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
Concept. $a = v\dfrac{dv}{dx}$.
Step 1. Differentiate. $\dfrac{dv}{dx} = \dfrac{6}{2\sqrt x} = \dfrac{3}{\sqrt x}$.
Step 2. Acceleration.
$$a = 6\sqrt x \cdot \frac{3}{\sqrt x} = 18\ \text{m/s}^2$$
The $\sqrt x$ cancels, so $a$ is constant (consistent with $v^2 = 36x = 2(18)x$).
Answer. $18$ m/s$^2$, constant (Option C).
Solution by Sreeraj P, M.Sc Physics