A point moves in the plane `xy` according to the law `x=at`, `y=at(1-alphat)`, where a and `alpha` are positive constants, and t is time. Find:
(a) the equation of the point's trajectory `y(x)`, plot this function,
(b) the velocity v and the acceleration w of the point as functions of time,
(c) the moment `t_0` at which the velocity vector forms an angle `pi//4` with the acceleration vector.

Answer

(a) `y=x-x^2alpha//a`; (b) `v=asqrt(1+(1-2alphat)^2)`, `w=2alphaa=const`; (c) `t_0=1//alpha`.

Explanation

We have
`x=at` and `y=at(1-alphat)` (1)
Hence, `y(x)` becomes,
`y=(ax)/(a)(1-(alphax)/(a))=x-(alpha)/(a)x^2` (parabola)
(b) Differenting Eq. (1) anwe get
`v_x=a` and `v_y=a(1-2alphat)` (2)
So `v=sqrt(v_x^2+v_y^2)=asqrt(1+(1-2alphat)^2)`
Diff. Eq. (2) with respect to time
`w_x=0` and `w_y=-2aalpha`
So, `w=sqrt(w_x^2+w_y^2)=2aalpha`
(c) From Eqs. (2) and (3)
We have `vecv=aveci+a(1-2alphat)vecj` and `vecw=2aalphavecj`
So, `cospi/4=1/sqrt2=(vecv*vecw)/(vw)=(-a(1-2alphat_0)2aalpha)/(asqrt(1+(1-2alphat_0)^2)2aalpha)`
On simplifying. `1-2alphat_0+-1`
As, `t_0!=0`, `t_0=1/alpha`