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(A) Let the velocity at point $A$ be $\vec{v}_1$ and at point $B$ be $\vec{v}_2$. Since the speed is constant,$|\vec{v}_1| = |\vec{v}_2| = v$.The chan

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$2 v \sin \frac{	heta}{2}$

Vedclass · Answer

(A) Let the velocity at point $A$ be $\vec{v}_1$ and at point $B$ be $\vec{v}_2$. Since the speed is constant,$|\vec{v}_1| = |\vec{v}_2| = v$.The change in velocity is given by $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$.The magnitude of the change in velocity for a particle moving in a circle with constant speed $v$ through an angle $	heta$ is given by the vector subtraction formula:$|\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2v^2 \cos 	heta}$$|\Delta \vec{v}| = \sqrt{2v^2(1 - \cos 	heta)}$Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$,we get:$|\Delta \vec{v}| = \sqrt{2v^2 \cdot 2 \sin^2 \frac{	heta}{2}}$$|\Delta \vec{v}| = \sqrt{4v^2 \sin^2 \frac{	heta}{2}}$$|\Delta \vec{v}| = 2v \sin \frac{	heta}{2}$

$A$ particle is moving on a circular path with a constant speed $v$. Its change of velocity as it moves from $A$ to $B$ in the figure is

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