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 $6s$.

Answer 1

Given $t=\sqrt{x}+3$ or $\sqrt{x}=(t-3)...(i)$
$\therefore x=(t-3{)}^2=t^2-6t+9...(ii)$
Differentiating this equation with respect to time, we get
velocity $v=\frac{dx}{dt}=2t-6....(iii)$
(a) $v=0$ when $2t-6=0$ or $t=3s$
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
$v_i=|v|=6m/s$
and at $t=6s$
$v_f=|v|=6m/s$
From work energy theorem,
Work done $=$ change in kinetic energy
$=\frac{1}{2}m\left[v_f^2-v_i^2\right]=\frac{1}{2}m\left[(6{)}^2-(6{)}^2\right]=0$