A particle is moving in a circle of radius $r$ with constant speed $v$ . The change in velocity in moving from $P$ to $Q$ is

2 v cos 20^{\circ}

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2 v sin 20^{\circ}

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2 v cos 40^{\circ}

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2 v sin 40^{\circ}

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Solution

The correct option is B $2 v sin 20^{\circ}$
Given : $|_{1} | = |_{2} | = v$
The angle between them is $40^{o}$
$| △ \to v |^{2} = v^{2} + v^{2} - 2 v^{2} cos 40^{\circ}$
$= 2 v^{2} [1 - (1 - 2 {sin}^{2} 20^{\circ}]$
$= 4 v^{2} {sin}^{2} 20^{\circ}$
or $| △ \to v | = 2 v sin 20^{\circ}$ .

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A particle is moving in a circle of radius $r$ with constant speed $v$ . The change in velocity in moving from $P$ to $Q$ is

The correct option is B $2 v sin 20^{\circ}$
Given : $|_{1} | = |_{2} | = v$
The angle between them is $40^{o}$
$| △ \to v |^{2} = v^{2} + v^{2} - 2 v^{2} cos 40^{\circ}$
$= 2 v^{2} [1 - (1 - 2 {sin}^{2} 20^{\circ}]$
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