An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by $\frac{dv}{dt} = - 2.5\sqrt{v}$ where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be

Question

Dr. Abubacker · Accepted Answer

(a) $\frac{dv}{dt} = - 2.5\ \sqrt{v}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{dv}{\sqrt{v}} = - 2.5dt\ $
$\Rightarrow \ \int_{6.25}^{0}v^{- 1/2}\ dv = - 2.5\ \int_{0}^{t}{dt}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \  - 2.5\lbrack t\rbrack_{0}^{t} = \left\lbrack 2v^{1/2} \right\rbrack_{6.25}^{0}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = 2\ s$

Q. An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by $\frac{dv}{dt} = - 2.5\sqrt{v}$ where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be

Explanation

Reviewed by Dr. Abubacker

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