Physics - Circular Motion Question with Solution | TestHub

Between A and B (a quarter circle), the displacement is $2R\sin 45^{\circ }=\sqrt{2}R$, so average velocity $=\frac{\sqrt{2}R}{t}$; the velocity vector turns through 90°, so $|\Delta \vec{v}|=\sqrt{v^2+v^2}$
$=\sqrt{2}v$ and average acceleration $=\frac{\sqrt{2}v}{t}$. The ratio is $\frac{v}{R}=\omega$ (Assertion true). The Reason is true (constant speed means constant $\omega$ ) but does not explain why the ratio equals $\omega$.