University Physics with Modern Physics 14th Edition Young TEST BANK

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1 Universi Phsics wih Modern Phsics 14h Ediion Young SOLUTIONS MANUAL Full clear download (no formaing errors) a: hps://esbankreal.com/download/universi-phsics-modern-phsics- 14h-ediion-oung-soluions-manual-/ Universi Phsics wih Modern Phsics 14h Ediion Young TEST BANK Full clear download (no formaing errors) a: hps://esbankreal.com/download/universi-phsics-modern-phsics- 14h-ediion-oung-es-bank-/ CHAPTER MOTION ALONG A STRAIGHT LINE Discussion Quesions Q.1 The speedomeer measures he magniude of he insananeous veloci, he speed. I does no measure veloci because i does no measure direcion. Q. Graph (d). The dos represen he insec s posiion as a funcion of ime. If he phoographs are aken a equal spaced ime inervals, hen he displacemen in successive inervals is increasing and his means he speed is increasing. Therefore, graphs (a) and (e) can be ruled ou. Graph (b) shows decreasing acceleraion so would correspond o he speed approaching a consan value, which is no wha he phoographs show. Graph (c) shows moion in he negaive -direcion, which is no he case. This leaves graph (d). This graph shows veloci in he posiive -direcion and increasing speed. This is consisen wih he phoographs. Q.3 The answer o he firs quesion is es. If he objec is iniiall moving and he acceleraion direcion is opposie o he veloci direcion, hen he objec slows down, sops for an insan and hen sars o move in he opposie direcion wih increasing speed. An eample is an objec hrown sraigh up ino he air. Gravi gives he objec a consan downward acceleraion. The objec ravels upward, sops a is maimum heigh and hen moves downward. The answer o he second quesion is no. Afer he firs reversal of he direcion of ravel he veloci and acceleraion are hen in he same direcion. The objec coninues moving in he second direcion wih increasing speed. Q.4 Average veloci equals insananeous veloci when he speed is consan and moion is in a sraigh line. Q.5 a) Yes. For an objec o be slowing down, all ha is required is ha he acceleraion be nonzero and for he veloci and acceleraion o be in opposie direcions. The magniude of he acceleraion deermines he rae a which he speed is changing. b) Yes. For an objec o be speeding up, all ha is required is ha he acceleraion be nonzero and for he veloci and acceleraion o be in he same direcion. The magniude of he acceleraion deermines he rae a which he speed is changing. Bu for an nonzero acceleraion he speed is increasing when he veloci and acceleraion are in he same direcion.

2 Q.6 Average veloci is he magniude of he displacemen divided b he ime inerval. Average speed is he disance raveled divided b he ime inerval. Displacemen equals he disance raveled when he moion is in he same direcion for he enire ime inerval, and herefore his is when average veloci equals average speed. Q.7 For he same ime inerval he have displacemens of equal magniude bu opposie direcions, so heir average velociies are in opposie direcions. One average veloci vecor is he negaive of he oher. Q.8 If in he ne ime inerval he second car had pulled ahead of he firs, hen he speed of he second car was greaer. The second car could also be observed o be alongside a pedesrian sanding a he curb, bu ha does no mean he pedesrian was speeding. Q.9 The answer o he firs quesion is no. Average veloci is displacemen divided b he ime inerval. If he displacemen is zero, hen he average veloci mus be zero. The answer o he second quesion is es. Zero displacemen means he objec has reurned o is saring poin, bu is speed a ha poin need no be zero. See Fig. DQ.9.

3 Figure DQ.9 Q.10 Zero acceleraion means consan veloci, so he veloci could be consan bu no zero. See Fig. DQ.10. An eample is a car raveling a consan speed in a sraigh line. Figure DQ.10 Q.11 No. Average acceleraion refers o an inerval of ime and if he veloci is zero hroughou ha inerval, he average acceleraion for ha ime inerval is zero. Bu es, ou can have zero veloci and nonzero acceleraion a one insan of ime. For eample, in Fig. DQ.11, v = 0 when he graph crosses he ime ais bu he acceleraion is he nonzero slope of he line. An eample is an objec hrown sraigh up ino he air. A is maimum heigh is veloci is zero bu is acceleraion is g downward. Figure DQ.11 Q.1 Yes. When he veloci and acceleraion are in opposie direcions he objec is slowing down. Q.13 (a) Two possible - graphs for he moion of he ruck are skeched in Fig. DQ.13. (b) Yes, he displacemen is 58 m and he ime inerval is 9.0 s, no maer wha pah he ruck akes beween 1 and. The average veloci is he displacemen divided b he ime inerval.

4 Figure DQ.13 Q.14 This is rue onl when he acceleraion is consan. The average veloci is defined o be he displacemen divided b he ime inerval. If he acceleraion is no consan, objecs can have he same iniial and final velociies bu differen displacemens and herefore differen average velociies. Q.15 I is greaer while he ball is being hrown. While being hrown, he ball acceleraes from res o veloci v 0 while raveling a disance less han our heigh. Afer i leaves our hand, i slows from v 0 o zero a he maimum heigh, while raveling a disance much greaer han our heigh. v v 0 Eq.(.13) sas ha a =. Larger ( 0 ) 0 means smaller a. Q.16 (a) Eq.(.13): v = v + a ( ). When an objec reurns o he release poin, = Eq.(.13) hen gives v = v 0 and v = ±v 0. (b) v = v 0 + a. A he highes poin v = 0, so up = v 0 / a. A he end of he moion, when he objec has reurned o he release poin, we have shown in (i) ha v = v 0, v v 0 v 0 so oal = = and oal a a = up. Q.17 The disance beween adjacen drops will increase. The drops have he downward acceleraion g = 9.8 m/s of a free-falling objec. Therefore, heir speed is coninuall increasing and he disance one drop ravels in each successive 1.0 s ime inerval increases. A given drop has fallen for 1.0 s longer han he ne drop released afer i, so he addiional disance i has fallen increases as he fall. Mahemaicall, le be he ime he second drop has fallen, so he firs drop has fallen for ime s. The disance beween hese wo drops hen is Δ = 1 g ( s) 1 g = 1 g (.0 s) s. The separaion Δ increases as increases. Q.18 Yes. Consider ver small ime inervals during which he acceleraion doesn have ime o change ver much, so can be assumed o be consan. Calculae Δv = a Δ 1, for a ver small ime inerval, saring a = 0. Then v 1 = v 0 + Δv. Since he acceleraion is assumed consan for he small ime inerval, v av, = (v 0 + v 1 ) / and Δ 1 = v av, Δ 1. Then he posiion a he end of he inerval is 1 = 0 + Δ 1. Repea he calculaion for he ne small ime inerval Δ : Δv = a Δ, v = v 1 + Δv, v av, = (v 1 + v ) /, Δ = v av, Δ. Repea for successive small ime inervals.

5 Q.19 In he absence of air resisance, he firs ball rises o is maimum heigh and hen reurns o he level of he op of he building. When i reurns o he heigh from which i was hrown, a he op of he building, i is moving downward wih speed v 0. The res of is moion is he same as for he second ball. (a) Since he las par of he moion of he firs ball sars wih i moving downward wih speed v 0 from he op of he building, he wo balls have he same speed jus before he reach he ground. (b) The second ball reaches he ground firs, since he firs ball has o move up and hen down before repeaing he moion of he second ball. (c) Displacemen is final posiion minus iniial posiion. Boh balls sar a he op of he building and end up a he ground. So he have he same displacemen. (d) The firs ball has raveled a greaer disance. Q.0 Le he + -direcion be eas. The average veloci is he displacemen divided b he ime inerval. The firs 10.0 m displacemen requires a ime of (10.0 m/s) / (3.00 m/s) = 40.0 s. The second 10.0 m displacemen requires a ime of (10.0 m/s) / (5.00 m/s) = 4.0 s. The average veloci is v = Δ 40.0 m av- = = 3.75 m/s. This is less han 4.00 m/s since ou spend more ime Δ 64.0 s running a 3.00 m/s han a 5.00 m/s. Q.1 A he highes poin he objec insananeousl has zero speed. Bu is veloci is coninuall changing, a a consan rae. Acceleraion measures he rae of change of veloci. For eample, in Fig..5b when he graph crosses he ime ais i sill has a consan slope ha corresponds o he acceleraion. Also noe he commens in par (d) of he soluion o Eample.7. Q. For an objec released from res and hen moving downward in free-fall, is downward 1 displacemen from is iniial posiion of 0 = 0 is given b = g. To increase b a facor of 3, increase b a facor of 3. You can also see his b leing Y be he original heigh, so Y = 1 gt. Le he new heigh be Y and he corresponding ime be T, so Y = 1 g (T ). Bu Y = 3Y = 3 ( 1 gt ), so 3 ( 1 gt ) = 1 g (T ) and T = 3T.

6 MOTION ALONG A STRAIGHT LINE.1. IDENTIFY: Δ = v av- Δ SET UP: We know he average veloci is 6.5 m/s. EXECUTE: Δ = v av- Δ = 5.0 m EVALUATE: In round numbers, 6 m/s 4 s = 4 m 5 m, so he answer is reasonable. Δ.. IDENTIFY: v av- = Δ SET UP: 13.5 das = s. A he release poin, = m m EXECUTE: (a) v av- = = = 4.4 m/s. Δ s (b) For he round rip, = 1 and Δ = 0. The average veloci is zero. EVALUATE: The average veloci for he rip from he nes o he release poin is posiive..3. IDENTIFY: Targe variable is he ime Δ i akes o make he rip in heav raffic. Use Eq. (.) ha relaes he average veloci o he displacemen and average ime. Δ SET UP: v av- = so Δ = v Δ av-δ and Δ = Δ. vav- EXECUTE: Use he informaion given for normal driving condiions o calculae he disance beween he wo ciies, where he ime is 1 h and 50 min, which is 110 min: Δ = v av- Δ = (105 km/h)(1 h/60 min)(110 min) = 19.5 km. Now use v av- for heav raffic o calculae Δ; Δ is he same as before: Δ = Δ = 19.5 km =.75 h = h and 45 min. v av- 70 km/h The addiional ime is ( h and 45 min) (1 h and 50 min) = (1 h and 105 min) (1 h and 50 min) = 55 min. EVALUATE: A he normal speed of 105 km/s he rip akes 110 min, bu a he reduced speed of 70 km/h i akes 165 min. So decreasing our average speed b abou 30% adds 55 min o he ime, which is 50% of 110 min. Thus a 30% reducion in speed leads o a 50% increase in ravel ime. This resul (perhaps surprising) occurs because he ime inerval is inversel proporional o he average speed, no direcl proporional o i. Δ.4. IDENTIFY: The average veloci is v av- =. Use he average speed for each segmen o find he ime Δ raveled in ha segmen. The average speed is he disance raveled divided b he ime. SET UP: The pos is 80 m wes of he pillar. The oal disance raveled is 00 m + 80 m = 480 m. EXECUTE: (a) The easward run akes ime 00 m = 40.0 s and he wesward run akes 5.0 m/s 80 m 480 m = 70.0 s. The average speed for he enire rip is = 4.4 m/s. 4.0 m/s s Δ 80 m (b) v av- = = = 0.73 m/s. The average veloci is direced wesward. Δ s Coprigh 016 Pearson Educaion, Inc. All righs reserved. This maerial is proeced under all coprigh laws as he currenl eis. No porion of his maerial ma be reproduced, in an form or b an means, wihou permission in wriing from he publisher. -1

7 - Chaper Moion Along a Sraigh Line - EVALUATE: The displacemen is much less han he disance raveled, and he magniude of he average veloci is much less han he average speed. The average speed for he enire rip has a value ha lies beween he average speed for he wo segmens..5. IDENTIFY: Given wo displacemens, we wan he average veloci and he average speed. Δ SET UP: The average veloci is v av- = and he average speed is jus he oal disance walked divided Δ b he oal ime o walk his disance. EXECUTE: (a) Le + be eas. Δ = 60.0 m 40.0 m = 0.0 m and Δ = 8.0 s s = 64.0 s. So v av- = Δ = 0.0 m = 0.31 m/s. Δ 64.0 s (b) average speed = 60.0 m m = 1.56 m/s 64.0 s EVALUATE: The average speed is much greaer han he average veloci because he oal disance walked is much greaer han he magniude of he displacemen vecor. Δ.6. IDENTIFY: The average veloci is v av- =. Use ( ) o find for each. Δ SET UP: (0) = 0, (.00 s) = 5.60 m, and (4.00 s) = 0.8 m 5.60 m 0 EXECUTE: (a) v av- = = +.80 m/s.00 s (b) v av- = 0.8 m 0 = +5.0 m/s 4.00 s (c) v av- = 0.8 m 5.60 m = m/s.00 s EVALUATE: The average veloci depends on he ime inerval being considered. Δ.7. (a) IDENTIFY: Calculae he average veloci using v av- =. Δ Δ SET UP: v av- = so use ( ) o find he displacemen Δ for his ime inerval. Δ EXECUTE: = 10.0 s: = 0 : = 0 = (.40 m/s )(10.0 s) (0.10 m/s 3 )(10.0 s) 3 = 40 m 10 m = 10 m. Δ 10 m Then v av- = = = 1.0 m/s. Δ 10.0 s d (b) IDENTIFY: Use v = o calculae v () and evaluae his epression a each specified. d SET UP: v = d = b 3c. d EXECUTE: (i) = 0 : v = 0 (ii) = 5.0 s: v = (.40 m/s )(5.0 s) 3(0.10 m/s 3 )(5.0 s) = 4.0 m/s 9.0 m/s = 15.0 m/s. (iii) = 10.0 s: v = (.40 m/s )(10.0 s) 3(0.10 m/s 3 )(10.0 s) = 48.0 m/s 36.0 m/s = 1.0 m/s. v = 0 ne when b 3c = 0 (c) IDENTIFY: Find he value of when v ( ) from par (b) is zero. SET UP: v = b 3c v = 0 a = 0. b (.40 m/s ) EXECUTE: b = 3c so = = = 13.3 s 3c 3(0.10 m/s 3 ) EVALUATE: v ( ) for his moion sas he car sars from res, speeds up, and hen slows down again. Coprigh 016 Coprigh Pearson 016 Educaion, Pearson Inc. Educaion, All righs Inc. reserved. All righs This reserved. maerial This proeced maerial under is proeced all coprigh under laws all coprigh as he currenl laws as he eis. currenl eis. No porion of No his porion maerial of ma his be maerial reproduced, ma be in reproduced, an form or in b an an form means, or b wihou an means, permission wihou in permission wriing from in he wriing publisher. from he publisher.

8 -3 Chaper Moion Along a Sraigh Line IDENTIFY: We know he posiion () of he bird as a funcion of ime and wan o find is insananeous veloci a a paricular ime. SET UP: The insananeous veloci is v () = d d = 8.0 m + (1.4 m/s) ( m/s 3 ) 3. d d EXECUTE: v () = d = 1.4 m/s (0.135 m/s 3 ). Evaluaing his a = 8.0 s gives v = 3.76 m/s. d EVALUATE: The acceleraion is no consan in his case. Δ.9. IDENTIFY: The average veloci is given b v av- =. We can find he displacemen Δ for each Δ consan veloci ime inerval. The average speed is he disance raveled divided b he ime. SET UP: For = 0 o =.0 s, v =.0 m/s. For =.0 s o = 3.0 s, v = 3.0 m/s. In par (b), v = 3.0 m/s for =.0 s o = 3.0 s. When he veloci is consan, Δ = v Δ. EXECUTE: (a) For = 0 o =.0 s, Δ = (.0 m/s)(.0 s) = 4.0 m. For =.0 s o = 3.0 s, Δ = (3.0 m/s)(1.0 s) = 3.0 m. For he firs 3.0 s, Δ = 4.0 m m = 7.0 m. The disance raveled is Δ 7.0 m also 7.0 m. The average veloci is v av- = = =.33 m/s. The average speed is also.33 m/s. Δ 3.0 s (b) For =.0 s o 3.0 s, Δ = ( 3.0 m/s)(1.0 s) = 3.0 m. For he firs 3.0 s, Δ = 4.0 m + ( 3.0 m) = +1.0 m. The ball ravels 4.0 m in he +-direcion and hen 3.0 m in he Δ 1.0 m -direcion, so he disance raveled is sill 7.0 m. v av- = = = 0.33 m/s. The average speed is Δ 3.0 s 7.00 m =.33 m/s s EVALUATE: When he moion is alwas in he same direcion, he displacemen and he disance raveled are equal and he average veloci has he same magniude as he average speed. When he moion changes direcion during he ime inerval, hose quaniies are differen..10. IDENTIFY and SET UP: The insananeous veloci is he slope of he angen o he versus graph. EXECUTE: (a) The veloci is zero where he graph is horizonal; poin IV. (b) The veloci is consan and posiive where he graph is a sraigh line wih posiive slope; poin I. (c) The veloci is consan and negaive where he graph is a sraigh line wih negaive slope; poin V. (d) The slope is posiive and increasing a poin II. (e) The slope is posiive and decreasing a poin III. EVALUATE: The sign of he veloci indicaes is direcion..11. IDENTIFY: Find he insananeous veloci of a car using a graph of is posiion as a funcion of ime. SET UP: The insananeous veloci a an poin is he slope of he versus graph a ha poin. Esimae he slope from he graph. EXECUTE: A: v = 6.7 m/s; B: v = 6.7 m/s; C: v = 0; D: v = 40.0 m/s; E: v = 40.0 m/s; F: v = 40.0 m/s; G: v = 0. EVALUATE: The sign of v shows he direcion he car is moving. v is consan when versus is a sraigh line..1. IDENTIFY: a av- = Δv Δ. a () is he slope of he v versus graph. SET UP: 60 km/h = 16.7 m/s 16.7 m/s m/s EXECUTE: (a) (i) a av- = = 1.7 m/s. (ii) a av- = = 1.7 m/s. 10 s 10 s (iii) Δv = 0 and a av- = 0. (iv) Δv = 0 and a av- = 0. (b) A = 0 s, v is consan and a = 0. A = 35 s, he graph of v versus is a sraigh line and a = a av- = 1.7 m/s. Coprigh 016 Coprigh Pearson 016 Educaion, Pearson Inc. Educaion, All righs Inc. reserved. All righs This reserved. maerial This proeced maerial under is proeced all coprigh under laws all coprigh as he currenl laws as he eis. currenl eis. No porion of No his porion maerial of ma his be maerial reproduced, ma be in reproduced, an form or in b an an form means, or b wihou an means, permission wihou in permission wriing from in he wriing publisher. from he publisher.

9 -4 Chaper Moion Along a Sraigh Line -4 EVALUATE: When a av- and v have he same sign he speed is increasing. When he have opposie signs, he speed is decreasing..13. IDENTIFY: The average acceleraion for a ime inerval Δ is given b a av- = Δv. Δ SET UP: Assume he car is moving in he + direcion. 1 mi/h = m/s, so 60 mi/h = 6.8 m/s, 00 mi/h = m/s and 53 mi/h = m/s. EXECUTE: (a) The graph of v versus is skeched in Figure.13. The graph is no a sraigh line, so he acceleraion is no consan. (b) (i) a = 6.8 m/s 0 = 1.8 m/s (ii) a = m/s 6.8 m/s = 3.50 m/s av- av-.1 s 0.0 s.1 s m/s m/s (iii) a av- = = m/s. The slope of he graph of v versus decreases as 53 s 0.0 s increases. This is consisen wih an average acceleraion ha decreases in magniude during each successive ime inerval. EVALUATE: The average acceleraion depends on he chosen ime inerval. For he inerval beween 0 and 53 s, a av- = m/s 0 =.13 m/s. 53 s Figure IDENTIFY: We know he veloci v() of he car as a funcion of ime and wan o find is acceleraion a he insan ha is veloci is 1.0 m/s. SET UP: We know ha v () = (0.860 m/s 3 ) and ha a ( ) = dv = d (0.860 m/s3 ) EXECUTE: a () = dv = (1.7 m/s 3 ). When v = 1.0 m/s, (0.860 m/s 3 ) = 1.0 m/s, which gives d = s. A his ime, a = 6.4 m/s. EVALUATE: The acceleraion of his car is no consan. d dv.15. IDENTIFY and SET UP: Use v = and a = o calculae v () and a (). d d EXECUTE: v = d =.00 cm/s (0.15 cm/s ) d a = dv = 0.15 cm/s d (a) A = 0, = 50.0 cm, v =.00 cm/s, a = 0.15 cm/s. (b) Se v = 0 and solve for : = 16.0 s. (c) Se = 50.0 cm and solve for. This gives = 0 and = 3.0 s. The urle reurns o he saring poin afer 3.0 s. (d) The urle is 10.0 cm from saring poin when = 60.0 cm or = 40.0 cm. d d. Coprigh 016 Coprigh Pearson 016 Educaion, Pearson Inc. Educaion, All righs Inc. reserved. All righs This reserved. maerial This proeced maerial under is proeced all coprigh under laws all coprigh as he currenl laws as he eis. currenl eis. No porion of No his porion maerial of ma his be maerial reproduced, ma be in reproduced, an form or in b an an form means, or b wihou an means, permission wihou in permission wriing from in he wriing publisher. from he publisher.

10 -5 Chaper Moion Along a Sraigh Line -5 Se = 60.0 cm and solve for : = 6.0 s and = 5.8 s. A = 6.0 s, v = +1.3 cm/s. A = 5.8 s, v = 1.3 cm/s. Se = 40.0 cm and solve for : = 36.4 s (oher roo o he quadraic equaion is negaive and hence nonphsical). A = 36.4 s, v =.55 cm/s. (e) The graphs are skeched in Figure.15. Figure.15 EVALUATE: The acceleraion is consan and negaive. v is linear in ime. I is iniiall posiive, decreases o zero, and hen becomes negaive wih increasing magniude. The urle iniiall moves farher awa from he origin bu hen sops and moves in he -direcion. Δv.16. IDENTIFY: Use aav- = Δ, wih Δ = 10 s in all cases. SET UP: v is negaive if he moion is o he lef. EXECUTE: (a) [(5.0 m/s) (15.0 m/s)]/(10 s) = 1.0 m/s (b) [( 15.0 m/s) ( 5.0 m/s)]/(10 s) = 1.0 m/s (c) [( 15.0 m/s) (+15.0 m/s)]/(10 s) = 3.0 m/s EVALUATE: In all cases, he negaive acceleraion indicaes an acceleraion o he lef. Δv.17. IDENTIFY: The average acceleraion is aav- =. Use v () o find v a each. The insananeous Δ dv acceleraion is a =. d SET UP: v (0) = 3.00 m/s and v (5.00 s) = 5.50 m/s. EXECUTE: (a) a = Δv = 5.50 m/s 3.00 m/s = m/s av- Δ 5.00 s (b) a = dv = (0.100 m/s 3 )() = (0.00 m/s 3 ). A = 0, a = 0. A = 5.00 s, a = 1.00 m/s. d (c) Graphs of v () and a () are given in Figure.17 (ne page). EVALUATE: a ( ) is he slope of v ( ) and increases as increases. The average acceleraion for = 0 o = 5.00 s equals he insananeous acceleraion a he midpoin of he ime inerval, =.50 s, since a ( ) is a linear funcion of. Coprigh 016 Coprigh Pearson 016 Educaion, Pearson Inc. Educaion, All righs Inc. reserved. All righs This reserved. maerial This proeced maerial under is proeced all coprigh under laws all coprigh as he currenl laws as he eis. currenl eis. No porion of No his porion maerial of ma his be maerial reproduced, ma be in reproduced, an form or in b an an form means, or b wihou an means, permission wihou in permission wriing from in he wriing publisher. from he publisher.

