he finl posiion is less hn he iniil posiion, he displcemen is negie. () The finl posiion is greer hn he iniil posiion, so he displcemen will be posiie. (b) The finl posiion is less hn he iniil posiion, so he displcemen will be negie. (c) The finl posiion is greer hn he iniil posiion, so he displcemen will be posiie. SOLUTION The displcemen is defined s Displcemen = x x, where x is he finl posiion nd x is he iniil posiion. The displcemens for he hree cses re: () Displcemen = 6. m. m = (b) Displcemen =. m 6. m = 4. m 4. m (c) Displcemen = 7. m (3. m) = +. m 3. SSM REASONING The erge speed is he disnce reled diided by he elpsed ime (Equion.). Since he erge speed nd disnce re known, we cn use his relion o find he ime. SOLUTION The ime i kes for he coninens o drif pr by 5 m is Disnce 5 m 4 Elpsed ime = 5 yr Aerge speed cm m 3 yr cm
5 46 KINEMATICS IN ONE DIMENSION 4. REASONING Since he erge speed of he impulse is equl o he disnce i rels diided by he elpsed ime (see Equion.), he elpsed ime is jus he disnce diided by he erge speed. SOLUTION The ime i kes for he impulse o rel from he foo o he brin is Disnce.8 m Time =.6 s (.) Aerge speed. m/s x x 5. REASONING According o Equion., he erge elociy is equl o he displcemen diided by he elpsed ime, nd he direcion of he erge elociy is he sme s h of he displcemen. The displcemen is equl o he difference beween he finl nd iniil posiions. x x SOLUTION Equion. gies he erge elociy s x x Therefore, he erge elociies for he hree cses re: () Aerge elociy = (6. m. m)/(.5 s) = +8. m/s (b) Aerge elociy = (. m 6. m)/(.5 s) = 8. m/s (c) Aerge elociy = [7. m (3. m)]/(.5 s) = +. m/s The lgebric sign of he nswer coneys he direcion in ech cse. 6. REASONING Disnce nd displcemen re differen physicl quniies. Disnce is sclr, nd displcemen is ecor. Disnce nd he mgniude of he displcemen, howeer, re boh mesured in unis of lengh. SOLUTION. The disnce reled is equl o hreefourhs of he circumference of he circulr lke. The circumference of circle is πr, where r is he rdius of he circle. Thus, he disnce d h he couple rels is r d.5 km 7.7 km
6 Chper Problems 47 b. The couple s displcemen is he hypoenuse of righ ringle wih sides equl o he rdius of he circle (see he drwing). The mgniude R of he displcemen cn be obined wih he id of he Pyhgoren heorem: R r r.5 km. km The ngle h he displcemen mkes wih due es is r n n () 45. norh of es r 7. REASONING AND SOLUTION In minues he sloh rels disnce of x s 6 s = s = (.37 m/s)( min) = 7 m min while he oroise rels disnce of x 6 s = = (.76 m/s)( min) = 55 m min The oroise goes frher hn he sloh by n moun h equls 55 m 7 m = 8 m 8. REASONING The younger (nd fser) runner should sr he rce fer he older runner, he dely being he difference beween he ime required for he older runner o complee he rce nd h for he younger runner. The ime for ech runner o complee he rce is equl o he disnce of he rce diided by he erge speed of h runner (see Equion.). SOLUTION The difference beween he imes for he wo runners o complee he rce is, where 5 8 Disnce Disnce nd (.) Aerge Speed Aerge Speed yrold 8yrold The difference beween hese wo imes (which is how much ler he younger runner should sr) is
7 48 KINEMATICS IN ONE DIMENSION Disnce 5 8 Aerge Speed Aerge Speed 5yrold 3 3 Disnce 8yrold. m. m 64 s 4.7 m/s 4.39 m/s 9. REASONING In order for he ber o cch he ouris oer he disnce d, he ber mus rech he cr he sme ime s he ouris. During he ime h i kes for he ouris o rech he cr, he ber mus rel ol disnce of d + 6 m. From Equion., ouris d () nd ber d 6 m () Equions () nd () cn be soled simulneously o find d. SOLUTION Soling Equion () for nd subsiuing ino Equion (), we find ber d 6 m ( d 6 m) d / d ouris ouris Soling for d yields: ber 6 m d ouris 6 m 6 m d ber 6. m/s 4. m/s ouris 5 m. REASONING AND SOLUTION Le wes be he posiie direcion. The erge elociy of he bckpcker is x x x x w e where w nd e w e w e w e Combining hese equions nd soling for x e (suppressing he unis) gies e.34 m/s /.447 m/s / w xw.34 m/s /.68 m/s 6.44 km xe.8 km /
8 Chper Problems 49 The disnce reled is he mgniude of x e, or.8 km.. REASONING AND SOLUTION. The ol displcemen reled by he bicyclis for he enire rip is equl o he sum of he displcemens reled during ech pr of he rip. The displcemen reled during ech pr of he rip is gien by Equion.:. Therefore, SSM x 6 s x (7. m/s)( min) 95 m min 6 s x (5. m/s)(36 min) m min 6 s x3 (3 m/s)(8. min) 6 m min The ol displcemen reled by he bicyclis during he enire rip is hen x 95 m m 6 m.67 m 4 b. The erge elociy cn be found from Equion.. 4 x.67 m min 6.74 m/s, due norh min 36 min 8. min 6 s. REASONING The definiion of erge elociy is gien by Equion. s Aerge elociy = Displcemen/(Elpsed ime). The displcemen in his expression is he ol displcemen, which is he sum of he displcemens for ech pr of he rip. Displcemen is ecor quniy, nd we mus be creful o ccoun for he fc h he displcemen in he firs pr of he rip is norh, while he displcemen in he second pr is souh. SOLUTION According o Equion., he displcemen for ech pr of he rip is he erge elociy for h pr imes he corresponding elpsed ime. Designing norh s he posiie direcion, we find for he ol displcemen h Displcemen = ( 7 m/s) Norh Norhwrd + (7 m/s) Souh Souhwrd where Norh nd Souh denoe, respeciely, he imes for ech pr of he rip. Noe h he minus sign indices direcion due souh. Noing h he ol elpsed ime is
