Question:
The distance moved by a particle travelling in straight line in $t$ seconds is given by $s=45 t+11 t^{2}-t^{3}$. The time taken by the particle to come to rest is
Solution:
(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,
$t=9 \mathrm{sec}$