For a particle in uniform circular motion , the acceleration ` vec(a)` at a point ` p ( R, theta )` on the circle of radiu ` R` is ( Here ` theta` is measured from the ` x- axis` )

Options

A. `-v^(2)/R cos theta hati + v^(2)R sin theta hatj`
B. `-v^(2)/R sin theta hati + v^(2)/R cos theta hatj`
C. `-v^(2)/R cos theta hati - v^(2)/R sin theta hatj`
D. `v^(2)/R hati + v^(2)/R hatj`

Answer

C

Explanation

`veca = v^(2)/r (-cos theta hati - sin theta hatj)`