For a particle in uniform circular motion the acceleration ` vec a ` at a point ` P ( R, theta )` on the circle of radius 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 hai - (v ^ 2 ) / ( R ) sin theta hatj `
D. ` (v^ 2 )/ (R ) hati + (v ^ 2 ) /( R ) hatj `

Answer

D

Explanation

For a particle in uniform circular motion, acceleration `a=v^(2)/R` is toward the centre. From figure, we have `overset(-)(a)=-a cos theta hati-a sin theta hatj rArr overset(-)a=-v^(2)/R cos theta hati-v^(2)/R sin theta hatj`