Question

An object,moving with a speed of $6.25m{s}^{-1}$, is decelerated at a rate given by $dv/dt=-2.5(v{)}^{0.5}$ Where $v$ is the instantaneous speed.Find the time taken by the object , to come to rest

Answer 1

Given that,

$\frac{dv}{dt}=-2.5\sqrt{v}$

Now

$\frac{dv}{dt}=-2.5\sqrt{v}$

$\frac{dv}{\sqrt{v}}=-2.5dt.....\left(I\right)$

On integrating of equation (I)

$\underset{6.25}{\overset{0}{\int }}{\left(v\right)}^{-\frac{1}{2}}dv=-2.5\underset{0}{\overset{t}{\int }}dt$

${\left[2v^{\frac{1}{2}}\right]}_{6.25}^0=-2.5{\left[t\right]}_0^t$

$-2\times \sqrt{6.25}=-2.5t$

$t=\frac{2\times -2.5}{-2.5}$

$t=2\sec$

Hence, the time taken by the object, to come to rest in $2$ sec