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


 Arron Kelley
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1 CHAPTER MOTION ALONG A STRAIGHT LINE Discussion Quesions Q. The speedomeer measures he magniude of he insananeous eloci, he speed. I does no measure eloci 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 inerals, hen he displacemen in successie inerals 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 alue, which is no wha he phoographs show. Graph (c) shows moion in he negaie direcion, which is no he case. This leaes graph (d). This graph shows eloci in he posiie 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 moing and he acceleraion direcion is opposie o he eloci direcion, hen he objec slows down, sops for an insan and hen sars o moe in he opposie direcion wih increasing speed. An eample is an objec hrown sraigh up ino he air. Grai gies he objec a consan downward acceleraion. The objec raels upward, sops a is maimum heigh and hen moes downward. The answer o he second quesion is no. Afer he firs reersal of he direcion of rael he eloci and acceleraion are hen in he same direcion. The objec coninues moing in he second direcion wih increasing speed. Q.4 Aerage eloci equals insananeous eloci 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 eloci 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 eloci 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 eloci and acceleraion are in he same direcion. Q.6 Aerage eloci is he magniude of he displacemen diided b he ime ineral. Aerage speed is he disance raeled diided b he ime ineral. Displacemen equals he disance raeled when he moion is in he same direcion for he enire ime ineral, and herefore his is when aerage eloci equals aerage speed. Q.7 For he same ime ineral he hae displacemens of equal magniude bu opposie direcions, so heir aerage elociies are in opposie direcions. One aerage eloci ecor is he negaie of he oher. Q.8 If in he ne ime ineral he second car had pulled ahead of he firs, hen he speed of he second car was greaer. The second car could also be obsered 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. Aerage eloci is displacemen diided b he ime ineral. If he displacemen is zero, hen he aerage eloci 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.
2 Figure DQ.9 Q. Zero acceleraion means consan eloci, so he eloci could be consan bu no zero. See Fig. DQ.. An eample is a car raeling a consan speed in a sraigh line. Figure DQ. Q. No. Aerage acceleraion refers o an ineral of ime and if he eloci is zero hroughou ha ineral, he aerage acceleraion for ha ime ineral is zero. Bu es, ou can hae zero eloci and nonzero acceleraion a one insan of ime. For eample, in Fig. DQ., = 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 eloci is zero bu is acceleraion is g downward. Figure DQ. Q. Yes. When he eloci and acceleraion are in opposie direcions he objec is slowing down. Q.3 (a) Two possible  graphs for he moion of he ruck are skeched in Fig. DQ.3. (b) Yes, he displacemen is 58 m and he ime ineral is 9. s, no maer wha pah he ruck akes beween and. The aerage eloci is he displacemen diided b he ime ineral.
3 Figure DQ.3 Q.4 This is rue onl when he acceleraion is consan. The aerage eloci is defined o be he displacemen diided b he ime ineral. If he acceleraion is no consan, objecs can hae he same iniial and final elociies bu differen displacemens and herefore differen aerage elociies. Q.5 I is greaer while he ball is being hrown. While being hrown, he ball acceleraes from res o eloci while raeling a disance less han our heigh. Afer i leaes our hand, i slows from o zero a he maimum heigh, while raeling a disance much greaer han our heigh. Eq.(.3) sas ha a =. Larger means smaller a. ( ) Q.6 (a) Eq.(.3): a = and = + ( ). When an objec reurns o he release poin, =. Eq.(.3) hen gies = ±. (b) = + a. A he highes poin =, so up = / a. A he end of he moion, when he objec has reurned o he release poin, we hae shown in (i) ha =, so oal = = and a a oal = up. Q.7 The disance beween adjacen drops will increase. The drops hae he downward acceleraion g = 9.8 m/s of a freefalling objec. Therefore, heir speed is coninuall increasing and he disance one drop raels in each successie. s ime ineral increases. A gien drop has fallen for. 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 = g( +. s) g = g (. s) +. s. The separaion Δ increases as increases. Q.8 Yes. Consider er small ime inerals during which he acceleraion doesn hae ime o change er much, so can be assumed o be consan. Calculae Δ = a Δ, for a er small ime ineral, saring a =. Then = +Δ. Since he acceleraion is assumed consan for he small ime ineral, a, = ( + ) / and Δ = a, Δ. Then he posiion a he end of he ineral is = +Δ. Repea he calculaion for he ne small ime ineral Δ : Δ = a Δ, = +Δ, a, = ( + ) /, Δ = a, Δ. Repea for successie small ime inerals.
4 Q.9 In he absence of air resisance, he firs ball rises o is maimum heigh and hen reurns o he leel 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 moing downward wih speed. 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 moing downward wih speed from he op of he building, he wo balls hae he same speed jus before he reach he ground. (b) The second ball reaches he ground firs, since he firs ball has o moe 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 hae he same displacemen. (d) The firs ball has raeled a greaer disance. Q. Le he + direcion be eas. The aerage eloci is he displacemen diided b he ime ineral. The firs. m displacemen requires a ime of (. m/s) / (3. m/s) = 4. s. The second. m displacemen requires a ime of (. m/s) / (5. m/s) = 4. s. The aerage Δ 4. m eloci is a = = = 3.75 m/s. This is less han 4. m/s since ou spend more ime Δ 64. s running a 3. m/s han a 5. m/s. Q. A he highes poin he objec insananeousl has zero speed. Bu is eloci is coninuall changing, a a consan rae. Acceleraion measures he rae of change of eloci. 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 moing downward in freefall, is downward displacemen from is iniial posiion of = is gien 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 = gt. Le he new heigh be Y and he corresponding ime be T, so = 3 = 3( ), so ( ) Y Y gt 3 gt = g( T ) and T = 3T. Y =. Bu g( T )
5 MOTION ALONG A STRAIGHT LINE.. IDENTIFY: Δ = aδ SET UP: We know he aerage eloci is 6.5 m/s. EXECUTE: Δ = a Δ = 5. m EVALUATE: In round numbers, 6 m/s 4 s = 4 m 5 m, so he answer is reasonable. Δ.. IDENTIFY: a = Δ 6 6 SET UP: 3. 5 das =. 66 s. A he release poin, =+ 55. m m a 6 EXECUTE: (a) = = = 4. 4 m/s. Δ. 66 s (b) For he round rip, = and Δ =. The aerage eloci is zero. EVALUATE: The aerage eloci for he rip from he nes o he release poin is posiie..3. IDENTIFY: Targe ariable is he ime Δ i akes o make he rip in hea raffic. Use Eq. (.) ha relaes he aerage eloci o he displacemen and aerage ime. Δ Δ SET UP: a = so Δ = aδ and Δ =. Δ a EXECUTE: Use he informaion gien for normal driing condiions o calculae he disance beween he wo ciies, where he ime is h and 5 min, which is min: Δ = a Δ = (5 km/h)( h/6 min)( min) = 9.5 km. Now use a for hea raffic o calculae Δ; Δ is he same as before: Δ 9.5 km Δ = = =.75 h = h and 45 min. 7 km/h a The addiional ime is ( h and 45 min) ( h and 5 min) = ( h and 5 min) ( h and 5 min) = 55 min. EVALUATE: A he normal speed of 5 km/s he rip akes min, bu a he reduced speed of 7 km/h i akes 65 min. So decreasing our aerage speed b abou 3% adds 55 min o he ime, which is 5% of min. Thus a 3% reducion in speed leads o a 5% increase in rael ime. This resul (perhaps surprising) occurs because he ime ineral is inersel proporional o he aerage speed, no direcl proporional o i. Δ.4. IDENTIFY: The aerage eloci is a =. Use he aerage speed for each segmen o find he ime Δ raeled in ha segmen. The aerage speed is he disance raeled diided b he ime. SET UP: The pos is 8 m wes of he pillar. The oal disance raeled is m + 8 m = 48 m. m EXECUTE: (a) The easward run akes ime 4 s 5. m/s =. and he wesward run akes 8 m 7 s. 4. m/s =. The aerage speed for he enire rip is 48 m 44 m/s.. s =. Δ 8 m (b) a = = = 73. m/s. The aerage eloci is direced wesward. Δ. s Coprigh 6 Pearson Educaion, Inc. All righs resered. 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. 
