AP Calculus BC Practice Quiz: Solving Motion Problems Using Parametric and Vector-Valued Functions

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 9 questions to check your progress.

Question 1 of 9

A particle moves in the xy-plane so that its position at any time t is given by the parametric equations x(t) = 3t^2 - 1 and y(t) = sin(πt). What is the velocity vector of the particle at time t = 2?

A)<12, π>B)<12, 0>C)<6, 0>D)<11, 1>

All Questions (9)

A) <12, π>

B) <12, 0>

C) <6, 0>

D) <11, 1>

Correct Answer: A

To find the velocity vector, we need to find the derivative of each component of the position vector r(t) = <x(t), y(t)>. The velocity vector is v(t) = <x'(t), y'(t)>. Here, x'(t) = 6t and y'(t) = πcos(πt). Evaluating at t = 2, we get x'(2) = 6(2) = 12 and y'(2) = πcos(2π) = π(1) = π. Therefore, the velocity vector at t = 2 is <12, π>. [cite: 3051]

A particle moves along a curve in the plane with velocity vector v(t) = <ln(t), 2t^3> for t > 0. What is the acceleration vector of the particle at time t = 1?

A) <1, 6>

B) <0, 6>

C) <1, 3>

D) <e, 2>

Correct Answer: A

The acceleration vector a(t) is the derivative of the velocity vector v(t). Given v(t) = <ln(t), 2t^3>, we find a(t) = v'(t) = <d/dt(ln(t)), d/dt(2t^3)> = <1/t, 6t^2>. To find the acceleration at t = 1, we substitute t = 1 into the expression for a(t): a(1) = <1/1, 6(1)^2> = <1, 6>. [cite: 3051]

A particle moves in the xy-plane with position given by x(t) = 4cos(t) and y(t) = 4sin(t) for t ≥ 0. What is the speed of the particle at time t = π/3?

A) 2

B) 4

C) 8

D) 2√3

Correct Answer: B

First, find the velocity vector v(t) = <x'(t), y'(t)>. Here, x'(t) = -4sin(t) and y'(t) = 4cos(t). The speed is the magnitude of the velocity vector, given by the formula speed = ||v(t)|| = sqrt((x'(t))^2 + (y'(t))^2). So, speed = sqrt((-4sin(t))^2 + (4cos(t))^2) = sqrt(16sin^2(t) + 16cos^2(t)) = sqrt(16(sin^2(t) + cos^2(t))) = sqrt(16) = 4. The speed is constant for all t. Therefore, at t = π/3, the speed is 4. [cite: 3051]

The velocity of a particle moving in the xy-plane is given by the vector v(t) = <3t^2, 1/(t+1)>. What is the displacement of the particle from time t = 0 to t = 2?

A) <8, ln(3)>

B) <12, ln(2)>

C) <8, 3/2>

D) <6, -1/9>

Correct Answer: A

The displacement of the particle is the net change in position, which is found by integrating the velocity vector over the time interval. Displacement = ∫[from 0 to 2] v(t) dt = <∫[from 0 to 2] 3t^2 dt, ∫[from 0 to 2] 1/(t+1) dt>. The x-component is ∫[from 0 to 2] 3t^2 dt = [t^3] from 0 to 2 = 2^3 - 0^3 = 8. The y-component is ∫[from 0 to 2] 1/(t+1) dt = [ln|t+1|] from 0 to 2 = ln(3) - ln(1) = ln(3). Thus, the displacement vector is <8, ln(3)>. [cite: 3052]

A particle moves in the plane with velocity vector v(t) = <t^2, sin(e^t)>. Which of the following integrals gives the total distance traveled by the particle from t = 0 to t = 2?

A) ∫[from 0 to 2] (t^2 + sin(e^t)) dt

B) ∫[from 0 to 2] sqrt(t^2 + sin^2(e^t)) dt

C) ∫[from 0 to 2] sqrt(4t^2 + e^(2t)cos^2(e^t)) dt

D) ∫[from 0 to 2] sqrt(t^4 + sin^2(e^t)) dt

Correct Answer: D

The total distance traveled by a particle over a time interval [a, b] is the integral of its speed. The speed is the magnitude of the velocity vector, ||v(t)||. Given v(t) = <x'(t), y'(t)> = <t^2, sin(e^t)>, the speed is sqrt((x'(t))^2 + (y'(t))^2) = sqrt((t^2)^2 + (sin(e^t))^2) = sqrt(t^4 + sin^2(e^t)). The total distance traveled from t = 0 to t = 2 is the definite integral of the speed over this interval, which is ∫[from 0 to 2] sqrt(t^4 + sin^2(e^t)) dt. [cite: 3053]

At time t = 1, a particle moving in the xy-plane is at the position (3, -2). If the velocity of the particle is given by v(t) = <2t, 6t^2 - 2>, what is the position of the particle at time t = 2?

