If displacement of a particle is given with the equation $s=\frac{1}{2}{t}^3-6t$ meters, then find the acceleration at time when the velocity tends to zero.

Given displacement equation of the particle is  $s=\frac{1}{2}{t}^3-6t$, as a function of time.

The formula of velocity $v=\frac{ds}{dt}m/s.$

Velocity $v=\frac{ds}{dt}=\left(\frac{3t^2}{2}-6\right)$m/sec

We need to find the acceleration of the particle when velocity tends to zero i.e.$v=0.$

The formula of acceleration of particle is $a=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}m/s^2$

Acceleration, $a=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}=\left(\frac{3}{2}2t\right)=\left(3t\right) m/s^2$

To find acceleration, when velocity $\to 0$

 $\Rightarrow \frac{3t^2}{2}-6=0\Rightarrow 3t^2=12$

$\Rightarrow t^2=4\Rightarrow t=\pm 2sec$.

As time cant be negative, the value of time is $t=2sec.$

Thus, acceleration of the particle when velocity tends to zero i.e. acceleration when $t=2 sec$ is

$\Rightarrow a=3\times 2=6m/s^2$

Therefore, acceleration of the particle when velocity tends to zero is $6$ $m/s^2.$