2005 ap calculus ab free-response questions (form b)€¦ · 2009 ap® calculus ab free-response...

2005 AP® CALCULUS AB FREE-RESPONSE QUESTIONS (Form B)

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4

3. A particle moves along the x-axis so that its velocity v at time t, for 0 5,t is given by 2ln 3 3 .v t t t The particle is at position 8x at time 0.t

(a) Find the acceleration of the particle at time 4.t

(b) Find all times t in the open interval 0 5t at which the particle changes direction. During which time intervals, for 0 5,t does the particle travel to the left?

(c) Find the position of the particle at time 2.t

(d) Find the average speed of the particle over the interval 0 2.t

WRITE ALL WORK IN THE TEST BOOKLET.

END OF PART A OF SECTION II

��

2005 AP® CALCULUS AB FREE-RESPONSE QUESTIONS

Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

GO ON TO THE NEXT PAGE.

6

5. A car is traveling on a straight road. For 0 24t� � seconds, the car’s velocity � �,v t in meters per second, is modeled by the piecewise-linear function defined by the graph above.

(a) Find � �24

0.v t dtÔ Using correct units, explain the meaning of � �

24

0.v t dtÔ

(b) For each of � �4v� and � �20 ,v� find the value or explain why it does not exist. Indicate units of measure.

(c) Let � �a t be the car’s acceleration at time t, in meters per second per second. For 0 24,t� � write a piecewise-defined function for � �.a t

(d) Find the average rate of change of v over the interval 8 20.t� � Does the Mean Value Theorem guarantee a value of c, for 8 20,c� � such that � �v c� is equal to this average rate of change? Why or why not?

WRITE ALL WORK IN THE TEST BOOKLET.


© 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).

7

t

(sec) 0 15 25 30 35 50 60

� �v t

� �ft sec –20 –30 –20 –14 –10 0 10

� �a t

� �2ft sec 1 5 2 1 2 4 2

6. A car travels on a straight track. During the time interval 0 60t� � seconds, the car’s velocity v, measured in

feet per second, and acceleration a, measured in feet per second per second, are continuous functions. The table above shows selected values of these functions.

(a) Using appropriate units, explain the meaning of � �60

30v t dtÔ in terms of the car’s motion. Approximate

� �60

30v t dtÔ using a trapezoidal approximation with the three subintervals determined by the table.

(b) Using appropriate units, explain the meaning of � �30

0a t dtÔ in terms of the car’s motion. Find the exact

value of � �30

0.a t dtÔ

(c) For 0 60,t� � must there be a time t when � � 5 ?v t � Justify your answer.

(d) For 0 60,t� � must there be a time t when � � 0 ?a t Justify your answer.

WRITE ALL WORK IN THE EXAM BOOKLET.

END OF EXAM




5

CALCULUS AB SECTION II, Part B

Time—45 minutes Number of problems—3

No calculator is allowed for these problems.

t (seconds) 0 10 20 30 40 50 60 70 80

� �v t (feet per second)

5 14 22 29 35 40 44 47 49

4. Rocket A has positive velocity � �v t after being launched upward from an initial height of 0 feet at time 0t

seconds. The velocity of the rocket is recorded for selected values of t over the interval 0 80t� � seconds, as shown in the table above.

(a) Find the average acceleration of rocket A over the time interval 0 80t� � seconds. Indicate units of measure.

(b) Using correct units, explain the meaning of � �70

10v t dtÔ in terms of the rocket’s flight. Use a midpoint

Riemann sum with 3 subintervals of equal length to approximate � �70

10.v t dtÔ

(c) Rocket B is launched upward with an acceleration of � � 31

a tt

�

feet per second per second. At time

0t seconds, the initial height of the rocket is 0 feet, and the initial velocity is 2 feet per second. Which of the two rockets is traveling faster at time 80t seconds? Explain your answer.

WRITE ALL WORK IN THE PINK EXAM BOOKLET.




-3-

2. A particle moves along the x-axis so that its velocity v at time 0�t is given by � � � �2sin .v t t The graph of v

is shown above for 0 5 .t p� � The position of the particle at time t is � �x t and its position at time 0 t is

� �0 5. x

(a) Find the acceleration of the particle at time 3. t

(b) Find the total distance traveled by the particle from time 0 t to 3. t

(c) Find the position of the particle at time 3. t

(d) For 0 5 ,t p� � find the time t at which the particle is farthest to the right. Explain your answer.





-5-




4. A particle moves along the x-axis with position at time t given by � � sin� tx t e t for 0 2 .p� �t

(a) Find the time t at which the particle is farthest to the left. Justify your answer.

(b) Find the value of the constant A for which � �x t satisfies the equation � � � � � � 0Ax t x t x t� � �� for 0 2 .p� �t

t (minutes) 0 2 5 7 11 12

� ��r t (feet per minute)

5.7 4.0 2.0 1.2 0.6 0.5

5. The volume of a spherical hot air balloon expands as the air inside the balloon is heated. The radius of the

balloon, in feet, is modeled by a twice-differentiable function r of time t, where t is measured in minutes. For 0 12,� �t the graph of r is concave down. The table above gives selected values of the rate of change, � �,�r t of the radius of the balloon over the time interval 0 12.� �t The radius of the balloon is 30 feet when 5. t

(Note: The volume of a sphere of radius r is given by 34 .3p V r )

(a) Estimate the radius of the balloon when 5.4 t using the tangent line approximation at 5. t Is your estimate greater than or less than the true value? Give a reason for your answer.

(b) Find the rate of change of the volume of the balloon with respect to time when 5. t Indicate units of measure.

(c) Use a right Riemann sum with the five subintervals indicated by the data in the table to approximate

� �12

0.�Ô r t dt Using correct units, explain the meaning of � �

12

0�Ô r t dt in terms of the radius of the balloon.

(d) Is your approximation in part (c) greater than or less than � �12

0?�Ô r t dt Give a reason for your answer.





-5-




4. A particle moves along the x-axis so that its velocity at time t, for 0 6,t is given by a differentiable function v whose graph is shown above. The velocity is 0 at 0,t 3,t and 5,t and the graph has horizontal tangents at 1t and 4.t The areas of the regions bounded by the t-axis and the graph of v on the intervals 0, 3 , 3, 5 , and 5, 6 are 8, 3, and 2, respectively. At time 0,t the particle is at 2.x

(a) For 0 6,t find both the time and the position of the particle when the particle is farthest to the left. Justify your answer.

(b) For how many values of t, where 0 6,t is the particle at 8 ?x Explain your reasoning.

(c) On the interval 2 3,t is the speed of the particle increasing or decreasing? Give a reason for your answer.

(d) During what time intervals, if any, is the acceleration of the particle negative? Justify your answer.



© 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.

-7-

t

(seconds) 0 8 20 25 32 40

� �v t (meters per second)

3 5 –10 –8 –4 7

6. The velocity of a particle moving along the x-axis is modeled by a differentiable function v, where the position x

is measured in meters, and time t is measured in seconds. Selected values of � �v t are given in the table above. The particle is at position 7x meters when 0t seconds.

(a) Estimate the acceleration of the particle at 36t seconds. Show the computations that lead to your answer. Indicate units of measure.


20v t dtÔ in the context of this problem. Use a trapezoidal sum

with the three subintervals indicated by the data in the table to approximate � �40

20.v t dtÔ

(c) For 0 40,t� � must the particle change direction in any of the subintervals indicated by the data in the table? If so, identify the subintervals and explain your reasoning. If not, explain why not.

(d) Suppose that the acceleration of the particle is positive for 0 8t� � seconds. Explain why the position of the particle at 8t seconds must be greater than 30x meters.


END OF EXAM


© 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.

GO ON TO THE NEXT PAGE.-2-

CALCULUS AB SECTION II, Part A


A graphing calculator is required for some problems or parts of problems.

1. Caren rides her bicycle along a straight road from home to school, starting at home at time 0t minutes andarriving at school at time 12t minutes. During the time interval 0 12t� � minutes, her velocity � �,v t inmiles per minute, is modeled by the piecewise-linear function whose graph is shown above.

(a) Find the acceleration of Caren’s bicycle at time 7.5t minutes. Indicate units of measure.


0v t dtÔ in terms of Caren’s trip. Find the value of

� �12

0.v t dtÔ

(c) Shortly after leaving home, Caren realizes she left her calculus homework at home, and she returns to get it. At what time does she turn around to go back home? Give a reason for your answer.

(d) Larry also rides his bicycle along a straight road from home to school in 12 minutes. His velocity is modeled

by the function w given by � � � �sin ,15 12

w t tp p where � �w t is in miles per minute for 0 12t� � minutes.

Who lives closer to school: Caren or Larry? Show the work that leads to your answer.


Calculator


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GO ON TO THE NEXT PAGE. -4-




4. A squirrel starts at building A at time 0 t and travels along a straight, horizontal wire connected to building B. For 0 18,� �t the squirrel’s velocity is modeled by the piecewise-linear function defined by the graph above.

(a) At what times in the interval 0 18,� �t if any, does the squirrel change direction? Give a reason for your answer.

(b) At what time in the interval 0 18� �t is the squirrel farthest from building A ? How far from building A is the squirrel at that time?

(c) Find the total distance the squirrel travels during the time interval 0 18.� �t

(d) Write expressions for the squirrel’s acceleration � �,a t velocity � �,v t and distance � �x t from building A that are valid for the time interval 7 10.� �t


AP® CALCULUS AB 2005 SCORING GUIDELINES (Form 8)

Question 3

A particle moves along the x-axis so that its velocity v at time t, for 0 St S 5, is given by

v(t) = In( t 2 - 3t + 3 ). The particle is at position x = 8 at time t = 0.

(a} Find the acceleration of the particle at time t = 4.

(b) Find all times t in the open interval 0 < t < 5 at which the particle changes direction. During which time intervals, for 0 :S t :S 5, does the particle travel to the left?

(c) Find the position of the particle at time t = 2.

( d) Find the average speed of the particle over the interval 0 :S t :S 2.

(a) a(4) = v'(4) = 1 7

(b) v(t) = 0

t2 - 3t + 3 =I

t 2 -3t + 2 = 0 (t-2) (t-1) = 0 t = 1, 2

v(t) > 0 for 0 < t <I

v( t) < 0 for 1 < t < 2

v(t) > 0 for 2 < t < 5

The particle changes direction when t =I and t = 2. The particle travels to the left when 1 < t < 2.

(c) s(t) = s(O) + J;1n(u 2 - 3u + 3) du

s(2) = 8 + Jo2 In(u2 - 3u + 3) du

= 8.368 or 8.369

1 J2 (d) 2 0 lv(t)j dt = 0.370 or 0.371

1: answer

{ I : sets v( t) = 0

3 : 1 : direction change at t = 1, 2

1 : interval with reason

{ 1: Jo2 In(u2 - 3u + 3) du

3 : 1 : handles initial condition

1: answer

2 : { 1 : integral 1: answer

Copyright© 2005 by College Board All rights reserved. \f;o;t ".:lnl".c>-ntr".lll "'"'11""'nC>hA"JIT'A .f"l''\Tn /fnr AP n-rn.f'""'"'"';"""Jllc\ "Jlnrl u1nrnr rnllP.n-Phrv.:ar.-1 r-Aml<::tnc.+nrla.ntc- ffAr AP c<t11AP.ntc ".:lnt'l TV.::.T'At1teo\

AP® CALCULUS AB 2005 SCORING GUIDELINES

Questions

I"( t)

(16. 201 A car is traveling on a straight road. For 0 :S t :S 24 seconds, the car's velocity v(t), in meters per second, is modeled by the piecewise-linear

function defined by the graph above.

