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)

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}$