and v y . The changes occur, respectively, because of the acceleration components a x and a y

Transcription

1 Week 3 Reciaion: Chaper3 : Problems: 1, 16, 9, 37, 41, A spacecraf is raveling wih a veloci of v0 = 5480 m/s along he + direcion. Two engines are urned on for a ime of 84 s. One engine gives he spacecraf an acceleraion in he + direcion of a = 1.0 m/s, while he oher gives i an acceleraion in he + direcion of a = 8.40 m/s. A he end of he firing, find (a) v and (b) v. REASONING The moion in he direcion occurs independenl of he moion in he direcion. The componens of he veloci change from heir iniial values of v 0 and v 0 o heir final values of v and v. The changes occur, respecivel, because of he acceleraion componens a and a. The final values can be deermined wih he aid of Equaions 3.3a and 3.3b. SOLUTION a. According o Equaion 3.3a, he componen of he veloci is ( v = v )( ) 0 + a= 5480 m/s m/s 84 s = 6490 m/s b. According o Equaion 3.3b, he componen of he veloci is ( v )( ) = v0 + a = 0 m/s m/s 84 s = 7070 m/s 16. A puck is moving on an air hocke able. Relaive o an, coordinae ssem a ime = 0 s, he componens of he puck s iniial veloci and acceleraion are v0 = +1.0 m/s and a = +.0 m/s. The componens of he puck s iniial veloci and acceleraion are v0 = +.0 m/s and a =.0 m/s. Find he magniude and direcion of he puck s veloci a a ime of = 0.50 s. Specif he direcion relaive o he + ais REASONING The magniude v of he puck s veloci is relaed o is and veloci componens (v and v ) b he Phagorean heorem (Equaion 1.7); v = v + v. The relaion v = v0 + a (Equaion 3.3a) ma be used o find v, since v 0, a, and are known. Likewise, he relaion v = v0 + a (Equaion 3.3b) ma be emploed o deermine v, since v 0, a, and are known. Once v and v are deermined, he angle θ ha he veloci makes wih respec o he ais can be found b using he 1 θ = an v / v. inverse angen funcion (Equaion 1.6); ( )

2 SOLUTION Using Equaions 3.3a and 3.3b, we find ha 0 0 v = v + a= m/s +.0 m/s 0.50 s = +.0 m/s v = v + a= +.0 m/s +.0 m/s 0.50 s = m/s The magniude v of he puck s veloci is ( ) ( ) v = v + v =.0 m/s m/s =. m/s The angle θ ha he veloci makes wih respec o he + ais is 1 v m/s = an = an = 7 above he + ais v.0 m/s θ 9. A major-league picher can hrow a baseball in ecess of 41.0 m/s. If a ball is hrown horizonall a his speed, how much will i drop b he ime i reaches a cacher who is 17.0 m awa from he poin of release? REASONING The verical displacemen of he ball depends on he ime ha i is in he air before being caugh. These variables depend on he -direcion daa, as indicaed in he able, where he + direcion is "up." -Direcion Daa a v v 0? 9.80 m/s 0 m/s? Since onl wo variables in he direcion are known, we canno deermine a his poin. Therefore, we eamine he daa in he direcion, where + is aken o be he direcion from he picher o he cacher. -Direcion Daa a v v m 0 m/s m/s? Since his able conains hree known variables, he ime can be evaluaed b using an equaion of kinemaics. Once he ime is known, i can hen be used wih he -direcion daa, along wih he appropriae equaion of kinemaics, o find he verical displacemen.

3 SOLUTION Using he -direcion daa, Equaion 3.5a can be emploed o find he ime ha he baseball is in he air: ( ) = v + a = v 0 0 since a = 0 m/s Solving for gives m = = = s v m/s 1 0 The displacemen in he direcion can now be evaluaed b using he -direcion daa able and he value of = s. Using Equaion 3.5b, we have = v + a = + = m/s s 9.80 m/s s m 0 The disance ha he ball drops is given b he magniude of his resul, so Disance = m. *37. ssm An airplane wih a speed of 97.5 m/s is climbing upward a an angle of 50.0 wih respec o he horizonal. When he plane s aliude is 73 m, he pilo releases a package. (a) Calculae he disance along he ground, measured from a poin direcl beneah he poin of release, o where he package his he earh. (b) Relaive o he ground, deermine he angle of he veloci vecor of he package jus before impac. SSM REASONING a. The drawing shows he iniial veloci v 0 of he package when i is released. The iniial speed of he package is 97.5 m/s. The componen of is displacemen along he ground is labeled as. The daa for he direcion are indicaed in he daa able below. v Direcion Daa a v v 0? 0 m/s +(97.5 m/s) cos 50.0 = +6.7 m/s Since onl wo variables are known, i is no possible o deermine from he daa in his able. A value for a hird variable is needed. We know ha he ime of fligh is he same for boh he and moions, so le s now look a he daa in he direcion.

