4 SCORING GUIDELINES (Form B) Quesion A es plane flies in a sraigh line wih (min) 5 1 15 5 5 4 posiive velociy v (), in miles per v ()(mpm) 7. 9. 9.5 7. 4.5.4.4 4. 7. minue a ime minues, where v is a differeniable funcion of. Seleced values of v () for 4 are shown in he able above. (a) Use a midpoin Riemann sum wih four subinervals of equal lengh and values from he able o 4 approximae v () d. Show he compuaions ha lead o your answer. Using correc unis, 4 explain he meaning of v () din erms of he plane s fligh. (b) Based on he values in he able, wha is he smalles number of insances a which he acceleraion of he plane could equal zero on he open inerval < < 4? Jusify your answer. 7 (c) The funcion f, defined by f() = 6 + cos( ) + sin ( ), is used o model he velociy of he 1 4 plane, in miles per minue, for 4. According o his model, wha is he acceleraion of he plane a =? Indicaes unis of measure. (d) According o he model f, given in par (c), wha is he average velociy of he plane, in miles per minue, over he ime inerval 4? (a) Midpoin Riemann sum is 1 [ v( 5) + v( 15) + v( 5) + v( 5) ] = 1 [ 9. + 7. +.4 + 4.] = 9 The inegral gives he oal disance in miles ha he plane flies during he 4 minues. : 1 : v( 5) + v( 15) + v( 5) + v( 5) 1 : answer 1 : meaning wih unis (b) By he Mean Value Theorem, v () = somewhere in he inerval (, 15 ) and somewhere in he inerval ( 5, ). Therefore he acceleraion will equal for a leas wo values of. 1 : wo insances : 1 : jusificaion (c) f ( ) =.47 or.48 miles per minue 1 : answer wih unis 1 4 (d) Average velociy = () 4 f d = 5.916 miles per minue : 1 : limis 1 : inegrand 1 : answer Copyrigh 4 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). 46 4
6 SCORING GUIDELINES (Form B) Quesion 6 (sec) v () ( f sec ) a () ( f sec ) 15 5 5 5 6 14 1 1 1 5 1 4 A car ravels on a sraigh rack. During he ime inerval 6 seconds, he car s velociy v, measured in fee per second, and acceleraion a, measured in fee per second per second, are coninuous funcions. The able above shows seleced values of hese funcions. (a) Using appropriae unis, explain he meaning of v () din erms of he car s moion. Approximae 6 v () dusing a rapezoidal approximaion wih he hree subinervals deermined by he able. 6 (b) Using appropriae unis, explain he meaning of a () din erms of he car s moion. Find he exac value of a () d. (c) For < < 6, mus here be a ime when v () = 5? Jusify your answer. (d) For < < 6, mus here be a ime when a () =? Jusify your answer. 6 (a) v () dis he disance in fee ha he car ravels from = sec o = 6 sec. Trapezoidal approximaion for v () 6 d: 1 1 1 A = ( 14 + 1) 5 + ( 1)( 15) + ( 1)( 1) = 185 f (b) a () dis he car s change in velociy in f/sec from = sec o = sec. a () d= v () d= v( ) v( ) = 14 ( ) = 6 f/sec (c) Yes. Since v( 5) = 1 < 5 < = v( 5 ), he IVT guaranees a in ( 5, 5 ) so ha v () = 5. (d) Yes. Since v( ) = v( 5 ), he MVT guaranees a in (, 5 ) so ha a () = v () =. Unis of f in (a) and f/sec in (b) : { 1 : explanaion 1 : value : { 1 : explanaion 1 : value 1 : v( 5) < 5 < v( 5) : 1 : Yes; refers o IVT or hypoheses 1 : v( ) = v( 5) : 1 : Yes; refers o MVT or hypoheses 1 : unis in (a) and (b) 6 The College Board. All righs reserved. Visi apcenral.collegeboard.com (for AP professionals) and www.collegeboard.com/apsudens (for AP sudens and parens). 7 48
6 SCORING GUIDELINES Quesion 4 (seconds) v () (fee per second) 1 4 5 6 7 8 5 14 9 5 4 44 47 49 Rocke A has posiive velociy v () afer being launched upward from an iniial heigh of fee a ime = seconds. The velociy of he rocke is recorded for seleced values of over he inerval 8 seconds, as shown in he able above. (a) Find he average acceleraion of rocke A over he ime inerval 8 seconds. Indicae unis of measure. 7 (b) Using correc unis, explain he meaning of v () din erms of he rocke s fligh. Use a midpoin Riemann sum wih subinervals of equal lengh o approximae v () d. 1 (c) Rocke B is launched upward wih an acceleraion of a () = fee per second per second. A ime + 1 = seconds, he iniial heigh of he rocke is fee, and he iniial velociy is fee per second. Which of he wo rockes is raveling faser a ime = 8 seconds? Explain your answer. 7 1 (a) Average acceleraion of rocke A is 1 : answer v( 8) v( ) 49 5 11 f sec = = 8 8 (b) Since he velociy is posiive, v () drepresens he disance, in fee, raveled by rocke A from = 1 seconds o = 7 seconds. 7 1 1 : explanaion : 1 : uses v( ), v( 4 ), v( 6) 1 : value A midpoin Riemann sum is [ v( ) + v( 4) + v( 6) ] = [ + 5 + 44] = f (c) Le vb () be he velociy of rocke B a ime. vb () = d = 6 + 1 + C + 1 = v ( ) = 6 + C B vb () = 6 + 1 4 v ( 8) = 5 > 49 = v( 8) B 4 : 1 : 6 + 1 1 : consan of inegraion 1 : uses iniial condiion 1 : finds vb ( 8 ), compares o v( 8 ), and draws a conclusion Rocke B is raveling faser a ime = 8 seconds. Unis of f sec in (a) and f in (b) 1 : unis in (a) and (b) 6 The College Board. All righs reserved. Visi apcenral.collegeboard.com (for AP professionals) and www.collegeboard.com/apsudens (for AP sudens and parens). 5 49
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 = LJ =IMAH > =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 J CH=FD Copyrigh by College Enrance Examinaion Board. All righs reserved. Advanced Placemen Program and AP are regisered rademarks of he College Enrance Examinaion Board. 4 JDHAA]DKFI^ FAHE@E?>AD=LEH IJ=HJI=JHECE HA=I=>AHA=JELA=N=@EL=KAI 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 97
SCORING GUIDELINES Quesion )>A?JLAI=CJDA N=NEIMEJDEEJE=FIEJE N 6DALA?EJOBJDA>A?J=JJEA J F EICELA >O L J IE J! = 9D=JEIJDA=??AAH=JEBJDA>A?J=JJEA J " > +IE@AHJDABMECJMIJ=JAAJI 5J=JAAJ1.H! J "#JDALA?EJOBJDA>A?JEI@A?HA=IEC 5J=JAAJ11.H! J "#JDAIFAA@BJDA>A?JEIE?HA=IEC )HAAEJDAHH>JDBJDAIAIJ=JAAJI?HHA?J.HA=?DIJ=JAAJFHLE@A=HA=IMDOEJEI?HHA?JHJ?HHA?J? 9D=JEIJDAJJ=@EIJ=?AJH=LAA@>OJDA>A?JLAHJDAJEAEJAHL= > J > " @ 9D=JEIJDAFIEJEBJDA>A?J=JJEA J " " = = " L= "?I!! Г HГ#! HГ# " $ =IMAH >! J "# =J L= J?I J!! 5J=JAAJ1EI?HHA?JIE?A=J 5J=JAAJ11EI?HHA?JIE?A LJ =@ =J?,EIJ=?A " LJ @J!&% 4!! NJ Г?I J! N N" '!"!!' LJ MDAJ! $ N!!''&$ N! ГN N" ГN! #!&% @ N" N LJ @J!"! 4!! NJ Г?I J! N" '!"! " Copyrigh by College Enrance Examinaion Board. All righs reserved. Advanced Placemen Program and AP are regisered 98 rademarks of he College Enrance Examinaion Board. 4 1?HHA?JMEJDHA=I! 11?HHA?J HA=IBH11 EEJIB=@"=EJACH= BLJ H LJ H KIAIN=@N"J?FKJA @EIJ=?A! D=@AI?D=CAB@EHA?JE=J IJK@AJ\IJKHECFEJ =IMAH EBE?HHA?JJKHECFEJH JKHECFEJ EJACH= =IMAH 4! NJ Г?I J +! =IMAH EB?IJ=JBEJACH=JE