6  Chaper EVALUATE: The displacemen is much less han he disance raeled, and he magniude of he aerage eloci is much less han he aerage speed. The aerage speed for he enire rip has a alue ha lies beween he aerage speed for he wo segmens..5. IDENTIFY: Gien wo displacemens, we wan he aerage eloci and he aerage speed. Δ SET UP: The aerage eloci is a = and he aerage speed is jus he oal disance walked diided Δ b he oal ime o walk his disance. EXECUTE: (a) Le + be eas. Δ = 6. m 4. m =. m and Δ = 8. s s = 64. s. So Δ. m a = = =. 3 m/s. Δ 64. s 6. m + 4. m (b) aerage speed = =. 56 m/s 64. s EVALUATE: The aerage speed is much greaer han he aerage eloci because he oal disance walked is much greaer han he magniude of he displacemen ecor. Δ.6. IDENTIFY: The aerage eloci is a =. Use () o find for each. Δ. =. and (4. s) =. 8 m SET UP: () =, ( s) 5 6 m, 5. 6 m EXECUTE: (a) a = =+ 8. m/s. s. 8 m (b) a = =+ 5. m/s 4. s. 8 m 5. 6 m (c) a = =+ 76. m/s. s EVALUATE: The aerage eloci depends on he ime ineral being considered. Δ.7. (a) IDENTIFY: Calculae he aerage eloci using a =. Δ Δ SET UP: a = so use ( ) o find he displacemen Δ for his ime ineral. Δ EXECUTE: = : = =. s: Then a 3 3 = (. 4 m/s )(. s) (. m/s )(. s) = 4 m m = m. Δ m = = =. m/s. Δ. s d (b) IDENTIFY: Use = o calculae ( ) and ealuae his epression a each specified. d d SET UP: = = b 3c. d EXECUTE: (i) = : = 3 (ii) = 5. s: = (. 4 m/s )(5. s) 3(. m/s )(5. s) = 4. m/s 9. m/s = 5. m/s. 3 (iii) =. s: = (4. m/s)(. s) 3(. m/s)(. s) = 48. m/s 36. m/s=. m/s. (c) IDENTIFY: Find he alue of when ( ) from par (b) is zero. SET UP: = b 3c = a =. = ne when b 3c = b (. 4 m/s ) EXECUTE: b= 3c so = = = 3. 3 s 3 3c 3(. m/s ) EVALUATE: () for his moion sas he car sars from res, speeds up, and hen slows down again. Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
7 Moion Along a Sraigh Line IDENTIFY: We know he posiion () of he bird as a funcion of ime and wan o find is insananeous eloci a a paricular ime. 3 3 d d 8. m + (. 4 m/s) (. 45 m/s ) SET UP: The insananeous eloci is () = =. d d d 3 EXECUTE: ( ) = =. 4 m/s (. 35 m/s ). Ealuaing his a = 8. s gies = 376. m/s. d EVALUATE: The acceleraion is no consan in his case. Δ.9. IDENTIFY: The aerage eloci is gien b a =. We can find he displacemen Δ for each Δ consan eloci ime ineral. The aerage speed is he disance raeled diided b he ime. SET UP: For = o =. s, =. m/s. For =. s o = 3. s, = 3. m/s. In par (b), = 3. m/sfor =. s o = 3. s. When he eloci is consan, Δ = Δ. EXECUTE: (a) For = o =. s, Δ = (. m/s)(. s) = 4. m. For =. s o = 3. s, Δ = (3. m/s)(. s) = 3. m. For he firs 3. s, Δ = 4. m + 3. m = 7. m. The disance raeled is Δ 7. m also 7. m. The aerage eloci is a = = = 33. m/s. The aerage speed is also.33 m/s. Δ 3. s (b) For =. s o 3. s, Δ = ( 3. m/s)(. s) = 3. m. For he firs 3. s, Δ = 4. m + ( 3. m) =+. m. The ball raels 4. m in he +direcion and hen 3. m in he Δ. m direcion, so he disance raeled is sill 7. m. a = = = 33. m/s. The aerage speed is Δ 3. s 7. m = 33. m/s. 3. s EVALUATE: When he moion is alwas in he same direcion, he displacemen and he disance raeled are equal and he aerage eloci has he same magniude as he aerage speed. When he moion changes direcion during he ime ineral, hose quaniies are differen... IDENTIFY and SET UP: The insananeous eloci is he slope of he angen o he ersus graph. EXECUTE: (a) The eloci is zero where he graph is horizonal; poin IV. (b) The eloci is consan and posiie where he graph is a sraigh line wih posiie slope; poin I. (c) The eloci is consan and negaie where he graph is a sraigh line wih negaie slope; poin V. (d) The slope is posiie and increasing a poin II. (e) The slope is posiie and decreasing a poin III. EVALUATE: The sign of he eloci indicaes is direcion... IDENTIFY: Find he insananeous eloci of a car using a graph of is posiion as a funcion of ime. SET UP: The insananeous eloci a an poin is he slope of he ersus graph a ha poin. Esimae he slope from he graph. EXECUTE: A: = 6. 7 m/s; B: = 6. 7 m/s; C: = ; D: = 4. m/s; E: = 4. m/s; F: = 4. m/s; G: =. EVALUATE: The sign of shows he direcion he car is moing. is consan when ersus is a sraigh line. Δ.. IDENTIFY: aa =. a() is he slope of he ersus graph. Δ SET UP: 6 km/h = 6. 7 m/s 6. 7 m/s 6. 7 m/s EXECUTE: (a) (i) aa = = 7. m/s. (ii) aa = = 7. m/s. s s (iii) Δ = and a a =. (i) Δ = and a a =. (b) A = s, is consan and a =. A = 35 s, he graph of ersus is a sraigh line and a = a = 7. m/s. a Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
