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A particle moves with deceleration along...

A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal acceleration are equal in moduli. At the initial moment `t=0` the speed of the particle equals `v_(0)`, then th speed of the particle as a function of the distance covered S will be

A

`v=v_(0) e^(-S//R)`

B

`v=v_(0)e^(S//R)`

C

`v=v_(0) e^(-R//S)`

D

`v=v_(0) e^(R//S)`

Text Solution

Verified by Experts

The correct Answer is:
A
Given `(dv)/(dt)=v^(2)/rrArr(dv)/(ds)=v^(2)/r, -underset(v_(0))overset(v)(int)1/vdv=underset(o)overset(s)(int)(ds)/r`
`rArr ln[v_(0)/v]=S/rrArr v_(0)/v=e^(S//r)`
`rArr v_(0)=ve^(S//r)rArrv=v_(0)e^(-S//r)`
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Knowledge Check

  • A body moves in a circular path of the radius R with deceleration so that at any moment of time its tangential and normal acceleration are equal in magnitude. At the initial moment t = 0 , the velocity of the body is v_(0) then the velocity of the body after it has moved S at any time will be

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