Q 11-02-038JEE MainMedium
The velocity of a particle moving along a straight line varies with its displacement as $v = 6\sqrt{x}$ (SI units). The acceleration of the particle is
Answer: (C) $18\ \text{m/s}^2$, constant
$$a = v\frac{dv}{dx} = 6\sqrt{x}\cdot\frac{6}{2\sqrt{x}} = 18\ \text{m/s}^2$$
The $\sqrt{x}$ cancels, so the acceleration is constant. This matches $v^2 = 36x = 2(18)x$, which is $v^2 = 2ax$ for uniform acceleration from rest.
Solution by Sreeraj P, M.Sc Physics