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A cyclist is moving on a circular path w...

A cyclist is moving on a circular path with constant speed V . What is the change in its velocity after it has desscribed an angle of `60^(@)`

A

Zero

B

`2v sin theta `

C

`2V sin 2 theta `

D

`2V sin ""(theta )/(2)`

Text Solution

Verified by Experts

The correct Answer is:
D

Let `V_(1) and V_(2) ` be the particle at A and B respectively .
AS this is a U.C.M `|V_(1) |=|V_(2)|=V`
` therefore Delta V=|V_(2) - v_(1) |=sqrt(v^(2)+v^(2)-2v . V cos theta )`
` = sqrt(2v ^(2) (1-cos theta))`
`= sqrt(2v^(2 ) . 2 sin ^(2) ((theta )/(2)))`
` sqrt(4v^(2) . sin ^(2)((theta)/(2)))`
` therefore Delta =2V sin ""(theta)/(2)`
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