8 4 Chaper EVALUATE: When a a and hae he same sign he speed is increasing. When he hae opposie signs, he speed is decreasing. Δ.3. IDENTIFY: The aerage acceleraion for a ime ineral Δ is gien b aa =. Δ SET UP: Assume he car is moing in he + direcion. mi/h =. 447 m/s, so 6 mi/h = 6. 8 m/s, mi/h = 89.4 m/s and 53 mi/h = 3. m/s. EXECUTE: (a) The graph of ersus is skeched in Figure.3. The graph is no a sraigh line, so he acceleraion is no consan m/s m/s 6. 8 m/s (b) (i) aa = =. 8 m/s (ii) aa = = 35. m/s. s. s. s 3. m/s m/s (iii) aa = = 78. m/s. The slope of he graph of ersus decreases as 53 s. s increases. This is consisen wih an aerage acceleraion ha decreases in magniude during each successie ime ineral. EVALUATE: The aerage acceleraion depends on he chosen ime ineral. For he ineral beween and 3. m/s 53 s, aa = = 3. m/s. 53 s Figure.3.4. IDENTIFY: We know he eloci () of he car as a funcion of ime and wan o find is acceleraion a he insan ha is eloci is. m/s. 3 SET UP: We know ha () = (.86 m/s 3 ) d d (. 86 m/s ) and ha a() = =. d d d 3 EXECUTE: a() = = (7. m/s). When =. m/s, (.86 m/s 3 ) =. m/s, which gies d = s. A his ime, a = 6. 4 m/s. EVALUATE: The acceleraion of his car is no consan. d d.5. IDENTIFY and SET UP: Use = and a = o calculae ( ) and a( ). d d d EXECUTE: = =. cm/s (. 5 cm/s ) d d a = = 5. cm/s d (a) A =, = 5. cm, =. cm/s, a = 5. cm/s. (b) Se = and sole for : = 6. s. (c) Se = 5. cm and sole for. This gies = and = 3. s. The urle reurns o he saring poin afer 3. s. (d) The urle is. cm from saring poin when = 6. cm or = 4. cm. Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
9 Moion Along a Sraigh Line 5 Se = 6. cm and sole for : = 6. s and = 5. 8 s. A = 6. s, =+ 3. cm/s. A = 5. 8 s, = 3. cm/s. Se = 4. cm and sole for : = s (oher roo o he quadraic equaion is negaie and hence nonphsical). A = s, = 55. cm/s. (e) The graphs are skeched in Figure.5. Figure.5 EXECUTE: (a) [(5. m/s) (5. m/s)]/( s) =. m/s (b)[( 5. m/s) ( 5. m/s)]/( s) =. m/s (c) [( 5. m/s) ( + 5. m/s)]/( s) = 3. m/s EVALUATE: The acceleraion is consan and negaie. is linear in ime. I is iniiall posiie, decreases o zero, and hen becomes negaie wih increasing magniude. The urle iniiall moes farher awa from he origin bu hen sops and moes in he direcion. Δ.6. IDENTIFY: Use aa =, wih Δ = s in all cases. Δ SET UP: is negaie if he moion is o he lef. EVALUATE: In all cases, he negaie acceleraion indicaes an acceleraion o he lef. Δ.7. IDENTIFY: The aerage acceleraion is aa =. Use ( ) o find a each. The insananeous Δ d acceleraion is a =. d SET UP: () = 3. m/s and (5. s) = 5. 5 m/s. Δ 5. 5 m/s 3. m/s EXECUTE: (a) aa = = = 5. m/s Δ 5. s d 3 3 (b) a = = (. m/s )( ) = (. m/s ). A =, a =. A = 5. s, a =. m/s. d (c) Graphs of ( ) and a( ) are gien in Figure.7 (ne page). EVALUATE: a() is he slope of ( ) and increases as increases. The aerage acceleraion for = o = 5. s equals he insananeous acceleraion a he midpoin of he ime ineral, =. 5 s, since a () is a linear funcion of. Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
10 6 Chaper Figure.7 d d.8. IDENTIFY: () = and a() = d d d n n SET UP: ( ) n = for n. d EXECUTE: (a) ( ) = (9. 6 m/s ) (. 6 m/s ) and a () = 96. m/s (3. m/s). Seing = gies = and =. s. A =, = 7. mand a = 9. 6 m/s. A =. s, = 5. m and a = m/s. (b) The graphs are gien in Figure.8. EVALUATE: For he enire ime ineral from = o =. s, he eloci is posiie and increases. While a is also posiie he speed increases and while a is negaie he speed decreases. Figure.8.9. IDENTIFY: Use he consan acceleraion equaions o find and a. (a) SET UP: The siuaion is skeched in Figure.9. Figure.9 = 7. m = 6. s = 5. m/s =? Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
11 Moion Along a Sraigh Line 7 + ( ) (7 m) EXECUTE: Use =, so = =. 5. m/s = 8.33 m/s. 6s. 5. m/s 5. m/s (b) Use = + a, so a = = =. m/s. 6s. EVALUATE: The aerage eloci is (7. m)/(6. s) =.7 m/s. The final eloci is larger han his, so he anelope mus be speeding up during he ime ineral; < 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. 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 = + a. (a) = 3(33 m/s) = 993 m/s, =, and a = 5g = 49. m/s. (b) = 5. s 993 m/s EXECUTE: (a) = + a and = = =. 3 s. Yes, he ime required is larger a 49. m/s han 5. s. (b) = + a = + (49. m/s )(5. s) = 45 m/s. EVALUATE: In 5. s he can onl reach abou /3 he speed of sound wihou blacking ou... IDENTIFY: For consan acceleraion, he sandard kinemaics equaions appl. SET UP: Assume he ball sars from res and moes in he + direcion. EXECUTE: (a) = 5. m, = 45. m/s and =. = + a( ) gies (45. m/s) a = = = 675 m/s. ( ) (. 5 m) + ( ) (. 5 m) (b) = gies = = = 667. s m/s 45. m/s EVALUATE: We could also use = + a o find = = =. 667 s which agrees wih a 675 m/s our preious resul. The acceleraion of he ball is er large... IDENTIFY: For consan acceleraion, he sandard kinemaics equaions appl. SET UP: Assume he ball moes in he + direcion. EXECUTE: (a) = m/s, = and = 3. ms. = + a gies m/s a = = = 44 m/s s m/s 3 (b) (3 = =. s) =. m. EVALUATE: We could also use = + a o calculae : 3 = (44 m/s )(3. s) =. m, which agrees wih our preious resul. The acceleraion of he ball is er large..3. IDENTIFY: Assume ha he acceleraion is consan and appl he consan acceleraion kinemaic equaions. Se a equal o is maimum allowed alue. SET UP: Le + be he direcion of he iniial eloci of he car. 5 km/h = 9. 7 m/s. a = 5 m/s. EXECUTE: = 9. 7 m/s. =. = + a( ) gies (9. 7 m/s) = = = 7. m. a ( 5 m/s ) EVALUATE: The car frame sops oer a shorer disance and has a larger magniude of acceleraion. Par of our.7 m sopping disance is he sopping disance of he car and par is how far ou moe relaie o he car while sopping. Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
