Q 11-03-024NEETJEE MainMedium
The position of a particle is $\vec{r} = (3t^2\,\hat{i} + 4t\,\hat{j})$ m, with $t$ in seconds. The angle between its velocity and its acceleration at $t = 1$ s is
Answer: (A) $\tan^{-1}\left(\dfrac{2}{3}\right)$
Concept. Differentiate the position to get $\vec v$ and $\vec a$.
Step 1. Vectors. $\vec v = 6t\hat i + 4\hat j$, $\vec a = 6\hat i$; at $t = 1$ s, $\vec v = 6\hat i + 4\hat j$.
Step 2. Angle. $\vec a$ lies along $\hat i$, so $\tan\theta = \dfrac46 = \dfrac23$.
Answer. $\tan^{-1}\left(\dfrac23\right)$ (Option A).
Solution by Sreeraj P, M.Sc Physics