9 5 KINEMATICS IN ONE DIMENSION Norh + Souh, we cn use Equion. o find he erge elociy for he enire rip s follows: Displcemen Aerge elociy = Elpsed ime 7 m/s 7 m/s Norh Norh Souh Souh Norh Souh = 7 m/s 7 m/s Norh Souh Norh Souh 3. Therefore, we he h 4 4 Norh Souh Bu nd Norh Souh Norh Souh 3 Aerge elociy = 7 m/s 7 m/s 6 m/s 4 4 The plus sign indices h he erge elociy for he enire rip poins norh. 3. REASONING AND SOLUTION The upper edge of he wll will dispper fer he rin hs reled he disnce d in he figure below..9 m A. m B d b A B The disnce d is equl o he lengh of he window plus he bse of he righ ringle of heigh.9 m. The bse of he ringle is gien by b =.9 m n = 4. m.9 m b Thus, d =. m + 4. m = 6. m. The ime required for he rin o rel 6. m is, from he definiion of erge speed, x 6. m 3. m/s. s
10 Chper Problems 5 4. REASONING AND SOLUTION Since = +, he ccelerion is gien by /. Since he direcion of rel is in he negie direcion hroughou he problem, ll elociies will be negie.. ( 9. m/s) ( 7. m/s).4 m/s 5. s Since he ccelerion is negie, i is in he sme direcion s he elociy nd he cr is speeding up. b. ( 3. m/s) ( 7. m/s).8 m/s 5. s Since he ccelerion is posiie, i is in he opposie direcion o he elociy nd he cr is slowing down or decelering. is defined by Equion.4 s he chnge in elociy diided by he elpsed ime. The chnge in elociy is equl o he finl elociy minus he iniil elociy. Therefore, he chnge in elociy, nd hence he ccelerion, is posiie if he finl elociy is greer hn he iniil elociy. The ccelerion is negie if he finl elociy is less hn he iniil elociy. The ccelerion is zero if he finl nd iniil elociies re he sme. 5. REASONING The erge ccelerion SOLUTION Equion.4 gies he erge ccelerion s. The iniil nd finl elociies re boh +8 m/s, since he elociy is consn. The erge ccelerion is = (8 m/s 8 m/s)/( ) = m/s b. The iniil elociy is +8 m/s, nd he finl elociy is 8 m/s. The erge ccelerion is = (8 m/s 8 m/s)/( s) = 4 m/s
11 5 KINEMATICS IN ONE DIMENSION 6. REASONING Alhough he plne follows cured, wodimensionl ph hrough spce, his cuses no difficuly here becuse he iniil nd finl elociies for his period re in opposie direcions. Thus, he problem is effeciely problem in one dimension only. Equion.4 reles he chnge in he plne s elociy o is erge ccelerion nd he elpsed ime =.6 yers. I will be conenien o coner he elpsed ime o seconds before clculing he erge ccelerion. SOLUTION. The ne chnge in he plne s elociy is he finl minus he iniil elociy: 8.5 km/s.9 km/s 39.4 km/s 39.4 km s m m/s km b. Alhough he plne s elociy chnges by lrge moun, he chnge occurs oer long ime inerl, so he erge ccelerion is likely o be smll. Expressed in seconds, he inerl is.6 yr Then he erge ccelerion is 365 d 4 h 6 min yr d h 6 s s min m/s 5.79 m/s s 4 7. REASONING Since he elociy nd ccelerion of he moorcycle poin in he sme direcion, heir numericl lues will he he sme lgebric sign. For conenience, we will choose hem o be posiie. The elociy, ccelerion, nd he ime re reled by Equion.4:. SOLUTION. Soling Equion.4 for we he b. Similrly, (+3 m/s) (+ m/s) +.5 m/s (+6 m/s) (+5 m/s) +.5 m/s 4. s 4. s
12 Chper Problems REASONING We cn use he definiion of erge ccelerion / (Equion.4) o find he spriner s finl elociy he end of he ccelerion phse, becuse her iniil elociy ( m/s, since she srs from res), her erge ccelerion, nd he ime inerl re known. SOLUTION. Since he spriner hs consn ccelerion, i is lso equl o her erge ccelerion, so.3 m/s Her elociy he end of he.s period is m/s.3 m/s. s.8 m/s b. Since her ccelerion is zero during he reminder of he rce, her elociy remins consn..8 m/s 9. REASONING When he elociy nd ccelerion ecors re in he sme direcion, he speed of he objec increses in ime. When he elociy nd ccelerion ecors re in opposie direcions, he speed of he objec decreses in ime. () The iniil elociy nd ccelerion re in he sme direcion, so he speed is incresing. (b) The iniil elociy nd ccelerion re in opposie direcions, so he speed is decresing. (c) The iniil elociy nd ccelerion re in opposie direcions, so he speed is decresing. (d) The iniil elociy nd ccelerion re in he sme direcion, so he speed is incresing. SOLUTION The finl elociy is reled o he iniil elociy, he ccelerion, nd he elpsed ime hrough Equion.4 ( = + ). The finl elociies nd speeds for he four moing objecs re:. = m/s + (3. m/s )(. s) = 8 m/s. The finl speed is 8 m/s. b. = m/s + (3. m/s )(. s) = 6. m/s. The finl speed is 6. m/s. c. = m/s + (3. m/s )(. s) = 6. m/s. The finl speed is 6. m/s. d. = m/s + (3. m/s )(. s) = 8 m/s. The finl speed is 8 m/s.. REASONING The fc h he emu is slowing down ells us h he ccelerion nd he elociy he opposie direcions. Furhermore, since he ccelerion remins he sme in boh prs of he moion, we cn deermine is lue from he firs pr of he moion nd hen use i in he second pr o deermine he bird s finl elociy he end of he ol 6.s ime inerl. SOLUTION
13 54 KINEMATICS IN ONE DIMENSION. The iniil elociy of he emu is direced due norh. Since he bird is slowing down, is ccelerion mus poin in he opposie direcion, or due souh. b. We ssume h due norh is he posiie direcion. Wih he d gien for he firs pr of he moion, Equion.4 shows h he erge ccelerion is.6 m/s 3. m/s.6 m/s 4. s s The negie lue for he ccelerion indices h i indeed poins due souh, which is he negie direcion. Soling Equion.4 for he finl elociy gies.6 m/s.6 m/s. s s 9.4 m/s Since his nswer is posiie, he bird s elociy fer n ddiionl. s is in he posiie direcion nd is 9.4 m/s, due norh.. REASONING AND SOLUTION The mgniude of he cr's ccelerion cn be found from Equion.4 ( = + ) s 6.8 m/s m/s 8.8 m/s 3.75 s. REASONING According o Equion.4, he erge ccelerion of he cr for he firs wele seconds fer he engine cus ou is f () nd he erge ccelerion of he cr during he nex six seconds is f f f () The elociy f of he cr he end of he iniil welesecond inerl cn be found by soling Equions () nd () simulneously.
