If a particle is moving such that the velocity acquired is proportional to the square root of the distance covered, then its acceleration is
Answer & explanation
Correct answer: option 1
Let v be the velocity of the particle when the distance covered is s. Then,
$v \propto \sqrt{s}$ [Given]
$\Rightarrow v=\lambda \sqrt{s}$
$\Rightarrow \frac{d v}{d s}=\frac{\lambda}{2 \sqrt{s}}$
$\Rightarrow v \frac{d v}{d s}=\frac{\lambda v}{2 \sqrt{s}}=\frac{\lambda^2}{2}$ = constant.
Hence, the acceleration is constant.