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A particle moves with decreasing speed a...

A particle moves with decreasing speed along the circle of radius R so that at any moment of time its tangential and centripetal accelerations are equal in magnitude. At the initial moment , t =0 its speed is u.

The time after which the speed of particle reduces to half of its initial value is

A

`(2R)/u`

B

`R/u`

C

`R/(2u)`

D

`(3R)/(2u)`

Text Solution

Verified by Experts

The correct Answer is:
B
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Knowledge Check

  • 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)`

  • 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

    A
    `v=v_(0)e^(-(2S)/(R))`
    B
    `v=v_(0)e^(-(S)/(R))`
    C
    `v=v_(0)e^(-SR)`
    D
    `v=v_(0)e^(-2SR)`

  • A particle moves with constant speed v along a circular path of radius r and completes the circle in time T. The acceleration of the particle is

    A
    `2pi v//T`
    B
    `2pi r//T`
    C
    `2pi r^(2)//T`
    D
    `2pi v^(2)//T`