Q 11-02-054JEE MainJEE Main 2024 (1 Feb, Shift 1)Easy
A particle is moving in one dimension (along the $x$ axis) under the action of a variable force. Its initial position was $16\ \text{m}$ right of the origin. The variation of its position $x$ with time $t$ is given as $x = -3t^3 + 18t^2 + 16t$, where $x$ is in m and $t$ is in s. The velocity of the particle when its acceleration becomes zero is ______ $\text{m s}^{-1}$.
Numerical value type. Enter your answer.
Answer: 52
$$v = \frac{dx}{dt} = -9t^2 + 36t + 16,\qquad a = \frac{dv}{dt} = -18t + 36$$
$a = 0$ at $t = 2\ \text{s}$:
$$v = -36 + 72 + 16 = 52\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics