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?

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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)#