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

If the equation for the displacement of a particle moving on a circle path is given by, theta = 2t^(2)+0.5 where, theta is in radian and t is in second, then the angular velocity of the particle after 2 s 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

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

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 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

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 seconds. Then the angular velocity and angular accleration of the particle at the end of 5 s will be

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^3)/60 - t/4 where theta is in radian and 't' is in seconds. Then the acceleration of the particle at the end 10 s will be