12 8 Chaper.4. 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 rael if he mainains his acceleraion during his ime. SET UP: Le + be he direcion he je raels and ake =. Wih consan acceleraion, he equaions = + a and = + + a boh appl. = 4g = 39. m/s, 4(33 m/s) 34 m/s, =. 34 m/s EXECUTE: (a) Soling = + a for gies = = = s. a 39. m/s (b) = + + a = (39. m/s )(33. 8 s) =. 4 m =. 4 km. 4 Coprigh 6 Pearson Educaion, Inc. All righs resered. 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. a = = and 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 6g, how far does he rael during his ime? SET UP: Le + be he direcion he person raels. = (he sops), a is negaie since i is opposie o he direcion of he moion, and = 36 ms = 3. 6 s. The equaions = + a and = + + a boh appl since he acceleraion is consan. EXECUTE: Soling = + a for gies = a. Then = + + a gies = a ( 588 m/s )(3 6 s) 38 cm =. =. EVALUATE: Noice ha we were no gien he iniial speed, bu we could find i: 3 a ( 588 m/s )(36 s) m/s 47 mph = = = =..6. IDENTIFY: In (a) he hip pad mus reduce he person s speed from. m/s o.3 m/s oer a disance of. cm, and we wan he acceleraion oer his disance, assuming consan acceleraion. In (b) we wan o find ou how long he acceleraion in (a) lass. SET UP: Le + be downward. =. m/s, = 3. m/s, and =. m. The equaions + = + a( ) and = appl for consan acceleraion. EXECUTE: (a) Soling = + a( ) for a gies (. 3 m/s) (. m/s) a = = = 58 m/s = 5. 9 g. ( ) (. m) + ( ) (. m) (b) = gies = = = ms. +. m/s +. 3 m/s EVALUATE: The acceleraion is er large, bu i onl lass for ms so i produces a small eloci change..7. IDENTIFY: We know he iniial and final elociies of he objec, and he disance oer which he eloci change occurs. From his we wan o find he magniude and duraion of he acceleraion of he objec. SET UP: The consanacceleraion kinemaics formulas appl. = + a ( ), where =, 3 = 5. m/s, and = 4. m. 3 (5. m/s) EXECUTE: (a) = + a( ) gies a = = = ( ) (4. m) = 3 5. m/s (b) = + a gies = = =.6 ms. 6 a 3. m/s m/s 3. g. EVALUATE: (c) The calculaed a is less han 45, g so he acceleraion required doesn rule ou his hpohesis..8. IDENTIFY: Appl consan acceleraion equaions o he moion of he car. SET UP: Le + be he direcion he car is moing. ( m/s) EXECUTE: (a) From = + a( ), wih =, a = 67m/s ( ) = ( m) =..
13 Moion Along a Sraigh Line 9 (b) Using Eq. (.4), = ( )/ = ( m)/( m/s) = s. (c) ( s)( m/s) = 4 m. EVALUATE: The aerage eloci of he car is half he consan speed of he raffic, so he raffic raels wice as far. Δ.9. IDENTIFY: The aerage acceleraion is aa =. For consan acceleraion, he sandard kinemaics Δ equaions appl. SET UP: Assume he rocke ship raels in he + direcion. 6 km/h = m/s and 6 km/h = 447. m/s.. min = 6. s Δ m/s EXECUTE: (a) (i) aa = = = m/s Δ 8. s 447. m/s 447. m/s (ii) aa = = 774. m/s 6. s 8. s m/s (b) (i) = 8. s, =, and = m/s. = = (8. s) = 79 m. (ii) Δ = 6. s 8. s = 5. s, = m/s, and = 447. m/s m/s m/s 4 = = (5. s) = 8. m. EVALUATE: When he acceleraion is consan he insananeous acceleraion hroughou he ime ineral equals he aerage acceleraion for ha ime ineral. We could hae calculaed he disance in par (a) as = + a (5 59 m/s )(8 s) 79 m, =.. = which agrees wih our preious calculaion..3. IDENTIFY: The acceleraion a is he slope of he graph of ersus. SET UP: The signs of and of a indicae heir direcions. EXECUTE: (a) Reading from he graph, a = 4. s, = 7. cm/s, o he righ and a = 7. s, = 3. cm/s, o he lef. (b) ersus is a sraigh line wih slope equal o 8. cm/s =. 6. s 3. cm/s, o he lef. I has his alue a all imes. (c) Since he acceleraion is consan,. 3 cm/s. = + a For = o 4.5 s, = =. For = o 7.5 s, (8 cm/s)(4 5 s) ( 3 cm/s )(4 5 s) 8 cm. (8 cm/s)(7 5 s) ( 3 cm/s )(7 5 s) 3 4 cm = =. (d) The graphs of a and ersus are gien in Figure.3. + EVALUATE: In par (c) we could hae insead used =. The acceleraion is consan and Figure.3 Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
14  Chaper.3. (a) IDENTIFY and SET UP: The acceleraion a a ime is he slope of he angen o he ersus cure a ime. EXECUTE: A = 3 s, he ersus cure is a horizonal sraigh line, wih zero slope. Thus a =. A = 7 s, he ersus cure is a sraighline segmen wih slope Thus a = 63. m/s. A = s he cure is again a sraighline segmen, now wih slope Thus a =. m/s. 45 m/s m/s 9 s 5 s 45 m/s 3 s 9 s 63 m/s =.. m/s =.. EVALUATE: a = when is consan, a > when is posiie and he speed is increasing, and a < when is posiie and he speed is decreasing. (b) IDENTIFY: Calculae he displacemen during he specified ime ineral. SET UP: We can use he consan acceleraion equaions onl for ime inerals 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 ineral = o = 5 s he acceleraion is consan and equal o zero. For he ime ineral = 5 s o = 9 s he acceleraion is consan and equal o 65. m/s. For he ineral = 9 s o = 3 s he acceleraion is consan and equal o. m/s. EXECUTE: During he firs 5 seconds he acceleraion is consan, so he consan acceleraion kinemaic formulas can be used. = m/s a = = 5 s =? = ( a = so no a erm) = ( m/s)(5 s) = m; his is he disance he officer raels in he firs 5 seconds. During