Q 11-02-056JEE MainJEE Main 2024 (1 Feb, Shift 2)Easy
A particle initially at rest starts moving from the reference point $x = 0$ along the $x$-axis, with velocity $v$ that varies as $v = 4\sqrt{x}\ \text{m s}^{-1}$. The acceleration of the particle is ______ $\text{m s}^{-2}$.
Numerical value type. Enter your answer.
Answer: 8
$$a = v\frac{dv}{dx} = 4\sqrt x\times\frac{4}{2\sqrt x} = 8\ \text{m s}^{-2}$$
Solution by Sreeraj P, M.Sc Physics