14 Chper Problems 55 SOLUTION Diiding Equion () by Equion (), we he Soling for f, we obin ( ) / ( ) ( ) / ( ) f f f f f f f ( / ) ( / ) f f f.5(. s)(+8. m/s) (6. s)( 36. m/s).5(. s) 6. s +3. m/s 3. REASONING AND SOLUTION Boh moorcycles he he sme elociy he end of he four second inerl. Now for moorcycle A nd = A + A = B + B for moorcycle B. Subrcion of hese equions nd rerrngemen gies A B = (4. m/s. m/s )(4 s) = 8. m/s The posiie resul indices h moorcycle A ws iniilly reling fser. 4. REASONING AND SOLUTION The erge ccelerion of he bskebll plyer is /, so 6. m/s x.5 s 4.5 m.5 s 5. SSM REASONING AND SOLUTION. The mgniude of he ccelerion cn be found from Equion.4 ( = + ) s 3. m/s m/s. s.5 m/s b. Similrly he mgniude of he ccelerion of he cr is
15 56 KINEMATICS IN ONE DIMENSION 4. m/s 38. m/s. s.5 m/s c. Assuming h he ccelerion is consn, he displcemen coered by he cr cn be found from Equion.9 ( = + x): (4. m/s) (38. m/s) x (.5 m/s ) 79 m Similrly, he displcemen reled by he jogger is (3. m/s) ( m/s) x (.5 m/s ) 3. m Therefore, he cr rels 79 m 3. m = 76 m furher hn he jogger. 6. REASONING The erge ccelerion is defined by Equion.4 s he chnge in elociy diided by he elpsed ime. We cn find he elpsed ime from his relion becuse he ccelerion nd he chnge in elociy re gien. SOLUTION. The ime h i kes for he VW Beele o chnge is elociy by n moun = is (nd noing h.447 m/s = mi/h).447 m / s 6. mi / h m / s mi / h.35 m / s.4 s b. From Equion.4, he ccelerion (in m/s ) of he drgser is.447 m / s 6. mi / h m / s mi / h.6 s s 44.7 m / s 7. REASONING We know he iniil nd finl elociies of he blood, s well s is displcemen. Therefore, Equion.9 x cn be used o find he ccelerion
16 Chper Problems 57 of he blood. The ime i kes for he blood o rech i finl elociy cn be found by using x Equion.7. SOLUTION. The ccelerion of he blood is 6 cm / s cm / s.7 cm / s x. cm b. The ime i kes for he blood, sring from cm/s, o rech finl elociy of +6 cm/s is x. cm.5 s cm / s + 6 cm / s 8. REASONING AND SOLUTION. From Equion.4, he definiion of erge ccelerion, he mgniude of he erge ccelerion of he skier is 8. m/s m/s 5. s.6 m/s b. Wih x represening he displcemen reled long he slope, Equion.7 gies: x ( ) (8. m/s m/s)(5. s) =. m 9. SSM REASONING AND SOLUTION The erge ccelerion of he plne cn be found by soling Equion.9 x for. Tking he direcion of moion s posiie, we he (+6. m/s) (+69 m/s) 3. m/s x (+75 m) The minus sign indices h he direcion of he ccelerion is opposie o he direcion of moion, nd he plne is slowing down. 3. REASONING A consn elociy he ime required for Secreri o run he finl mile is gien by Equion. s he displcemen (+69 m) diided by he elociy. The cul ime required for Secreri o run he finl mile cn be deermined from Equion.8, since
17 58 KINEMATICS IN ONE DIMENSION he iniil elociy, he ccelerion, nd he displcemen re gien. I is he difference beween hese wo resuls for he ime h we seek. SOLUTION According o Equion., wih he ssumpion h he iniil ime is = s, he run ime consn elociy is x 69 m 97.4 s 6.58 m/s Soling Equion.8 x for he ime shows h 4 x.5 m/s 6.58 m/s 6.58 m/s 4.5 m/s 69 m 94. s We he ignored he negie roo s being unphysicl. The ccelerion llowed Secreri o run he ls mile in ime h ws fser by 97.4 s 94. s.8 s 3. SSM REASONING The cr hs n iniil elociy of = +5. m/s, so iniilly i is moing o he righ, which is he posiie direcion. I eenully reches poin where he displcemen is x = +.5 m, nd i begins o moe o he lef. This mus men h he cr comes o momenry hl his poin (finl elociy is = m/s), before beginning o moe o he lef. In oher words, he cr is decelering, nd is ccelerion mus poin opposie o he elociy, or o he lef. Thus, he ccelerion is negie. Since he iniil elociy, he finl elociy, nd he displcemen re known, Equion.9 x cn be used o deermine he ccelerion. SOLUTION Soling Equion.9 for he ccelerion shows h m/s 5. m/s. m/s x.5 m 3. REASONING A ime boh rockes reurn o heir sring poins nd he displcemen of zero. This occurs, becuse ech rocke is decelering during he firs hlf of is journey.
