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(D) The velocity of the particle at any point is tangent to the circular path.Let the velocity at point $A$ be $\vec{v}_A$ and at point $B$ be $\vec{v

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$v$

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(D) The velocity of the particle at any point is tangent to the circular path.Let the velocity at point $A$ be $\vec{v}_A$ and at point $B$ be $\vec{v}_B$.The magnitude of both velocities is $v$,so $|\vec{v}_A| = |\vec{v}_B| = v$.The angle between the radii $OA$ and $OB$ is $	heta = 60^{\circ}$.The angle between the velocity vectors $\vec{v}_A$ and $\vec{v}_B$ is also $60^{\circ}$ because the velocity is always perpendicular to the radius.The magnitude of the change in velocity is given by $|\Delta \vec{v}| = |\vec{v}_B - \vec{v}_A|$.Using the formula for the magnitude of the difference of two vectors of equal magnitude $v$ with an angle $	heta$ between them:$|\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2v^2 \cos 	heta} = \sqrt{2v^2(1 - \cos 	heta)} = \sqrt{2v^2(2 \sin^2(	heta/2))} = 2v \sin(	heta/2)$.Substituting $	heta = 60^{\circ}$:$|\Delta \vec{v}| = 2v \sin(60^{\circ}/2) = 2v \sin(30^{\circ}) = 2v 	imes (1/2) = v$.Therefore,the magnitude of the change in velocity is $v$.

$A$ particle is moving in a circle of radius $r$ with center at $O$,with a constant speed $v$. The magnitude of the change in velocity when moving from point $A$ to point $B$ is

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