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) \(7.365\)