Answers o Een Numbered Problems Chper. () 7 m s, 6 m s (b) 8 5 yr 4.. m ih 6. () 5. m s (b).5 m s (c).5 m s (d) 3.33 m s (e) 8. ().3 min (b) 64 mi..3 h. ().3 s (b) 3 m 4..8 mi wes of he flgpole 6. (b) 4. m/s, 4. m/s, 4. m/s (c) = 7. m s, much less hn he resuls of (b) 8. () 5.4 fs, 55. fs, 55.5 fs, 57.4 fs (b).598 fs..75 m s. (),.6 m s,.8 m s (b),.6 m s, 4..33 m s 6. () 6.6 m s (b) 8. ().5 s (b).448 m s.9 m s (c) 3. s 3. () 35 s (b) 6 m s 3. (). s (b) No, he minimum disnce o sop =. km 34. () 5.5 km (b).8 m s, 4.6 m s,.8 m s; 38.7 m s 36. () 7 m (b).49 m s 38. 9. s 4. Ide () is no rue unless he ccelerion is zero. Ide (b) is rue for ll consn lues of ccelerion. 7
8 CHAPTER 4. 96 m 44. () 5 m (b).4 s 3 46. Hrdwood Floor: =. m s, =.4m s Crpeed Floor: = 3.9 m s, = 7.m s 48. () 9.8 m s (b) 4.9 m 5. ().3 s (b) 33 m s 5. () 4. m s,. m s (b).8 m 54. () 4.53 s (b) 4. m s 56. () 5 4.9 m s (b) 4 3.57 s (c).8 m 58..3 s 6. 3. m s 6. () 3. s (b) 5. m s (c) 3.4 m s, 34.8 m s 64. (). s (b) m s (c).3 s 66..5 s 68. 3 ~ m s, ssumes he bll drops.5 m nd compresses. cm upon hiing he floor 7. Yes, her mximum ccelerion is more hn sufficien.
Moion in One Dimension 9 35. Displcem en = 85. km h h + 3 km 6..5 () x = 49.6 + 3 km = 8 km (b) ( + ) ( + ) Displcem en 49.6 3 km Aerge elociy elpsed im e 35. 5. +. h 6. 63.4 km h. m.6 (). s (b) (c) (d) 5. m 4. s 5. m. m 4. s. s 5. m s.5 m s 5. m 5. m 7. s 4. s x x (e) = 8. s.5 m s 3.33 m s x(m) 8 6 4 4 6 3 4 5 6 7 8 (s) 4. m.7 (). s (b) (c). m 4. s 4. m 5. s. s + 4. m s.5 m s. m s x(m) 4 3 3 4 5 (s) (d) 5. s. The ol ime for he rip is = +. min= +.367 h, where is he ime spen reling 89.5 km h. Thus, he disnce reled is ( 89.5 km h) ( 77.8 km h)(.367 h) x= = = + or, 89.5 km h = 77.8 km h + 8.5 km
CHAPTER From which, =.44 h for ol ime of = +.367 h =.8 h Therefore, = ( 77.8 km h)(.8 h) x = = 8 km.3 The mximum ime o complee he rip is oldisnce 6 m km h = = = 3. s required erge speed 5 km h.78 m s The ime spen in he firs hlf of he rip is hlfdisnce 8 m km h = = =.5 s ( ) 3 km h.78 m s Thus, he mximum ime h cn be spen on he second hlf of he rip is = - = 3. s.5 s=.5 s, nd he required erge speed on he second hlf is ( ) hlfdisnce 8 m km h 76. m s =.5 s.78 m s 74 km h. From =, we he ( ) 6 55 m ih.447 m s = = =.6 m s m ih 3.7 s. () From = o = 5. s, (m/s) From 5. s = 5. s o = 5 s, 8. m s 8. m s = = 5 s 5. s nd from = o = s, 8. m s 8. m s = = s.6 m s.8 m s 8 6 4 4 6 8 5 5 (s) (b) A =. s, he slope of he ngen line o he cure is. A = s, he slope of
Moion in One Dimension he ngen line is.6 m s, nd = 8 s, he slope of he ngen line is..3 () The erge ccelerion cn be found from he cure, nd is lue will be 6 m s. s 8. m s 8 (b) The insnneous ccelerion =.5 s equls he slope of he ngen line o he cure h 4 ime. This slope is bou m s...5..5. (s).3 () The ime for he ruck o rech m s is found from = + s (m/s) 6 m s s. m s The ol ime is ol = s+ s+ 5. s= 35 s (b) The disnce reled during he firs s is + m s = = s = m ( x) ( ) The disnce reled during he nex s (wih = ) is = + = m s s + = 4 m ( x) ( ) The disnce reled in he ls 5. s is m s+ x = 3 3 3 = 5. s = 5 m The ol displcemen is hen + + x= x = m + 4 m + 5 m = 55 m 3 nd he erge elociy for he ol moion is gien by
