AP Calculus AB 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals - MCQs - Exam Style Questions
Calc-Ok Question
(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) \(7.365\)
No-Calc Question
An object moves along a straight line so that at any time \(t\) its acceleration is \(a(t)=6t\). At time \(t=0\), the object’s velocity is \(10\) and the object’s position is \(7\). What is the object’s position at time \(t=2\)?
(A) \(22\)
(B) \(27\)
(C) \(28\)
(D) \(35\)
▶️ Answer/Explanation
Velocity: \(v(t)=v(0)+\displaystyle\int_{0}^{t}6s\,ds=10+3t^{2}\).
Position: \(x(t)=x(0)+\displaystyle\int_{0}^{t}v(s)\,ds=7+\int_{0}^{t}\!\big(3s^{2}+10\big)\,ds=7+\Big(s^{3}+10s\Big)\Big|_{0}^{t}\).
\(x(2)=7+\big(2^{3}+10\times 2\big)=7+(8+20)=35\).
✅ Answer: (D) \(35\)