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?

1 Answer
Jun 4, 2016

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