he ineral = 5 s o 9 s he acceleraion is again consan. The consan acceleraion formulas can be applied o his 4second ineral. I is conenien o resar our clock so he ineral sars a ime = and ends a ime = 4 s. (Noe ha he acceleraion is no consan oer he enire = o = 9 s ineral.) = m/s a = 65. m/s = 4 s = m =? = + a = +. = + = ( m/s)(4 s) (6 5 m/s )(4 s) 8 m 5 m 3 m. Thus + 3 m = m + 3 m = 3 m. A = 9 s he officer is a = 3 m, so she has raeled 3 m in he firs 9 seconds. During he ineral = 9 s o = 3 s he acceleraion is again consan. The consan acceleraion formulas can be applied for his 4second ineral bu no for he whole = o = 3 s ineral. To use he equaions resar our clock so his ineral begins a ime = and ends a ime = 4 s. = 45 m/s (a he sar of his ime ineral) a =. m/s 4 s = + a = = 3 m =? (45 m/s)(4 s) ( m/s )(4 s) 8 m 89 6 m 9 4 m. = +. =. =. Thus = m = 3 m m = 3 m. A = 3 s he officer is a = 3 m, so she has raeled 3 m in he firs 3 seconds. EVALUATE: The eloci is alwas posiie so he displacemen is alwas posiie and displacemen and disance raeled are he same. The aerage eloci for ime ineral Δ is a =Δ/ Δ. For = o 5 s, a = m/s. For = o 9 s, a = 6 m/s. For = o 3 s, a = 5 m/s. These resuls are consisen wih he figure in he ebook..3. IDENTIFY: () is he slope of he ersus graph. Car B moes wih consan speed and zero acceleraion. Car A moes wih posiie acceleraion; assume he acceleraion is consan. Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
15 Moion Along a Sraigh Line  SET UP: For car B, is posiie and a =. For car A, a is posiie and increases wih. EXECUTE: (a) The moion diagrams for he cars are gien in Figure.3a. (b) The wo cars hae he same posiion a imes when heir  graphs cross. The figure in he problem shows his occurs a approimael = s and = 3 s. (c) The graphs of ersus for each car are skeched in Figure.3b. (d) The cars hae he same eloci when heir  graphs hae he same slope. This occurs a approimael = s. (e) Car A passes car B when A moes aboe B in he  graph. This happens a = 3 s. (f) Car B passes car A when B moes aboe A in he  graph. This happens a = s. EVALUATE: When a =, he graph of ersus is a horizonal line. When a is posiie, he graph of ersus is a sraigh line wih posiie slope. Figure IDENTIFY: For consan acceleraion, he kinemaics formulas appl. We can use he oal displacemen ande final eloci 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 = + a( ) applies. EXECUTE: Firs look a he moion oer 6.8 m. We use he following numbers: =, = 6.8 m, and = 3.8 /s. Soling he aboe equaion for a gies a =.6 m/s. Now look a he moion oer he 3.4 m using =, a =.6 m/s and = 3.4 m. Soling he same equaion, bu his ime for, gies =.69 m/s. EVALUATE: Een hough he block has raeled 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 ehicle. The ruck passes he car when he are a he same a he same >. SET UP: The ruck has a =. The car has =. Le + be in he direcion of moion of he ehicles. Boh ehicles sar a =. The car has a C =.8 m/s. The ruck has =. m/s. EXECUTE: (a) = + a gies T = T and C = a C. Seing T = C gies = and T (. m/s) T = a C, so = = = 4.9 s. A his, T = (. m/s)(4.9 s) = 86 m and ac.8 m/s = (3. m/s )(4.9 s) = 86 m. The car and ruck hae each raeled 86 m. (b) A = 4.9 s, he car has = + a = (.8 m/s )(4.9 s) = 4 m/s. (c) T = T and C = a C. The  graph of he moion for each ehicle is skeched in Figure.34a. (d) =. = a. The  T T C C graph for each ehicle is skeched in Figure.34b (ne page). EVALUATE: When he car oerakes he ruck is speed is wice ha of he ruck. Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
16  Chaper Figure IDENTIFY: Appl he consan acceleraion equaions o he moion of he flea. Afer he flea leaes he ground, a = g, downward. Take he origin a he ground and he posiie direcion o be upward. (a) SET UP: A he maimum heigh =. = = 44. m a = 98. m/s =? = + a ( ) EXECUTE: a = ( ) = ( 9. 8 m/s )(. 44 m) =. 94 m/s (b) SET UP: When he flea has reurned o he ground =. = =+. 94 m/s = + a a = 98. m/s =? (. 94 m/s) EXECUTE: Wih = his gies = = = 6. s. a 9. 8 m/s EVALUATE: We can use = + a o show ha wih =. 94 m/s, = afer.3 s..36. IDENTIFY: The rock has a consan downward acceleraion of 9.8 m/s. We know is iniial eloci and posiion and is final posiion. SET UP: We can use he kinemaics formulas for consan acceleraion. EXECUTE: (a) = 3 m, =. m/s, a = 9. 8 m/s. The kinemaics formulas gie Coprigh 6 Pearson Educaion, Inc. All righs resered. 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. = + a( ) = (. m/s) + ( 9. 8 m/s )( 3 m) = 3.74 m/s, so he speed is 3.7 m/s m/s. m/s (b) = + a and = = = 5.59 s. a 9. 8 m/s EVALUATE: The erical eloci in par (a) is negaie because he rock is moing downward, bu he speed is alwas posiie. The 5.59 s is he oal ime in he air..37. IDENTIFY: The pin has a consan downward acceleraion of 9.8 m/s and reurns o is iniial posiion. SET UP: We can use he kinemaics formulas for consan acceleraion. EXECUTE: The kinemaics formulas gie = + a. We know ha =, so (8. m/s) = = =+ 67. s. a 9. 8 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 9.8 m/s. We know he iniial eloci of he pu and he disance i raels.