4 -Direcion Daa a v v 0 73 m 9.80 m/s +(97.5 m/s) sin 50.0 = m/s? Noe ha he displacemen of he package poins from is iniial posiion oward he ground, so is value is negaive, i.e., = 73 m. The daa in his able, along wih he appropriae equaion of kinemaics, can be used o find he ime of fligh. This value for can, in urn, be used in conjuncion wih he -direcion daa o deermine. b. The drawing a he righ shows he veloci of he package jus before impac. The angle ha he veloci makes wih respec o he ground can be found from he inverse 1 angen funcion as θ = an ( v / v ). Once he ime has been found in par (a), he values of v and v can be deermined from he daa in he ables and he appropriae equaions of kinemaics. v SOLUTION a. To deermine he ime ha he package is in he air, we 1 will use Equaion 3.5b ( = v + a 0 ) and he daa in he -direcion daa able. Solving his quadraic equaion for he ime ields v + v θ + 1 ( a ) ( )( ) 1 v ± v 4 a 0 0 = ( 74.7 m/s) ( 74.7 m/s) 4( 1 )( 9.80 m/s )( 73 m) ± = = 6.78 s and.0 s 9.80 m/s 1 ( )( ) We discard he firs soluion, since i is a negaive value and, hence, unrealisic. The displacemen can be found using =.0 s, he daa in he -direcion daa able, and Equaion 3.5a:

5 b. The angle θ ha he veloci of he package makes wih respec o he ground is 1 given b θ an ( v / v ) =. Since here is no acceleraion in he direcion (a = 0 m/s ), v is he same as v 0, so ha v = v 0 = +6.7 m/s. Equaion 3.3b can be emploed wih he -direcion daa o find v : Therefore, v = v + a= m/s m/s.0 s = 141 m/s 0 v m/s = an = an = 66.0 v m/s θ where he minus sign indicaes ha he angle is 66.0 below he horizonal. *41. mmh A soccer plaer kicks he ball oward a goal ha is 16.8 m in fron of him. The ball leaves his foo a a speed of 16.0 m/s and an angle of 8.0 above he ground. Find he speed of he ball when he goalie caches i in fron of he ne. REASONING The speed v of he soccer ball jus before he goalie caches i is given b v v v = +, where v and v are he and componens of he final veloci of he ball. The daa for his problem are (he + direcion is from he kicker o he goalie, and he + direcion is he up direcion): -Direcion Daa a v v m 0 m/s? +(16.0 m/s) cos 8.0 = m/s -Direcion Daa a v v m/s? +(16.0 m/s) sin 8.0 = m/s Since here is no acceleraion in he direcion (a = 0 m/s ), v remains he same as v 0, so v = v 0 = m/s. The ime ha he soccer ball is in he air can be found from he -direcion daa, since hree of he variables are known. Wih his value for he ime and he -direcion daa, he componen of he final veloci can be deermined.

6 SOLUTION Since a = 0 m/s, he ime can be calculaed from Equaion 3.5a as m = = = 1.19 s. The value for v v m/s can now be found b using Equaion 0 3.3b wih his value of he ime and he -direcion daa: v = v + a= m/s m/s 1.19 s = 4.15 m/s 0 The speed of he ball jus as i reaches he goalie is ( ) ( ) v = v + v = m/s m/s = 14.7 m/s 71. ssm Muliple-Concep Eample 4 provides useful background for his problem. A diver runs horizonall wih a speed of 1.0 m/s off a plaform ha is 10.0 m above he waer. Wha is his speed jus before sriking he waer? SSM REASONING Once he diver is airborne, he moves in he direcion wih consan veloci while his moion in he direcion is acceleraed (a he acceleraion due o gravi). Therefore, he magniude of he componen of his veloci remains consan a 1.0 m/s for all imes. The magniude of he componen of he diver's veloci afer he has fallen hrough a verical displacemen can be deermined from Equaion 3.6b: v = v + a. Since he diver runs off 0 he plaform horizonall, v 0 = 0 m/s. Once he and componens of he veloci are known for a paricular verical displacemen, he speed of he diver can be obained from v = v + v. SOLUTION For convenience, we will ake downward as he posiive direcion. Afer he diver has fallen 10.0 m, he componen of his veloci is, from Equaion 3.6b, v = v + a = 0 + (9.80 m / s )( m) = m / s 0 Therefore, v v + v (1.0 m / s) + ( 14.0 m / s) = 14.1 m / s