Finding the expression for the x-component of velocity (vectors?)

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SUMMARY

The discussion centers on finding the x-component of velocity for a particle described by the position vector r=(ct^2-2dt^3)i+(4ct^2-dt^3)j, where c and d are positive constants. The correct approach involves deriving the velocity vector v=(2ct-6dt^2)i+(8ct-3d^2)j and determining when the particle moves solely in the x-direction by setting the y-component of velocity (Vy) to zero. The solution requires solving for time t and substituting back into the x-component of velocity (Vx) to find the correct expression.

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ashhlyn
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Homework Statement


a particle's position is the vector r=(ct^2-2dt^3)i+(4ct^2-dt^3)j where c and d are positive constants. find the expression for the x-component of the velocity (for time t>0) when the particle is moving in the x-direction. you should express your answer in terms of variables c and d.

Homework Equations


(vector)r=(ct^2-2dt^3)i+(4ct^2-dt^3)j
is the only given equation

The Attempt at a Solution


I'm confused on what this problem is asking
I can derive the position vector to get v=(2ct-6dt^2)i+(8ct-3d^2)j but what is the question asking? I thought the x-component would be 2ct-6dt^2 but that was wrong. thank you much for any help

EDIT: when I put the wrong answer in the hint is "first fine the velocity vector and use this to determine the times when the particle is traveling in the x direction)
 
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ashhlyn said:
when the particle is moving in the x-direction
I.e., at a time when there is no other component to the velocity.
 
haruspex said:
I.e., at a time when there is no other component to the velocity.
ahh thank you! Set Vy equal to zero, solve for t, and put back into Vx. I got the right answers thanks much
 

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