A) (4, 8)

B) (6, 10)

C) (6, 12)

D) (7, 6)

Correct Answer: B

The position at time t=2 can be found by adding the initial position at t=1 to the displacement from t=1 to t=2. The position vector is p(t) = p(1) + ∫[from 1 to 2] v(t) dt. The x-coordinate is x(2) = x(1) + ∫[from 1 to 2] 2t dt = 3 + [t^2] from 1 to 2 = 3 + (2^2 - 1^2) = 3 + 3 = 6. The y-coordinate is y(2) = y(1) + ∫[from 1 to 2] (6t^2 - 2) dt = -2 + [2t^3 - 2t] from 1 to 2 = -2 + [(2(2)^3 - 2(2)) - (2(1)^3 - 2(1))] = -2 + [(16 - 4) - (2 - 2)] = -2 + 12 = 10. The position at t=2 is (6, 10). [cite: 3050, 3052]

A particle moves in the xy-plane with its position given by the parametric equations x(t) = t^3 - 3t and y(t) = t^2 - 4t + 1. For what value of t > 0 is the tangent line to the particle's path horizontal?

A) t = 1

B) t = √3

C) t = 2

D) t = 4

Correct Answer: C

A horizontal tangent line occurs when the vertical component of velocity, dy/dt, is zero, and the horizontal component, dx/dt, is not zero. First, find dy/dt: y'(t) = 2t - 4. Set y'(t) = 0 to find potential times: 2t - 4 = 0, which gives t = 2. Now, we must check that dx/dt is not zero at t = 2. Find dx/dt: x'(t) = 3t^2 - 3. At t = 2, x'(2) = 3(2)^2 - 3 = 12 - 3 = 9. Since x'(2) ≠ 0, the tangent line is horizontal at t = 2. [cite: 3050, 3051]

The position of a particle moving in the plane is given by r(t) = <t^3, t^2>. At time t = 1, which of the following statements is true?

A) The particle is slowing down.

B) The particle is speeding up.

C) The particle's speed is constant.

D) The particle is at rest.

Correct Answer: B

To determine if the particle is speeding up or slowing down, we can examine the sign of the dot product of the velocity and acceleration vectors, v(t) · a(t). First, find the velocity and acceleration vectors. v(t) = r'(t) = <3t^2, 2t>. a(t) = v'(t) = <6t, 2>. Now, evaluate both at t = 1: v(1) = <3(1)^2, 2(1)> = <3, 2>. a(1) = <6(1), 2> = <6, 2>. Calculate the dot product: v(1) · a(1) = (3)(6) + (2)(2) = 18 + 4 = 22. Since the dot product is positive, the angle between the velocity and acceleration vectors is acute, which means the particle is speeding up. [cite: 3051]

A particle moves in the xy-plane with velocity v(t) = <x'(t), y'(t)>. If the particle is at the point (5, 1) at time t = 2, which expression represents the y-coordinate of the particle's position at time t = 5?

A) 5 + ∫[from 2 to 5] x'(t) dt

B) ∫[from 2 to 5] y'(t) dt

C) 1 + ∫[from 2 to 5] y'(t) dt

D) 1 + ∫[from 2 to 5] sqrt((x'(t))^2 + (y'(t))^2) dt

Correct Answer: C

According to the Fundamental Theorem of Calculus, the final position is the initial position plus the accumulated change (displacement). The displacement in the y-direction from t=2 to t=5 is given by the definite integral of the y-component of velocity, ∫[from 2 to 5] y'(t) dt. To find the y-coordinate at t=5, we add this displacement to the y-coordinate at t=2. Therefore, y(5) = y(2) + ∫[from 2 to 5] y'(t) dt. Since y(2) = 1, the expression is 1 + ∫[from 2 to 5] y'(t) dt. [cite: 3052]