A particle starts from rest and moves on a curve with constant angular acceleration of `3.0 rad//s^(2)`. An observer starts his stopwatch at a certain instant and recond that the particle covers an angular span of `120` rad at the end of `4^(th)` second. How long the particle has moved when the observer started his stopwatch?

Explanation

Let the instants when the particles starts moving and the observer starts his stopwatch, are `t_(0)=0` to `t=t_(1)`. Denoting angular positions and angular velocity at the instant `t=t_(1)` by `theta_(1)` and `omega_(1)` and the angular position at the instant `t_(2)=t_(1)+4s` by `theta_(2)`, we can express the angular span covered during the interval from eq.
`theta-theta_(o)=omega_(o)t+1/2alphat^(2) rarr theta_(2)-theta_(1)=omega_(1)(t_(2)-t_(1))+1/2alpha(t_(2)-t_(1))^(2)`
Substituting values `theta_(1), theta_(2), t_(1)` and `t_(2)`, we have `omega_(1)=24 rad//s`
From eq. `omega=omega_(0)+alphat`, we have `omega_(1)=omega_(0)+alphat_(1)`
Now substituting `omega_(0)=0, omega_(1)=24` and `alpha=3 rad//s^(2)`, we have `t_(1)=8.0 s`