A particle moves along a straight line OX. At a time t (in seconds), the distance .$x = 40 + 12t - t^3$ Then the displacement of the particle  before coming to rest is 

Distance travelled by the particle is $x = 40 + 12t - t^3$

We know that, speed is defined as the rate of change of distance, i.e. $v = \frac{dx}{dt}$

$\therefore$                     $v=\frac{d}{dt}\left(40+12t-t^3\right)=0+12-3t^2$

But final velocity, v = 0

$\therefore$              $\mathrm{𝟚}-3t^2=0\Rightarrow t^2=\frac{12}{3}=4$

$\Rightarrow$                       t = 2s

Hence, distance travelled by the particle before coming to rest is given by

               x = 40 + 12(2) - (2)³ = 40 + 24 - 8 = 64 - 8 = 56 m