11 -6 Chaper Moion Along a Sraigh Line -6 Figure IDENTIFY: v () = d and a ( ) = d d SET UP: ( n ) = n n 1 for n 1. d dv d EXECUTE: (a) v () = (9.60 m/s ) (0.600 m/s 6 ) 5 and a () = 9.60 m/s (3.00 m/s 6 ) 4. Seing v = 0 gives = 0 and =.00 s. A = 0, =.17 m and a = 9.60 m/s. A =.00 s, = 15.0 m and a = 38.4 m/s. (b) The graphs are given in Figure.18. EVALUATE: For he enire ime inerval from = 0 o =.00 s, he veloci v is posiive and increases. While a is also posiive he speed increases and while a is negaive he speed decreases. Figure IDENTIFY: Use he consan acceleraion equaions o find v 0 and a. (a) SET UP: The siuaion is skeched in Figure.19. Figure.19 0 = 70.0 m = 6.00 s v = 15.0 m/s v 0 =? Coprigh 016 Coprigh Pearson 016 Educaion, Pearson Inc. Educaion, All righs Inc. reserved. All righs This reserved. maerial This proeced maerial under is proeced all coprigh under laws all coprigh as he currenl laws as he eis. currenl eis. No porion of No his porion maerial of ma his be maerial reproduced, ma be in reproduced, an form or in b an an form means, or b wihou an means, permission wihou in permission wriing from in he wriing publisher. from he publisher.

12 -7 Chaper Moion Along a Sraigh Line -7 EXECUTE: Use = v + v 0 ( ), so v 0 (70.0 m) = v = 15.0 m/s = 8.33 m/s s v v m/s 5.0 m/s (b) Use v = v 0 + a, so a = = = 1.11 m/s s EVALUATE: The average veloci is (70.0 m)/(6.00 s) = 11.7 m/s. The final veloci is larger han his, so he anelope mus be speeding up during he ime inerval; v 0 < v and a > IDENTIFY: In (a) find he ime o reach he speed of sound wih an acceleraion of 5g, and in (b) find his speed a he end of 5.0 s if he has an acceleraion of 5g. SET UP: Le + be in his direcion of moion and assume consan acceleraion of 5g so he sandard kinemaics equaions appl so v = v 0 + a. (a) v = 3(331 m/s) = 993 m/s, v 0 = 0, and a = 5g = 49.0 m/s. (b) = 5.0 s v v m/s 0 EXECUTE: (a) v = v 0 + a and = = = 0.3 s. Yes, he ime required is larger a 49.0 m/s han 5.0 s. (b) v = v + a = 0 + (49.0 m/s 0 )(5.0 s) = 45 m/s. EVALUATE: In 5.0 s he can onl reach abou /3 he speed of sound wihou blacking ou..1. IDENTIFY: For consan acceleraion, he sandard kinemaics equaions appl. SET UP: Assume he ball sars from res and moves in he + -direcion. EXECUTE: (a) = 1.50 m, v = 45.0 m/s and v = 0. v = v + a ( ) gives v v 0 (45.0 m/s) a = = = 675 m/s. ( 0 ) (1.50 m) (b) = v + v 0 ( gives = 0 ) (1.50 m) 0 = = s v 0 + v 45.0 m/s v 45.0 m/s EVALUATE: We could also use v = v 0 + a o find = = = s which agrees wih a 675 m/s our previous resul. The acceleraion of he ball is ver large... IDENTIFY: For consan acceleraion, he sandard kinemaics equaions appl. SET UP: Assume he ball moves in he + direcion. EXECUTE: (a) v = m/s, v 0 = 0 and = 30.0 ms. v = v 0 + a gives a = v v 0 = m/s 0 = 440 m/s s (b) = v + v m/s = ( s) = 1.10 m. EVALUATE: We could also use = v + 1 a o calculae : = (440 m/s )( s) = 1.10 m, which agrees wih our previous resul. The acceleraion of he ball is ver large..3. IDENTIFY: Assume ha he acceleraion is consan and appl he consan acceleraion kinemaic equaions. Se a equal o is maimum allowed value. SET UP: Le + be he direcion of he iniial veloci of he car. a = 50 m/s. 105 km/h = 9.17 m/s. EXECUTE: v 0 = 9.17 m/s. v = 0. v = v 0 + a ( 0 ) gives v v 0 (9.17 m/s) 0 0 = = = 1.70 m. a ( 50 m/s ) EVALUATE: The car frame sops over a shorer disance and has a larger magniude of acceleraion. Par Coprigh 016 Coprigh Pearson 016 Educaion, Pearson Inc. Educaion, All righs Inc. reserved. All righs This reserved. maerial This proeced maerial under is proeced all coprigh under laws all coprigh as he currenl laws as he eis. currenl eis. No porion of No his porion maerial of ma his be maerial reproduced, ma be in reproduced, an form or in b an an form means, or b wihou an means, permission wihou in permission wriing from in he wriing publisher. from he publisher.

13 -8 Chaper Moion Along a Sraigh Line -8 of our 1.70 m sopping disance is he sopping disance of he car and par is how far ou move relaive o he car while sopping. Coprigh 016 Coprigh Pearson 016 Educaion, Pearson Inc. Educaion, All righs Inc. reserved. All righs This reserved. maerial This proeced maerial under is proeced all coprigh under laws all coprigh as he currenl laws as he eis. currenl eis. No porion of No his porion maerial of ma his be maerial reproduced, ma be in reproduced, an form or in b an an form means, or b wihou an means, permission wihou in permission wriing from in he wriing publisher. from he publisher.

