Question

A point object is moving on the cartesian coordinate plane according to: `vecr(t)=b^2thatu_x+(ct^3−q_0)hatu_y`. Determine: a) The equation of the trajectory of object on the cartesian plane b) The magnitude and direction of the velocity?

Answer 1

trajectory `y = (c/(b^6))x^3-q_0`
velocity magnitude `norm (vec v(t)) = sqrt(b^4+9c^2t^4)`
velocity normalized direction `(vec v(t))/(norm (vec v(t)) ) = {{b^2,3c t^2}}/sqrt(b^4+9c^2t^4)`

Explanation:

Making `vec r(t) = {x(t),y(t)}` we have

`{x(t),y(t)} = {b^2t, ct^3-q_0}`

which is the so called movement parametric equation.
Eliminating `t` we get

`y = (c/(b^6))x^3-q_0`

which is the trajectory equation.

Velocity determination is done differentiating `vec r(t)` regarding `t` so

`vec v(t) = {b^2,3c t^2}`

so `norm (vec v(t)) = sqrt(b^4+9c^2t^4)` is the velocity magnitude and

`(vec v(t))/(norm (vec v(t)) ) = {{b^2,3c t^2}}/sqrt(b^4+9c^2t^4)` is the normalized direction