Question

The displacement $x$ of a particle moving in one dimension, under the action of a constant force is related to the time $t$ by the equation $t=\sqrt{x}+3$ where $x$ is in metre and $t$ in second. Find
(a) the displacement of the particle when its velocity is zero, and
(b) the work done by the force in the first $6 s$.

Answer 1

Given $t=\sqrt{x}+3 $ or $ \sqrt{x}=(t-3)\,\,\,... (i)$
$\therefore x=(t-3)^{2}=t^{2}-6 t+9\,\,\, ...(ii)$
Differentiating this equation with respect to time, we get
velocity $v=\frac{d x}{d t}=2 t-6\,\,\, ....(iii)$
(a) $v=0$ when $2 t-6=0$ or $t=3 s$
Substituting in Eq. (i), we get
$\sqrt{x}=0 \text { or } x=0$
i.e. displacement of particle when velocity is zero is also zero.
(b) From Eq. (iii), speed of particle at $t=0$ is
$v_{i}=|v|=6\, m / s$
and at $t=6\, s$
$v_{f}=|v|=6\, m / s$
From work energy theorem,
Work done $=$ change in kinetic energy
$=\frac{1}{2} m\left[v_{f}^{2}-v_{i}^{2}\right]=\frac{1}{2} m\left[(6)^{2}-(6)^{2}\right]=0$