CHAPTER Quck Quzzes (a) 00 yd (b) 0 (c) 0 (a) False The car may be slowng down, so ha he drecon o s acceleraon s oppose he drecon o s elocy (b) True I he elocy s n he drecon chosen as negae, a pose acceleraon causes a decrease n speed (c) True For an accelerang parcle o sop a all, he elocy and acceleraon mus hae oppose sgns, so ha he speed s decreasng I hs s he case, he parcle wll eenually come o res I he acceleraon remans consan, howeer, he parcle mus begn o moe agan, oppose o he drecon o s orgnal elocy I he parcle comes o res and hen says a res, he acceleraon has become zero a he momen he moon sops Ths s he case or a brakng car he acceleraon s negae and goes o zero as he car comes o res 3 The elocy-me graph (a) has a consan slope, ndcang a consan acceleraon, whch s represened by acceleraon-me graph (e) Graph (b) represens an objec whose speed always ncreases, and does so a an eer ncreasng rae Thus, he acceleraon mus be ncreasng, and he acceleraon-me graph ha bes ndcaes hs s (d) Graph (c) depcs an objec ha rs has a elocy ha ncreases a a consan rae, whch means s acceleraon s consan The moon hen changes o one a consan speed, ndcang ha he acceleraon o he objec becomes zero Thus, he bes mach o hs suaon s graph () 4 (b) Accordng o graph b, here are some nsans n me when he objec s smulaneously a wo deren x-coordnaes Ths s physcally mpossble 5 (a) The blue graph o Fgure 4b bes shows he puck s poson as a uncon o me As seen n Fgure 4a, he dsance he puck has raeled grows a an ncreasng rae or approxmaely hree me nerals, grows a a seady rae or abou our me nerals, and hen grows a a dmnshng rae or he las wo nerals (b) The red graph o Fgure 4c bes llusraes he speed (dsance raeled per me neral) o he puck as a uncon o me I shows he puck ganng speed or approxmaely hree me nerals, mong a consan speed or abou our me nerals, hen slowng o res durng he las wo nerals (c) The green graph o Fgure 4d bes shows he puck s acceleraon as a uncon o me The puck gans elocy (pose acceleraon) or approxmaely hree me nerals, moes a consan elocy (zero acceleraon) or abou our me nerals, and hen loses elocy (negae acceleraon) or roughly he las wo me nerals 9
6 (c) The acceleraon o he ball remans consan whle s n he ar The magnude o s acceleraon s he ree-all acceleraon, g = 980 m/s 7 (c) As raels upward, s speed decreases by 980 m/s durng each second o s moon When reaches he peak o s moon, s speed becomes zero As he ball moes downward, s speed ncreases by 980 m/s each second 8 (a) and () The rs jumper wll always be mong wh a hgher elocy han he second Thus, n a gen me neral, he rs jumper coers more dsance han he second Thus, he separaon dsance beween hem ncreases A any gen nsan o me, he eloces o he jumpers are denely deren, because one had a head sar In a me neral aer hs nsan, howeer, each jumper ncreases hs or her elocy by he same amoun, because hey hae he same acceleraon Thus, he derence n eloces says he same 0
Problem Soluons Dsances raeled are = = 800 km h 0500 h = 400 km = = 00 km h 000 h = 00 km = = 400 km h 0750 h = 300 km 3 3 3 Thus, he oal dsance raeled s x = 400 + 00 + 300 km = 900 km, and he elapsed me s = 0500 h + 000 h + 0750 h + 050 h = 70 h (a) 900 km = = = 70 h 59 km h (b) x = 900 km (see aboe) (a) 0 m yr = = = 7 yr 38 356 0 s or n parcularly wndy mes 7 0 m s, 00 m yr = = = 7 yr 38 356 0 s 6 0 m s (b) The me requred mus hae been 3 3 x 3 0 m 609 m 0 mm = = = 0 mm yr m m 8 5 0 yr 3 (a) Boa A requres 0 h o cross he lake and 0 h o reurn, oal me 0 h Boa B requres 0 h o cross he lake a whch me he race s oer Boa A wns, beng 60 km ahead o B when he race ends (b) Aerage elocy s he ne dsplacemen o he boa dded by he oal elapsed me The wnnng boa s back where sared, s dsplacemen hus beng zero, yeldng an aerage elocy o zero 4 (a) 0 m yr = = = 7 4000 yr 38 356 0 s - 5 0 m s