18 Chper Problems 59 Howeer, rocke A hs smller iniil elociy hn rocke B. Therefore, in order for rocke B o decelere nd reurn o is poin of origin in he sme ime s rocke A, rocke B mus he decelerion wih greer mgniude hn h for rocke A. Since we know h he displcemen of ech rocke is zero ime, since boh iniil elociies re gien, nd since we seek informion bou he ccelerion, we begin our soluion wih Equion.8, for i conins jus hese ribles. SOLUTION Applying Equion.8 o ech rocke gies A A A x A A A A A A B B B x B B B B B B The ime for ech rocke is he sme, so h we cn eque he wo expressions for, wih he resul h A B A B or Soling for B gies B A A B A B A 5 m/s B 86 m/s m/s 58 m/s As expeced, he mgniude of he ccelerion for rocke B is greer hn h for rocke A. 33. REASONING The sopping disnce is he sum of wo prs. Firs, here is he disnce he cr rels. m/s before he brkes re pplied. According o Equion., his disnce is he mgniude of he displcemen nd is he mgniude of he elociy imes he ime. Second, here is he disnce he cr rels while i deceleres s he brkes re pplied. This disnce is gien by Equion.9, since he iniil elociy, he ccelerion, nd he finl elociy ( m/s when he cr comes o sop) re gien. SOLUTION Wih he ssumpion h he iniil posiion of he cr is x = m, Equion. gies he firs conribuion o he sopping disnce s x x. m/s.53 s
19 6 KINEMATICS IN ONE DIMENSION Soling Equion.9 x for x shows h he second pr of he sopping disnce is x m/s. m/s 7. m/s Here, he ccelerion is ssigned negie lue, becuse we he ssumed h he cr is reling in he posiie direcion, nd i is decelering. Since i is decelering, is ccelerion poins opposie o is elociy. The sopping disnce, hen, is m/s. m/s xsopping x x. m/s.53 s 39. m 7. m/s 34. REASONING The enering cr minins consn ccelerion of = 6. m/s from he ime i srs from res in he pi re unil i cches he oher cr, bu i is conenien o sepre is moion ino wo inerls. During he firs inerl, lsing = 4. s, i cceleres from res o he elociy wih which i eners he min speedwy. This elociy is found from Equion.4, wih = m/s, =, =, nd = : () The second inerl begins when he enering cr eners he min speedwy wih elociy, nd ends when i cches up wih he oher cr, which rels wih consn elociy = 7. m/s. Since boh crs begin nd end he inerl sidebyside, hey boh undergo he sme displcemen x during his inerl. The displcemen of ech cr is gien by Equion.8 x. For he ccelering cr, =, nd =, so x () For he oher cr, = nd = m/s, nd so Equion.8 yields x (3) SOLUTION The displcemen during he second inerl is no required, so equing he righ hnd sides of Equions () nd (3) elimines x, leing n equion h my be soled for he elpsed ime, which is now he only unknown quniy:
20 Chper Problems 6 (4) Subsiuing Equion () for ino Equion (4), we find h 7. m/s 6. m/s 4. s 6. m/s 5 s 35. REASONING The drwing shows he wo knighs, iniilly sepred by he displcemen d, reling owrd ech oher. A ny momen, Sir George s displcemen is x G nd h of Sir Alfred is x A. When hey mee, heir displcemens re he sme, so x G = x A. Sring poin for Sir George + Sring poin for Sir Alfred x G x A d According o Equion.8, Sir George's displcemen s funcion of ime is x m/s () G,G G G G where we he used he fc h Sir George srs from res,g = m/s. Since Sir Alfred srs from res x = d = s, we cn wrie his displcemen s (gin, employing Equion.8) x d d m/s d () A,A A A A
21 6 KINEMATICS IN ONE DIMENSION Soling Equion for xg / G nd subsiuing his expression ino Equion yields xg xg xa d A d A (3) G G Noing h x A = x G when he wo riders collide, we see h Equion 3 becomes xg xg d A G d Soling his equion for x G gies xg A G SOLUTION Sir George s ccelerion is posiie G.3 m/s since he srs from res nd moes o he righ (he posiie direcion). Sir Alfred s ccelerion is negie A.. m/s since he srs from res nd moes o he lef (he negie direcion). The displcemen of Sir George is, hen, x d 88. m 5.8 m G A. m/s G +.3 m/s 36. REASONING The plyers collide when hey he he sme x coordine relie o common origin. For conenience, we will plce he origin he sring poin of he firs plyer. From Equion.8, he x coordine of ech plyer is gien by x x d d where d = +48 m is he iniil posiion of he second plyer. When x x, he plyers collide ime. SOLUTION. Equing Equions () nd () when, we he d
22 Chper Problems 63 We noe h = +.5 m/s, while =.3 m/s, since he firs plyer cceleres in he +x direcion nd he second plyer in he x direcion. Soling for, we he b. From Equion (), d 48 m s.5 m/s.3 m/s x.5 m/s s 3. m 37. REASONING A consn elociy he ime required for he firs cr o rel o he nex exi is gien by Equion. s he mgniude of he displcemen (.5 3 m) diided by he mgniude of he elociy. This is lso he rel ime for he second cr o rech he nex exi. The ccelerion for he second cr cn be deermined from Equion.8, since he iniil elociy, he displcemen, nd he ime re known. This equion pplies, becuse he ccelerion is consn. SOLUTION According o Equion., wih he ssumpion h he iniil ime is = s, he ime for he firs cr o rech he nex exi consn elociy is x.5 m 76 s 33 m/s Remembering h he iniil elociy of he second cr is zero, we cn sole Equion.8 x for he ccelerion o show h 3 x (.5 m) (76 s) 3.87 m/s Since he second cr s speed is incresing, his ccelerion mus be in he sme direcion s he elociy. 38. REASONING Le he ol disnce beween he firs nd hird sign be equl o d. Then, he ime A is gien by d d d A ()
23 64 KINEMATICS IN ONE DIMENSION Equion.7 x cn be wrien s x /, so h B d d d () SOLUTION Diiding Equion () by Equion () nd suppressing unis for conenience, we obin B A REASONING Becuse he cr is reling in he +x direcion nd decelering, is ccelerion is negie: =.7 m/s. The finl elociy for he inerl is gien ( = +4.5 m/s), s well s he elpsed ime ( = 3. s). Boh he cr s displcemen x nd is iniil elociy he insn brking begins re unknown. Compre he lis of known kinemic quniies (,, ) o he equions of kinemics for consn ccelerion: (Equion.4), x (Equion.7), x (Equion.8), nd x (Equion.9). None of hese four equions conins ll hree known quniies nd he desired displcemen x, nd ech of hem conins he iniil elociy. Since he iniil elociy is neiher known nor requesed, we cn combine wo kinemic equions o elimine i, leing n equion in which x is he only unknown quniy. SOLUTION For he firs sep, sole Equion.4 for : () Subsiuing he expression for in Equion () ino Equion.8 x yields n expression for he cr s displcemen solely in erms of he known quniies,, nd : x Subsiue he known lues of,, nd ino Equion (): x ()
