JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Kinematics

The speed $v$ is constant, so the instantaneous velocity has magnitude $v$. On turning through $90^{\circ }$ the particle covers a quarter of the circle, an arc length $\frac{\pi R}{2}$, so the time taken is $t=\frac{\pi R}{2v}$. Its displacement is the chord joining the start and end points, of length $R\sqrt{2}$, so the average velocity is $⟨v⟩=\frac{R\sqrt{2}}{\pi R/2v}=\frac{2\sqrt{2}v}{\pi }$. Hence $\frac{v}{⟨v⟩}=\frac{\pi }{2\sqrt{2}}$, and comparing with $\pi :x\sqrt{2}$ gives $x=2$.