AP CALCULUS AB 17 SCORING GUIDELINES
16 SCORING GUIDELINES Quesion For, a paricle moves along he x-axis. The velociy of he paricle a ime is given by v ( ) = 1 + sin. The paricle is a posiion x = a ime. = (a) A ime =, is he paricle speeding up or slowing down? (b) Find all imes in he inerval < < when he paricle changes direcion. Jusify your answer. (c) Find he posiion of he paricle a ime =. (d) Find he oal disance he paricle ravels from ime = o ime =. (a) v( ) =.978716 > v ( ) = 1.16 < : conclusion wih reason The paricle is slowing down since he velociy and acceleraion have differen signs. (b) v ( ) = =.7768 v ( ) changes from posiive o negaive a =.77. Therefore, he paricle changes direcion a his ime. (c) x( ) = x( ) + v( ) d = + ( 5.8157) =.815 { 1 : =.77 : 1 : jusificaion 1 : inegral : 1 : uses iniial condiion 1 : answer = : { 1 : inegral (d) Disance v( ) d = 5.1 1 : answer 16 The College Board. Visi he College Board on he Web: www.collegeboard.org.
1 SCORING GUIDELINES Quesion A paricle moves along a sraigh line. For 5, he velociy of he paricle is given by ( ) ( ) 6 5 v = + +, and he posiion of he paricle is given by s ( ). I is known ha s ( ) = 1. (a) Find all values of in he inerval for which he speed of he paricle is. (b) Wrie an expression involving an inegral ha gives he posiion s ( ). Use his expression o find he posiion of he paricle a ime = 5. (c) Find all imes in he inerval 5 a which he paricle changes direcion. Jusify your answer. (d) Is he speed of he paricle increasing or decreasing a ime =? Give a reason for your answer. (a) Solve v ( ) = on. =.18 (or.17) and =.7 1 : considers : { 1 : answer v ( ) = (b) s ( ) = 1 + vx ( ) dx 5 s ( 5) = 1 + v( x) dx = 9.7 : 1 : s ( ) 1 : s( 5) (c) v ( ) = when =.56,.17756 v ( ) changes sign from negaive o posiive a ime =.56. v ( ) changes sign from posiive o negaive a ime =.17756. : { 1 : considers v ( ) = : answers wih jusificaion Therefore, he paricle changes direcion a ime =.56 and ime =.18 (or.17). (d) v ( ) = 11.75758 <, a( ) = v ( ) =.9571 < : conclusion wih reason The speed is increasing a ime = because velociy and acceleraion have he same sign. 1 The College Board. Visi he College Board on he Web: www.collegeboard.org.
1 SCORING GUIDELINES Quesion 6 For 1, a paricle moves along he x-axis. The velociy of he paricle a ime is given by v () = cos ( ). The paricle is a posiion x = a ime =. 6 (a) For 1, when is he paricle moving o he lef? (b) Wrie, bu do no evaluae, an inegral expression ha gives he oal disance raveled by he paricle from ime = o ime = 6. (c) Find he acceleraion of he paricle a ime. Is he speed of he paricle increasing, decreasing, or neiher a ime =? Explain your reasoning. (d) Find he posiion of he paricle a ime =. 6 (a) () ( ) v = cos = =, 9 The paricle is moving o he lef when v () <. This occurs when < < 9. 1 : considers v () = : 1 : inerval 6 (b) v () d (c) a () = sin ( ) a v 6 6 sin = < 6 1 ( ) = ( ) 1 cos = < ( ) = ( ) 1 : answer 1 : a () : : conclusion wih reason The speed is increasing a ime =, because velociy and acceleraion have he same sign. (d) x( ) = + cos ( ) 6 6 = + sin ( 6 ) 6 = + ( ) d sin 6 = + = + : 1 : aniderivaive 1 : uses iniial condiion 1 : answer 1 The College Board. Visi he College Board on he Web: www.collegeboard.org.
11 SCORING GUIDELINES Quesion 1 For 6, a paricle is moving along he x-axis. The paricle s posiion, x( ), is no explicily given. v = sin e + 1. The acceleraion of he paricle is given by The velociy of he paricle is given by () ( ) 1 () cos( ) a = e e and x ( ) =. (a) Is he speed of he paricle increasing or decreasing a ime = 5.5? Give a reason for your answer. (b) Find he average velociy of he paricle for he ime period 6. (c) Find he oal disance raveled by he paricle from ime = o = 6. (d) For 6, he paricle changes direcion exacly once. Find he posiion of he paricle a ha ime. (a) v ( 5.5) =.57, a ( 5.5) = 1.5851 : conclusion wih reason The speed is increasing a ime = 5.5, because velociy and acceleraion have he same sign. 1 6 (b) Average velociy = v () d= 1.99 6 : { 1 : inegral 1 : answer 6 : { 1 : inegral (c) Disance = v () d= 1.57 1 : answer (d) v ( ) = when = 5.1955. Le b = 5.1955. v ( ) changes sign from posiive o negaive a ime = b. b xb ( ) = + v ( ) d= 1.1 or 1.15 : 1 : considers v ( ) = 1 : inegral 1 : answer 11 The College Board. Visi he College Board on he Web: www.collegeboard.org.
1 SCORING GUIDELINES (Form B) Quesion 6 Two paricles move along he x-axis. For 6, he posiion of paricle P a ime is given by p() = cos ( ), while he posiion of paricle R a ime is given by r () = 6 + 9+. (a) For 6, find all imes during which paricle R is moving o he righ. (b) For 6, find all imes during which he wo paricles ravel in opposie direcions. (c) Find he acceleraion of paricle P a ime =. Is paricle P speeding up, slowing down, or doing neiher a ime =? Explain your reasoning. (d) Wrie, bu do no evaluae, an expression for he average disance beween he wo paricles on he inerval 1. (a) r () = 1 + 9 = ( 1)( ) r () = when = 1 and = r () > for < < 1 and < < 6 r () < for 1< < 1 : r () : 1 : answer Therefore R is moving o he righ for < < 1 and < < 6. (b) p () = sin( ) p () = when = and = p () < for < < p () > for < < 6 Therefore he paricles ravel in opposie direcions for < < 1 and < <. : 1 : p () 1 : sign analysis for 1 : answer p () (c) p () = cos ( ) p ( ) ( ) ( ) = cos = > 8 p ( ) < Therefore paricle P is slowing down a ime =. 1 : p ( ) : 1 : answer wih reason 1 (d) p() r() d 1 : { 1 : inegrand 1 : limis and consan 1 The College Board. Visi he College Board on he Web: www.collegeboard.com.
7 SCORING GUIDELINES (Form B) Quesion A paricle moves along he x-axis so ha is velociy v a ime is given by v () = sin ( ). The graph of v is shown above for 5. The posiion of he paricle a ime is x( ) and is posiion a ime = is x ( ) = 5. (a) Find he acceleraion of he paricle a ime =. (b) Find he oal disance raveled by he paricle from ime = o =. (c) Find he posiion of he paricle a ime =. (d) For 5, find he ime a which he paricle is farhes o he righ. Explain your answer. (a) a( ) = v ( ) = 6cos9 = 5.66 or 5.67 1 : a ( ) (b) Disance = v () d= 1.7 OR For < <, v ( ) = when = = 1.775 and = =.566 x ( ) = 5 x x ( ) () = 5 + v d = 5.898 ( ) () = 5 + v d = 5.1 x( ) = 5 + v( ) d = 5.7756 ( ) ( ) ( ) ( ) ( ) ( ) x x + x x + x x = 1. 7 : { 1 : seup 1 : answer 1 : inegrand (c) x( ) = 5 + v( ) d = 5.77 or 5.77 : 1 : uses x( ) = 5 1 : answer (d) The paricle s righmos posiion occurs a ime = = 1.77. The paricle changes from moving righ o moving lef a hose imes for which v ( ) = wih v ( ) changing from posiive o negaive, namely a =,, 5 ( = 1.77,.7,.96 ). T Using x( T) = 5 + v( ) d, he paricle s posiions a he imes i changes from righward o lefward movemen are: T: 5 xt ( ): 5 5.895 5.788 5.75 The paricle is farhes o he righ when T =. : 1 : ses v () = 1 : answer 1 : reason 7 The College Board. All righs reserved. Visi apcenral.collegeboard.com (for AP professionals) and www.collegeboard.com/apsudens (for sudens and parens).
7 SCORING GUIDELINES Quesion A paricle moves along he x-axis wih posiion a ime given by x() = e sin for. (a) Find he ime a which he paricle is farhes o he lef. Jusify your answer. (b) Find he value of he consan A for which x() saisfies he equaion Ax () + x () + x() = for < <. (a) () sin x = e + e cos = e ( cos sin ) x () = when cos = sin. Therefore, x () = on 5 for = and =. The candidaes for he absolue minimum are a 5 =,,, and. 5 : : x () 1 : ses x () = 1 : answer 1 : jusificaion x() e sin ( ) = 5 e e ( ) > ( ) sin 5 5 sin < e sin ( ) = The paricle is farhes o he lef when = 5. (b) x () = e ( cos sin ) + e ( sin cos ) = e cos Ax () + x () + x() ( ) ( ) = A e cos + e cos sin + e sin = ( A + 1) e cos = : : x () 1 : subsiues x (), x (), and x() ino Ax () + x () + x() 1 : answer Therefore, A = 1. 7 The College Board. All righs reserved. Visi apcenral.collegeboard.com (for AP professionals) and www.collegeboard.com/apsudens (for sudens and parens).
5 SCORING GUIDELINES (Form B) Quesion A paricle moves along he x-axis so ha is velociy v a ime, for 5, is given by () ( ) v = ln +. The paricle is a posiion x = 8 a ime =. (a) Find he acceleraion of he paricle a ime =. (b) Find all imes in he open inerval < < 5 a which he paricle changes direcion. During which ime inervals, for 5, does he paricle ravel o he lef? (c) Find he posiion of he paricle a ime =. (d) Find he average speed of he paricle over he inerval. 5 (a) a( ) = v ( ) = 1 : answer 7 (b) v () = + = 1 + = ( ) ( 1) = = 1, 1 : ses v () = : 1 : direcion change a = 1, 1 : inerval wih reason v () > for < < 1 v () < for 1 < < v () > for < < 5 The paricle changes direcion when = 1 and =. The paricle ravels o he lef when 1 < <. (c) () = ( ) + ln( + ) ( ) = 8 + ln( + ) s s u u du s u u du = 8.68 or 8.69 : ( ) 1 : ln u u + du 1 : handles iniial condiion 1 : answer 1 (d) () v d=.7 or.71 1 : inegral : 1 : answer Copyrigh 5 by College Board. All righs reserved. Visi apcenral.collegeboard.com (for AP professionals) and www.collegeboard.com/apsudens (for AP sudens and parens).
SCORING GUIDELINES Quesion A paricle moves along he y-axis so ha is velociy v a ime A ime =, he paricle is a y = 1. (Noe: an 1 x = arcan x ) 1 is given by v () = ( e) 1 an. (a) Find he acceleraion of he paricle a ime =. (b) Is he speed of he paricle increasing or decreasing a ime =? Give a reason for your answer. (c) Find he ime a which he paricle reaches is highes poin. Jusify your answer. (d) Find he posiion of he paricle a ime =. Is he paricle moving oward he origin or away from he origin a ime =? Jusify your answer. (a) a( ) = v ( ) =.1 or.1 1 : answer (b) v ( ) =.6 Speed is increasing since a ( ) < and v ( ) <. 1 : answer wih reason 1 v = when ( e ) (c) () an = 1 = ln ( an() 1 ) =. is he only criical value for y. v () > for < < ln( an() 1 ) v () < for > ln ( an () 1 ) 1 : ses v () = : 1 : idenifies =. as a candidae 1 : jusifies absolue maximum y () has an absolue maximum a =.. (d) y( ) = 1 + v( ) d = 1.6 or 1.61 The paricle is moving away from he origin since v ( ) < and y ( ) <. : 1 : v () d 1 : handles iniial condiion 1 : value of y( ) 1 : answer wih reason Copyrigh by College Enrance Examinaion Board. All righs reserved. Visi apcenral.collegeboard.com (for AP professionals) and www.collegeboard.com/apsudens (for AP sudens and parens).
SCORING GUIDELINES (Form B) Quesion A paricle moves along he x-axis wih velociy a ime given by v( ) = 1 + e1. (a) Find he acceleraion of he paricle a ime =. (b) Is he speed of he paricle increasing a ime =? Give a reason for your answer. (c) Find all values of a which he paricle changes direcion. Jusify your answer. (d) Find he oal disance raveled by he paricle over he ime inerval. 1 (a) a () = v() = e a() = e : 1 : v( ) 1 : a() (b) a () < v() = 1 + e < Speed is increasing since v () < and a () <. 1 : answer wih reason 1 (c) v () = when 1 = e, so = 1. v () > for < 1 and v () < for > 1. Therefore, he paricle changes direcion a = 1. : 1 : solves v ( ) = o ge = 1 1 : jusifies change in direcion a = 1 (d) Disance = v () d 1 1 1+ + 1 1 1 1 1 1 e e 1 1 1 e e 1 1 = ( ) + ( + ) = ( ) ( ) e d e d = ( + ) + ( + ) : 1 : limis 1 : inegrand 1 : anidiffereniaion 1 : evaluaion = e + 1 e OR OR 1 () = x e x() = e x (1) = x() = e Disance = ( x(1) x() ) + ( x(1) x() ) ( + e) + 1+ e = ( ) = e + 1 e 1 : any aniderivaive 1 : evaluaes x ( ) when =, 1, : 1 : evaluaes disance beween poins 1 : evaluaes oal disance Copyrigh by College Enrance Examinaion Board. All righs reserved. Available a apcenral.collegeboard.com. 5
SCORING GUIDELINES Quesion A paricle moves along he x-axis so ha is velociy a ime is given by v( ) = ( + 1) sin. A ime =, he paricle is a posiion x = 1. (a) Find he acceleraion of he paricle a ime =. Is he speed of he paricle increasing a =? Why or why no? (b) Find all imes in he open inerval < < when he paricle changes direcion. Jusify your answer. (c) Find he oal disance raveled by he paricle from ime = unil ime =. (d) During he ime inerval, wha is he greaes disance beween he paricle and he origin? Show he work ha leads o your answer. (a) a() = v() = 1.587 or 1.588 v () = sin() < Speed is decreasing since a () > and v () <. : 1: a() 1 : speed decreasing wih reason (b) v () = when = = or.56 or.57 Since v () < for < < and v () > for < <, he paricle changes direcions a =. : 1: = only 1 : jusificaion (c) Disance = v () d =. or. : 1: limis 1: inegrand 1: answer (d) v ( ) d=.65 x( ) = x() + v( ) d =.65 Since he oal disance from = o = is., he paricle is sill o he lef of he origin a =. Hence he greaes disance from he origin is.65. 1 : ± (disance paricle ravels : while velociy is negaive) 1: answer Copyrigh by College Enrance Examinaion Board. All righs reserved. Available a apcenral.collegeboard.com.
SCORING GUIDELINES (Form B) Quesion )F=HJE?ALAI=CJDAN=NEIIJD=JEJILA?EJOL=J=OJEAJBH > J > $ EICELA>O IEJ LJ A Г)JJEAJJDAF=HJE?AEI=JJDAHECE = JDA=NAIFHLE@A@IAJ?DJDACH=FDBLJ BH > J > $ >,KHECMD=JEJAHL=IBJEAEIJDAF=HJE?ALECJJDAABJ/ELA=HA=IBHOKH =IMAH?.E@JDAJJ=@EIJ=?AJH=LAA@>OJDAF=HJE?ABHJJJ" @ 1IJDAHA=OJEAJ J > $ =JMDE?DJDAF=HJE?AHAJKHIJJDAHECEKIJEBOOKH =IMAH = LJ CH=FD JDHAA]DKFI^ FAHE@E?>AD=LEH J IJ=HJI=JHECE HA=I=>AHA=JELA=N=@EL=KAI > =HJE?AEILECJJDAABJMDA LJ EA A IE?! " =@# $ H " LJ IEJ LJ A @J#" Г JHJ N LJ @J $#$ N" " J LJ @J '$%$$ N ГN N" ГN #" @ 6DAHAEIIK?DJEA>A?=KIA 6 LJ @J BH=6 EJAHL=I! Г A=?DEIIECHE?HHA?JEJAHL= HA=I EEJIB=@"=EJACH=B LJ H LJ H KIAIN=@N"J?FKJA@EIJ=?A! D=@AI?D=CAB@EHA?JE=JIJK@AJ\I JKHECFEJ =IMAH JAEBE?HHA?JJKHECFEJ IK?DJEA HA=I Copyrigh by College Enrance Examinaion Board. All righs reserved. Advanced Placemen Program and AP are regisered rademarks of he College Enrance Examinaion Board.
AB{1 1999 1. A paricle moves along he y{axis wih velociy given by v() = sin( )for. (a) In which direcion (up or down) is he paricle moving a ime =1:5? Why? (b) Find he acceleraion of he paricle a ime = 1:5. Is he velociy of he paricle increasing a =1:5? Why orwhyno? (c) Given ha y() is he posiion of he paricle a ime and ha y() =, nd y(). (d) Find he oal disance raveled by he paricle from =o =. (a) v(1:5)=1:5 sin(1:5 )=1:167 1: answer and reason Up, because v(1:5) > (b) a() =v () = sin + cos a(1:5) = v (1:5) = ;:8 or ;:9 ( 1: a(1:5) 1: conclusion and reason No v is decreasing a 1.5 because v (1:5) < Z (c) y() = Z = v() d sin cos d = ; + C 8 Z 1: y() = v() d >< 1: y() =; 1 cos + C >: 1: y() y() = = ; 1 + C =) C = 7 y() =; 1 cos + 7 y() = ; 1 cos + 7 =:86 or :87 (d) disance = or Z jv()j d =1:17 v() = sin = = or = p 1:77 y() = y( p )= y() = :86 or :87 [y( p ) ; y()] + [y( p ) ; y()] =1:17 or 1:17 8 >< >: 1: limis of and on an inegral of v() orjv()j or uses y() and y() o compue disance 1: handles change of direcion a suden's urning poin 1: answer /1 if incorrec urning poin