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(B) The change in velocity $\Delta \vec{v}$ is given by the vector difference $\vec{v}_2 - \vec{v}_1$.Since the speed is constant,the magnitude of vel

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$\sqrt{2}v$

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(B) The change in velocity $\Delta \vec{v}$ is given by the vector difference $\vec{v}_2 - \vec{v}_1$.Since the speed is constant,the magnitude of velocity is $|\vec{v}_1| = |\vec{v}_2| = v$.The angle between the velocity vectors is equal to the angle subtended at the center,which is $	heta = 90^{\circ}$.The magnitude of the change in velocity is given by the formula $|\Delta \vec{v}| = 2v \sin(	heta/2)$.Substituting $	heta = 90^{\circ}$:$|\Delta \vec{v}| = 2v \sin(90^{\circ}/2) = 2v \sin(45^{\circ})$.Since $\sin(45^{\circ}) = 1/\sqrt{2}$,we get:$|\Delta \vec{v}| = 2v 	imes (1/\sqrt{2}) = \sqrt{2}v$.

$A$ particle moves from point $P_1$ to $P_2$ along a circular path with constant speed $v$. If the angle subtended by the arc $P_1P_2$ at the center is $90^{\circ}$,what is the change in velocity?

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