Question

A particle moves along a straight line. Its position at any instant is given by $x=32t-\frac{8t^3}{3}$ where $x$ is in $\text{m}$ and $t$ in $\text{s}$. The acceleration of the particle at the instant when particle is at rest will be

• A. $-16{\text{m/s}}^2$
• B. $-32{\text{m/s}}^2$
• C. $16{\text{m/s}}^2$
• D. $32{\text{m/s}}^2$

Answer 1

The correct option is B $-32{\text{m/s}}^2$

Given, $x=32t-\frac{8t^3}{3}$
So, $v=\frac{dx}{dt}=32-8t^2$ ... (1)
$\Rightarrow 32-8t^2=0$ (particle is at rest)
$\Rightarrow t=2\text{s}$ ... (2)
Now, from (1)
$a=\frac{dv}{dt}=-16t$
$\Rightarrow a=-16\times 2$ [from (2)]
$\Rightarrow a=-32{\text{m/s}}^2$