Q 11-03-055JEE MainJEE Main 2025 (24 Jan, Shift 2)Easy
The position vector of a moving body at any instant of time is given as $\vec r = (5t^2\hat i - 5t\hat j)\ \text{m}$. The magnitude and direction of velocity at $t = 2\ \text{s}$ is
Answer: (D) $5\sqrt{17}\ \text{m/s}$, making an angle of $\tan^{-1}4$ with $-$ve $Y$ axis
$$\vec v = \frac{d\vec r}{dt} = 10t\,\hat i - 5\hat j \Rightarrow \vec v(2) = 20\hat i - 5\hat j$$
$$|\vec v| = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17}\ \text{m/s}$$
The velocity points right and down. Its angle $\alpha$ with the negative $Y$ axis satisfies $\tan\alpha = \dfrac{20}{5} = 4$, so $\alpha = \tan^{-1}4$ with the $-$ve $Y$ axis.
Solution by Sreeraj P, M.Sc Physics