m day (b) = = 4 = day 38 864 0 s -6 7 0 m s 350 Dsplacemen = 850 km h h + 30 km 600 x = 496 + 30 km = 80 km 5 (a) (b) ( + ) ( 350 + 50) Dsplacemen 496 30 km Aerage elocy = = = elapsed me + 00 h 600 634 km h 6 (a) 00 m = = = 00 s 500 m s x(m) (b) (c) (d) (e) 500 m = = = 5 m s 400 s 500 m 00 m = = = 400 s 00 s 500 m 500 m = = = 700 s 400 s x x 0 0 800 s 0 = = = = 50 m s 333 m s 0 0 8 6 4 0 4 6 3 4 5 6 7 8 (s) 7 (a) (b) (c) x x 40 m 0 0 s 0 0, = = = x x 0 m 0 4 0 0,4 = = = x x 40 s 0 40 m 40 s 5,5 = = = + 40 m s 05 m s 0 m s x(m) 4 3 3 4 5 (s) (d) x x 0 0 50 s 5 0 0,5 = = = 0
8 (a) The me or a car o make he rp s = Thus, he derence n he mes or he wo cars o complee he same 0 mle rp s 0 m 0 m 60 mn = = = = 3 mn 55 m h 70 m h h (b) When he aser car has a 50 mn lead, s ahead by a dsance equal o ha raeled by he slower car n a me o 50 mn Ths dsance s gen by x = ( ) = ( 55 m h)( 5 mn) The aser car pulls ahead o he slower car a a rae o: = 70 m h 55 m h = 5 m h Thus, he me requred or o ge dsance relae x ahead s: 55 m h 5 mn = = = 55 mn 50 m h relae Fnally, he dsance he aser car has raeled durng hs me s h x = = ( 70 m h )( 55 mn ) = 60 mn 64 m 9 (a) To correspond o only orward moon, he graph mus always hae a pose slope Graphs o hs ype are a, e, and (b) For only backward moon, he slope o he graph mus always be negae Only graph c s o hs ype (c) The slope o he graph mus be consan n order o correspond o consan elocy Graphs lke hs are d and (d) The graph correspondng o he larges consan elocy s he one wh he larges consan slope Ths s graph (e) To correspond o no moon (e, zero elocy), he slope o he graph mus hae a consan alue o zero Ths s graph d 3
0 The dsance raeled by he space shule n one orb s 4 π π Earh s radus + 00 mles = 3963 + 00 m = 6 0 m Thus, he requred me s 4 6 0 m = 9 800 m h 3 h The oal me or he rp s = + 0 mn = + 0367 h, where s he me spen raelng a 895 km h Thus, he dsance raeled s ( 895 km h ) ( 778 km h )( 0367 h ) x = = = + or, 895 km h = 778 km h + 85 km From whch, = 44 h or a oal me o = + 0367 h = 80 h Thereore, x= = ( 778 km h)( 80 h) = 8 km (a) A he end o he race, he orose has been mong or me and he hare or a me 0 mn = 0 s The speed o he orose s = 000 m s, and he speed o he hare s h = 0 = 0 m s The orose raels dsance x, whch s 00 m larger han he dsance x h raeled by he hare Hence, x = xh + 00 m, whch = 0 s + 00 m or becomes h ( ) 000 m s = 0 m s 0 s + 00 m Ths ges he me o he race as = (b) x ( 000 m s)( 3 0 s ) = = = 3 m 3 0 s 4
3 The maxmum me o complee he rp s oal dsance 600 m km h = = =30 s requred aerage speed 50 km h 078 m s The me spen n he rs hal o he rp s hal dsance 800 m km h = = 5 s 30 km h = 078 m s Thus, he maxmum me ha can be spen on he second hal o he rp s = - = 30 s 5 s = 05 s, and he requred aerage speed on he second hal s hal dsance 800 m km h = = = 76 m s = 05 s 078 m s 74 km h 4 Choose a coordnae axs wh he orgn a he lagpole and eas as he pose drecon Then, usng x = x + + a wh a = 0 or each runner, he x-coordnae o each runner a me s B x = 40 m + 60 m h and x = 30 m + 50 m h A When he runners mee, x A = x B or 40 m ( 60 m h) 30 m ( 50 m h) + = + 70 m Ths ges he elapsed me when hey mee as = = 064 h A hs me, 0 m h x A = x = 08 m Thus, hey mee 08 m wes o he lagpole B 5 (a) (b) (c) 00 m 0 = = 500 m s 00 s 0 ( ) 500 00 m = = 50 m s 400 00 s ( ) ( ) ( ) 500 500 m = = 0 500 400 s x(m) 0 8 6 4 0 4 6 3 4 5 6 7 8 (s) 5