SCORING GUIDELINES (Form B) Quesion 4 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 () < e v() = 1 + < 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 ) e d e d 1 1 1 1 e e 1 = ( 1 1+ e) + ( + e 1 1) 4 : 1 : limis 1 : inegrand 1 : anidiffereniaion 1 : evaluaion = e + 1 e OR OR 1 () = x e x() = e x (1) = e x() = Disance = ( x(1) x() ) + ( x(1) x() ) = ( + e) + ( 1+ e ) = e + 1 e 4 : 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. 99 5
SCORING GUIDELINES Quesion A paricle moves along he x-axis so ha is velociy a ime is given by A ime =, he paricle is a posiion x = 1. v( ) = ( + 1) sin. (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 = 4. or 4.4 : 1: limis 1: inegrand 1: answer (d) v () d=.65 x( ) = x() + v( ) d =.65 Since he oal disance from = o = is : 1: ± (disance paricle ravels while velociy is negaive) 1 : answer 4.4, he paricle is sill o he lef of he origin a =. Hence he greaes disance from he origin is.65. Copyrigh by College Enrance Examinaion Board. All righs reserved. Available a apcenral.collegeboard.com. 1
4 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 (a) Find he acceleraion of he paricle a ime =. 1 x = arcan x ) 1 is given by v () = ( e) 1 an. (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 ( ) =.46 Speed is increasing since a ( ) < and v ( ) <. 1 : answer wih reason 1 v = when ( e ) (c) () an = 1 = ln ( an () 1 ) =.44 is he only criical value for y. v () > for < < ln ( an () 1 ) v () < for > ln ( an () 1 ) 1 : ses v () = : 1 : idenifies =.44 as a candidae 1 : jusifies absolue maximum y () has an absolue maximum a =.44. (d) y( ) = 1+ v( ) d = 1.6 or 1.61 The paricle is moving away from he origin since v ( ) < and y ( ) <. 4 : 1 : v () d 1 : handles iniial condiion 1 : value of y( ) 1 : answer wih reason Copyrigh 4 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). 11 4
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 = 4. (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( 4) = v ( 4) = 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). 1 4
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 5.466 or 5.467 (b) Disance v d 1.7 OR For, v when 1.7745 and.566 x 5 x x 5 v d 5.8948 5 v d 5.441 x 5 v d 5.7756 x x x x x x 1. 7 1 : a : 1 : seup 1 : answer 1 : inegrand (c) x 5 v d 5.77 or 5.774 : 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 xt 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). 1
7 SCORING GUIDELINES Quesion 4 A paricle moves along he x-axis wih posiion a ime given by x() = e sin for π. (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 < < π. (a) () sin x = e + e cos = e ( cos sin ) x () = when cos = sin. Therefore, x () = on π 5 π π for = and =. 4 4 The candidaes for he absolue minimum are a π 5π,,, and π. 4 4 5 : : x () 1 : ses x () = 1 : answer 1: jusificaion x() e sin ( ) = π 4 5π 4 e e π π ( 4 ) > π ( ) 4 sin 5π 4 5 sin < 4 π e π sin ( π ) = The paricle is farhes o he lef when = 5 π. 4 (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 = 4 : : 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). 14