at $t = 0$ a particle starts at rest and moves along a line in such a way that at time $t$ its acceleration…

Explanation:

Step1: Find the velocity function

Since acceleration (a(t) = 24t^{2}), and (v(t)=\int a(t)dt). [ \begin{align*} v(t)&=\int 24t^{2}dt\ &=24\times\frac{t^{3}}{3}+C\ & = 8t^{3}+C \end{align*} ] Given that the particle starts at rest, so (v(0) = 0). Substituting (t = 0) and (v(0)=0) into (v(t)=8t^{3}+C), we get (C = 0). So (v(t)=8t^{3}).

Step2: Find the position function

Since (s(t)=\int v(t)dt), and (v(t)=8t^{3}). [ \begin{align*} s(t)&=\int 8t^{3}dt\ &=8\times\frac{t^{4}}{4}+K\ &=2t^{4}+K \end{align*} ] Assume (s(0) = 0) (initial position), then (K = 0). So (s(t)=2t^{4}).

Step3: Calculate the displacement from (t = 0) to (t = 2)

We use (s(2)-s(0)). [ \begin{align*} s(2)-s(0)&=2\times2^{4}-2\times0^{4}\ &=2\times16-0\ &=32 \end{align*} ]

Answer:

A. 32