A particle is performing circular motion of radius 1 m. Its speed is `v=(2t^(2)) m//s`. What will be magnitude of its acceleration at `t=1s`.

A particle is moving in a circle of radius 1 m with speed varying with time as v=(2t)m//s . In first 2 s

A particle is moving in a circle of radius 1 m with speed varying with time as v=(2t)m//s . In first 2 s

A particle performs simple harmonic motion about O with amolitude A and time period T . The magnitude of its acceleration at t=(T)/(8) s after the particle reaches the extreme position would be

A particle performs simple harmonic motion about O with amolitude A and time period T . The magnitude of its acceleration at t=(T)/(8) s after the particle reaches the extreme position would be

A particle describes uniform circular motion in a circle of radius 2 m, with the angular speed of 2 rad s^(-1) . The magnitude of the change in its velocity in pi/2 s is

A particle is moving in a circle of radius 1 m with speed varying with time as v = (2t) m/s. In first 2 sec:

A particle is moving on a circular path of radius 1 m with 2 m/s. If speed starts increasing at a rate of 2m//s^(2) , then acceleration of particle is

A particle is moving on a circular path of radius 1 m with 2 m/s. If speed starts increasing at a rate of 2m//s^(2) , then acceleration of particle is

A body of mass 1 kg starts moving from rest t = 0 in a circular path of radius 8 m. Its kinetic energy varies with time as k = 2t^(2) J then magnitude of centripetal acceleration ( in m//s^(2) ) at t = 2s is.