24 Chper Problems 65 x 4.5 m/s 3. s.7 m/s 3. s 5.7 m Noe: Equion () cn lso be obined by combining Equion () wih Equion.7 x, or, wih more effor, by combining Equion () wih Equion.9 x. 4. REASONING AND SOLUTION As he plne deceleres hrough he inersecion, i coers ol disnce equl o he lengh of he plne plus he widh of he inersecion, so x = 59.7 m + 5. m = 84.7 m The speed of he plne s i eners he inersecion cn be found from Equion.9. Soling Equion.9 for gies x (45. m) ( 5.7 m/s )(84.7 m) 54.7 m/s The ime required o rerse he inersecion cn hen be found from Equion.4. Soling Equion.4 for gies 45. m/s 54.7 m/s =.7 s 5.7 m/s 4. SSM REASONING As he rin psses hrough he crossing, is moion is described by Equions.4 ( = + ) nd.7 x ( ), which cn be rerrnged o gie nd These cn be soled simulneously o obin he speed when he rin reches he end of he crossing. Once is known, Equion.4 cn be used o find he ime required for he rin o rech speed of 3 m/s. x SOLUTION Adding he boe equions nd soling for, we obin x (. m) (.6 m/s )(.4 s). m/s.4 s
25 66 KINEMATICS IN ONE DIMENSION The moion from he end of he crossing unil he locomoie reches speed of 3 m/s requires ime 3 m/s. m/s 4 s.6 m/s 4. REASONING Since he cr is moing wih consn elociy, he displcemen of he cr in ime cn be found from Equion.8 wih = m/s nd equl o he elociy of he cr: xcr cr. Since he rin srs from res wih consn ccelerion, he displcemen of he rin in ime is gien by Equion.8 wih = m/s: rin x A ime, when he cr jus reches he fron of he rin, xcr Lrin x, where rin is he lengh of he rin. Thus, ime, rin L rin cr rin rin L () A ime, when he cr is gin he rer of he rin, xcr x. Thus, ime rin cr rin () Equions () nd () cn be soled simulneously for he speed of he cr ccelerion of he rin. rin SOLUTION. Soling Equion () for we he rin cr nd he rin cr (3) Subsiuing his expression for ino Equion () nd soling for rin, we he cr cr Lrin 9 m 4 s (4 s) 8 s 3 m/s b. Direc subsiuion ino Equion (3) gies he ccelerion of he rin:
26 Chper Problems 67 cr (3 m/s) rin.93 m/s 8 s 43. REASONING AND SOLUTION When ir resisnce is negleced, free fll condiions re pplicble. The finl speed cn be found from Equion.9; SSM y where is zero since he sun mn flls from res. If he origin is chosen he op of he hoel nd he upwrd direcion is posiie, hen he displcemen is y = 99.4 m. Soling for, we he y ( 9.8 m/s )( 99.4 m) 44. m/s The speed impc is he mgniude of his resul or 44. m/s. 44. REASONING Becuse here is no effec due o ir resisnce, he rock is in free fll from is lunch unil i his he ground, so h he ccelerion of he rock is lwys 9.8 m/s, ssuming upwrd o be he posiie direcion. In (), we will consider he inerl beginning lunch nd ending. s ler. In (b), we will consider he inerl beginning lunch nd ending 5. s ler. Since he displcemen isn required, Equion.4 suffices o sole boh prs of he problem. The sone slows down s i rises, so we expec he speed in () o be lrger hn 5 m/s. The speed in (b) could be smller hn 5 m/s (he rock does no rech is mximum heigh) or lrger hn 5 m/s (he rock reches is mximum heigh nd flls bck down below is heigh he.s poin). SOLUTION. For he inerl from lunch o =. s, he finl elociy is = 5 m/s, he ccelerion is = 9.8 m/s, nd he iniil elociy is o be found. Soling Equion.4 for gies 5 m/s ( 9.8 m/s )(. s) 35 m/s Therefore, lunch, Speed 35 m/s b. Now we consider he inerl from lunch o = 5. s. The iniil elociy is h found in pr (), = 35 m/s. The finl elociy is
27 68 KINEMATICS IN ONE DIMENSION 35 m/s ( 9.8 m/s )(5. s) = 4 m/s (.4) Insnneous speed is he mgniude of he insnneous elociy, so we drop he minus sign nd find h Speed 4 m/s 45. REASONING AND SOLUTION In ime he crd will undergo ericl displcemen y gien by y = where = 9.8 m/s. When = 6. ms = 6. s, he displcemen of he crd is.8 m, nd he disnce is he mgniude of his lue or d.8 m. Similrly, when = ms, d.7 m, nd when = 8 ms, d3.6 m. 46. REASONING Assuming h ir resisnce cn be negleced, he ccelerion is he sme for boh he upwrd nd downwrd prs, nmely 9.8 m/s (upwrd is he posiie direcion). Moreoer, he displcemen is y = m, since he finl nd iniil posiions of he bll re he sme. The ime is gien s = 8. s. Therefore, we my use Equion.8 y o find he iniil elociy of he bll. SOLUTION Soling Equion.8 y for he iniil elociy gies y m 9.8 m/s 8. s 39 m/s 8. s
28 Chper Problems REASONING AND SOLUTION The figure he righ shows he phs ken by he pelles fired from gun A nd gun B. The wo phs differ by he exr disnce coered by he pelle from gun A s i rises nd flls bck o he edge of he cliff. When i flls bck o he edge of he cliff, he pelle from gun A will he he sme speed s he pelle fired from gun B, s Concepul Exmple 5 discusses. Therefore, he fligh ime of pelle A will be greer hn h of B by he moun of ime h i kes for pelle A o coer he exr disnce. A B The ime required for pelle A o reurn o he cliff edge fer being fired cn be found from Equion.4: = +. If "up" is ken s he posiie direcion hen = +3. m/s nd = 3. m/s. Soling Equion.4 for gies = = ( 3. m/s) ( 3. m/s) = 9.8 m/s 6. s Noice h his resul is independen of he heigh of he cliff. 48. REASONING The iniil elociy nd he elpsed ime re gien in he problem. Since he rock reurns o he sme plce from which i ws hrown, is displcemen is zero (y = m). Using his informion, we cn employ Equion.8 y o deermine he ccelerion due o griy. SOLUTION Soling Equion.8 for he ccelerion yields y m 5 m / s. s. s.5 m / s 49. SSM REASONING The iniil elociy of he compss is +.5 m/s. The iniil posiion of he compss is 3. m nd is finl posiion is m when i srikes he ground. The displcemen of he compss is he finl posiion minus he iniil posiion, or y = 3. m. As he compss flls o he ground, is ccelerion is he ccelerion due o griy, = 9.8 m/s. Equion.8 y cn be used o find how much ime elpses before he compss his he ground. SOLUTION Sring wih Equion.8, we use he qudric equion o find he elpsed ime.
29 7 KINEMATICS IN ONE DIMENSION.5 m/s.5 m/s m/s 3. m m/s y There re wo soluions o his qudric equion, = nd =.568 s. The second soluion, being negie ime, is discrded..8 s 5. REASONING The iniil speed of he bll cn be deermined from Equion.9 y. Once he iniil speed of he bll is known, Equion.9 cn be used second ime o deermine he heigh boe he lunch poin when he speed of he bll hs decresed o one hlf of is iniil lue. SOLUTION When he bll hs reched is mximum heigh, is elociy is zero. If we ke upwrd s he posiie direcion, we he from Equion.9 h y m/s ( 9.8 m/s )(6 m) 8 m/s When he speed of he bll hs decresed o one hlf of is iniil lue,, nd Equion.9 gies ( ) (8 m/s) y m 4 ( 9.8 m/s ) 4 5. REASONING AND SOLUTION. y.8 m/s 9.8 m/s 3. m 7.9 m/s The minus is chosen, since he dier is now moing down. Hence, 7.9 m/s. b. The dier's elociy is zero his highes poin. The posiion of he dier relie o he bord is.8 m/s y.7 m 9.8 m/s The posiion boe he wer is 3. m +.7 m = 3. m.
