Q 11-02-114JEE MainJEE Main 2021 (27 Aug, Shift 1)Easy
If the velocity of a body related to displacement $x$ is given by $v = \sqrt{5000 + 24x}$ m s$^{-1}$, then the acceleration of the body is ______ m s$^{-2}$.
Numerical value type. Enter your answer.
Answer: 12
$v^2 = 5000 + 24x$. Differentiating: $2v\dfrac{dv}{dt} = 24\dfrac{dx}{dt} = 24v$, so
$$a = \frac{dv}{dt} = 12\ \text{m s}^{-2}$$
Solution by Sreeraj P, M.Sc Physics