A particle moves along a cirvle of radius 'R' with speed varies as `v = a_(0)t`, where `a_(0)` is a positive constant. Then the angle between the velocity vector and the acceleration vertor of the particle when it has covered one fourth of the circle is

Options

A. `tan^(-1) ((pi)/(4))`
B. `tan^(-1) ((pi)/(2))`
C. `tan^(-1) (pi)`
D. `tan^(-1)(2pi)`

Answer

C

Explanation

Given `v = a_(0)t`
`a_(t) = (dv)/(dt) = a_(0)`
Now, `v^(2) = u^(2) + 2a_(t)S`
`= 0 + 2a_(0) (piR)/(2)` ,
`v^(2) = pia_(0)R`
So, `a_(n) = (v^(2))/(R ) = pia_(0)`
The angle between the velocity vector and the acceleration vector is
`phi = tan^(-1) ((a_(n))/(a_(t))) = tan^(-1)((pia_(0))/(a_(0))) = tan^(-1)(pi)`