tangential acceleration in nonuniform circular motion is const.if we apply eqn s=ut+1/2 at^2,s represents distance covered along circumference or displacement?
If the tangential acceleration in the circular motion is constant in module (its direction is changing because the trajectory is circular) then the equation
`s = v*t +a*t^2/2`
represents the distance covered along the circle.
To demonstrate this take an infinitesimal displacement along the same circle
If the circular motion has a tangential acceleration uniform in module, it means the angular acceleration `epsilon` is constant as a vector over time. For angles `theta` on the circle the following equation is valid
`theta =omega*t +epsilon*t^2/2`
By differentiating we get
`d(theta) = omega +epsilon*t*dt`
Multiplying with R we obtain
`ds =R*d(theta) = omega*R +epsion*R*t*dt =v +a*t*dt`
By integration from `s=0` to `s` (from angles `theta=0` to `theta` ) we obtain the total distance covered along the circle (not the displacement vector).
`s =R*theta = v*t +a*t^2/2`
`R*theta` (`theta` is measured in radians) represents the distance along the circle since `R*2pi` represents the circumference of the entire circle.
For some reason the ds diferrential above does not appear like it should:
`ds =R*d(theta) =omega*R +epsilon*R*t*dt =v + a*t*dt`