Question

The motion of a particle along a stright line is described by the equation $x=8+12t-t^3$ where, $x$ is in meter and $t$ in second. The retardation of the particle when its velocity beocmes zero, is

• A. $12{\text{ms}}^{-2}$
• B. $\text{Zero}$
• C. $6{\text{ms}}^{-2}$
• D. $24{\text{ms}}^{-2}$

Answer 1

The correct option is A $12{\text{ms}}^{-2}$

Given:
$x\left(t\right)=8+12t-t^3$
$\Rightarrow v\left(t\right)=\frac{dx}{dt}=12-3t^2$
When velocity of particle becomes zero, then time is given by
$12-3t^2=0\Rightarrow t=2\text{s}$
Now,
$a\left(t\right)=\frac{dv}{dt}=-6t$
$\Rightarrow a_{|t=2\text{s}}=-6\times 2=-12{\text{m/s}}^2$
$\therefore \text{Retardation is}12{\text{m/s}}^2$