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?

Question

1658 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-08T05:50:17

Let #(x,y)# represents the cartesian coordinate of the point having position vector at t th instant

#vecr(t)=b^2thatu_x+(ct^3-q_0)hatu_y....(1)#

Then
#x=b^2t and y= ct^3-q_0#

#:.y=c(x/b^2)^3-q_0#

On simplification we get

#b^6y=cx^3-q_0b^6#
#=>cx^3-b^6y-q_0b^6=0-># is the eqution of trajectory.

Now differentiating (1) w.r.to.t we get velocity

#vecv(t)=(d(r(t)))/(dt)=b^2hatu_x+3ct^2hatu_y#
Its magnitude

#abs(vecv(t))=sqrt(b^4+9c^2t^4)#

The direction of the velocity.
If it is directed at an angle #theta(t)# with the x-axis at t th instant then

#theta(t)=tan^-1((3ct^2)/b^2)#