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

(b) $\int_{6.25}^{0}\frac{dv}{\sqrt{v}} = - 2.5\int_{0}^{t}{dt}$
$\left| 2\sqrt{v} \right|_{6.25}^{0} = - 2.5t$
$2\sqrt{6.25} = 2.5t$
$t = 2sec$

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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