CHAPTER 55 m 6 m s 35 s ol.37 A he end of he ccelerion period, he elociy is = + = +.5 m s 5. s = 7.5 m s This is lso he iniil elociy for he brking period. () Afer brking, (b) The ol disnce reled is = + = 7.5 m s+. m s 3. s =.5 m s x= x + x = + ccel brke ccel brke + 7.5 m s 7.5 m s+.5 m s x = ( 5. s) + ( 3. s) = 3 m.4 The ime he Thunderbird spends slowing down is ( ) 5 m = ( ) + + 7.5m s 6.99 s The ime required o regin speed fer he pi sop is ( ) 35 m = ( ) + 7.5m s+ 9.79 s Thus, he ol elpsed ime before he Thunderbird is bck up o speed is = + 5. s+ = 6.99 s+ 5. s+ 9.79 s=.8 s During his ime, he Mercedes hs reled ( consn speed) disnce x = = 7.5m s.8 s = 558 m M nd he Thunderbird hs fllen behind disnce d=m x = 558 m 5m 35m = 958 m.4 The cr is disnce d from he dog nd hs iniil elociy when he brkes re pplied giing i consn ccelerion.
Moion in One Dimension 3 + Apply = = obin o he enire rip (for which x= d+ 4.m, = s,nd = ) o d+ 4.m + s = or d+ 4.m = () 5.s Then, pplying = + ( x) o he enire rip yields ( d 4. m ) = + +. d+ 4.m Subsiue for from Equion () o find h = () 5 s Finlly, pply x= + o he firs 8. s of he rip (for which x d = ). This gies d ( 8.s) ( 64s ) = + (3) Subsiue Equions () nd () ino Equion (3) nd sole for he hree unknowns,, nd d o find h = m s, =.m s, nd d= 96 m.44 () For he upwrd fligh of he rrow, = + m s, finl elociy is =. Thus, ( y) = + yields 9.8 m s = g=, nd he ( y) ( ) m s mx 9.8 m s 5 m (b) The ime for he upwrd fligh is ( y) ( y) 5 m mx mx up = o + m s+ up.s y= y = 5 m, =, nd = 9.8 m s. Thus, For he downwrd fligh, mx y= + gies dow n ( y) ( ) fligh is = + =. s+.s=.4 s ol dow n dow n 5m.s nd he ol ime of he 9.8 m s.49 () When i reches heigh of 5 m, he speed of he rocke is
4 CHAPTER = + y = 5. m s +. m s 5 m = 55.7 m s Afer he engines sop, he rocke coninues moing upwrd wih n iniil elociy of = 55.7 m s nd ccelerion = g= 9.8 m s. When he rocke reches mximum heigh, =. The displcemen of he rocke boe he poin where he engines sopped (h is, boe he 5 m leel) is ( ) 55.7 m s y 58 m 9.8 m s The mximum heigh boe ground h he rocke reches is hen gien by h mx = 5 m + 58 m= 38 m (b) The ol ime of he upwrd moion of he rocke is he sum of wo inerls. The firs is he ime for he rocke o go from = 5. m s he ground o elociy of = 55.7 m s n liude of 5 m. This ime is gien by ( y) ( ) ( + ).84 s 55.7 5. m s 5 m The second inerl is he ime o rise 58 m sring wih = 55.7 m s nd ending wih =. This ime is ( y) ( ) 5.67 s +55.7 m s 58 m The ol ime of he upwrd fligh is hen up = + =.84 + 5.67 s= 8.5 s (c) The ime for he rocke o fll 38 m bck o he ground, wih = nd ccelerion = g= 9.8 m s, is found from y= + s dow n ( y) ( ) 38 m 7.93 s 9.8 m s so he ol ime of he fligh is fligh = up + dow n = 8.5+ 7.93 s= 6.4 s