"" ( "". 20) § 20 ___ ....,

"' E~ 1;

JM JM (a) Find 0 v(t) dt. Using correct units, explain the meaning of 0 v(t) dt. g=.; ~ Q. 10 >~

i:: 5 E (b) For each of v'( 4) and v'(20), find the value or explain why it does not

exist. Indicate units of measure. 0 4 8 12 16 20 24

Time (seconds)

(c) Let a(t) be the car's acceleration at time t, in meters per second per second. For 0 < t < 24, write a

piecewise-defined function for a( t).

(d) Find the average rate of change of v over the interval 8 :S t :S 20. Does the Mean Value Theorem guarantee a value of c, for 8 < c < 20, such that v'(c) is equal to this average rate of change? Why or why not?

(a) J:4v(t) dt = ~ (4)(20) + (12)(20) + ~ (8)(20) = 360

The car travels 360 meters in these 24 seconds.

(b) v'( 4) does not exist because

lim (v(t)- v( 4)) = 5 * 0 = lim (v(t)- v( 4)). t~4- t - 4 t~4+ t - 4

'( 20) 20 - o 5 I 2 v = = -- m sec 16- 24 2

(c) { 5

a(t) ~ -~ if0<t<4

if 4 < t < 16

if 16 < t < 24

a(t) does not exist at t = 4 and t = 16.

(d) The average rate of change of v on [8, 20) is

v(20) - v(8) _ _ 1 / 2 20 - 8 - 6 m sec .

No, the Mean Value Theorem does not apply to v on [8, 20] because v is not differentiable at t = 16.

2 . { l: value · 1 : meaning with units

{ 1 : v'( 4) does not exist, with explanation

3 : 1 : v'(20)

1 : units

{ 5

2 : 1 : finds the values 5, 0, - 2 1 : identifies constants with correct intervals

2 : { 1 : average rate of change of v on [ 8, 20] 1 : answer with explanation

Copyright © 2005 by College Board. All rights reserved. .. .. ' ..

Ap® CALCULUS AB 2006 SCORING GUIDELINES (Form B)

Question 6

t 0 15 25 30 35 50 60

(sec)

v(t) -20 -30 -20 -14 -10 0 10

(ft/sec)

a(t)

(ft/ sec2 ) 1 5 2 1 2 4 2

A car travels on a straight track. During the time interval 0 ~ t ~ 60 seconds, the car's velocity v, measured in feet per second, and acceleration a, measured in feet per second per second, are continuous functions. The table above shows selected values of these functions.

(a) Using appropriate units, explain the meaning of J 60 lv(t)I dt in tenns of the car's motion. Approximate 30

J60 lv(t)I dt using a trapezoidal approximation with the three subintervals detennined by the table.

30

I 30 (b) Using appropriate units, explain the meaning of 0 a(t) dt in tenns of the car's motion. Find the exact value

J30 of 0 a(t) dt.

(c) For 0 < t < 60, must there be a time t when v(t) = -5? Justify your answer.

(d) For 0 < t < 60, must there be a time t when a(t) = 0? Justify your answer.

f 60 (a) lv(t)I dt is the distance in feet that the car travels

30

from t = 30 sec to t = 60 sec.

Trapezoidal approximation for J60lv(t)I dt: 30

A= ~ (14 + 10}5 + ~ (10)(15) + ~ (10)(10) = 185 ft

J30 (b) 0 a( t) dt is the car's change in velocity in ft/sec from

t = 0 sec to t = 30 sec.

J30 J30 , 0 a(t) dt = 0 v (t) dt = v(30) -v(O)

= -14 -(-20) = 6 ft/sec

(c) Yes. Since v(35) = -10 < -5 < 0 = v(50), the IVT

guarantees at in (35, 50) so that v(t) = -5.

