A particle is moving on a circular path of ra | vedclass

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Question

$\vec{v}_{i}=v \cos 	heta \hat{i}+v \sin 	heta \hat{j}$

$\vec{v}_{f}=v \hat{i}$

$ 	herefore  \Delta \vec{v}=\vec{v}_{f}-\vec{v}_{i}$

Vedclass · Accepted Answer

$2v\, sin\, 20^o$

Vedclass · Answer

$\vec{v}_{i}=v \cos 	heta \hat{i}+v \sin 	heta \hat{j}$

$\vec{v}_{f}=v \hat{i}$

$ 	herefore  \Delta \vec{v}=\vec{v}_{f}-\vec{v}_{i}$

$=v(1-\cos 	heta) \hat{i}+v \sin 	heta \hat{j}$

or $\quad \Delta \mathrm{v}=\mathrm{v} \sqrt{(1-\cos 	heta)^{2}+(\sin 	heta)^{2}}$

${=v \sqrt{1-2 \cos 	heta+\cos ^{2} 	heta+\sin ^{2} 	heta}}$

${=v \sqrt{2(1-\cos 	heta)}}$

${=v \sqrt{2\left[1-\left(1-2 \sin ^{2} \frac{	heta}{2}ight)ight]}}$

${v=\sqrt{2 	imes 2 \sin ^{2} \frac{	heta}{2}}}$

${=2 v \sin \frac{	heta}{2}}$

$\Rightarrow \Delta \mathrm{v}=2 \mathrm{v} \sin 20^{\circ} \quad\left(\because \quad 	heta=40^{\circ}ight)$

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

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