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

Options

A. 8 rad/sec
B. 12 rad/sec
C. 24 rad/sec
D. 36 rad/sec

Answer

C

Explanation

`omega= (d theta)/(dt)= (d)/(dt) (2t^(3) + 0.5) = 6t^(2)`
at t= 2s, `omega = 6 xx (2)^(2) = 24` rad/s