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 $v(t)=\int a(t)dt$, then $v(t)=\int 24t^{2}dt = 8t^{3}+C$. Given that the particle starts at rest ($v(0) = 0$), substituting $t = 0$ and $v(0)=0$ into $v(t)$ gives $C = 0$. So $v(t)=8t^{3}$.

Step2: Find the position - function

Since position $s(t)=\int v(t)dt$, then $s(t)=\int 8t^{3}dt=2t^{4}+D$. When $t = 0$, assume the initial position $s(0)=0$, so $D = 0$. Thus $s(t)=2t^{4}$.

Step3: Calculate the displacement in the first 2 seconds

We want to find the displacement from $t = 0$ to $t = 2$. Using the fundamental theorem of calculus, $s(2)-s(0)$. Substitute $t = 2$ into $s(t)$: $s(2)=2\times2^{4}=2\times16 = 32$ and $s(0)=0$. So the particle moves 32 feet in the first 2 seconds.

Answer:

A. 32