The motion of a particle along a straight line is described by equation
$$x = 8 + 12t - {t^3}$$
where, $$x$$ is in metre and $$t$$ in $$sec.$$  The retardation of the particle when its velocity becomes zero, is

Solution :
Double differentiation of displacement equation gives acceleration and single differentiation gives velocity of the body.
Given, $$x = 8 + 12t - {t^3}$$
We know $$v = \frac{{dx}}{{dt}}$$  and acceleration $$a = \frac{{dv}}{{dt}}$$
$$\eqalign{ & {\text{So,}}\,v = 12 - 3{t^2}\,{\text{and}}\,a = - 6t \cr & {\text{At}}\,t = 2s \cr & v = 0\,{\text{and}}\,a = - 6 \times 2 \cr & a = - 12\,m/{s^2} \cr} $$
So, retardation of the particle $$ = 12\,m/{s^2}.$$