17 Moion Along a Sraigh Line 3 SET UP: We can use he kinemaics formulas for consan acceleraion. EXECUTE: (a) = 9.5 m/s and = 3.6 m, which gies = + a( ) = (9. 5 m/s) + ( 9. 8 m/s )(3. 6 m) = m/s m/s 9. 5 m/s (b) = = = 57. s a 98. m/s EVALUATE: The pu is sopped b he ceiling, no b grai..39. IDENTIFY: A ball on Mars ha is hi direcl upward reurns o he same leel in 8.5 s wih a consan downward acceleraion of.379g. How high did i go and how fas was i iniiall raeling upward? SET UP: Take + upward. = a he maimum heigh. a =. 379g = 3. 7 m/s. The consanacceleraion formulas = + a and = + + a boh appl. EXECUTE: Consider he moion from he maimum heigh back o he iniial leel. For his moion = and = 4. 5 s. = + + a ( 3 7 m/s )(4 5 s) 33 5 m. =.. =. The ball wen 33.5 m aboe is original posiion. (b) Consider he moion from jus afer i was hi o he maimum heigh. For his moion = and = 45. s. = + a gies = a = ( 3. 7 m/s )(4. 5 s) = 5. 8 m/s. (c) The graphs are skeched in Figure.39. Figure.39 EVALUATE: The answers can be checked seeral was. For eample, =, = 5. 8 m/s, and (5. 8 m/s) a = 37. m/s in = + a( ) gies = = = m, a ( 3. 7 m/s ) agrees wih he heigh calculaed in (a)..4. IDENTIFY: Appl consan acceleraion equaions o he moion of he lander. SET UP: Le + be downward. Since he lander is in freefall, a =+ 6. m/s. EXECUTE: = 8. m/s, = 5. m, a =+ 6. m/s in = + a ( ) = (. 8 m/s) + (. 6 m/s )(5. m) = 4. m/s. = + a ( ) gies which EVALUATE: The same descen on earh would resul in a final speed of 9.9 m/s, since he acceleraion due o grai on earh is much larger han on he moon..4. 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 = and a = 9. 8 m/s. Le d be he disance he meersick falls. EXECUTE: (a) d = + a gies d = (4. 9 m/s ) and = m/s. 76 m (b) = = 9. s 49m/s. EVALUATE: The reacion ime is proporional o he square of he disance he sick falls. Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
18 4 Chaper.4. IDENTIFY: Appl consan acceleraion equaions o he erical moion of he brick. SET UP: Le + be downward. a = 98. m/s EXECUTE: (a) =, =.9 s, a = 9. 8 m/s. = + a (9 8 m/s )(.9 s) 7.7 m. =. = The building is 7.7 m all. (b) = + a = + (9. 8 m/s )(.9 s) = 8.6 m/s (c) The graphs of a, EVALUATE: We could use eiher and ersus are gien in Figure.4. Take = a he ground. + = or = + a( ) o check our resuls. Figure IDENTIFY: When he onl force is grai he acceleraion is 98. m/s, downward. There are wo inerals of consan acceleraion and he consan acceleraion equaions appl during each of hese inerals. SET UP: Le + be upward. Le = a he launch pad. The final eloci for he firs phase of he moion is he iniial eloci for he freefall phase. EXECUTE: (a) Find he eloci when he engines cu off. = 55 m, a = 5. m/s, =. = + a( ) gies = (. 5 m/s )(55 m) = m/s. Now consider he moion from engine cuoff o maimum heigh: = 55 m, = m/s, = (a he maimum heigh), a = 9. 8 m/s. = + a( ) gies (48. 6 m/s) = = = m and = m + 55 m = 646 m. a ( 9. 8 m/s ) (b) Consider he moion from engine failure unil jus before he rocke srikes he ground: = 55 m, a = 98. m/s, = m/s. = + a ( ) gies = (48. 6 m/s) + ( 9. 8 m/s )( 55 m) = m/s. Then m/s m/s 6 4 s. = = =. a 9. 8 m/s = + a gies (c) Find he ime from blasoff unil engine failure: = 55 m, =, a =+ 5. m/s. ( ) (55 m) = + a gies = = =. 6 s. The rocke srikes he launch pad a 5. m/s 6. s+ 64. s= 38. s afer blasoff. The acceleraion a is 9. 8 m/s from =. 6 s o 38 s.. so he graph of ersus is a sraigh line wih posiie slope of + 5. m/s from = o =. 6 s. I is = + a applies during each consan acceleraion segmen, 5. m/s during he blasoff phase and wih negaie slope of 9. 8 m/s afer engine failure. During each phase = + a. The sign of a deermines he curaure of ( ). A = 38. s he rocke has reurned o =. The graphs are skeched in Figure.43. Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
19 Moion Along a Sraigh Line 5 EVALUATE: In par (b) we could hae found he ime from us o aoid soling for from a quadraic equaion. = + a finding firs allows, Figure IDENTIFY: Appl consan acceleraion equaions o he erical moion of he sandbag. SET UP: Take + upward. a = 9. 8 m/s. The iniial eloci of he sandbag equals he eloci of he balloon, so =+ 5. m/s. When he balloon reaches he ground, = 4. m. A is maimum heigh he sandbag has =. EXECUTE: (a) = 5. s: = + a (5 m/s)( 5 s) ( 9 8 m/s )( 5 s) 94 m. = =. The sandbag is 4.9 m aboe he ground. = + a =+ 5. m/s + ( 9. 8 m/s )(. 5 s) =. 55 m/s. =. s: = (5. m/s)(. s) + ( 9. 8 m/s )(. s) =. m. The sandbag is 4. m aboe he ground. = + a =+ 5. m/s + ( 9. 8 m/s )(. s) = 4. 8 m/s. (b) = 4. m, = 5. m/s, a = 9. 8 m/s. = + a gies. =.. 4 m (5 m/s) (4 9 m/s ). ( ) (4 9 m/s ) (5 m/s) 4 m... = and = 5. ± ( 5). 4(49)(. 4). s = (5. ± 9). s. mus be posiie, so = 3. 4 s. 98. (c) = + a =+ 5. m/s + ( 9. 8 m/s )(3. 4 s) = 8. 4 m/s (d) = 5. m/s, a = 98. m/s, =. = + a( ) gies (5. m/s) = = = 8. m. The maimum heigh is 4.3 m aboe he ground. a ( 9. 8 m/s ) (e) The graphs of a,, and ersus are gien in Figure.44. Take = a he ground. EVALUATE: The sandbag iniiall raels upward wih decreasing eloci and hen moes downward wih increasing speed. Figure.44 Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
