Question

A particle moving with an initial velocity $um{s}^{-1}$ is retarded by a force at the rate of $a=-k\sqrt{v}$ . where $K$ is a positive constant and $v$ is the instantaneous velocity. The particle comes to rest in a time given by

• A. $\frac{2\sqrt{u}}{k}$
• B. $k\sqrt{u}$
• C. $\frac{\sqrt{u}}{k}$
• D. $\frac{\sqrt{u}}{2k}$

Answer 1

Answer: (A)

Given, $a=-k\sqrt{v}$
or $a=\frac{dv}{dt}=-k\sqrt{v}[\because \frac{dx}{dt}=v]$
or $dv=-k\sqrt{v}dt$
or $\frac{dv}{\sqrt{v}}=-kdt$
Integration will yield
$\underset{u}{\overset{v}{\int }}\frac{dv}{\sqrt{v}}=\underset{0}{\overset{t}{\int }}-kdt$
${\left[2\sqrt{v}\right]}_u^v=-k{\left[t\right]}_0^t$
$2\left(\sqrt{v}-\sqrt{u}\right)=-kt$
$\therefore$ Particle comes to rest, so final velocity $v=0$
and $-2\sqrt{u}=-kt$
$\Rightarrow t=\frac{2\sqrt{u}}{k}$