A radius vector of a particle varies with time t as `r=at(1-alphat)`, where a is a constant vector and `alpha` is a positive factor. Find:
(a) the velocity v and the acceleration w of the particles as functions of time,
(b) the time interval `Deltat` taken by the particle to return to the initial points, and the distance s covered during that time.

Answer

(a) `v=a(1-2alphat)`, `w=-2alphaa=const`; (b) `Deltat=1//alpha`, `s=a//2alpha`.

Explanation

we have `vecr=vecat(1-alphat)`
So, `vecv=(vec(dr))/(dt)=veca(1-2alphat)`
and `vecw=(dvecv)/(dt)=-2alphaveca`
(b) From the equation
`vecr=vecat(1-alphat)`,
`vecr=0`, at `t=0` and also at `t=Deltat=1/alpha`
So, the sought time `Deltat=1/alpha`
As `vecv=veca(1-2alphat)`
So, `v=|vecv|={:(a(1-2alphat)),(a(2alphat-1)):}}{:("for t"le(1)/(2alpha)),("for t" gt (1)/(2alpha)):}`
Hence, the sought distance
`s=int vdt=underset0 overset(1//2alpha)int=a(1-2alphat)dt+underset(1//2alpha)overset(1//alpha)inta(2alphat-1)dt`
Simplifying, we get, `s=(a)/(2alpha)`