Moion in One Dimension 5.5 () The keys he ccelerion cugh.5 s ler. Thus, = g= -9.8 m s from he relese poin unil hey re y= + gies ( + ) ( ) y 4. m 9.8 m s.5 s +. m s.5 s or =. m s upw rd (b) The elociy of he keys jus before he cch ws = + =. m s+ 9.8 m s.5 s = 4.68 m s or = 4.68 m s downwrd.56 We ssume h he bulle is cylinder which slows down jus s he fron end pushes pr wood fibers. () The ccelerion is ( ) 8 m s 4 m s. m 5 4.9 m s (b) The erge elociy s he fron of he bulle psses hrough he bord is + ( ) = = 35 m s nd he ol ime of conc wih he bord is bord ( ) ( ) bord Lbulle. m. m = + = + = 35 m s 8 m s bord 4 3.57 s = +, wih =, gies he required hickness s (c) From ( x) ( 4 m s) 5 ( ) x 4.9 m s.8 m.57 The flling bll moes disnce of ( 5 m h) before hey mee, where h is he heigh boe he ground where hey mee. Apply 5 m ( h) = g, or h 5 m y= +, wih g = o obin = g ()
6 CHAPTER Applying y= + o he rising bll gies h g = 5 m s () Combining equions () nd () gies 5 m = =.6 s 5 m s.59 () When eiher bll reches he ground, is ne displcemen is y = 9.6 m Applying y= + o he moion of he firs bll gies 9.6 m = ( 4.7 m s) + ( 9.8 m s ) which hs posiie soluion of =. s. Similrly, pplying his relion o he moion of he second bll gies 9.6 m = ( + 4.7 m s) + ( 9.8 m s ) which hs single posiie soluion of = 4. s. Thus, he difference in he ime of fligh for he wo blls is = = 4.. s= 3. s (b) When he blls srike he ground, heir elociies re: nd = g = 4.7 m s 9.8 m s. s = 4.5 m s = g =+ 4.7 m s 9.8 m s 4. s = 4.5 m s (c) A =.8 s, he displcemen of ech bll from he blcony is: y = y = g = i 4.7 m s.8 s 4.9 m s.8 s y = y = g = + i 4.7 m s.8 s 4.9 m s.8 s These yield y= 4.9 m nd y =+ 8.6 m. Therefore he disnce sepring he wo blls his ime is d= y y= 8.6 m 4.9 m = 3.5 m
Moion in One Dimension 7.6 When Khy hs been moing for seconds, Sn s elpsed ime is +. s. A his ime, he displcemens of he wo crs re x Khy Khy ( ) = ( ) + Khy = + ( 4.9 m s ) nd ( x ) = ( ) + ( +. s) = + ( 3.5 m s )( +. s) Sn Sn Sn =, or () When Khy oerkes Sn, ( x) ( x) Khy ( 4.9 m s ) = ( 3.5 m s )( +. s) Sn which gies he ime s = 5.46 s (b) Khy s displcemen his ime is = Khy 4.9 m s 5.46 s = 73. m ( x) (c) A his ime, he elociies of he wo crs re = + = + 4.9 m s 5.46 s = 6.7 m s Khy Khy Khy nd ( ) ( ) Sn = + Sn +. s = + 3.5 m s 6.46 s =.6 m s Sn.63 () The displcemen x of he sled during he ime h i hs ccelerion =+ 4 fs is: x ( ) ( fs ) = + = or x ( fs ) A he end of ime, he sled hd chieed elociy of = + = + ( 4fs ) or ( 4 fs ) = () = () The displcemen of he sled while moing consn elociy for ime is x ( = = 4 fs ) or x ( 4 fs ) = (3) I is known h x+ x = 7 5 f, nd subsiuions from Equions () nd (3) gie: fs + 4f s = 75 f
8 CHAPTER or + = 875 s (4) Also, i is known h: + = 9 s (5) Soling Equions (4) nd (5) simulneously yields = 5.s nd = 85 s (b) From Equion () boe, = 4 fs = 4fs 5.s = fs (c) The displcemen x3 of he sled s i comes o res (wih ccelerion 3 = fs ) is: ( fs) ( ) x f 3 3 fs Thus, he ol displcemen for he rip (mesured from he sring poin) is xol = x+ x + x3 = 7 5 f+ f = 8 5 f (d) The ime required o come o res from elociy (wih ccelerion 3) is fs 3 3 fs s so he durion of he enire rip is: = + + 3 = 5.s+ 85s+ s= s ol.67 The ime required for he womn o fll 3. m, sring from res, is found from y= + s 3. m = + ( 9.8 m s ), giing =.78 s () Wih he horse moing wih consn elociy of. m s, he horizonl disnce x= =. m s.78 s = 7.8 m is horse (b) The required ime is =.78 s s clculed boe.