AP Calculus AB 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration - MCQs - Exam Style Questions
No-Calc Question
A particle moves along the \(x\)-axis so that for \(t>0\) its acceleration is \(a(t)=\dfrac{4}{t^{2}}\). Which of the following could be an expression for the position \(x(t)\) of the particle at time \(t\)?
(A) \(x(t)=-\dfrac{8}{t^{3}}\)
(B) \(x(t)=-\dfrac{4}{t}\)
(C) \(x(t)=-4\ln t\)
(D) \(x(t)=\dfrac{24}{t^{4}}\)
(B) \(x(t)=-\dfrac{4}{t}\)
(C) \(x(t)=-4\ln t\)
(D) \(x(t)=\dfrac{24}{t^{4}}\)
▶️ Answer/Explanation
Integrate \(a(t)=4t^{-2}\): \(v(t)=\int 4t^{-2}\,dt=-4t^{-1}+C_{1}\).
Integrate again: \(x(t)=\int\!\big(-4t^{-1}+C_{1}\big)dt=-4\ln t+C_{1}t+C_{2}\).
A valid form (ignoring the linear/constant terms) is \(-4\ln t\).
✅ Answer: (C)
Integrate again: \(x(t)=\int\!\big(-4t^{-1}+C_{1}\big)dt=-4\ln t+C_{1}t+C_{2}\).
A valid form (ignoring the linear/constant terms) is \(-4\ln t\).
✅ Answer: (C)
No-Calc Question
A particle moves along the \(y\)-axis so that its position is \(y(t)=\dfrac{1}{3}t^{3}-2t^{2}+3t+4\) for \(t\ge 0\). What is the position when it changes from moving down to moving up?
(A) \(0\)
(B) \(4\)
(C) \(\dfrac{14}{3}\)
(D) \(\dfrac{16}{3}\)
(B) \(4\)
(C) \(\dfrac{14}{3}\)
(D) \(\dfrac{16}{3}\)
▶️ Answer/Explanation
Moving down \(\Leftrightarrow y'(t)<0\); moving up \(\Leftrightarrow y'(t)>0\).
\(y'(t)=t^{2}-4t+3=(t-1)(t-3)\) changes from negative to positive at \(t=3\).
Position: \(y(3)=\dfrac{1}{3}(27)-2(9)+3(3)+4=9-18+9+4=4\).
✅ Answer: (B)
\(y'(t)=t^{2}-4t+3=(t-1)(t-3)\) changes from negative to positive at \(t=3\).
Position: \(y(3)=\dfrac{1}{3}(27)-2(9)+3(3)+4=9-18+9+4=4\).
✅ Answer: (B)