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A particle moves in a circular path such...

A particle moves in a circular path such that its speed v varies with distance s as `v=alphasqrt(s)` where `alpha` is a positive constant. If the acceleration of the particle after traversing a distance s is `[alpha^2sqrt(x+s^2/R^2)]` find x.

A

`alpha^(2) sqrt((1)/(4)-(s^(2))/(R^(2)))`

B

`alpha^(2) sqrt((1)/(4)+(s^(2))/(R^(2)))`

C

`alpha sqrt((1)/(2)+(s^(2))/(R^(2)))`

D

`alpha^(2) sqrt((1)/(2)+(s^(2))/(R^(2)))`

Text Solution

Verified by Experts

The correct Answer is:
B
`a= sqrt(a_(t)^(2)+a_(t)^(2))= sqrt(((dv)/(dt))^(2)+((v^(2))/(R ))^(2))`
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