(d) 0 500 m = = 800 700 s ( ) 500 m s 6 (a) A ew ypcal alues are (s) x(m) 00 575 00 60 300 353 400 680 500 9 600 9 (b) We wll use a 0400 s neral cenered a = 400 s We nd a = 380 s, x = 60 m and a = 40 s, x = 766 m Thereore, 64 m = = = 40 m s 0400 s x(m) 00 80 60 40 0 00 80 60 40 0 0 000 00 00 300 400 500 600 (s) Usng a me neral o 000 s, we nd he correspondng alues o be: a = 390 s, 80 m x = 640 m and a = 40 s, x = 7 m Thus, = = = 40 m s 000 s For a me neral o 000 s, he alues are: a = 395 s, x = 660 m, and a 40 m = 405 s, x = 70 m Thereore, = = = 40 m s 000 s 680 m (c) A = 4 00 s, x = 680 m Thus, or he rs 400 s, = = = 400 s alue s much less han he nsananeous elocy a = 400 s 70 m s Ths 7 (a) (b) (c) x x 40 m 0 s 0 0 s 0 s = 0 = = = 050 s 0 s 30 s x x 5 s 0 s 60 m = = = 5 s 0 s 5 s x x 40 s 5 s 0 = = = 0 40 s 5 s 5 s 40 m s 40 m s x(m) 4 3 3 4 5 (s) (d) x x + 0 m 50 s 40 s = = 4 5 s 50 s 40 s 0 s = 0 m s 6
8 (a) The aerage speed durng a me neral s dsance raeled = Durng he rs quarer mle segmen, Secreara s aerage speed was 050 m 30 5 s 5 s = = = 54 s 356 m h Durng he second quarer mle segmen, 30 40 s = = 550 s 374 m h For he hrd quarer mle o he race, 30 38 s 3 = = 555 s 377 m h, and durng he nal quarer mle, 30 30 s 4 = = 574 s 390 m h (b) Assumng ha = 4 and recognzng ha = 0, he aerage acceleraon durng he race was 574 s 0 a = = = oal elapsed me 5+40+38+30 s 0598 s 9 0 600 m s a = = = 50 s 0 400 m s The negae sgn n he aboe resul ndcaes ha he acceleraon s n he negae x drecon 0 ( 80 ms) ( 00 ms) a = = = 0 s 3 5 0 m s From a =, we hae ( ) 60 55 m h 0447 m s = = a 060 m s = m h 37 s 7
(a) From = 0 o = 50 s, (m/s) 0 0 a = = = 0 50 s 0 From = 50 s o = 5 s, 80 m s 80 m s a = = 5 s 50 s and rom = 0 o = 0 s, 6 m s, 8 6 4 4 6 8 5 0 5 0 (s) 80 m s 80 m s a = = 0 s 0 080 m s (b) A = 0 s, he slope o he angen lne o he cure s 0 A = 0 s, he slope o he angen lne s 6 m s, and a = 8 s, he slope o he angen lne s 0 3 (a) The aerage acceleraon can be ound rom he cure, and s alue wll be 6 m s a = = = 0 s 80 m s (b) The nsananeous acceleraon a = 5 s equals he slope o he angen lne o he cure a ha me Ths slope s abou m s 4 The dsplacemen whle comng o res s 3 x = 0 km = 0 0 m The nal speed s 300 km h = and he nal speed s 0 Thereore, rom a( x) (m/s) 6 8 4 0 000 050 00 50 00 (s) = +, ( x) 3 0 300 km h 078 m s a = = = 0 0 m km h 90 m s 5 From a( x) a = 3 = +, we hae ( 097 0 m s) 0 a( 0 m ) 5 74 0 m s whch s = + so ha 4 79 0 mes g! 8
6 (a) + x = = 80m s + = becomes 400 m ( 850 s), (b) whch yelds = 66 m s 80 m s 66 m s a = = = 850 s 0448 m s 7 Suppose he unknown acceleraon s consan as a car nally mong a = 350 m h comes o a = 0 sop n x = 400 We nd s acceleraon rom = + a x ( x) 0 350 m h 47 s a = = = 33 s 400 m h Now consder a car mong a = 700 m h and soppng o = 0 wh a = 33 s From he same equaon s soppng dsance s ( ) 0 700 m h 47 s x = = a 33 s m h = 60 8 (a) From he denon o acceleraon, we hae 0 40 m s a = = = 50 s 80 m s (b) From x = + a, he dsplacemen s x = ( 40 m s)( 50 s) + ( 80 m s )( 50 s) = 00 m 9 (a) Wh 0 km h =, a( x) = + yelds ( x) 0 km h 0 078 m s a = = = 40 m km h 3 m s 9
(b) The requred me s 0 km h 0 078 m s = = a 3 m s = km h 44 s 30 (a) The me or he ruck o reach 0 m s s ound rom = + a as 0 m s 0 = = = 0 s a 0 m s The oal me s oal = 0 s+ 0 s+ 50 s= 35 s (b) The dsance raeled durng he rs 0 s s 0+ 0 m s = = 0 s = 00 m ( x) The dsance raeled durng he nex 0 s (wh a = 0) s = + = 0 m s 0 s + 0 = 400 m ( x) ( ) a The dsance raeled n he las 50 s s 0 m s+0 = 3 3 3= 50 s = 50 m ( x) The oal dsplacemen s hen + + x = x = 00 m + 400 m + 50 m=550 m 3 and he aerage elocy or he oal moon s gen by 550 m = = = 6 m s 35 s oal 3 (a) Usng = 894 s x = + a wh 0 = ges 400 m 0 ( 00 m s ) = +, yeldng (b) From = + a, wh 0 =, we nd 0 ( 00 m s )( 894 s) = + = 894 m s 30
