The displacement of a particle travelling in straight line in $t$ seconds is given by $S=45t+11t^2-t^3$, Find the time where the particle comes to rest.

Given $S=45t+11t^2-t^3$

Velocity $\left(V\right)=\frac{ds}{dt}$

$=45\left(1\right)+11\left(2t\right)-3t^2\Rightarrow V=45+22t-3t^2$

Partical comes to rest

$V=0;45+22t-3t^2=0$

$3t^2-22t-45=0\Rightarrow 3t^2-27t+St-45=0$

$3t(t-9)+5(t-9)=0\Rightarrow (3t+5)(t-9)=0$

$t=\frac{-5}{3},9$

$\therefore t=9\sec$