The distance moved by a particle travelling in straight line

(a) $9 \mathrm{sec}$

$s=45 t+11 t^{2}-t^{3}$

$\Rightarrow \frac{d s}{d t}=45+22 t-3 t^{2}$

According to the question,

$3 t^{2}-22 t-45=0$

$\Rightarrow 3 t^{2}-27 t+5 t-45=0$

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

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

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

As time can't be negative,

$t=9 \mathrm{sec}$