14 -9 Chaper Moion Along a Sraigh Line IDENTIFY: In (a) we wan he ime o reach Mach 4 wih an acceleraion of 4g, and in (b) we wan o know how far he can ravel if he mainains his acceleraion during his ime. SET UP: Le + be he direcion he je ravels and ake 0 = 0. Wih consan acceleraion, he equaions v = v 0 + a v 0 = and = + v + 1 a boh appl. a = 4g = 39. m/s, v = 4(331 m/s) = 134 m/s, and 0 0 EXECUTE: (a) Solving v = v 0 + a for gives = v v 0 a EXECUTE: Solving v = v 0 + a for v 0 gives v 0 = a. Then = 0 + v a gives m/s 0 = = 33.8 s. 39. m/s (b) = 0 + v 0 + a = (39. m/s )(33.8 s) =.4 10 m =.4 km. EVALUATE: The answer in (a) is abou ½ min, so if he waned o reach Mach 4 an sooner han ha, he would be in danger of blacking ou..5. IDENTIFY: If a person comes o a sop in 36 ms while slowing down wih an acceleraion of 60g, how far does he ravel during his ime? SET UP: Le + be he direcion he person ravels. v = 0 (he sops), a is negaive since i is opposie o he direcion of he moion, and = 36 ms = s. The equaions v = v + a and = + v + 1 a boh appl since he acceleraion is consan. 1 1 = a = ( 588 m/s )( s) = 38 cm. 0 EVALUATE: Noice ha we were no given he iniial speed, bu we could find i: 3 v 0 = a = ( 588 m/s )(36 10 s) = 1 m/s = 47 mph..6. IDENTIFY: In (a) he hip pad mus reduce he person s speed from.0 m/s o 1.3 m/s over a disance of.0 cm, and we wan he acceleraion over his disance, assuming consan acceleraion. In (b) we wan o find ou how long he acceleraion in (a) lass. SET UP: Le + be downward. v 0 =.0 m/s, v = 1.3 m/s, and 0 = 0.00 m. The equaions v 0 + v v = v 0 + a ( 0 ) and 0 = appl for consan acceleraion. EXECUTE: (a) Solving v = v + a ( ) for a gives 0 0 v v (1.3 m/s) (.0 m/s) a = 0 = = 58 m/s = 5.9g. ( 0 ) (0.00 m) v (b) 0 + v ( ) (0.00 m) 0 = gives = = = 1 ms. 0 v + v.0 m/s m/s 0 EVALUATE: The acceleraion is ver large, bu i onl lass for 1 ms so i produces a small veloci change..7. IDENTIFY: We know he iniial and final velociies of he objec, and he disance over which he veloci change occurs. From his we wan o find he magniude and duraion of he acceleraion of he objec. SET UP: The consan-acceleraion kinemaics formulas appl. v = v + a ( ), where v = 0, v = m/s, and = 4.0 m. 0 0 v v ( m/s) EXECUTE: (a) v = v 0 + a ( ) gives a = = = m/s = g. ( 0 ) (4.0 m) v v m/s 0 (b) v = v 0 + a gives = = = 1.6 ms. a m/s EVALUATE: (c) The calculaed a is less han 450,000 g so he acceleraion required doesn rule ou his hpohesis. Coprigh 016 Coprigh Pearson 016 Educaion, Pearson Inc. Educaion, All righs Inc. reserved. All righs This reserved. maerial This proeced maerial under is proeced all coprigh under laws all coprigh as he currenl laws as he eis. currenl eis. No porion of No his porion maerial of ma his be maerial reproduced, ma be in reproduced, an form or in b an an form means, or b wihou an means, permission wihou in permission wriing from in he wriing publisher. from he publisher.

15 Chaper Moion Along a Sraigh Line IDENTIFY: Appl consan acceleraion equaions o he moion of he car. -10 SET UP: Le + be he direcion he car is moving. v (0 m/s) EXECUTE: (a) From v = v + a ( ), wih v = 0, a = = = 1.67 m/s ( 0 ) (10 m) Coprigh 016 Coprigh Pearson 016 Educaion, Pearson Inc. Educaion, All righs Inc. reserved. All righs This reserved. maerial This proeced maerial under is proeced all coprigh under laws all coprigh as he currenl laws as he eis. currenl eis. No porion of No his porion maerial of ma his be maerial reproduced, ma be in reproduced, an form or in b an an form means, or b wihou an means, permission wihou in permission wriing from in he wriing publisher. from he publisher.

16 -11 Chaper Moion Along a Sraigh Line -11 (b) Using Eq. (.14), = ( 0 )/v = (10 m)/(0 m/s) = 1 s. (c) (1 s)(0 m/s) = 40 m. EVALUATE: The average veloci of he car is half he consan speed of he raffic, so he raffic ravels wice as far. Δv.9. IDENTIFY: The average acceleraion is a av- =. For consan acceleraion, he sandard kinemaics Δ equaions appl. SET UP: Assume he rocke ship ravels in he + direcion. 161 km/h = 44.7 m/s and 1610 km/h = 447. m/s min = 60.0 s EXECUTE: (a) (i) a = Δv = 44.7 m/s 0 = 5.59 m/s av- Δ 8.00 s (ii) a av- = 447.m/s 44.7 m/s = 7.74 m/s 60.0 s 8.00 s v 0 + v m/s (b) (i) = 8.00 s, v 0 = 0, and v = 44.7 m/s. 0 = = (8.00 s) = 179 m. (ii) Δ = 60.0 s 8.00 s = 5.0 s, v 0 = 44.7 m/s, and v = 447. m/s. = v + v m/s m/s = (5.0 s) = m. EVALUATE: When he acceleraion is consan he insananeous acceleraion hroughou he ime inerval equals he average acceleraion for ha ime inerval. We could have calculaed he disance in par (a) as = v 0 + a = (5.59 m/s )(8.00 s) = 179 m, which agrees wih our previous calculaion..30. IDENTIFY: The acceleraion a is he slope of he graph of v versus. SET UP: The signs of v and of a indicae heir direcions. EXECUTE: (a) Reading from he graph, a = 4.0 s, v =.7 cm/s, o he righ and a = 7.0 s, v = 1.3 cm/s, o he lef. 8.0cm/s (b) v versus is a sraigh line wih slope = 1.3 cm/s. The acceleraion is consan and 6.0 s equal o 1.3 cm/s, o he lef. I has his value a all imes. (c) Since he acceleraion is consan, = v + 1 a 0 0. For = 0 o 4.5 s, 1 0 = (8.0 cm/s)(4.5 s) + ( 1.3 cm/s )(4.5 s) =.8 cm. For = 0 o 7.5 s, 1 0 = (8.0 cm/s)(7.5 s) + ( 1.3 cm/s )(7.5 s) = 3.4 cm (d) The graphs of a and versus are given in Figure.30. EVALUATE: In par (c) we could have insead used = v + v 0 0. Figure.30

