Q 11-02-081JEE MainJEE Main 2023 (15 Apr, Shift 1)Easy
The position of a particle related to time is given by $x=(5t^2-4t+5)$ m. The magnitude of velocity of the particle at $t=2$ s will be
Answer: (D) $16\ \text{m s}^{-1}$
$v=10t-4=16\ \text{m s}^{-1}$ at $t=2$ s.
Solution by Sreeraj P, M.Sc Physics