AP CALCULUS AB 27 SCORING GUIDELINES (Form B) Question 2 A particle moves along the x-axis so that its velocity v at time 2 t is given by vt () = sin ( t ). The graph of v is shown above for t 5 π. The position of the particle at time t is x( t ) and its position at time t = is x ( ) = 5. (a) Find the acceleration of the particle at time t = 3. (b) Find the total distance traveled by the particle from time t = to t = 3. (c) Find the position of the particle at time t = 3. (d) For t 5 π, find the time t at which the particle is farthest to the right. Explain your answer. (a) a( 3) = v ( 3) = 6cos9 = 5.466 or 5.467 1 : a ( 3) (b) Distance = vt () dt= 1.72 3 OR For < t < 3, vt ( ) = when t = π = 1.77245 and t = 2π = 2.5663 x ( ) = 5 x x ( ) () π = 5 + v t dt = 5.89483 π 2π ( ) () 2π = 5 + v t dt = 5.4341 3 x( 3) = 5 + v( t) dt = 5.77356 ( π ) ( ) ( π ) ( π ) ( ) ( π ) x x + x 2 x + x 3 x 2 = 1. 72 2 : { 1 : setup 1 : answer 1 : integrand 3 2 (c) x( 3) = 5 + v( t) dt = 5.773 or 5.774 3 : 1 : uses x( ) = 5 1 : answer (d) The particle s rightmost position occurs at time t = π = 1.772. The particle changes from moving right to moving left at those times t for which vt ( ) = with vt ( ) changing from positive to negative, namely at t = π, 3 π, 5π ( t = 1.772, 3.7, 3.963 ). T Using x( T) = 5 + v( t) dt, the particle s positions at the times it changes from rightward to leftward movement are: T: π 3π 5π xt ( ): 5 5.895 5.788 5.752 The particle is farthest to the right when T = π. 3 : 1 : sets vt () = 1 : answer 1 : reason 27 The College Board. All rights reserved.
27 The College Board. All rights reserved.
27 The College Board. All rights reserved.
27 The College Board. All rights reserved.
27 The College Board. All rights reserved.
27 The College Board. All rights reserved.
27 The College Board. All rights reserved.
AP CALCULUS AB 27 SCORING COMMENTARY (Form B) Question 2 Sample: 2A Score: 9 The student earned all 9 points. Sample: 2B Score: 6 The student earned 6 points: no points in part (a), 2 points in part (b), 3 points in part (c), and 1 point in part (d). Correct work is presented in parts (b) and (c). In part (a) the student did not earn the first point because the derivative of vt ( ) is incorrect. The student could have used the graphing calculator to find the numerical derivative. In part (d) the student does not set vt ( ) =, so the first point was not earned. The answer point was earned but not the reason point since the student does not explicitly rule out the other times for which vt ( ) =. Sample: 2C Score: 3 The student earned 3 points: no points in part (a), no points in part (b), 2 points in part (c), and 1 point in part (d). In part (a) the derivative of vt ( ) is correct, but the student makes an error when evaluating the acceleration at t = 3. In part (b) the student integrates the velocity to find displacement instead of integrating the speed to find distance traveled. In this case, since the particle changes direction on the interval from t = to t = 3, displacement is not the same as distance traveled. In part (c) the student has a correct integrand and uses x ( ) = 5, which earned the first 2 points. The student attempts to find the antiderivative of vt ( ) but did not earn the last point. In part (d) the student does not set vt ( ) =, so the first point was not earned. The answer point was earned but not the reason point since the student does not explicitly rule out the other times when vt ( ) =. 27 The College Board. All rights reserved.