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) $\tan^{-1}\left(\dfrac{2}{3}\right)$
$$\vec{v} = \frac{d\vec{r}}{dt} = 6t\,\hat{i} + 4\hat{j}, \qquad \vec{a} = 6\hat{i}$$
At $t = 1$ s, $\vec{v} = 6\hat{i} + 4\hat{j}$. The acceleration lies along $\hat{i}$, so the angle is the angle of $\vec{v}$ with the $x$-axis:
$$\tan\theta = \frac{4}{6} = \frac{2}{3}$$
Solution by Sreeraj P, M.Sc Physics