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Standard XII

Physics

Subtraction of a Vector

A particle is...

Question

A particle is moving on a circular path of radius r with uniform velocity v. The change in velocity when the particle moves from P to Q is $(∠ P O Q = 40^{\circ})$

$2 v c o s 40^{\circ}$

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$2 v s i n 40^{\circ}$

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$2 v s i n 20^{\circ}$

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$2 v c o s 20^{\circ}$

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Solution

The correct option is C
$2 v s i n 20^{\circ}$

Change in velocity $= 2 v s i n (\frac{θ}{2}) = 2 v s i n 20^{\circ}$

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Q. 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 :-

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A particle is moving on a circular path of radius r with uniform velocity v. The change in velocity when the particle moves from P to Q is (∠POQ=40∘)

The correct option is C 2vsin20∘ Change in velocity =2v sin(θ2)=2v sin 20∘

A particle is moving on a circular path of radius r with uniform velocity v. The change in velocity when the particle moves from P to Q is $(∠ P O Q = 40^{\circ})$

The correct option is C
$2 v s i n 20^{\circ}$

Change in velocity $= 2 v s i n (\frac{θ}{2}) = 2 v s i n 20^{\circ}$