Q 11-03-066JEE MainJEE Main 2024 (27 Jan, Shift 1)Easy
Position of an ant ($S$ in metres) moving in the Y–Z plane is given by $\vec S = 2t^2\,\hat j + 5\,\hat k$ (where $t$ is in second). The magnitude and direction of velocity of the ant at $t = 1\ \text{s}$ will be:
Answer: (D) $4\ \text{m s}^{-1}$ in $y$-direction
$$\vec v = \frac{d\vec S}{dt} = 4t\,\hat j$$
At $t = 1\ \text{s}$, $\vec v = 4\,\hat j\ \text{m s}^{-1}$: magnitude $4\ \text{m s}^{-1}$ along the $y$-direction.
Solution by Sreeraj P, M.Sc Physics