0 00 m s 3 (a) The me requred o sop s = = = a 500 m s 00 s (b) The mnmum dsance needed o sop he plane s + 0+ 00 m s x = = = ( 00 s) = 000 m = 00 km Thus, he plane canno sop n 08 km 33 Usng a( x) = +, wh = 0 and = 60 m h, yelds ( x) 0 60 m h 0447 m s a = = = 00 m m h 36 m s 34 The elocy a he end o he rs neral s = + a = 0 + (77 m s) 50 s = 46 m s Ths s also he consan elocy durng he second neral and he nal elocy or he hrd neral (a) From x = + a, he oal dsplacemen s x = x + x + x 3 = 0 + ( 77 m s ) ( 50 s) + ( 46 m s)( 3 s) + 0 + + ( 46 m s)( 439 s) ( 947 m s ) ( 439 s) or 3 3 x = 3 m + 5 0 m + 9 m = 55 0 m = 55 km 3
(b) ( ) 3 m = = = 08 m s, 50 s 3 ( x) 5 0 m = = = 46 m s, 3 s 3 ( ) 3 9 m = = = 08 m s, and he aerage elocy or he 439 s 3 oal rp s oal 3 x 55 0 m = = = 387 m s 50+3+439 s oal 35 Usng he unormly acceleraed moon equaon yelds x = ( 0 m s)( 40 s) + ( 0 m s )( 40 s) = 0, whch s obously wrong x = + a or he ull 40 s neral The source o he error s ound by compung he me requred or he ran o come o 0 0 m s res Ths me s = = = 0 s Thus, he ran s slowng down or he a 0 m s rs 0 s and s a res or he las 0 s o he 40 s neral The acceleraon s no consan durng he ull 40 s I s, howeer, consan durng he rs 0 s as he ran slows o res Applcaon o x = + a o hs neral ges he soppng dsance as x = ( 0 m s)( 0 s) + ( 0 m s )( 0 s) = 00 m 36 = 400 m h = 79 m s and = 0 (a) To nd he dsance raeled, we use + 79 m s + 0 x = = = ( 0 s ) = 07 m (b) The acceleraon s 79 m s 0 a = = = 0 s 49 m s 3
37 A he end o he acceleraon perod, he elocy s = + a = 0 + 5 m s 50 s = 75 m s Ths s also he nal elocy or he brakng perod (a) Aer brakng, a 75 m s ( 0 m s )( 30 s) (b) The oal dsance raeled s = + = + = 5 m s x = x + x = + accel brake accel brake 0+ 75 ms 75 ms+ 5 ms x = ( 50 s) + ( 30 s) = 3 m 38 The nal elocy o he ran s = 84 km h and he nal elocy s = 64 km h The me requred or he 400 m ran o pass he crossng s ound rom + x = = as ( x) 0400 km 3600 s = = = 9 s + 84 + 64 km h h 39 (a) Take = 0 a he me when he player sars o chase hs opponen A hs me, he opponen s 36 m n ron o he player A me > 0, he dsplacemens o he players rom her nal posons are xplayer = ( ) + aplayer = 0+ ( 40 m s player ), () x = + a = m s + 0 () opponen and, opponen ( ) opponen When he players are sde-by-sde, x = x + 36 m (3) player opponen From Equaons (), (), and (3), we nd 60 s 8 s = 0 whch has soluons o = s and =+ 8 s Snce he me mus be greaer han zero, we mus choose = 8 s as he proper answer player = + player player = + = (b) x ( ) a 0 ( 40 m s )( 8 s) 3 0 m 33
40 Takng = 0 when he car sars aer he ruck, he dsplacemens o he ehcles rom her nal posons a me > 0 are: ( xcar = + acar = 0+ 5 m s car ), x = + a = 40 km h + 0 ruck and ruck ( ) ruck When he car oerakes he ruck, xcar = xruck, or ( 5 m s ) = ( 40 km h ) Ths has a soluon = 0 descrbng he nal suaon and a second soluon 078 m s = ( 40 km h ) 5 m s = km h 89 s The dsance he car has raeled beore cachng he ruck s x ( 5 m s )( 89 s ) car =+ = 99 m 4 The dsance he car raels a consan elocy, 0, durng he reacon me s x = 0 r The me or he car o come o res, rom nal elocy, aer he brakes are appled 0 0 0 s = = = and he dsance raeled durng hs brakng perod s a a a + 0 + 0 0 0 x = = = = a a Thus, he oal dsance raeled beore comng o a sop s 0 sop s = x + x = 0 0 r a 34
0 4 (a) I a car s a dsance ssop = 0 r (See he soluon o Problem 4) rom he a nersecon o lengh s when he lgh urns yellow, he dsance he car mus rael beore he lgh urns red s x = s + s = +s a 0 sop 0 r Assume he drer does no accelerae n an aemp o bea he lgh (an exremely dangerous pracce!) The me he lgh should reman yellow s hen he me requred or he car o rael dsance x a consan elocy Ths s 0 lgh 0r 0 a+ s = = = 0 0 s 0 r a + 0 (b) Wh s = 6 m, 0 = 60 km h, a = 0 m s, and = s, r lgh 60 km h 078 m s 6 m km h = s + = 0 m s km h 60 km h 078 m s 6 s 43 (a) From a( y) = + wh = 0, we hae ( y) ( ) ( ) 0 50 m s = = = 39 m max a 980 m s (b) The me o reach he hghes pon s up 0 50 m s = = = 55 s a 980 m s (c) The me requred or he ball o all 39 m, sarng rom res, s ound rom y = ( 0) + a as ( y) ( ) 39 m = = = 55 s a 980 m s (d) The elocy o he ball when reurns o he orgnal leel (55 s aer sars o all rom res) s = + a = 0+ 980 m s 55 s = 50 m s 35
44 From a( y) = +, wh = 0, = 9 000 km h, and y =+ 8 m, ( y) 9 0 km h 0 078 m s a = = 8 m km h 4 = 6 8 0 m s 45 Assume he whales are raelng sragh upward as hey leae he waer Then = + a y, wh = 0 when y =+ 75 m, ges = a y = 0 98 m s 75 m = m s 46 Use y = + a, wh = 0, a= 980 m s, and y = 760 m o nd ( y) ( ) 760 m = = = 394 s a 980 m s 47 (a) Aer 00 s, he elocy o he malbag s ( ) a bag = + = 50 m s + 980 m s 00 s = m s The negae sgn ells ha he bag s mong downward and he magnude o he elocy ges he speed as m s (b) The dsplacemen o he malbag aer 00 s s ( y) + m s + 50 m s = ( 00 s) 6 m bag = = Durng hs me, he helcoper, mong downward wh consan elocy, undergoes a dsplacemen o = + = 5 m s 00 s + 0 = 300 m coper ( y) a Durng hs he end, he malbag s 00 s, boh he malbag and he helcoper are mong downward A 6 m 300 m= 96 m below he helcoper 36
(c) Here, ( ) ( ) 50 m s bag = =+ and coper 00 s, he speed o he malbag s a = 0 a bag = 980 m s whle coper Aer ( ) bag m m m = 50 + 980 00 s = 8 s s s = m 8 s In hs case, he helcoper rses 300 m durng he 00 s neral whle he malbag has a dsplacemen o 8 m s + 50 m s = 00 s 66 m bag = ( y) rom he release pon Thus, he separaon beween he wo a he end o 300 m ( 66 m ) = 96 m 00 s s 48 (a) Consder he relaon y = + a wh a= g When he ball s a he hrowers hand, he dsplacemen y s zero, or 0 = g Ths equaon has wo soluons, = 0 whch corresponds o when he ball was hrown, and = g correspondng o when he ball s caugh Thereore, he ball s caugh a = 00 s, he nal elocy mus hae been ( 980 m s )( 00 s) g = = = 980 m s (b) From a( y) = +, wh = 0 a he maxmum hegh, ( y) ( ) ( ) 0 980 m s = = = 490 m max a 980 m s 37
49 (a) When reaches a hegh o 50 m, he speed o he rocke s = + a y = 500 m s + 00 m s 50 m = 557 m s Aer he engnes sop, he rocke connues mong upward wh an nal elocy o = 557 m s and acceleraon a= g= 980 m s When he rocke reaches maxmum hegh, = 0 The dsplacemen o he rocke aboe he pon where he engnes sopped (e, aboe he 50 m leel) s ( ) ( ) 0 557 m s y = = = 58 m a 980 m s The maxmum hegh aboe ground ha he rocke reaches s hen gen by h = 50 m + 58 m= 308 m max (b) The oal me o he upward moon o he rocke s he sum o wo nerals The rs s he me or he rocke o go rom = 500 m s a he ground o a elocy o = 557 m s a an alude o 50 m Ths me s gen by ( y ) 50 ( m) = = = ( 557 + 500 ) m s 84 s The second neral s he me o rse 58 m sarng wh wh = 0 Ths me s = 557 m s and endng ( y) 58 ( m) 567 s = = = 0+557 m s The oal me o he upward lgh s hen up = + = 84 + 567 s = 85 s 38
(c) The me or he rocke o all 308 m back o he ground, wh a= g= 980 m s, s ound rom y = + a as down ( y) ( ) 308 m = = = 793 s, a 980 m s = 0 and acceleraon so he oal me o he lgh s lgh = up + down = 85 + 793 s = 64 s 50 (a) The camera alls 50 m wh a ree-all acceleraon, sarng wh = 0 m s Is elocy when reaches he ground s = + a y = 0 ms + 980 ms 50 m = 33 ms The me o reach he ground s gen by 33 ms 0 ms = = = 3 s a 980 m s (b) Ths elocy was ound o be = 33 m s n par (a) aboe 5 (a) The keys hae acceleraon caugh 50 s laer Thus, a = g= -980 m s rom he release pon unl hey are y = + a ges ( + ) ( ) y a 400 m 980 m s 50 s = = =+ 00 m s, 50 s or = 00 m s upward (b) The elocy o he keys jus beore he cach was = + a = 00 m s + 980 m s 50 s = 468 m s, or 468 m s downward 39
5 In hs case, = +, y = + a yelds 300 m ( 800 m s) ( 980 m s ) or 490 + 800 s + 300 s = 0 Usng he quadrac ormula o sole or he me ges 800 s ± 800 s 4 490 300 s = 490 Snce he me when he ball reaches he ground mus be pose, we use only he pose soluon o nd = 79 s 53 Durng he 0600 s requred or he rg o pass compleely ono he brdge, he ron bumper o he racor moes a dsance equal o he lengh o he rg a consan elocy o = 00 km h Thereor he lengh o he rg s Lrg km 078 m s = = 00 ( 0600 s) 67 m h = km h Whle some par o he rg s on he brdge, he ron bumper moes a dsance x = L + L = 400 m + 67 m Wh a consan elocy o = 00 km h, he me or brdge rg hs o occur s Lbrdge + Lrg 400 m + 67 m km h = = = 00 km h 078 m s 50 s 54 (a) From = + x = + a, we hae 00 m ( 300 m s) ( 350 m s ) Ths reduces o 350 + 600 s + 00 s = 0, and he quadrac ormula ges ( 600 s) ( 600 s) 4( 350)( 00 s ) ± = 350 The desred me s he smaller soluon o = 453 s The larger soluon o = 6 s s he me when he boa would pass he buoy mong backwards, assumng mananed a consan acceleraon 40
(b) The elocy o he boa when rs reaches he buoy s = + a = 300 m s+ 350 m s 453 s = 4 m s 55 (a) The acceleraon o he bulle s ( ) ( 300 m/ s) ( 400 m/ s) a = = = 000 m 5 350 0 m s (b) The me o conac wh he board s ( ) 300 400 m s = = = 5 a 350 0 m s 4 86 0 s 56 We assume ha he bulle s a cylnder whch slows down jus as he ron end pushes apar wood bers (a) The acceleraon s ( ) ( 80 m s) ( 40 m s) a = = = 000 m 5-490 0 m s (b) The aerage elocy as he ron o he bulle passes hrough he board s + board = = 350 m s and he oal me o conac wh he board s ( ) L 000 m 0000 m board bulle = + = + = 350 m s 80 m s board 4 357 0 s (c) From a( x) = +, wh = 0, ges he requred hckness as 0 ( 40 m s) 5 ( ) x = = = a 490 0 m s 080 m 4
57 The allng ball moes a dsance o ( 5 m h) beore hey mee, where h s he hegh aboe he ground where hey mee Apply ( 5 m h) = 0 g, or h 5 m y = + a, wh a g = o oban = g () Applyng y = + a o he rsng ball ges h ( 5 m s) g = () Combnng equaons () and () ges 5 m = = 060 s 5 m s 58 The dsance requred o sop he car aer he brakes are appled s m 47 s 0 350 h m h = = = 47 sop a 900 s ( x) ( ) Thus, he deer s no o be h, he maxmum dsance he car can rael beore he brakes are appled s gen by ( x) ( x) = 00 = 00 47 = 530 beore sop Beore he brakes are appled, he consan speed o he car s 350 m/h Thus, he me requred or o rael 530, and hence he maxmum allowed reacon me, s ( ) r max ( ) 530 beore = = = 03 s m 47 s 350 h m h 4
59 (a) When eher ball reaches he ground, s ne dsplacemen s y = 96 m Applyng y = + a o he moon o he rs ball ges 96 m = ( 47 m s) + ( 980 m s ) whch has a pose soluon o = 00 s Smlarly, applyng hs relaon o he moon o he second ball ges 96 m = ( + 47 m s) + ( 980 m s ) whch has a sngle pose soluon o = 400 s Thus, he derence n he me o lgh or he wo balls s = = ( 400 00 ) s= 300 s (b) When he balls srke he ground, her eloces are: and = g = 47 m s 980 m s 00 s = 45 m s, = g =+ 47 m s 980 m s 400 s = 45 m s (c) A = 0800 s, he dsplacemen o each ball rom he balcony s: y = y 0 = g = 47 m s 0800 s 490 m s 0800 s, y = y 0 = g = + 47 m s 0800 s 490 m s 0800 s These yeld wo balls a hs me s y = 49 m and y =+ 86 m Thereore he dsance separang he d= y y = 86 m 49 m = 35 m 43