20 6 Chaper.45. IDENTIFY: Use he consan acceleraion equaions o calculae a and. (a) SET UP: = + a = 4 m/s, =, = 9. s, a =? 4 m/s EXECUTE: a = = = 49 m/s. 9 s (b) a / g = (49 m/s )/(9. 8 m/s ) = 5. 4 (c) = + a = + (49 m/s )(. 9 s) = m (d) SET UP: Calculae he acceleraion, assuming i is consan: =. 4 s, = 83 m/s, = (sops), a =? = + a 83 m/s EXECUTE: a = = = m/s 4s. a / g = ( m/s )/(9. 8 m/s ) =. 6; a =. 6g If he acceleraion while he sled is sopping is consan hen he magniude of he acceleraion is onl.6g. Bu if he acceleraion is no consan i is cerainl possible ha a some poin he insananeous acceleraion could be as large as 4g. 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 freefall and has a consan downward acceleraion of magniude 9. 8 m/s. Appl he consan acceleraion equaions o he moion of he egg. SET UP: Take + o be upward. A he maimum heigh, =. EXECUTE: (a) = 3. m, = 5. s, a = 9. 8 m/s. = + a gies 3 m = a ( 9 8 m/s )(5 s) 8 5 m/s. =... =+. 5. s (b) = m/s, = (a he maimum heigh), a = 9. 8 m/s. = + a( ) gies (8. 5 m/s) = = = 7. 5 m. a ( 9. 8 m/s ) (c) A he maimum heigh =. (d) The acceleraion is consan and equal o 98. m/s, downward, a all poins in he moion, including a he maimum heigh. (e) The graphs are skeched in Figure m/s EVALUATE: The ime for he egg o reach is maimum heigh is = = =. 89 s. The a 98. m/s egg has reurned o he leel of he cornice afer 3.78 s and afer 5. s i has raeled downward from he cornice for. s. Figure.46 Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
21 Moion Along a Sraigh Line IDENTIFY: We can aoid soling for he common heigh b considering he relaion beween heigh, ime of fall, and acceleraion due o grai, and seing up a raio inoling ime of fall and acceleraion due o grai. SET UP: Le g En be he acceleraion due o grai 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 = h. = EXECUTE: = + a gies h= g and h= g Combining hese wo equaions gies E En En. E 75. s En = g = (9. 8 m/s ) 868 m/s. =. En 8. 6 s ge = genen and g EVALUATE: The acceleraion due o grai is inersel proporional o he square of he ime of fall..48. IDENTIFY: Since air resisance is ignored, he boulder is in freefall and has a consan downward acceleraion of magniude 9. 8 m/s. Appl he consan acceleraion equaions o he moion of he boulder. SET UP: Take + o be upward. EXECUTE: (a) =+ 4. m/s, =+. m/s, a = 9. 8 m/s. = + a gies. m/s 4. m/s = = =+. 4 s. a 98. m/s. m/s 4. m/s (b) =. m/s. = = =+ 6. s. a 9. 8 m/s (c) =, =+ 4. m/s, a = 9. 8 m/s. = + a gies = and (4. m/s) = = = s. a 98. m/s 4. m/s (d) =, =+ 4. m/s, a = 9. 8 m/s. = + a gies = = = 48. s. a 98. m/s (e) The acceleraion is 98. m/s, downward, a all poins in he moion. (f) The graphs are skeched in Figure.48. EVALUATE: = 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. 4 s < 4. 8 s < 6. s. The boulder is going upward unil i reaches is maimum heigh and afer he maimum heigh i is raeling downward. Figure IDENTIFY: The rock has a consan downward acceleraion of 9.8 m/s. The consanacceleraion kinemaics formulas appl. SET UP: The formulas = + + a and = + a( ) boh appl. Call + upward. Firs find he iniial eloci and hen he final speed. EXECUTE: (a) 6. s afer i is hrown, he rock is back a is original heigh, so = a ha insan. Using a = 9.8 m/s and = 6. s, he equaion = + + a gies = 9.4 m/s. When he rock Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
22 8 Chaper reaches he waer, = 8. m. The equaion = + a( ) gies = 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..5. IDENTIFY: The acceleraion is no consan, so we mus use calculus insead of he sandard kinemaics formulas. SET UP: The general calculus formulas are = + a d and = +. d Firs inegrae a o find (), and hen inegrae ha o find (). 3 EXECUTE: Find (): ( ) = + a d = + (.3 m/s )(5. s ) d. Carring ou he inegral and puing in he numbers gies () = 8. m/s (.3 m/s 3 )[(5. s) /]. Now use his resul o find (). 3 = + d = + 8. m/s (.3 m/s )((5. s) ) d, which gies = + (8. m/s) (.3 m/s 3 )[(7.5 s) 3 /6)]. Using = 4. m and =. s, we ge = 47.3 m. EVALUATE: The sandard kinemaics formulas appl onl when he acceleraion is consan..5. IDENTIFY: The acceleraion is no consan, bu we know how i aries wih ime. We can use he definiions of insananeous eloci and posiion o find he rocke s posiion and speed. SET UP: The basic definiions of eloci and posiion are () = + a d and d. EXECUTE: (a) ( ) = a d = (. 8 m/s ) d = (. 4 m/s ) ( 4 m/s ) ( 4667 m/s ). = d =. d =. For s, =. = 467 m. 3 3 (b) = 35 m so ( m/s ) = 35 m and = s. A his ime =.. = 3 ( 4 m/s )(8 864 s) m/s. = EVALUATE: The ime in par (b) is less han. s, so he gien formulas are alid..5. IDENTIFY: The acceleraion is no consan so he consan acceleraion equaions canno be used. Insead, use = + a d and = + d. Use he alues of and of a SET UP: EXECUTE: (a) n n+ d, for n. =. s o ealuae and. = n + 3 = + αd = + α = + (. 6 m/s ). = 5. m/s when =. s gies = 44. m/s. Then, a =. s, (b) 3 α α 6 =. +.. = m/s ( 6 m/s )( s) 6 8 m/s. = + ( + ) d = =. 4 m + (4. 4 m/s)(. s) + (. m/s )(. s) =. 8 m. =. s, 6 = 6. m a =. s gies = 4. m. Then, a (c) () = 4. m + (44. m/s) + (. m/s). ( ) = 4. 4 m/s + (. 6 m/s ). a () = (m/s).. The graphs are skeched in Figure.5. d d EVALUATE: We can erif ha a = and =. d d Figure.5 Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
