The distance time formula for the motion of a particle along a straight line $S=t^3–9t^2+24t–18$
then find when and where the velocity is zero.

Given, $S=t^3–9t^2+24t–18$

Velocity $\left(V\right)=\frac{ds}{dt}=3t^2-18t+24$

Velocity $\left(V\right)=0$

$\Rightarrow 3t^2-18t+24=0$

$t^2-6t+8=0\Rightarrow t^2-4t-2t+8=0$

$t(t-4)-2(t-4)=0\Rightarrow (t-2)(t-4)=0$

$\text{ At }t=2\text{ or }4;\text{ If }t=2$

$S=(2{)}^3-9(2{)}^2+24\left(2\right)-18$

$=8-36+48-18=56-54=2$

if $t=4;S=(4{)}^3-9(4{)}^2+24\left(4\right)-18$

$=64-144+96-18$

$=160-162=-2$ (not possible)

$\therefore t=2\sec \text{ and }S=2$ units the velocity becomes zero.