(d) Yes. Since v(O) = v(25), the MVTguarantees at in

(0, 25) so that a(t) = v'(t) = 0.

Units of ft in (a) and ft/sec in (b)

2 : { 1 : explanation I : value

2 : { I : explanation I : value

{ 1 : v(35) < -5 < v(50) 2·

· 1 : Yes; refers to IVT or hypotheses

{ 1 : v(O) = v(25)

2· · 1 : Yes; refers to MVT or hypotheses

1 : units in (a) and (b)

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7


Question 4

t 0 10 20 30 40 50 60 70 80 (seconds)

v(t) (feet per second)

5 14 22 29 35 40 44 47 49

Rocket A has positive velocity v(t) after being launched upward from an initial height ofO feet at time t = 0

seconds. The velocity of the rocket is recorded for selected values oft over the interval 0 :s; t :s; 80 seconds, as shown in the table above. (a) Find the average acceleration ofrocketA over the time interval 0 :s; t :s; 80 seconds. Indicate units of

measure.

J70 (b) Using correct units, explain the meaning of v(t) dt in terms of the rocket's flight. Use a midpoint

IO

J70 Riemann sum with 3 subintervals of equal length to approximate v(t) dt.

IO

(c) Rocket Bis launched upward with an acceleration of a(t) = ~ feet per second per second. At time vt + 1

t = 0 seconds, the initial height of the rocket is 0 feet, and the initial velocity is 2 feet per second. Which of the two rockets is traveling faster at time t = 80 seconds? Explain your answer.

(a) Average acceleration of rocket A is

v{80) - v(O} _ 49 - 5 _ 11 ft/ 2

80 - 0 - 80 - 20 sec

J70 (b) Since the velocity is positive, v(t) dt represents the

IO

distance, in feet, traveled by rocket A from t = I 0 seconds to t = 70 seconds.

A midpoint Riemann sum is 20[v(20) + v( 40) + v(60)]

= 20(22 + 35 + 44] = 2020 ft

( c) Let v B ( t) be the velocity ofrocket B attime t.

Vs(t) = f ~ dt = 6.Jt+i + C vt +I

2=v8 (0)=6+C

Vs(t) = 6.Jt+i - 4

v8 (80) = 50 > 49 = v(80)

Rocket B is traveling faster at time t = 80 seconds.

Units of ft/ sec2 in (a) and ft in (b)

1: answer

{ I : explanation

3 : 1 : uses v(20), v( 40), v( 60)

1 : value

1: 6.ft+f 1 : constant of integration

4 : 1 : uses initial condition

1 : finds v8 (80), compares to v(80),

and draws a conclusion


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"

AP® CALCULUS AB 2007 SCORING GWDELINES (Form B)

Question 2

A particle moves along the x-axis so that its velocity v at time 1·( n t ~ 0 isgivenby v(t) = sin(t2 ). Thegraphof vis shown above

for 0 ~ t ~ J5ii. The position of the particle at time t is x( t) and

its position at time t = 0 is x(O) = 5.

(a) Find the acceleration of the particle at time t = 3.

(b) Find the total distance traveled by the particle from time t = 0 to t = 3.

(c) Find the position of the particle at time t = 3.

(d) For 0 :-;;; t :-;;; &, find the time t at which the particle

is farthest to the right. Explain your answer.

(a) a(3) = v'(3) = 6cos 9 = -5.466 or -5.467

(b) Distance = J :lv(t)I dt = 1.702

OR For 0 < t < 3, v(t) = 0 when t = ..fii = 1.77245 and

t = Jfi = 2.50663 x(O) = 5

J ..[jf x(..fii) = 5 + 0 v(t) dt = 5.89483

J ,,ff; x(.&) = 5 + 0 v(t) dt = 5.43041

x(3) = 5 + J: v(t) dt = 5.77356

()

-I

lx(..fii)- x(O)j + lx(Jfi)- x(..fii)I + jx(3)- x(Jfi)I = 1.702

(c) x(3) = 5 + J: v(t) dt = 5.773 or 5.774

( d) The particle's rightmost position occurs at time t = ..fii = 1. 772.

The particle changes from moving right to moving left at those times t for which v(t) = 0 with v(t) changing from positive to negative, namely at

t =Ji,&, J5ii (t = 1.772, 3.070, 3.963).

Using x(T) = 5 + J: v(t) dt, the particle's positions at the times it

changes from rightward to leftward movement are: T: 0 Ji .J3ii .J5ii

x(T): 5 5.895 5.788 5.752

The particle is farthest to the right when T = Ji.

© 2007 The College Board. All rights reserved.

I : a(3)

2 : {I : setup I : answer

{ {

I : integrand

3 : 2 I : uses x(O) = 5

I: answer

{ I : sets v(t) = 0

3: I: answer

I : reason


Question 4

A particle moves along the x-axis with position at time t given by x(t) = e-t sin t for 0 :5: t :5: 2JC.

(a) Find the time t at which the particle is farthest to the left. Justify your answer.

(b) Find the value of the constant A for which x(t) satisfies the equation Ax•(t) + x'(t) + x(t) = 0

for 0 < t < 2Jl.

(a) x' ( t) = -e -t sin t + e -t cos t = e -t (cos t - sin t)

x'(t) = 0 when cost= sint. Therefore, x'(t) = 0 on

0 < t < 2- £or t = Jl and t = 5JC --'~ 4 4·

The candidates for the absolute minimum are at Jl 5Jl

t = 0, 4 , 4 , and 2JC.

t x(t)

0 e0 sin(O) = 0 1t Jl

e 4 sin(~)> 0 4

5Jl 5Jr -4 . (5Jl) 0

4 e sm 4 <

2Jl e-2/f sin (2K) = 0

The particle is farthest to the left when t = 5:.

(b) x'(t) = -e-t (cost - sin t) + e-t (-sin t - cost)

= -2e-t cost

Ax•(t) + x'(t) + x(t)

= A ( -2e -t cos t) + e -t (cos t - sin t) + e-t sin t

= (-2A + I)e-t cost

=0

1 Therefore, A= 2 .

{

2: x'(t) 1: sets x'(t) = 0

5: 1: answer

I : justification

{

2:x•(t)

I: substitutes x•(t), x'(t), and x(t) 4 : into Ax'(t) + x'(t) + x(t)

I: answer

© 2007 The College Board. All rights reserved. Visit apcentraLcollegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).


Question 4

v( /)

Grnph of r

A particle moves along the x-axis so that its velocity at time t, for 0 ~ t ~ 6, is given by a differentiable function v whose graph is shown above. The velocity is 0 at t = 0, t = 3, and t = 5, and the graph has horizontal tangents at t = 1 and t = 4. The areas of the regions bounded by the t-axis and the graph of v on the intervals [O, 3], [3, 5], and [5, 6] are 8, 3, and 2, respectively. At time t = 0, the particle is at x = -2.

(a) For 0 ~ t ~ 6, find both the time and the position of the particle when the particle is farthest to the left. Justify your answer.

(b) For how many values of t, where 0 5 t 5 6, is the particle at x = -8 ? Explain your reasoning.

( c) On the interval 2 < t < 3, is the speed of the particle increasing or decreasing? Give a reason for your answer.

( d) During what time intervals, if any, is the acceleration of the particle negative? Justify your answer.

(a) Since v(t) < 0 for 0 < t < 3 and 5 < t < 6, and v(t) > 0

for 3 < t < 5, we consider t = 3 and t = 6.

x(3) = -2 + J: v(t) dt = -2 - 8 = -10

x(6) = -2 + J: v(t) dt = -2 - 8 + 3 - 2 = -9

Therefore, the particle is farthest left at time t = 3 when its position is x( 3) = -10.

