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?

Solution :
First $$X$$ by $$X$$ differentiating displacement equation we get velocity of the body, since body comes to rest so velocity becomes zero. Now by putting the value of time $$t$$ in displacement equation we get the distance travelled by the body when it comes to rest.
Distance travelled by the particle is $$x = 40 + 12t - {t^3}$$
We know that, velocity is the rate of change of distance i.e. $$v = \frac{{dx}}{{dt}}.$$
$$\eqalign{ & \therefore v = \frac{d}{{dt}}\left( {40 + 12t - {t^3}} \right) \cr & = 0 + 12 - 3{t^2} \cr} $$
but final velocity $$v = 0$$
$$\eqalign{ & \therefore 12 - 3{t^2} = 0 \cr & {\text{or}}\,{t^2} = \frac{{12}}{3} = 4 \cr & {\text{or}}\,t = 2\,s \cr} $$
Hence, distance travelled by the particle before coming to rest is given by
$$\eqalign{ & x = 40 + 12\left( 2 \right) - {\left( 2 \right)^3} \cr & = 40 + 24 - 8 \cr & = 64 - 8 \cr & = 56\,m \cr} $$