Q 11-02-066JEE MainJEE Main 2024 (29 Jan, Shift 2)Easy
A particle is moving in a straight line. The variation of position $x$ as a function of time $t$ is given as $x = (t^3 - 6t^2 + 20t + 15)\ \text{m}$. The velocity of the body when its acceleration becomes zero is:
Answer: (B) $8\ \text{m s}^{-1}$
$$v = 3t^2 - 12t + 20,\qquad a = 6t - 12$$
$a = 0$ at $t = 2\ \text{s}$:
$$v = 12 - 24 + 20 = 8\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics