Practice a College Board-style free response question on straight-line motion — velocity, acceleration, and displacement. Write your response, then reveal the model answer to see exactly what earns each point.
Free Response Question · Unit 4 · Contextual Applications
A particle moves along a horizontal line with position given by s(t) = t³ − 6t² + 9t for t ≥ 0, where s is in meters and t is in seconds.
A
Find the velocity v(t) and the acceleration a(t). Find the velocity at t = 2 seconds, including units.
✓ Model answer
v(t) = s′(t) = 3t² − 12t + 9, and a(t) = v′(t) = 6t − 12. At t = 2: v(2) = 3(4) − 24 + 9 = −3 meters per second.
Why it scores: Correct v(t) and a(t) by differentiating, correct value v(2) = −3, and units (m/s). Missing units or a differentiation slip loses a point.
B
At t = 2, is the particle speeding up or slowing down? Justify your answer.
✓ Model answer
At t = 2, v(2) = −3 (negative) and a(2) = 6(2) − 12 = 0. Because the acceleration is 0, the particle is neither speeding up nor slowing down at that instant. (More generally, the particle is speeding up when v and a share a sign and slowing down when they have opposite signs.)
Why it scores: Compares the signs of v and a with correct values, and justifies the conclusion. A bare answer without the sign comparison earns no justification point.
C
Find all times t ≥ 0 when the particle is at rest, and find the total displacement of the particle on the interval [0, 4].
✓ Model answer
At rest when v(t) = 0: 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3) = 0, so t = 1 and t = 3 seconds. Displacement on [0, 4] = s(4) − s(0). s(4) = 64 − 96 + 36 = 4 and s(0) = 0, so the displacement is 4 meters.
Why it scores: Solves v(t) = 0 for both times, and computes displacement as s(4) − s(0) = 4 m. Confusing displacement with total distance would lose credit here.
How to score points on AP Calculus FRQs
Differentiate for motion: v = s′, a = v′; keep track of signs.
Justify speeding up/slowing down by comparing the signs of v and a.
Displacement is s(b) − s(a); total distance is different (needs direction changes).
Always include units when interpreting a rate in context.
Answer the verb. "Find" needs the computation; "justify" needs supporting reasoning.