17 -1 Chaper Moion Along a Sraigh Line (a) IDENTIFY and SET UP: The acceleraion a curve a ime. a ime is he slope of he angen o he v versus EXECUTE: A = 3 s, he v versus curve is a horizonal sraigh line, wih zero slope. Thus a = 0. A = 7 s, he v versus curve is a sraigh-line segmen wih slope Thus a = 6.3 m/s. A = 11 s he curve is again a sraigh-line segmen, now wih slope Thus a = 11. m/s m/s 0 m/s 9 s 5 s = 6.3 m/s m/s = 11. m/s. 13 s 9 s EVALUATE: a = 0 when v is consan, a > 0 when v is posiive and he speed is increasing, and a < 0 when v is posiive and he speed is decreasing. (b) IDENTIFY: Calculae he displacemen during he specified ime inerval. SET UP: We can use he consan acceleraion equaions onl for ime inervals during which he acceleraion is consan. If necessar, break he moion up ino consan acceleraion segmens and appl he consan acceleraion equaions for each segmen. For he ime inerval = 0 o = 5 s he acceleraion is consan and equal o zero. For he ime inerval = 5 s o = 9 s he acceleraion is consan and equal o 6.5 m/s. For he inerval = 9 s o = 13 s he acceleraion is consan and equal o 11. m/s. EXECUTE: During he firs 5 seconds he acceleraion is consan, so he consan acceleraion kinemaic formulas can be used. v 0 = 0 m/s a = 0 = 5 s 0 =? 1 0 = v 0 (a = 0 so no a erm) 0 = (0 m/s)(5 s) = 100 m; his is he disance he officer ravels in he firs 5 seconds. During he inerval = 5 s o 9 s he acceleraion is again consan. The consan acceleraion formulas can be applied o his 4-second inerval. I is convenien o resar our clock so he inerval sars a ime = 0 and ends a ime = 4 s. (Noe ha he acceleraion is no consan over he enire = 0 o = 9 s inerval.) v 0 = 0 m/s a = 6.5 m/s = 4 s 0 = 100 m 0 =? = v + 1 a 1 0 = (0 m/s)(4 s) + (6.5 m/s )(4 s) = 80 m + 50 m = 130 m. Thus m = 100 m m = 30 m. A = 9 s he officer is a = 30 m, so she has raveled 30 m in he firs 9 seconds. During he inerval = 9 s o = 13 s he acceleraion is again consan. The consan acceleraion formulas can be applied for his 4-second inerval bu no for he whole = 0 o = 13 s inerval. To use he equaions resar our clock so his inerval begins a ime = 0 and ends a ime = 4 s. v 0 = 45 m/s (a he sar of his ime inerval) a = 11. m/s = 4 s 0 = 30 m 0 =? = v + 1 a = (45 m/s)(4 s) + ( 11. m/s )(4 s) = 180 m 89.6 m = 90.4 m. Thus = m = 30 m m = 30 m. A = 13 s he officer is a = 30 m, so she has raveled 30 m in he firs 13 seconds. EVALUATE: The veloci v is alwas posiive so he displacemen is alwas posiive and displacemen and disance raveled are he same. The average veloci for ime inerval Δ is v av- = Δ/Δ. For = 0 o 5 s, v av- = 0 m/s. For = 0 o 9 s, v av- = 6 m/s. For = 0 o 13 s, v av- = 5 m/s. These resuls are consisen wih he figure in he ebook..3. IDENTIFY: v ( ) is he slope of he versus graph. Car B moves wih consan speed and zero acceleraion. Car A moves wih posiive acceleraion; assume he acceleraion is consan.

18 -13 Chaper Moion Along a Sraigh Line -13 SET UP: For car B, v is posiive and a = 0. For car A, a is posiive and v increases wih. EXECUTE: (a) The moion diagrams for he cars are given in Figure.3a. (b) The wo cars have he same posiion a imes when heir - graphs cross. The figure in he problem shows his occurs a approimael = 1 s and = 3 s. (c) The graphs of v versus for each car are skeched in Figure.3b. (d) The cars have he same veloci when heir - graphs have he same slope. This occurs a approimael = s. (e) Car A passes car B when A moves above B in he - graph. This happens a = 3 s. (f) Car B passes car A when B moves above A in he - graph. This happens a = 1 s. EVALUATE: When a = 0, he graph of v versus is a horizonal line. When a v versus is a sraigh line wih posiive slope. is posiive, he graph of Figure IDENTIFY: For consan acceleraion, he kinemaics formulas appl. We can use he oal displacemen ande final veloci o calculae he acceleraion and hen use he acceleraion and shorer disance o find he speed. SET UP: Take + o be down he incline, so he moion is in he + direcion. The formula v = v + a( ) applies. 0 0 EXECUTE: Firs look a he moion over 6.80 m. We use he following numbers: v 0 = 0, 0 = 6.80 m, and v = 3.80 /s. Solving he above equaion for a gives a = 1.06 m/s. Now look a he moion over he 3.40 m using v 0 = 0, a = 1.06 m/s and 0 = 3.40 m. Solving he same equaion, bu his ime for v, gives v =.69 m/s. EVALUATE: Even hough he block has raveled half wa down he incline, is speed is no half of is speed a he boom..34. IDENTIFY: Appl he consan acceleraion equaions o he moion of each vehicle. The ruck passes he car when he are a he same a he same > 0. SET UP: The ruck has a = 0. The car has v 0 = 0. Le + be in he direcion of moion of he vehicles. Boh vehicles sar a = 0. The car has a =.80 m/s. The ruck has v = 0.0 m/s. 0 C EXECUTE: (a) = v + 1 a gives = v and = 1 a 0 0. Seing = gives = 0 and T 0T C C T C v 0T = 1 a C, so = v 0T (0.0 m/s) = = 14.9 s. A his, a C.80 m/s T = (0.0 m/s)(14.9 s) = 86 m and 1 = (3.0 m/s )(14.9 s) = 86 m. The car and ruck have each raveled 86 m. (b) A = 14.9 s, he car has v = v + a = (.80 m/s )(14.9 s) = 40 m/s. 0 1 (c) T = v 0T and C = a C. The - graph of he moion for each vehicle is skeched in Figure.34a. (d) v T = v 0T. v C = a C. The v - graph for each vehicle is skeched in Figure.34b (ne page). EVALUATE: When he car overakes he ruck is speed is wice ha of he ruck.

19 -14 Chaper Moion Along a Sraigh Line -14 Figure IDENTIFY: Appl he consan acceleraion equaions o he moion of he flea. Afer he flea leaves he ground, a = g, downward. Take he origin a he ground and he posiive direcion o be upward. (a) SET UP: A he maimum heigh v = 0. v = 0 0 = m a = 9.80 m/s v 0 =? v = v 0 + a ( 0 ) EXECUTE: v = a ( ) = ( 9.80 m/s )(0.440 m) =.94 m/s 0 0 (b) SET UP: When he flea has reurned o he ground 0 = 0 v 0 = +.94 m/s a = 9.80 m/s =? 0 = v a v 0 (.94 m/s) 0 = 0. EXECUTE: Wih 0 = 0 his gives = a = = s m/s EVALUATE: We can use v = v 0 + a o show ha wih v 0 =.94 m/s, v = 0 afer s..36. IDENTIFY: The rock has a consan downward acceleraion of 9.80 m/s. We know is iniial veloci and posiion and is final posiion. SET UP: We can use he kinemaics formulas for consan acceleraion. EXECUTE: (a) 0 = 30 m, v 0 =.0 m/s, a = 9.80 m/s. The kinemaics formulas give v = v 0 + a ( 0 ) = (.0 m/s) + ( 9.80 m/s )( 30 m) = 3.74 m/s, so he speed is 3.7 m/s. (b) v = v 0 + a and = v v 0 a = 3.74 m/s.0 m/s 9.80 m/s = 5.59 s. EVALUATE: The verical veloci in par (a) is negaive because he rock is moving downward, bu he speed is alwas posiive. The 5.59 s is he oal ime in he air..37. IDENTIFY: The pin has a consan downward acceleraion of 9.80 m/s and reurns o is iniial posiion. SET UP: We can use he kinemaics formulas for consan acceleraion. EXECUTE: The kinemaics formulas give = v + 1 a 0 0. We know ha 0 = 0, so v 0 (8.0 m/s) = = = s. a 9.80 m/s EVALUATE: I akes he pin half his ime o reach is highes poin and he remainder of he ime o reurn..38. IDENTIFY: The pu has a consan downward acceleraion of 9.80 m/s. We know he iniial veloci of

20 -15 Chaper Moion Along a Sraigh Line -15 he pu and he disance i ravels.

