A particle starts moving at t = 0 in a circle of radius `R = 2m` with constant angular acceleration of `a - 3 "rad/sec"^(2)`. Initial angular speed of the particle is 1 rad/sec. At time `t_(0)` the angle between the acceleration vector and the velocity vector of the particle is `37^(@)`.
What is the magnitude of the acceleration of the particle at `t_(0)` ?

Options

A. `6 m//s^(2)`
B. `4.5 m//s^(2)`
C. `7.5 m//s^(2)`
D. `5 m//s^(2)`

Answer

C

Explanation

`a (15)/(2) m//s^(2) = 7.5 m//s^(2)`