30 Chper Problems 7 5. REASONING Equion.9 y boe he lunch poin, where he finl speed is cn be used o deermine he mximum heigh m / s. Howeer, we will need o know he iniil speed, which cn be deermined i Equion.9 nd he fc h y 4. m (ssuming upwrd o be he posiie direcion). SOLUTION When he bll hs reched is mximum heigh, we he, so h Equion.9 becomes y y mx or m / s mx or mx when m / s nd y y y () Using Equion.9 nd he fc h when y 4. m (ssuming upwrd o be he posiie direcion), we find h 4. m y or 4. m or 3 / 4 Subsiuing Equion () ino Equion () gies y mx 4. m / 3 / m () 53. SSM REASONING AND SOLUTION Since he blloon is relesed from res, is iniil elociy is zero. The ime required o fll hrough ericl displcemen y cn be found from Equion.8 y wih m/s. Assuming upwrd o be he posiie direcion, we find y ( 6. m). s 9.8 m/s 54. REASONING Equion.9 y cn be used o find ou how fr boe he cliff's edge he pelle would he gone if he gun hd been fired srigh upwrd, proided h we cn deermine he iniil speed impred o he pelle by he gun. This iniil speed cn be found by pplying Equion.9 o he downwrd moion of he pelle described in he problem semen. SOLUTION If we ssume h upwrd is he posiie direcion, he iniil speed of he pelle is, from Equion.9, y ( 7 m/s) ( 9.8 m/s )( 5 m).9 m/s
31 7 KINEMATICS IN ONE DIMENSION Equion.9 cn gin be used o find he mximum heigh of he pelle if i were fired srigh up. A is mximum heigh, = m/s, nd Equion.9 gies (.9 m/s) y m ( 9.8 m/s ) 55. REASONING The displcemen y of he dier is equl o her erge elociy muliplied by he ime, or. Since he dier hs consn ccelerion (he ccelerion due o y griy), her erge elociy is equl o, where nd re, respeciely, he iniil nd finl elociies. Thus, ccording o Equion.7, he displcemen of he dier is y (.7) The finl elociy nd he ime in his expression re known, bu he iniil elociy is no. To deermine her elociy he beginning of he.s period (her iniil elociy), we urn o her ccelerion. The ccelerion is defined by Equion.4 s he chnge in her elociy,. Soling his equion for he iniil, diided by he elpsed ime : elociy yields / Subsiuing his relion for ino Equion.7, we obin y SOLUTION The dier s ccelerion is h due o griy, or = 9.8 m/s. The ccelerion is negie becuse i poins downwrd, nd his direcion is he negie direcion. The displcemen of he dier during he ls. s of he die is y. m/s. s 9.8 m/s. s 5.6 m The displcemen of he dier is negie becuse she is moing downwrd. 56. REASONING The bll is iniilly in free fll, hen collides wih he pemen nd rebounds, which pus i ino free fll gin, unil cugh by he boy. We don he enough informion o nlyze is collision wih he pemen, bu we re only sked o clcule he ime i spends in he ir, undergoing freefll moion. The moion cn be conenienly diided ino hree inerls: from relese (h = 9.5 m) o impc, from impc o he second highes poin (h = 5.7 m), nd from he second highes poin o h3 =. m boe he pemen. For ech of he inerls, he ccelerion is h due o griy. For he firs nd ls inerl, he
32 Chper Problems 73 bll s iniil elociy is zero, so he ime o fll gien disnce cn be found from Equion.8 y. The second inerl begins he pemen nd ends h, so he iniil elociy isn zero. Howeer, he symmery of freefll moion is such h i kes he bll s much ime o rise from he ground o mximum heigh h s i would ke for bll dropped from h o fll o he pemen, so we cn gin use Equion.8 o find he durion of he second inerl. SOLUTION Tking upwrd s he posiie direcion, we he = 9.8 m/s for he ccelerion in ech of he hree inerls. Furhermore, he iniil elociy for ech of he inerls is = m/s. Remember, we re using symmery o re he second inerl s if he bll were dropped from res heigh of 5.7 m nd fell o he pemen. Using Equion.8 y, wih = m/s, we cn sole for he ime o find h y Applying his resul o ech inerl gies he ol ime s ol = ( 9.5 m ) 9.8 m/s + ( 5.7 m ) 9.8 m/s + ( ) é ë m . m ù û 9.8 m/s = 3.43 s s inerl nd inerl 3 rd inerl Noe h he displcemen y for ech inerl is negie, becuse upwrd hs been designed s he posiie direcion. 57. REASONING To clcule he speed of he rf, i is necessry o deermine he disnce i rels nd he ime inerl oer which he moion occurs. The speed is he disnce diided by he ime, ccording o Equion.. The disnce is 7. m 4. m = 3. m. The ime is he ime i kes for he sone o fll, which cn be obined from Equion.8 y, since he displcemen y, he iniil elociy, nd he ccelerion re known. SOLUTION During he ime h i kes he sone o fll, he rf rels disnce of 7. m 4. m = 3. m, nd ccording o Equion., is speed is 3. m speed
33 74 KINEMATICS IN ONE DIMENSION The sone flls downwrd for disnce of 75. m, so is displcemen is y = 75. m, where he downwrd direcion is ken o be he negie direcion. Equion.8 cn be used o find he ime of fll. Seing = m/s, nd soling Equion.8 for he ime, we he Therefore, he speed of he rf is y 75. m 3.9s 9.8 m/s 3. m speed = 3.9 s.767 m/s 58. REASONING The sone h is hrown upwrd loses speed on he wy up. The sone h is hrown downwrd gins speed on he wy down. The sones cross phs below he poin h corresponds o hlf he heigh of he cliff. To see why, consider where hey would cross phs if hey ech minined heir iniil speed s hey moed. Then, hey would cross phs excly he hlfwy poin. Howeer, he sone reling upwrd begins immediely o lose speed, while he sone reling downwrd immediely gins speed. Thus, he upwrd moing sone rels more slowly hn he downwrd moing sone. Consequenly, he sone hrown downwrd hs reled frher when i reches he crossing poin hn he sone hrown upwrd. The iniil elociy is known for boh sones, s is he ccelerion due o griy. In ddiion, we know h he crossing poin he sones re he sme plce he sme ime. Furhermore, he posiion of ech sone is specified by is displcemen y from is sring poin. The equion of kinemics h reles he ribles,, nd y is Equion.8 y, nd we will use i in our soluion. In using his equion, we will ssume upwrd o be he posiie direcion. SOLUTION Applying Equion.8 o ech sone, we he y up = up + Upwrd moing sone nd y down = down + Downwrd moing sone
