Q 11-02-097JEE MainJEE Main 2022 (25 Jul, Shift 2)Easy
A particle is moving in a straight line such that its velocity is increasing at $5\ \text{m s}^{-1}$ per metre. The acceleration of the particle is ______ $\text{m s}^{-2}$ at a point where its velocity is $20\ \text{m s}^{-1}$.
Numerical value type. Enter your answer.
Answer: 100
$\dfrac{dv}{dx} = 5\ \text{s}^{-1}$, so $a = v\dfrac{dv}{dx} = 20\times5 = 100\ \text{m s}^{-2}$.
Solution by Sreeraj P, M.Sc Physics