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
Concept. The deceleration depends on the speed, $\dfrac{dv}{dt}=-2.5\sqrt v$, so it is not constant. Separate the variables (all $v$ on one side, $t$ on the other) and integrate from the initial speed to zero.
Step 1. Separate.
$$\frac{dv}{\sqrt v}=-2.5\,dt$$
Step 2. Integrate from $v=6.25$ m/s at $t=0$ to $v=0$ at time $t$.
$$\int_{6.25}^{0}v^{-1/2}\,dv=-2.5\int_0^t dt\ \Rightarrow\ \Big[2\sqrt v\Big]_{6.25}^{0}=-2.5\,t$$
Step 3. Evaluate. $\sqrt{6.25}=2.5$:
$$0-2(2.5)=-2.5\,t\ \Rightarrow\ t=\frac{5}{2.5}=2\ \text{s}$$
Answer. $2$ s (Option B). In general, $\dfrac{dv}{dt}=-k\sqrt v$ stops the body in $t=\dfrac{2\sqrt{u}}{k}$.
Solution by Sreeraj P, M.Sc Physics