34 Chper Problems 75 In hese expressions is he ime i kes for eiher sone o rech he crossing poin, nd is he ccelerion due o griy. Noe h y up is he displcemen of he upwrd y down moing sone boe he bse of he cliff, y down is he H displcemen of he downwrd moing sone below he op y of he cliff, nd H is he displcemen of he cliffop boe up he bse of he cliff, s he drwing shows. The disnces boe nd below he crossing poin mus dd o equl he heigh of he cliff, so we he y y H up down where he minus sign ppers becuse he displcemen y down poins in he negie direcion. Subsiuing he wo expressions for y up nd y down ino his equion gies up down H This equion cn be soled for o show h he rel ime o he crossing poin is H up down Subsiuing his resul ino he expression from Equion.8 for y up gies up up H H yup up down up down 6. m 6. m 9. m/s 9.8 m/s 9. m/s 9. m/s 9. m/s 9. m/s.46 m Thus, he crossing is loced disnce of.46 m boe he bse of he cliff, which is below he hlfwy poin of 3. m, s expeced. 59. SSM REASONING AND SOLUTION. We cn use Equion.9 o obin he speed cquired s she flls hrough he disnce H. Tking downwrd s he posiie direcion, we find
35 76 KINEMATICS IN ONE DIMENSION y m/s H or H To cquire speed of wice his lue or H, she mus fll n ddiionl disnce H. According o Equion.9 y, we he H H H or 4 H H H The ccelerion due o griy cn be elimined lgebriclly from his resul, giing 4H H H or H 3H b. In he preious clculion he ccelerion due o griy ws elimined lgebriclly. Thus, lue oher hn 9.8 m/s would no he ffeced he nswer o pr (). 6. REASONING When he rrows rech heir mximum heighs, hey come insnneously o hl, nd he finl speed of ech rrow is zero. Using his fc, we will be ble o deermine he ime i kes for ech rrow o rech is mximum heigh. Knowing his ime for he second rrow will llow us o deermine is iniil speed lunch. SOLUTION The ime required for he firs rrow o rech is mximum heigh cn be deermined from Equion.4 ( = + ). Tking upwrd s he posiie direcion, we he m/s 5. m/s.55 s 9.8 m/s Noe h he second rrow is sho. s fer he firs rrow. Therefore, since boh rrows rech heir mximum heigh he sme ime, he second rrow reches is mximum heigh.55 s. s =.35 s fer being fired. The iniil speed of he second rrow cn hen be found from Equion.4: m/s ( 9.8 m/s )(.35 s) 3. m/s 6. SSM REASONING Once he mn sees he block, he mn mus ge ou of he wy in he ime i kes for he block o fll hrough n ddiionl. m. The elociy of he block he insn h he mn looks up cn be deermined from Equion.9. Once he elociy is known h insn, Equion.8 cn be used o find he ime required for he block o fll hrough he ddiionl disnce.
36 Chper Problems 77 SOLUTION When he mn firs noices he block, i is 4. m boe he ground nd is displcemen from he sring poin is y 4. m 53. m. Is elociy is gien by Equion.9 y. Since he block is moing down, is elociy hs negie lue, y m/s ( 9.8 m/s )(4. m 53. m) 7.7 m/s The block hen flls he ddiionl. m o he leel of he mn's hed in ime which sisfies Equion.8: y where y =. m nd 7.7 m/s. Thus, is he soluion o he qudric equion where he unis he been suppressed for breiy. From he qudric formul, we obin 7.7 (7.7) 4(4.9)(.) (4.9).4 s or 6. s The negie soluion cn be rejeced s nonphysicl, nd he ime i kes for he block o rech he leel of he mn is.4 s. 6. REASONING Once is fuel is gone, he rocke is in free fll, so is moion consiss of wo inerls of consn bu differen ccelerion. We will ke upwrd s he posiie direcion. From lunch o engine burnou, he ccelerion is = +86. m/s, nd he rocke s displcemen is y. Is elociy he end of he burn,, is lso he iniil elociy for he second porion of is fligh: engine burnou o mximum liude. During his second porion, he rocke slows down wih he ccelerion of griy = 9.8 m/s nd undergoes n ddiionl displcemen of y in reching is mximum heigh. Is mximum liude is he sum of hese wo ericl displcemens: h = y + y. SOLUTION Firs we consider he ime period =.7 s from he igniion of he engine unil he fuel is gone. The rocke cceleres from = m/s o =, rising displcemen y, s gien by Equion.8 y : m/s y ()
37 78 KINEMATICS IN ONE DIMENSION Equion.4 gies is elociy he insn he fuel runs ou: m/s () From h momen onwrd, he second pr of he rocke s moion is free fll ( = 9.8 m/s ). I kes ime for he rocke s elociy o decrese from = o = m/s is mximum liude. We sole Equion.9 y o find is upwrd displcemen y during his ime: y y m/s or Subsiuing for from Equion (), we find for y h y (3) Using Equions () nd (3), we find h he rocke s mximum liude, relie o he ground, is h y y Using he lues gien, we find h 86. m/s h 86. m/s.7 s m 9.8 m/s 63. REASONING To find he iniil elociy, of he second sone, we will employ Equion.8, y +. In his expression is he ime h he second sone is in, he ir, nd i is equl o he ime h he firs sone is in he ir minus he ime 3. i kes for he firs sone o fll 3. m: 3. We cn find nd 3. by pplying Equion.8 o he firs sone. SOLUTION To find he iniil elociy, of he second sone, we employ Equion.8,, y. Soling his equion for, yields
38 Chper Problems 79, y The ime for he firs sone o srike he ground cn be obined from Equion.8,, + y his equion for, we he. Noing h, = m/s since he sone is dropped from res nd soling y 5. m.75 s 9.8 m/s () Noe h he sone is flling down, so is displcemen is negie (y = 5. m). Also, is ccelerion is h due o griy, so = 9.8 m/s. The ime 3. for he firs sone o fll 3. m cn lso be obined from Equion : y 3. m.88 s 9.8 m/s 3. The ime h he second sone is in he ir is The iniil elociy of he second sone is, s.88 s.94 s y 5. m 9.8 m/s.94 s m/s.94 s 64. REASONING We ssume h downwrd is he posiie direcion. The ile flls from res, so is iniil elociy is zero. The ile flls hrough displcemen y in going from he roof op o he op of he window. I is lue for y h we seek, nd i cn be obined from y (Equion.9). In his expression = m/s, nd is he ccelerion due window o griy. The elociy window he op of he window is no gien, bu i cn be obined from he ime of. s h i kes he ile o pss he window. SOLUTION Soling Equion.9 for y nd using he fc h = m/s gies
