The position vector of the particle is `r(t)= a cos omega t hat(i) + a sin omega t hat(j)`, where a and `omega` are real constants of suitable dimensions. The acceleration is

Options

A. perpendlcular to the velocity
B. parallel to the velocity
C. direction away from the origin
D. perpendicular to the position vector

Answer

A

Explanation

Given that, `r(t)= a cos omega t hat(i) + a sin omega t hat(j)`
`because v= (dr(t))/(dt)= - a omega sin omega t hat(i) + a omega cos omega t hat(j)`
`a= (dv)/(dt)= - a omega^(2) cos omega t hat(i)-a omega^(2) sin omega t hat(j)`
`because a.v= (-a omega sin omega t hat(i)+ a omega cos omega t hat(j)) (-a omega^(2) cos omega t hat(i)- a omega^(2) sin omega t hat(j))`
`a.v = a^(2) omega^(3) sin omega t cos omega t- a^(2) omega^(3) sin omega t cos omega t`
`rArr a.v= 0 [ because a.v |a||v| cos theta]`
Above result implies that acceleration is perpendicular to velocity