60 We do no know eher he nal elocy nor he nal elocy (e, elocy jus beore mpac) or he ruck Wha we do know s ha he ruck skds 64 m n 40 s whle accelerang a 560 m s We hae yeld = + a and + x = = Appled o he moon o he ruck, hese = a or = 560 m s 40 s = 35 m s, () and 64 m + = = = 97 m s () 40 s Addng equaons () and () ges he elocy a momen o mpac as = 35 + 97 m s, or = 30 m s 6 When Kahy has been mong or seconds, San s elapsed me s + 00 s A hs me, he dsplacemens o he wo cars are x = + a = +, 0 ( 490 m s Kahy K K ) = S + San S + = + + and ( x) a ( 00 s) 0 ( 350 m s )( 00 s) (a) When Kahy oerakes San, ( x) ( x) =, or Kahy ( 490 m s ) = ( 350 m s )( + 00 s) whch ges he me as = 546 s (b) Kahy s dsplacemen a hs me s San = Kahy 490 m s 546 s = 730 m ( x) 44
(c) A hs me, he eloces o he wo cars are ( ) 0 ( 490 m s )( 546 s K ak ) Kahy = + = + = 67 m s, and ( ) a ( 00 s) 0 ( 350 m s )( 646 s) = + + = + = 6 m s S S San 6 (a) The elocy wh whch he rs sone hs he waer s m m = + a( y) = 00 + 980 ( 500 m) = 34 s s m s The me or hs sone o h he waer s 34 m s 00 m s = = = 300 s a -980 m s (b) Snce hey h smulaneously, he second sone whch s released 00 s laer, wll h he waer aer an lgh me o 00 s Thus, y a 500 m 980 m s 00 s = = = 5 m s 00 s (c) From par (a), he nal elocy o he rs sone s = 34 m s The nal elocy o he second sone s = + a = 5 m s + 980 m s 00 s = 348 m s 63 (a) A 500 m aboe he surace, he elocy o he asronau s gen by = + a y = 0 + 00 m s 500 m = + 447 m s (b) The bol begns ree-all wh an nal elocy o =+ 447 m s a 500 m aboe he surace When reaches he surace, Thus, he bol hs 63 s aer s released y = + a ges 500 m = + 447 m s + 67 m s whch has a pose soluon o 63 s = 45
(c) When reaches he surace, he elocy o he bol s = + a =+ 447 m s + 67 m s 63 s = 606 m s (d) The dsplacemen o he asronau whle he bol s allng s Thereore, he alude o he asronau when he bol hs s y = + a = 447 m s 63 s + 00 m s 63 s = 680 m y = 500 m + 680 m = 730 m (e) The elocy o he asronau when he bol hs s = + a =+ 447 m s + + 00 m s 63 s = + 7 m s 64 (a) From y = + a wh = 0, we hae ( y) ( ) 3 m = = = s a 980 m s (b) The nal elocy s 0 ( 980 m s m s )( s) = + = m s (c) The me ake or he sound o he mpac o reach he specaor s y 3 m sound = = = 68 0 s, sound 340 m s so he oal elapsed me s oal = s + 68 0 s 3 s 65 (a) Snce he sound has consan elocy, he dsance raeled s ( 00 s)( 50 s) x = sound = = 3 55 0 (b) The plane raels hs dsance n a me o 50 s + 0 s = 5 s, so s elocy mus be 3 x 55 0 plane = = = 37 0 s 5 s 46
(c) The me he lgh ook o reach he obserer was lgh 3 x 55 0 m s = = = 8 300 0 m s 38 s lgh 6 56 0 s Durng hs me he plane would only rael a dsance o 000 66 The oal me or he sae o reach he ground s ound rom y = + a wh 0 = as oal ( y) ( ) 50 m = = = 6 s a 980 m s The me o all he rs een meers s ound smlarly: ( y) ( ) 50 m = = = 75 s a 980 m s 5 The me Wle E Coyoe has o reach saey s = oal 5 = 6 s 75 s = 0509 s 67 The me requred or he woman o all 300 m, sarng rom res, s ound rom y = + a as 300 m = 0 + ( 980 m s ), gng = 078 s (a) Wh he horse mong wh consan elocy o 00 m s, he horzonal dsance s ( 00 m s)( 078 s) x = horse = = 78 m (b) The requred me s = 078 s as calculaed aboe 47
68 Assume ha he ball alls 5 m, rom res, beore ouchng he ground Furher, assume ha aer conac wh he ground he ball moes wh consan acceleraon or an addonal 050 cm (hence compressng he ball) beore comng o res Wh he rs assumpon, a( y) ouches he ground as = + ges he elocy o he ball when rs = + a y = 0 + 980 m s 5 m = 54 m s Then, usng he second assumpon, he acceleraon whle comng o res s ound o be ( y) ( ) ( 3 ) 0 54 m s a = = = 9 0 m s 50 0 m 3, or 3 ~0 m s 48
