Question

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

• A. Proportional to $s^2$
• B. proportional to $\frac{1}{s^2}$
• C. Proportional to $\frac{1}{s}$
• D. A constant

Answer 1

Answer: (D)

A constant

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