A point moves in the plane `xy` according to the law `x=a sin omegat`, `y=a(1-cos omega t)`, where a and `omega` are positive constants. Find:
(a) the distance s traversed by the point during the time `tau`,
(b) the angle between the point's velocity and acceleration vectors.

Answer

(a) `s=aomegatau`; (b) `pi//2`.

Explanation

Differentiating motion law: `x=asin omegat`, `y=a(1-cosomegat)`, with respect to time, `v_x=aomegacosomegat`, `v_y=aomegasin omegat`
So, `vecv=aomegacosomegatveci+aomegasin omegatvecj` (1)
and `v=aomega=Const.` (2)
Differentiating Eq. (1) with respect to time
`vecw=(dvecv)/(dt)=-aomega^2sinomegat veci+aomega^2cos omegatvecj` (3)
(a) The distance s traversed by the point during the time `tau` is given by
`s=underset0oversettauintvdt=underset0oversettauintaomegadt=aomegatau` (using 2)
(b) Taking inner product of `vecv` and `vecw`
We get, `vecv.vecw=(a omega cos omegat veci+aomegasin omegatvecj)*(aomega^2sinomegat(-i)+aomega^2cosomegat-vecj)`
So, `vecv*vecw=-a^2omega^2sinomegatcosomegat+a^2omega^3sinomegatcosomegat=0`
Thus, `vecv_|_vecw`, i.e., the angle between velocity vector and acceleration vector equals `pi/2`.