(b) The particle moves continuously and monotonically from x(O) = -2 to x(3) = -10. Similarly, the particle moves

continuously and monotonically from x( 3) = -10 to

x(5) = -7 and also from x(5) = -7 to x(6) = -9.

By the Intermediate Value Theorem, there are three values oft for which the particle is at x(t) = -8.

( c) The speed is decreasing on the interval 2 < t < 3 since on this interval v < 0 and v is increasing.

(d) The acceleration is negative on the intervals 0 < t < I and 4 < t < 6 since velocity is decreasing on these intervals.

{

1 : identifies t : 3 as a candidate

3: I : considers J 0 v(t) dt

1 : conclusion

3 . and t = 6 {

1 : positions at t = 3, t = 5,

· I : description of motion

l : conclusion

1 : answer with reason

{ 1: answer

2 : 1 : justification

© 2008 The College Board. All rights reserved.

AP9 CALCULUS AB 2009 SCORING GUIDELINES (Form B)

Question 6

t 0 8 20 25 32 40 (seconds)

v(t) 3 5

(meters per second) -10 -8 -4 7

The velocity of a particle moving along the x-axis is modeled by a differentiable function v, where the position x is measured in meters, and time t is measured in seconds. Selected values of v(t) are given in the

table above. The particle is at position x = 7 meters when t = 0 seconds.

(a) Estimate the acceleration of the particle at t = 36 seconds. Show the computations that lead to your answer. Indicate units of measure.

(b) Using correct units, explain the meaning of J 40 v(t) dt in the context ofthis problem. Use a 20

J 40 trapezoidal sum with the three subintervals indicated by the data in the table to approximate v( t) dt.

20

(c) For 0 ~ t ~ 40, must the particle change direction in any of the subintervals indicated by the data in the table? If so, identify the subintervals and explain your reasoning. If not, explain why not.

(d) Suppose that the acceleration of the particle is positive for 0 < t < 8 seconds. Explain why the position of the particle at t = 8 seconds must be greater than x = 30 meters.

, v( 40) - v( 3 2) 11 / 2 (a) a(36) = v (36)"" = - meters sec 40 - 32 8

J 40 (b) v( t) dt is the particle's change in position in meters from time

20

t = 20 seconds to time t = 40 seconds.

J 40 v(t) dt"" v(20) + v(25) . 5 + v(25) + v(32) _ 7 + v(32) + v( 40) _ 8 20 2 2 2

= -75 meters

(c) v(8) > 0 and v(20) < 0

v(32) < 0 and v( 40) > 0

Therefore, the particle changes direction in the intervals 8 < t < 20 and 32 < t < 40.

(d) Since v'(t) = a(t) > 0 for 0 < t < 8, v(t) ~ 3 on this interval.

Therefore, x(8) = x(O) + J: v(t) dt ~ 7 + 8 -3 > 30.

© 2009 The Colleae Board. All riahts reserved.


1: answer

{ 1: meaning of J: v(t) dt

3 : 2 : trapezoidal

approximation

2 . { 1 : answer · 1 : explanation

{ 1 : v'(t) = a(t)

2 : 1 : explanation of x(8) > 30

APe CALCULUS AB 2009 SCORING GmDELINES

v(t)

0.3

0.2

0.1

0

-0.1

j_ ~ I ~

1 2~ 3 j_ ~v -0.2

Question 1

t N \ I

5 6 7 8 9 lO 11 12

Caren rides her bicycle along a straight road from home to school, starting at home at time t = 0 minutes and arriving at school at time t = 12 minutes. During the time interval 0 ~ t ~ 12 minutes, her velocity v( t), in miles per minute, is modeled by the piecewise-linear function whose graph is shown above.