Answers o Een Numbered Concepual Quesons Yes Zero elocy means ha he objec s a res I he objec also has zero acceleraon, he elocy s no changng and he objec wll connue o be a res 4 You can gnore he me or he lghnng o reach you because lgh raels a he speed o 3 x 0 8 m/s, a speed so as ha n our day-o-day aces s essenally nne 6 The aerage elocy o an objec s dened as he dsplacemen o he objec dded by he me neral durng whch he dsplacemen occurred I he aerage elocy s zero, he dsplacemen mus also be zero 8 In Fgure (b), he mages are equally spaced showng ha he objec moed he same dsance n each me neral Hence, he elocy s consan n (b) In Fgure (c,) he mages are arher apar or each successe me neral The objec s mong oward he rgh and speedng up Ths means ha he acceleraon s pose n (c) In Fgure (a), The rs our mages show an ncreasng dsance raeled each me neral and hereore a pose acceleraon Howeer, aer he ourh mage, he spacng s decreasng showng ha he objec s now slowng down (or has negae acceleraon) 0 Veloces are equal only boh magnude and drecon are he same These objecs are mong n deren drecons, so he eloces are no he same The rule o humb assumes consan elocy I he car(s) moe wh consan acceleraon, he elocy would connually be changng Ths would mean he dsance beween he cars would connually hae o change or he rule o humb o be ald, whch could requre a slowng down, whch would mply a change n he alue o he acceleraon 4 (a) The car s mong o he eas and ncreasng n speed (b) The car s mong o he eas bu slowng n speed (c) The car s mong o he eas a consan speed (d) The car s mong o he wes bu slowng n speed (e) The car s mong o he wes and speedng up () The car s mong o he wes a consan speed (g) The car sars rom res and begns o speed up oward he eas (h) The car sars rom res and begns o speed up oward he wes 6 The balls speed up and slow down a he same rae, bu he dsances o rael are deren n deren me nerals For example, he rsng ball sars a a hgh speed and slows down Ths means ha raels a longer dsance n, say, he rs second han does he slower mong dropped ball Thus, he balls wll mee aboe he mdway pon 8 (a) Successe mages on he lm wll be separaed by a consan dsance he ball has consan elocy 49
(b) Sarng a he rgh-mos mage, he mages wll be geng closer ogeher as one moes oward he le (c) Sarng a he rgh-mos mage, he mages wll be geng arher apar as one moes oward he le (d) As one moes rom le o rgh, he balls wll rs ge arher apar n each successe mage, hen closer ogeher when he ball begns o slow down 50
Answers o Een Numbered Problems (a), 7 0 m s 8 (b) 5 0 yr 6 0 m s 4 (a) (b) 5 0 m s 6 7 0 m s 6 (a) 500 m s (b) 5 m s (c) 50 m s (d) 333 m s (e) 0 8 (a) 3 mn (b) 64 m 0 3 h (a) 3 0 s (b) 3 m 4 08 m wes o he lagpole 6 (b) 40 m/s, 40 m/s, 40 m/s (c) = 70 m s, much less han he resuls o (b) 8 (a) 54 s, 550 s, 555 s, 574 s (b) 0598 s 0 3 5 0 m s (a) 0, 6 m s, 080 m s (b) 0, 6 m s, 0 4 90 m s 6 (a) 66 m s (b) 0448 m s 8 (a) 80 m s (b) 00 m 30 (a) 35 s (b) 6 m s 3 (a) 00 s (b) No, he mnmum dsance o sop = 00 km 34 (a) 55 km (b) 08 m s, 46 m s, 08 m s ; 387 m s 36 (a) 07 m, (b) 49 m s 38 9 s 40 89 s, 99 m 5
4 (b) 6 s 44 6 8 0 m s 46 394 s 48 (a) 980 m s (b) 490 m 50 (a) 3 s (b) 33 m s 5 79 s 54 (a) 453 s (b) 4 m s 56 (a) 5 4 490 0 m s (b) 357 0 s (c) 080 m 58 03 s 60 30 m s 6 (a) 300 s (b) 5 m s (c) 34 m s, 348 m s 64 (a) s (b) m s (c) 3 s 66 05 s 68 3 ~0 m s, assumes he ball drops 5 m and compresses 05 cm upon hng he loor 5