Question

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 instantaneous speed. The time taken by the object, to come to rest, would be:
(AIEEE 2011)

• A. 1 s
• B. 2 s
• C. 4 s
• D. 8 s

Answer 1

The correct option is B 2 s

${\int }_{6.25}^0\frac{dv}{\sqrt{v}}=-2.5{\int }_0^{t}dt\Rightarrow |2\sqrt{v}{|}_{6.25}^0=-2.5t$
$\Rightarrow 2\sqrt{6.25}=2.5t\Rightarrow t=2s$