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 metre and $t$ in second. Find the acceleration of the particle at the instant when particle is at rest.

• A. $-16m/s^2$
• B. $-32m/s^2$
• C. $32m/s^2$
• D. $16m/s^2$

Answer 1

The correct option is B $-32m/s^2$

$v=\frac{dx}{dt}=32-8t^2$
Particle at rest means $v=0$, which occurs at time, $t=\sqrt{\frac{32}{8}}=2sec$

Thus, we are asked to find acceleration at $t=2$ second.

Now, $a=\frac{dv}{dt}=0-16t=-16t$

At, $t=2$, $a=-32m/s^2$