21 -16 Chaper Moion Along a Sraigh Line -16 SET UP: We can use he kinemaics formulas for consan acceleraion. EXECUTE: (a) v 0 = 9.50 m/s and 0 = 3.60 m, which gives v = v 0 + a ( 0 ) = (9.50 m/s) + ( 9.80 m/s )(3.60 m) = 4.44 m/s v v 4.44 m/s 9.50 m/s (b) = 0 = = s a 9.8 m/s 0 0 EVALUATE: The pu is sopped b he ceiling, no b gravi..39. IDENTIFY: A ball on Mars ha is hi direcl upward reurns o he same level in 8.5 s wih a consan downward acceleraion of 0.379g. How high did i go and how fas was i iniiall raveling upward? SET UP: Take + upward. v = 0 a he maimum heigh. a = 0.379g = 3.71 m/s. The consanacceleraion formulas v = v 0 + a and = + v + 1 a boh appl. 0 0 EXECUTE: Consider he moion from he maimum heigh back o he iniial level. For his moion 1 1 v 0 = 0 and = 4.5 s. = 0 + v 0 + a = ( 3.71 m/s )(4.5 s) = 33.5 m. The ball wen 33.5 m above is original posiion. (b) Consider he moion from jus afer i was hi o he maimum heigh. For his moion v = 0 and = 4.5 s. v = v + a gives v = a = ( 3.71 m/s )(4.5 s) = 15.8 m/s. (c) The graphs are skeched in Figure.39. Figure.39 EVALUATE: The answers can be checked several was. For eample, v = 0, v 0 = 15.8 m/s, and EXECUTE: v 0 = 0.8 m/s, 0 = 5.0 m, a = +1.6 m/s in v = v v a = 3.71 m/s in v = v + a ( ) gives v 0 0 (15.8 m/s) = = = 33.6 m, which a ( 3.71 m/s ) agrees wih he heigh calculaed in (a)..40. IDENTIFY: Appl consan acceleraion equaions o he moion of he lander. SET UP: Le + be downward. Since he lander is in free-fall, a = +1.6 m/s. v = v 0 + a ( 0 ) = (0.8 m/s) + (1.6 m/s )(5.0 m) = 4.1 m/s. + a ( ) gives EVALUATE: The same descen on earh would resul in a final speed of 9.9 m/s, since he acceleraion due o gravi on earh is much larger han on he moon..41. IDENTIFY: Appl consan acceleraion equaions o he moion of he meersick. The ime he meersick falls is our reacion ime. SET UP: Le + be downward. The meer sick has v = 0 and a = 9.80 m/s. Le d be he disance he meersick falls. EXECUTE: (a) = v + 1 a gives d = (4.90 m/s ) and = 0 0 d m/s m (b) = = s 4.90 m/s EVALUATE: The reacion ime is proporional o he square of he disance he sick falls.

22 -17 Chaper Moion Along a Sraigh Line IDENTIFY: Appl consan acceleraion equaions o he verical moion of he brick. SET UP: Le + be downward. a = 9.80 m/s EXECUTE: (a) v = 0, = 1.90 s, a = 9.80 m/s. = v + 1 a = 1 (9.80 m/s )(1.90 s) = 17.7 m. 0 The building is 17.7 m all. (b) v = v + a = 0 + (9.80 m/s 0 )(1.90 s) = 18.6 m/s 0 0 (c) The graphs of a, v and versus are given in Figure.4. Take = 0 a he ground. EVALUATE: We could use eiher v 0 + v = or v = v + a ( ) o check our resuls Figure IDENTIFY: When he onl force is gravi he acceleraion is 9.80 m/s, downward. There are wo inervals of consan acceleraion and he consan acceleraion equaions appl during each of hese inervals. SET UP: Le + be upward. Le = 0 a he launch pad. The final veloci for he firs phase of he moion is he iniial veloci for he free-fall phase. EXECUTE: (a) Find he veloci when he engines cu off. = 55 m, a =.5 m/s, v = 0. v = v a ( ) gives v = (.5 m/s )(55 m) = 48.6 m/s. 0 0 Now consider he moion from engine cu-off o maimum heigh: 0 = 55 m, v 0 = m/s, v = 0 (a he maimum heigh), a = 9.80 m/s. v = v + a ( ) gives 0 0 v v 0 0 (48.6 m/s) 0 = = = 11 m and = 11 m + 55 m = 646 m. a ( 9.80 m/s ) (b) Consider he moion from engine failure unil jus before he rocke srikes he ground: 0 = 55 m, a = 9.80 m/s, v 0 = m/s. v = v 0 + a ( 0 ) gives v = (48.6 m/s) + ( 9.80 m/s )( 55 m) = 11 m/s. Then v = v + a gives v v 11 m/s 48.6 m/s = 0 = = 16.4 s. a 9.80 m/s 0 (c) Find he ime from blas-off unil engine failure: 0 = 55 m, v 0 = 0, a = +.5 m/s. 1 ( 0 ) (55 m) 0 = v 0 + a gives = = = 1.6 s. The rocke srikes he launch pad a.5 m/s 1.6 s s = 38.0 s afer blas-off. The acceleraion a is +.5 m/s from = 0 o = 1.6 s. I is 9.80 m/s from = 1.6 s o 38.0 s. v = v + a applies during each consan acceleraion segmen, 0 so he graph of v versus is a sraigh line wih posiive slope of.5 m/s during he blas-off phase and wih negaive slope of 9.80 m/s afer engine failure. During each phase = v + 1 a. The 0 0

23 -18 Chaper Moion Along a Sraigh Line -18 sign of a deermines he curvaure of ( ). A = 38.0 s he rocke has reurned o = 0. The graphs are skeched in Figure.43.

24 -19 Chaper Moion Along a Sraigh Line -19 EVALUATE: In par (b) we could have found he ime from = v + 1 a, finding v firs allows us o avoid solving for from a quadraic equaion. 0 0 Figure IDENTIFY: Appl consan acceleraion equaions o he verical moion of he sandbag. SET UP: Take + upward. a = 9.80 m/s. The iniial veloci of he sandbag equals he veloci of ( he balloon, so v 0 = m/s. When he balloon reaches he ground, maimum heigh he sandbag has v = 0. 0 = 40.0 m. A is EXECUTE: (a) = 0.50 s: = v + 1 a = (5.00 m/s)(0.50 s) + 1 ( 9.80 m/s )(0.50 s) 0 0 = 0.94 m. The sandbag is 40.9 m above he ground. v = v + a = m/s + ( 9.80 m/s )(0.50 s) =.55 m/s. 0 = 1.00 s: = (5.00 m/s)(1.00 s) + 1 ( 9.80 m/s )(1.00 s) = 0.10 m. The sandbag is 40.1 m above 0 he ground. v = v + a = m/s + ( 9.80 m/s )(1.00 s) = 4.80 m/s. 0 1 (b) 0 = 40.0 m, v 0 = 5.00 m/s, a = 9.80 m/s. 0 = v 0 + a gives 40.0 m = (5.00 m/s) (4.90 m/s ). (4.90 m/s ) (5.00 m/s) 40.0 m = 0 and 1 = 5.00 ± ( 5.00) 4(4.90)( 40.0) 9.80 ) s = (0.51 ±.90) s. mus be posiive, so = 3.41 s. (c) v = v + a = m/s + ( 9.80 m/s 0 )(3.41 s) = 8.4 m/s (d) v 0 = 5.00 m/s, a = 9.80 m/s, v = 0. v = v 0 + a ( 0 ) gives v v 0 0 (5.00 m/s) 0 = = = 1.8 m. The maimum heigh is 41.3 m above he ground. a ( 9.80 m/s ) (e) The graphs of a, v, and versus are given in Figure.44. Take = 0 a he ground. EVALUATE: The sandbag iniiall ravels upward wih decreasing veloci and hen moves downward wih increasing speed. Figure.44