23 Moion Along a Sraigh Line (a) IDENTIFY: Inegrae a( ) o find ( ) and hen inegrae ( ) o find (). SET UP: = + a d, a = A B wih 3 = + = + 3 EXECUTE: ( A B ) d A B A res a = sas ha =, so = A B = (. 5 m/s ) (. m/s ) 3 3 = (. 75 m/s ) (. 4 m/s ) SET UP: EXECUTE: d 3 A =. 5 m/s and = + ( A B ) d = + A B A he origin a = sas ha =, so = A B = (. 5 m/s ) (. m/s ) = (. 5 m/s ) (. m/s ) B =. m/s. d d EVALUATE: We can check our resuls b using hem o erif ha () = and a() =. d d d (b) IDENTIFY and SET UP: A ime, when is a maimum, d =. (Since a =, he maimum d d eloci is when a =. For earlier imes a is posiie so is sill increasing. For laer imes a is negaie and is decreasing.) d EXECUTE: a = = so A B = d One roo is =, bu a his ime = and no a maimum. 3 A 5. m/s The oher roo is = = =. 5 s 4 B. m/s A his ime = (. 75 m/s ) (. 4 m/s ) gies = (. 75 m/s )(. 5 s) (. 4 m/s )(. 5 s) = 7. m/s 78. m/s = 39. m/s. EVALUATE: For <. 5 s, a > and is increasing. For >. 5 s, a < and is decreasing..54. IDENTIFY: a () is he slope of he ersus graph and he disance raeled is he area under he ersus graph. SET UP: The ersus graph can be approimaed b he graph skeched in Figure.54 (ne page). EXECUTE: (a) Slope = a= for. 3 ms. (b) hma = Area under  graph ATriangle + ARecangle (. 3 ms)(33 cm/s) + (. 5 ms. 3 ms)(33cm/s) 5. cm 33 cm/s 5 (c) a = slope of  graph. a(. 5 ms) a(. ms) =. cm/s. 3. ms a (. 5 ms) = because he slope is zero. 3 (d) h = area under  graph. h(. 5 ms) ATriangle = (. 5 ms)(33 cm/s) = 8. 3 cm. h(. ms) ATriangle = (. ms)( cm/s) = 5. cm. h(. 5 ms) ATriangle + ARecangle = (. 3 ms)(33 cm/s) + (. ms)(33 cm/s) =. cm. Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
24  Chaper EVALUATE: The acceleraion is consan unil =. 3 ms, and hen i is zero. g = 98 cm/s. The acceleraion during he firs.3 ms is much larger han his and grai can be negleced for he porion of he jump ha we are considering. Figure IDENTIFY: The spriner s acceleraion is consan for he firs. s bu zero afer ha, so i is no consan oer he enire race. We need o break up he race ino segmens. + SET UP: When he acceleraion is consan, he formula = applies. The aerage Δ eloci is a =. Δ + +. m/s EXECUTE: (a) = = (. s) =. m. (b) (i) 4. m a. m/s so ime a consan speed is 4. s. The oal ime is 6. s, so Δ 5. m a = = = 833. m/s. Δ 6. s (ii) He runs 9. m a. m/s so he ime a consan speed is 9. s. The oal ime is. s, so m a = = 99. m/s.. s (iii) He runs 9 m a. m/s so ime a consan speed is 9. s. His oal ime is. s, so m a = = 9. 5 m/s.. s EVALUATE: His aerage eloci keeps increasing because he is running more and more of he race a his op speed..56. IDENTIFY: We know he erical posiion of he lander as a funcion of ime and wan o use his o find is eloci iniiall and jus before i his he lunar surface. d SET UP: B definiion, () =, so we can find as a funcion of ime and hen ealuae i for he d desired cases. d EXECUTE: (a) () = = c+ d. A =, () = c= 6. m/s. The iniial eloci is 6. m/s d downward. (b) () = sas b c+ d =. The quadraic formula sas = s ± s. I reaches he surface a =. 9 s. A his ime, = 6. m/s + (. 5 m/s )(. 9 s) = 5. 5 m/s. EVALUATE: The gien formula for () is of he form = + + a. For par (a), = c = 6 m/s..57. IDENTIFY: In ime S he Swaes rael a disance d = SS and in ime P he Pwaes rael a disance d = P P. SET UP: S = P + 33s EXECUTE: d d = + 33s. d = 33s and d = 5 km. S 3.5 km/s 6.5 km/s P Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
25 Moion Along a Sraigh Line  EVALUATE: The imes of rael for each wae are S = 7s and P = 38 s..58. IDENTIFY: The brick has a consan downward acceleraion, so we can use he usual kinemaics formulas. We know ha i falls 4. m in. s, bu we do no know which second ha is. We wan o find ou how far i falls in he ne.s ineral. SET UP: Le he + direcion be downward. The final eloci a he end of he firs.s ineral will be he iniial eloci for he second.s ineral. a = 9.8 m/s and he formula = + a applies. EXECUTE: (a) Firs find he iniial speed a he beginning of he firs.s ineral. Appling he aboe formula wih a = 9.8 m/s, =. s, and = 4. m, we ge = 35. m/s. A he end of his.s ineral, he eloci is = 35. m/s + (9.8 m/s )(. s) = 44.9 m/s. This is for he ne.s ineral. Using = + a wih his iniial eloci gies = 49.8 m. EVALUATE: The disance he brick falls during he second.s ineral is greaer han during he firs.s ineral, which i mus be since he brick is acceleraing downward. Δ.59. IDENTIFY: The aerage eloci is a =. Δ SET UP: Le + be upward. m 63 m EXECUTE: (a) a = = 97 m/s 4.75 s m (b) a = = 69 m/s 5.9 s 63 m EVALUATE: For he firs.5 s of he fligh, a = = 54.8 m/s. When he eloci isn.5 s consan he aerage eloci depends on he ime ineral chosen. In his moion he eloci is increasing..6. IDENTIFY: Use consan acceleraion equaions o find for each segmen of he moion. SET UP: Le + be he direcion he rain is raeling. EXECUTE: = o 4. s: = + a (.6 m/s )(4. s) 57 m. = = A = 4. s, he speed is = + a = (.6 m/s )(4. s) =.4 m/s. In he ne 7. s, a = and = = (.4 m/s)(7. s) = 568 m. For he ineral during which he rain is slowing down, =.4 m/s, a = 3.5 m/s and =. (.4 m/s) = + a( ) gies = = = 7 m. a ( 3.5 m/s ) The oal disance raeled is 57 m m + 7 m = 8 m. EVALUATE: The acceleraion is no consan for he enire moion, bu i does consis of consan acceleraion segmens, and we can use consan acceleraion equaions for each segmen..6. IDENTIFY: When he graph of ersus is a sraigh line he acceleraion is consan, so his moion consiss of wo consan acceleraion segmens and he consan acceleraion equaions can be used for each segmen. Since is alwas posiie he moion is alwas in he + direcion and he oal disance moed equals he magniude of he displacemen. The acceleraion a is he slope of he ersus graph. SET UP: For he = o =. s segmen, = 4. m/s and =. m/s. For he =. s o. s segmen, =. m/s and = m/s +. m/s EXECUTE: (a) For = o =. s, = = (. s) = 8. m.. m/s + For =. s o =. s, = (. s) =. m. (b) = 8. m +. m = 9. m The oal disance raeled is 9. m. Coprigh 6 Pearson Educaion, Inc. All righs resered. 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.
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