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A particle moves along a circle of ...

A particle moves along a circle of radius `((20)/(pi ))` metre with constant tangential acceleration .If the velocity of the particle is 40 m/s at the end of second revolution , after the revolution has began , then the tangential acceleration .If the velocity of the particle is 40 m/s at the end of second revolution , after the revolution has began , then the tangential acceleration is

A

`5m//s^(2)`

B

`10 m//s^(2)`

C

`15 m//s^(2)`

D

`20 m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`v^(2) = u ^(2) + 2as = 2 ar theta ( :' s = r theta ) `
` therefore a= (v^(2))/(2 r theta )`
` therefore a =(40xx40 )/(2xx (20)/(pi) xx4pi)=10 m//s^(2)`
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