A radius vector of a point A relative to the origin varies with time t as `r=ati-bt^2j`, where a and b are positive constants, and I and j are the unit vectors of the x and y axes. Find:
(a) the equation of the point's trajectory `y(x)`, plot this function,
(b) the time dependence of the velocity v and acceleration w vectors, as well as of the moduli of these quantities,
(c) the time dependence of the angle `alpha` between the vectors w and v,
(d) the mean velocity vector averaged over the first t seconds of motion, and the modulus of this vector.

Answer

(a) `y=-x^2b//a^2`; (b) `v=ai-2btj`, `w=-2bj`, `v=sqrt(a^2+4b^2t^2)`, `w=2b`; (c) `tan alpha=a//2bt`; (d) `<>=ai-btj`, `|<>|=sqrt(a^2+b^2t^2)`.

Explanation

As `vecr=atvecl-bt^2vecj`
So, `x=at`, `y=-bt^2`
and therefore `y=(-bx^2)/(a^2)`
which is Eq. of a parabola, whose graph is shown in the Fig.
(b) As `vevr=atvecl-bt^2vecj`
`vecv=(dvecr)/(dt)=aveci-2btvecj` (1)
So, `v=sqrt(a^2(-2bt)^2)=sqrt(a^2+4b^2t^2)`
Diff. Eq. (1) w.r.t. time, we get
`vecw=(dvecv)/(dt)=-2bvecj`
So, `|vecw|=w=2b`
(c) `cos alpha=(vecv*vecw)/(vw)=((1veci-2btvecj)*(-2bvecj))/((sqrt(a^2+4b^2t^2))2b)`
or, `cos alpha=(2bt)/(sqrt(a^2+4b^2t^2)`
so, `tan alpha=(a)/(2bt)`
or, `alpha=tan^-1((a)/(2bt))`
(d) The mean velocity vector
`lt vecv gt =(int vecvdt)/(int dt)=(underset0oversettint(aveci-2btvecj)dt)/(t)=aveci-btvecj`
Hence, `|lt vecv gt|=sqrt(a^2+(-bt)^2)=sqrt(a^2+b^2t^2)`