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(B) In uniform circular motion,the speed $v$ remains constant,but the direction of velocity changes.Let the velocity at point $P$ be $\vec{v}_1$ and a

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$2v \sin 20^\circ$

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(B) In uniform circular motion,the speed $v$ remains constant,but the direction of velocity changes.Let the velocity at point $P$ be $\vec{v}_1$ and at point $Q$ be $\vec{v}_2$.The magnitude of both velocities is $|\vec{v}_1| = |\vec{v}_2| = v$.The angle between the velocity vectors $\vec{v}_1$ and $\vec{v}_2$ is equal to the angle subtended by the arc $PQ$ at the center,which is $	heta = 40^\circ$.The change in velocity is given by $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$.The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2v^2 \cos 	heta} = \sqrt{2v^2(1 - \cos 	heta)} = \sqrt{2v^2(2 \sin^2(	heta/2))} = 2v \sin(	heta/2)$.Substituting $	heta = 40^\circ$,we get $|\Delta \vec{v}| = 2v \sin(40^\circ/2) = 2v \sin 20^\circ$.

$A$ particle is moving on a circular path of radius $r$ with uniform speed $v$. The change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = 40^\circ)$.

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