`A` point moves in the plane `xy` according to the law, `x=a sin omegat,y=a(1-cosomegat)` Answer the following question taking `a` and `omega` as positive constant
The magnitude of the velocity of the point as a function of time is

Options

A. `asqrt(1+(1-alphat)^(2))`
B. `asqrt(1+(1-2alphat)^(2))`
C. `2asqrt(1+(1-alphat)^(2))`
D. `2asqrt(1+(1-2alphat)^(2))`

Answer

B

Explanation

`x=a sin omegat rArr V_(x)=aomega cos omegat rArra_(x)=aomega^(2) sin omegat`
`y=a(1-cos omegat)rArr V_(y) =aomega sin omegat`
`rArr a_(y)=aomega^(2)cos omegat`
Again `x^(2)+(y-a)^(2)=a^(2)`
So the motion is a circular motion with centre at `(0,a)`and radius of `a`.We can also realise that speed at any instant `V=sqrt(V_(x)^(2)+V_(y)^(2))=aomega=` Constant.So distance travelled in a time `T=aomegaT`As motion is uniform circular motion velocity is always perpendicular to accelaration.