Question

A particle moving with an initial velocity $u\, ms^{-1}$ 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)

$\frac{2\sqrt u}{k}$

Answer 2

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
$\int\limits_{u}^{v} \frac{dv}{\sqrt{v}} = \int\limits_{0}^{t} -kdt$
$\left[2\sqrt{v}\right]_{u}^{v} = -k\left[t\right]_{0}^{t}$
$2\left(\sqrt{v} -\sqrt{u}\right) = -kt$
$\therefore$ Particle comes to rest, so final velocity $v =0$
and $-2\sqrt{u} = -kt$
$\Rightarrow t = \frac{2\sqrt{u}}{k}$