Q 11-03-099JEE MainJEE Main 2023 (6 Apr, Shift 2)Medium
As shown in the figure, a particle is moving with constant speed $\pi\ \text{m s}^{-1}$. Considering its motion from $A$ to $B$, the magnitude of the average velocity is
Answer: (C) $1.5\sqrt3\ \text{m s}^{-1}$
Displacement (chord) $=2r\sin60^\circ=\sqrt3r$. Time $=\dfrac{\text{arc}}{v}=\dfrac{2\pi r/3}{\pi}=\dfrac{2r}{3}$.
$$v_{avg}=\frac{\sqrt3r}{2r/3}=1.5\sqrt3\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics