Home / AP Calculus AB 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals – MCQs

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) \(7.365\)

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) \(35\)

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