A particle moves, along a straight line OX. At a time t (in second), the distance x (in metre) of the particle from O is given by $x=40+12t-t^3.$ How long would the particle travel before coming to rest?
 

$x=40+12t-t^3$
$\therefore$Velocity $v=\frac{dx}{dt}=12-3t^2$
When particle comes to rest, $\frac{dx}{dt}=v=0$
$\therefore 12-3t^2=0\Rightarrow 3t^2=12\Rightarrow t=2$seconds
Distance travelled by the particle before coming to rest is
$$\begin{gathered} {\int }_0^{x}ds={\int }_0^{2}vdt \\ s={\int }_0^2\left(12-3t^2\right)dt={\left[12t-\frac{3t^2}{3}\right]}_0^2 \\ =12\times 2-8=16m \end{gathered}$$
Hence, the correct answer is option (a).