The position vector of a particle moving in a plane is given by ` r =a cos omega t hati + b sin omega I hatj, ` where `hati and hatj ` are the unit vectors along the rectangular axes C and Y : a,b, and `omega` are constants and t is time. The acceleration of the particle is directed along the vector

Options

A. `- a hati+ b hatj`
B. ` b hati + a hatj`
C. `-r`
D. `(dr)/(dt)`

Answer

C

Explanation

Given, position vector,
`r =a cos omega t hati + b sin omega hatij" "…(i)`
On differentiating both sides w.r.t.t, we get
Velocity, `(dr)/(dt) =v=-a omega t sin omega r hati+ b omega cos omega t hatj`
Again, on differentiating w.r.t.t, we get acceleration `(dv)/(dt) =a=-a omega ^(2) cos omega t hati- b omega ^(2) sin omega t hatj`
`a =- omega ^(2) (a cos omega t hati + b sin omega t hatj)`
From Eq. (i), we get `a =- omega ^(2) (r) =- omega ^(2) r`
Hence, acceleration is along `-r.`