(a) Find the acceleration of Caren's bicycle at time t = 7.5 minutes. Indicate units of measure.

J 12 (b) Using correct units, explain the meaning of 0 lv(t)I dt in terms of Caren's trip. Find the value

J 12 of 0 lv(t)I dt.

( c) Shortly after leaving home, Caren realizes she left her calculus homework at home, and she returns to get it. At what time does she tum around to go back home? Give a reason for your answer.

(d) Larry also rides his bicycle along a straight road from home to school in 12 minutes. His velocity is

modeled by the function w given by w( t) = ~ sin ( ~ t), where w( t) is in miles per minute for

0 ~ t ~ 12 minutes. Who lives closer to school: Caren or Larry? Show the work that leads to your answer.

(a) a(7.5) = v'(7.5) = v(S~ = ;(7) = -0.l miles/minute2

J 12 (b) 0 lv(t)I dt is the total distance, in miles, that Caren rode

during the 12 minutes from t = 0 to t = 12.

J 12 J 2 f 4 J 12 0 lv(t)I dt = 0 v(t) dt - 2 v(t) dt + 4 v(t) dt

= 0.2 + 0.2 + 1.4 = 1.8 miles

( c) Caren turns around to go back home at time t = 2 minutes. This is the time at which her velocity changes from positive to negative.

J 12 (d) 0 w(t) dt = 1.6; Larry lives 1.6 miles from school.

J 12

0 v(t) dt = 1.4; Caren lives 1.4 miles from school.

Therefore, Caren lives closer to school.

2 : { 1 : an~wer 1: umts

2 . { 1 : meaning of integral · 1 : value of integral

2 : { 1 : answer 1 : reason

2: Larry's distance from school

1 : integral

3: 1 : value

1 : Caren's distance from school

and conclusion

© 2009 The Colleae Board. All riahts reserved.

AP9 CALCULUS AB 2010 SCORING GUIDELINES (Form B)

Question 4

A squirrel starts at building A at time t = 0 and travels along a straight wire connected to building B. For 0 :s: t :s: 18, the <2.20> r7, 20)

20 .-----..... squirrel· s velocity is modeled by the piecewise-linear function defined by the graph above. 10

-JO (10.-IO)

(a) At what times in the interval 0 < t < 18, if any, does the squirrel change direction? Give a reason for your answer.

(b) At what time in the interval 0 :s: t :s: 18 is the squirrel farthest from building A ? How far from building A is the squirrel at this time?

Grapbof v

(c) Find the total distance the squirrel travels during the time interval 0 :s: t :s: 18.

(16, 10)

(14.-10)

(d) Write expressions for the squirrel's acceleration a(t), velocity v(t), and distance x(t) from building A that are valid for the time interval 7 < t < 10.

(a) The squirrel changes direction whenever its velocity changes sign. This occurs at t = 9 and t = 15.

(b) Velocity is 0 at t = 0, t = 9, and t = 15.

t osition at time t 0 0

9 9 + 5 -20=140 2

15 140 - 6 + 4 · 10 = 90 2

18 90+ 3 ; 2 -10=115

The squirrel is farthest from building A at time t = 9; its greatest distance from the building is 140.

J 18 (c) The total distance traveled is 0 lv(t)I dt = 140 + 50 + 25 = 215.

(d) For 7 < t < 10 a(t) = 20 - (-lO) = -10 ' 7 -10

v(t) = 20 - lO(t - 7) = -lOt + 90

x(7) = 1 + 5 - 20 = 120 2

x(t) = x(7) + J; (-lOu + 90) du

= 120 + (-5u 2 + 90u )I::~

= -5t2 + 90t - 265

2 . { 1 : t-values · 1 : explanation

2 : { 1 : identifies candidates 1: answers

1: answer

{ 1 : a(t)

4: 1 : v(t)

2: x(t)

2005 ap calculus ab free-response questions (form b)€¦ · 2009 ap® calculus ab free-response...

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