The angular displacement of a particle performing circular motion is
`theta=(t^(3))/(60)-(t)/(4)`
where ` theta` is in radian and 't' is in second .Then the angular velocity and angular acceleraion of a particle at the end of 5 s will be

The angular displacement of a particle performing circular motion is theta=(t^(4))/(60)-(t)/(4) where theta is radian and 't' is in seconds. Then the acceleration of a particle at the end of 10 s will be

A particle is at rest, It starts rotating about a fixed point. Its angle of rotation (theta) with time (t) is given by the relation : theta = (6t^3)/(15) - (t^2)/(2) where theta is in radian and t is seconds. Find the angular velocity and angular acceleration of a particle at the end of 6 second.

A particle starts rotating from rest according to the formuls, theta=((3t^3)/20 ) - ((t^2)/3)) where theta is in radian and t is second. Find the angular velocity to and angular acceleration a at the end of 5 seconds.

The displacement of a particle moving in a circular path is given by theta = 3 t^(2) + 0.8 , where theta is in radian and t is in seconds . The angular velocity of the particle at t=3 sec . Is

IF the equation for the displancement of a particle moving on a circular path is given by theta=2t^3+0.5 , where theta is in radius and t is in seconds, then the angular velocity of the particle at t=2 s is

The infinitesimal angular displacement of a particle performing circular motion is a vetor because its obeys

If the equation for the displacement of a particle moving in a circular path is given by (theta)=2t^(3)+0.5 , where theta is in radians and t in seconds, then the angular velocity of particle after 2 s from its start is

The infinitesimally angular displacement delta theta in uniform circular motion is

A particle starts rotating from rest. Its angular displacement is expressed by the following equation theta=0.025t^(2)-0.1t where theta is in radian and t is in seconds. The angular acceleration of the particle is