Question

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

• A. 4 s
• B. 8 s
• C. 1 s
• D. 2 s

Answer 1

The correct option is D 2 s

$\frac{dv}{dt}=-2\cdot 5\sqrt{v}\Rightarrow \frac{dv}{\sqrt{v}}=-2\cdot 5dt$

$\text{Initial speed}6.25m/s\text{is decelerated to}$

$\text{rest position. So on integrating}\to$

${\int }_{6.25}^0{v}^{-1/2}dv=-2\cdot 5{\int }_0^{t}dt$

$\Rightarrow 2{\left[v^{-1/2}\right]}_{6.25}^0=-2\cdot 5t$

$\Rightarrow t=\frac{-2(6.25{)}^{1/2}}{-2.5}=2s$

$\text{Option}\left(D\right)\text{is the correct answer.}$