Q 11-02-014JEE MainAIEEE 2011Top questionMedium
An object, moving with a speed of $6.25$ m/s, is decelerated at a rate given by $\dfrac{dv}{dt} = -2.5\sqrt{v}$, where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be
Answer: (B) $2$ s
Separate the variables and integrate from $v = 6.25$ m/s to $v = 0$:
$$\frac{dv}{\sqrt{v}} = -2.5\,dt \;\Rightarrow\; \int_{6.25}^{0} v^{-1/2}\,dv = -2.5\int_0^t dt$$
$$\Big[2\sqrt{v}\Big]_{6.25}^{0} = -2.5\,t \;\Rightarrow\; 0 - 2(2.5) = -2.5\,t$$
$$t = 2\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics