AP Calculus AB 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals - MCQs - Exam Style Questions
Calc-Ok Question
The velocity of a particle moving along the \(x\)-axis is given by \(v(t)=4\sqrt{t+1}\cos t\) for time \(t\ge 0\). At time \(t=0\), the particle is at position \(x=3\). What is the position of the particle at time \(t=2\) ?
(A) \(-1.431\)
(B) \(0.569\)
(C) \(4.365\)
(D) \(7.365\)
▶️ Answer/Explanation
Position function: \(x(t)=x(0)+\displaystyle\int_{0}^{t} v(s)\,ds\).
Here \(v(t)=4\sqrt{t+1}\cos t\).
So \(x(2)=3+\displaystyle\int_{0}^{2} 4\sqrt{t+1}\cos t\,dt\).
Numerical evaluation: \(\displaystyle\int_{0}^{2} 4\sqrt{t+1}\cos t\,dt\approx 4.3648\).
Add initial position: \(3+4.3648\approx 7.365\).
✅ Answer: (D) \(7.365\)
▶️ Answer/Explanation
Solution
Acceleration \( a(t) = 24t^2 \).
Velocity: \( v(t) = \int 24t^2 \, dt = 8t^3 + C_1 \). Since \( v(0) = 0 \), \( C_1 = 0 \), so \( v(t) = 8t^3 \).
Position: \( s(t) = \int 8t^3 \, dt = 2t^4 + C_2 \). Assume \( s(0) = 0 \), so \( C_2 = 0 \), thus \( s(t) = 2t^4 \).
Distance from \( t = 0 \) to \( t = 2 \): \( s(2) – s(0) = 2(2)^4 – 2(0)^4 = 32 – 0 = 32 \text{ feet} \).
✅ Answer: A