Question

If a particle moves in such a way that the displacement is proportional to the square of the velocity acquired, then its acceleration is

• A. proportional to s²
• B. proportional to √s
• C. a constant
• D. proportional to $\frac{1}{s}$

Answer 1

The Correct Option is C

Explanation:
According to question, $s\propto v^2$$\Rightarrow v=k\sqrt{s}$Now, acceleration $=\frac{dv}{dt}=k\frac{1}{2\sqrt{s}}\cdot \frac{ds}{dt}$$=k\frac{1}{2\sqrt{s}}\cdot k\sqrt{s}(\because v=\frac{ds}{dt}=k\sqrt{s})$$=\frac{k^2}{2}=\mathrm{a}$ constant