AP Calculus BC 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration- MCQs - Exam Style Questions
Calc-Ok Question
The velocity of a particle moving along the \(x\)-axis is \(v(t)=\dfrac{1-t^{3}}{1+t}\) for \(t\ge0\). Which statement is true at \(t=1.5\)?
(A) Acceleration is negative, and the speed is decreasing
(B) Acceleration is negative, and the speed is increasing
(C) Acceleration is positive, and the speed is decreasing
(D) Acceleration is positive, and the speed is increasing
(B) Acceleration is negative, and the speed is increasing
(C) Acceleration is positive, and the speed is decreasing
(D) Acceleration is positive, and the speed is increasing
▶️ Answer/Explanation
Detailed solution
Velocity at \(t=1.5\): \(\displaystyle v(1.5)=\frac{1-1.5^{3}}{1+1.5}=\frac{-2.375}{2.5}\approx-0.95<0\).
Acceleration \(a(t)=v'(t)\). Using quotient rule with \(v(t)=\dfrac{1-t^{3}}{1+t}\):
\(\displaystyle v'(t)=\frac{(-3t^{2})(1+t)-(1-t^{3})}{(1+t)^{2}} =\frac{-3t^{2}-2t^{3}-1}{(1+t)^{2}}.\)
At \(t=1.5\), numerator \(=-3(1.5)^{2}-2(1.5)^{3}-1=-14.5<0\) so \(a(1.5)<0\).
Speed \(|v|\) increases when \(v\) and \(a\) have the same sign. Here \(v<0\) and \(a<0\) ⇒ speed is increasing.
✅ Answer: (B)