AP Calculus BC 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals - MCQs - Exam Style Questions
Question
The velocity of a particle moving along the \(x\)-axis is \( v(t)=t^{2}-9-4\sin(t^{2}-9) \) for \(t\ge0\). At what time \(t\), for \(0\le t\le 3\), is the particle farthest to the left?
(B) \(2.009\)
(C) \(2.554\)
(D) \(2.772\)
▶️ Answer/Explanation
Correct answer: (C) \(2.554\)
Farthest left \( \Longleftrightarrow \) minimum position \(x(t)\) on \([0,3]\). This occurs at endpoints or where \(v(t)=x'(t)=0\) with sign change \(- \to +\).
Solve \( v(t)=0 \): \[ t^{2}-9-4\sin(t^{2}-9)=0. \] Let \(u=t^{2}-9\) (so \(u\in[-9,0]\)). We need \(u-4\sin u=0\), i.e. \(u=4\sin u\). On \([-9,0]\) there are two solutions: \[ u=0 \;(\Rightarrow t=3),\qquad u\approx -2.470 \;(\Rightarrow t\approx \sqrt{9-2.470}\approx 2.554). \] Sign check: \[ v(2)=2^{2}-9-4\sin(4-9)=-5-4\sin(-5)\lt 0, \] \[ v(2.7)=2.7^{2}-9-4\sin(7.29-9)\approx -1.71-4\sin(-1.71)\gt 0, \] so \(v\) changes \(-\to+\) at \(t\approx 2.554\) (local minimum of \(x\)). At \(t=3\), \(v(3)=0\) but \(v'(3)=2\cdot3\!\left[1-4\cos(0)\right]=-18\lt0\) (no \(-\to+\) change). Therefore the particle is farthest left at \[ \boxed{t\approx 2.554}. \]
Calc-Ok Question
(B) \(0.845\)
(C) \(1.448\)
(D) \(2.887\)
▶️ Answer/Explanation
Total distance \(=\displaystyle \int_{0}^{2}\!|\sin(t^{2})|\,dt\).
Sign change at \(t=\sqrt{\pi}\approx1.772\); split or evaluate numerically.
Calculator result:
\(\displaystyle \int_{0}^{2}|\sin(t^{2})|\,dt\approx1.448.\)
✅ Answer: (C)