25 -0 Chaper Moion Along a Sraigh Line IDENTIFY: Use he consan acceleraion equaions o calculae a and 0. (a) SET UP: v = v 0 + a v = 4 m/s, v 0 = 0, = s, a =? EXECUTE: a = v v 0 = 4 m/s 0 = 49 m/s s (b) a /g = (49 m/s )/(9.80 m/s ) = (c) 0 = v 0 + a = 0 + (49 m/s )(0.900 s) = 101 m (d) SET UP: Calculae he acceleraion, assuming i is consan: = 1.40 s, v 0 = 83 m/s, v = 0 (sops), a =? v = v 0 + a EXECUTE: a = v v 0 = 0 83 m/s = 0 m/s 1.40 s a /g = ( 0 m/s )/(9.80 m/s ) = 0.6; a = 0.6g If he acceleraion while he sled is sopping is consan hen he magniude of he acceleraion is onl 0.6g. Bu if he acceleraion is no consan i is cerainl possible ha a some poin he insananeous acceleraion could be as large as 40g. EVALUATE: I is reasonable ha for his moion he acceleraion is much larger han g..46. IDENTIFY: Since air resisance is ignored, he egg is in free-fall and has a consan downward acceleraion of magniude 9.80 m/s. Appl he consan acceleraion equaions o he moion of he egg. SET UP: Take + o be upward. A he maimum heigh, v = 0. 1 EXECUTE: (a) 0 = 30.0 m, = 5.00 s, a = 9.80 m/s. 0 = v 0 + a gives v = 0 1 a = 30.0 m 1 ( 9.80 m/s 0 )(5.00 s) = m/s s (b) v 0 = m/s, v = 0 (a he maimum heigh), a = 9.80 m/s. v = v 0 + a ( 0 ) gives v v 0 0 (18.5 m/s) 0 = = = 17.5 m. a ( 9.80 m/s ) (c) A he maimum heigh v = 0. (d) The acceleraion is consan and equal o 9.80 m/s, downward, a all poins in he moion, including a he maimum heigh. (e) The graphs are skeched in Figure.46. v v 18.5 m/s EVALUATE: The ime for he egg o reach is maimum heigh is = 0 = = 1.89 s. The a 9.8 m/s egg has reurned o he level of he cornice afer 3.78 s and afer 5.00 s i has raveled downward from he cornice for 1. s. Figure.46

26 -1 Chaper Moion Along a Sraigh Line IDENTIFY: We can avoid solving for he common heigh b considering he relaion beween heigh, ime of fall, and acceleraion due o gravi, and seing up a raio involving ime of fall and acceleraion due o gravi. SET UP: Le g En be he acceleraion due o gravi on Enceladus and le g be his quani on earh. Le h be he common heigh from which he objec is dropped. Le + be downward, so 0 = h. v 0 = 0 EXECUTE: g = g E En En = v + 1 a gives h = 1 g and h = 1 g. Combining hese wo equaions gives 0 0 E En En and g 0 0 v 0 = = (40.0 m/s) = s. a 9.80 m/s E = g = (9.80 m/s ) 1.75 s = m/s. En En 18.6 s EVALUATE: The acceleraion due o gravi is inversel proporional o he square of he ime of fall..48. IDENTIFY: Since air resisance is ignored, he boulder is in free-fall and has a consan downward acceleraion of magniude 9.80 m/s. Appl he consan acceleraion equaions o he moion of he boulder. SET UP: Take + o be upward. EXECUTE: (a) v = m/s, v = +0.0 m/s, a = 9.80 m/s. v = v + a v v 0.0 m/s 40.0 m/s = 0 = = +.04 s. a 9.80 m/s v v m/s 40.0 m/s (b) v = 0.0 m/s. = = = +6.1 s. a 9.80 m/s gives 1 (c) 0 = 0, v 0 = m/s, a = 9.80 m/s. 0 = v 0 + a gives = 0 and v v m/s (d) v = 0, v 0 = m/s, a = 9.80 m/s. v = v 0 + a gives = a = 9.80 m/s = 4.08 s. (e) The acceleraion is 9.80 m/s, downward, a all poins in he moion. (f) The graphs are skeched in Figure.48. EVALUATE: v = 0 a he maimum heigh. The ime o reach he maimum heigh is half he oal ime in he air, so he answer in par (d) is half he answer in par (c). Also noe ha.04 s < 4.08 s < 6.1 s. The boulder is going upward unil i reaches is maimum heigh and afer he maimum heigh i is raveling downward. Figure IDENTIFY: The rock has a consan downward acceleraion of 9.80 m/s. The consan-acceleraion kinemaics formulas appl. SET UP: The formulas = + v + 1 a and v = v + a ( ) boh appl. Call + upward Firs find he iniial veloci and hen he final speed. EXECUTE: (a) 6.00 s afer i is hrown, he rock is back a is original heigh, so = 0 a ha insan. Using a = 9.80 m/s and = 6.00 s, he equaion = + v + 1 a gives v 0 = 9.4 m/s. When he rock 0 0

27 - Chaper Moion Along a Sraigh Line - reaches he waer, 0 = 8.0 m. The equaion v = v + a ( ) gives v = 37.6 m/s, so is speed is 37.6 m/s. EVALUATE: The final speed is greaer han he iniial speed because he rock acceleraed on is wa down below he bridge..50. IDENTIFY: The acceleraion is no consan, so we mus use calculus insead of he sandard kinemaics formulas SET UP: The general calculus formulas are v = v a d v(), and hen inegrae ha o find (). and = v d. Firs inegrae a o find EXECUTE: Find v(): v ( ) = v + a d = v + (0.030 m/s 3 )(15.0 s )d. Carring ou he inegral and puing in he numbers gives v () = 8.00 m/s (0.030 m/s 3 )[(15.0 s) /]. Now use his resul o find (). = v d = m/s (0.030 m/s )((15.0 s) ) 3 d, which gives = 0 + (8.00 m/s) (0.030 m/s 3 )[(7.50 s) 3 /6)]. Using 0 = 14.0 m and = 10.0 s, we ge = 47.3 m. EVALUATE: The sandard kinemaics formulas appl onl when he acceleraion is consan..51. IDENTIFY: The acceleraion is no consan, bu we know how i varies wih ime. We can use he definiions of insananeous veloci and posiion o find he rocke s posiion and speed. SET UP: The basic definiions of veloci and posiion are v ( ) = v a d and EXECUTE: (a) v () = a d = (.80 m/s 3 )d = (1.40 m/s 3 ) = 0 v d = 0 (1.40 m/s ) d = ( m/s ). For = 10.0 s, 0 = 467 m. 3 3 (b) 0 = 35 m so ( m/s ) = 35 m and = s. A his ime v = (1.40 m/s 3 )(8.864 s) = 110 m/s. 0 = 0 v d. EVALUATE: The ime in par (b) is less han 10.0 s, so he given formulas are valid..5. IDENTIFY: The acceleraion is no consan so he consan acceleraion equaions canno be used. Insead, use v = v a d 1 SET UP: n d = n +1, for n 0. n + 1 EXECUTE: (a) v = v and = v d. Use he values of v and of a = 1.0 s o evaluae v 0 and 0. + αd = v + 1 α = v + (0.60 m/s 3 ). v = 5.0 m/s when = 1.0 s gives v 0 = 4.4 m/s. Then, a =.0 s, v = 4.4 m/s + (0.60 m/s )(.0 s) = 6.8 m/s. (b) = (v 0 + α )d = 0 + v α. = 6.0 m a = 1.0 s gives 0 = 1.4 m. Then, a =.0 s, = 1.4 m + (4.4 m/s)(.0 s) + 1 (1. m/s 3 )(.0 s) 3 = 11.8 m. (c) () = 1.4 m + (4.4 m/s) + (0.0 m/s 3 ) 3. v () = 4.4 m/s + (0.60 m/s 3 ). a ( ) = (1.0 m/s 3 ). The graphs are skeched in Figure.5. EVALUATE: We can verif ha a dv = d d and v =. d

Q2.4 Average velocity equals instantaneous velocity when the speed is constant and motion is in a straight line.

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