39 8 KINEMATICS IN ONE DIMENSION window window y () The ile rels n ddiionl displcemen y window =.6 m in rersing he window in ime =. s. These d cn be used in window window y (Equion.8) o find he elociy window he op of he window. Soling Equion.8 for window gies window y.6 m 9.8 m/s. s window 7. m/s. s Using his lue for window in Equion (), we obin window 7. m/s y.5 m 9.8 m/s 65. SSM REASONING The slope of srighline segmen in posiionersusime grph is he erge elociy. The lgebric sign of he erge elociy, herefore, corresponds o he sign of he slope. SOLUTION. The slope, nd hence he erge elociy, is posiie for segmens A nd C, negie for segmen B, nd zero for segmen D. b. In he gien posiionersusime grph, we find he slopes of he four srighline segmens o be.5 km km A 6.3 km/h. h h B C.5 km.5 km 3.8 km/h.4 h. h.75 km.5 km.63 km/h.8 h.4 h.75 km.75 km D km/h. h.8 h
40 Chper Problems REASONING On posiionersusime grph, he elociy is he slope. Since he objec s elociy is consn nd i moes in he +x direcion, he grph will be srigh line wih posiie slope, beginning x = 6 m when = s. A = 8 s, is posiion should be x = 6 m + 48 m = +3 m. Once he grph is consruced, he objec s elociy is found by x clculing he slope of he grph:. SOLUTION The posiionersusime grph for he moion is s follows: Posiion x (m) Time (s) The objec s displcemen is +48 m, nd he elpsed ime is 8 s, so is elociy is x 48 m.7 m/s 8 s 67. REASONING AND SOLUTION The erge ccelerion for ech segmen is he slope of h segmen. 4 m/s m/s A.9 m/s s s B 4 m/s m/s m/s 48 s s 8 m/s m/s C 3.3 m/s 6 s 48 s 68. REASONING The erge elociy for ech segmen is he slope of he line for h segmen. SOLUTION Tking he direcion of moion s posiie, we he from he grph for segmens A, B, nd C,
41 8 KINEMATICS IN ONE DIMENSION A B. km 4. km. km/h.5 h. h. km. km. km/h.5 h.5 h 4. km. km C 4 km/h 3. h.5 h 69. REASONING The slope of he posiionime grph is he elociy of he bus. Ech of he hree segmens of he grph is srigh line, so he bus hs differen consn elociy for ech pr of he rip: A, B, nd C. The slope of ech segmen my be clculed from Equion. x, where Δx is he difference beween he finl nd iniil posiions of he bus nd Δ is he elpsed ime during ech segmen. The erge ccelerion of he bus is he chnge in is elociy diided by he elpsed ime, s in Equion.4. The rip lss from = h (he iniil insn on he grph) o = 3.5 h (he finl insn on he grph), so he ol elpsed ime is Δ = 3.5 h. The iniil elociy of he bus is is elociy =, which is is consn elociy for segmen A: = A. Similrly, he elociy of he bus he ls insn of segmen C is is finl elociy for he rip: = C. x SOLUTION In using Equion. o clcule he slopes of segmens A nd C, ny displcemen Δx wihin segmen my be chosen, so long s he corresponding elpsed ime Δ is used in he clculion. If he full displcemens for ech segmen re chosen, hen Apply hese resuls o Equion.4: C A xa 4 km km 4 km/h. h h A xc 7 km 33 km 5 km/h 3.5 h. h C 5 km/h 4 km/h 8.3 km/h 3.5 h 7. REASONING The runner is he posiion x = m when ime = s; he finish line is m wy. During ech ensecond segmen, he runner hs consn elociy nd runs
42 Chper Problems 83 hlf he remining disnce o he finish line. The following ble shows he firs four segmens of he moion: Time Inerl = s =. s =. s =. s =. s = 3. s = 3. s = 4. s Chnge in Posiion x = m x = 5. m x = 5. m x = 5. m + 5. m = 75. m x = 75. m x = 75. m +.5 m = 87.5 m x = 87.5 m x = 87.5 m m = 93.8 m This d cn be used o consruc he posiionime grph. Since he runner hs consn elociy during ech ensecond segmen, we cn find he elociy during ech segmen from he slope of he posiionime grph for h segmen. SOLUTION. The following figure shows he posiionime grph for he firs fory seconds.. Posiion x (m) Time (s) b. The slope of ech segmen of he posiionime grph is clculed s follows:
43 84 KINEMATICS IN ONE DIMENSION x 5. m. m. s o. s 5. m/s. s s x 75. m 5. m. s o. s.5 m/s. s. s x 87.5 m 75. m. s o 3. s.5 m/s 3. s. s 3. s o 4. s Therefore, he elociyime grph is: x 93.8 m 87.5 m.65 m/s 4. s 3. s 5. Velociy (m/s) Velociy (m/s) Time (s) SSM REASONING The wo runners sr one hundred meers pr nd run owrd ech oher. Ech runs en meers during he firs second nd, during ech second herefer, ech runner runs niney percen of he disnce he rn in he preious second. While he elociy of ech runner chnges from second o second, i remins consn during ny one second. SOLUTION The following ble shows he disnce coered during ech second for one of he runners, nd he posiion he end of ech second (ssuming h he begins he origin) for he firs eigh seconds. Time (s) Disnce coered (m) Posiion x (m)
44 Chper Problems The following grph is he posiionime grph consruced from he d in he ble boe.. Since he wo runners re running owrd ech oher in excly he sme wy, hey will mee hlfwy beween heir respecie sring poins. Th is, hey will mee x = 5. m. According o he grph, herefore, his posiion corresponds o ime of 6.6 s. b. Since he runners collide during he seenh second, he speed he insn of collision cn be found by king he slope of he posiionime grph for he seenh second. The speed of eiher runner in he inerl from = 6. s o = 7. s is x 5.6 m m 5.3 m/s 7. s 6. s Therefore, he momen of collision, he speed of eiher runner is 5.3 m/s.
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