Q70 / 90 · JEE Main 2011 · Physics
An object, moving with a speed of $6.25 \mathrm{~m} / \mathrm{s}$, is decelerated at a rate given by :
$$
\frac{\mathrm{dv}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}
$$
where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be:
- Correct answerA$2 \mathrm{~s}$
- B$4 \mathrm{~s}$
- C$8 \mathrm{~s}$
- D$1 \mathrm{~s}$
Solution
$$
\frac{d v}{d t}=-2.5 \sqrt{v}
$$
Integrating the above equation.
$$
\Rightarrow 2 \sqrt{v}=-2.5 \mathrm{t}+\mathrm{C}
$$
at $\mathrm{t}=0, \mathrm{v}=6.25 \Rightarrow \mathrm{C}=5$
at $v=0 \Rightarrow t=\frac{5}{2.5}=2 \mathrm{~s}$
Hence, the correct answer is option: 'A'.