Velocity and acceleration from a position equation

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The discussion revolves around finding the velocity and acceleration of a particle defined by the position equation x = y^2/6, moving at a constant velocity of 3 in/s in the y-direction. Participants suggest differentiating the position equation with respect to time to determine the velocity in the x-direction. It is noted that since Vy is constant, the acceleration in the y-direction is zero, simplifying the analysis to focus on the x-direction. The conversation highlights the importance of expressing y as a function of time to effectively calculate the required values. The thread concludes with acknowledgment of the need for a clearer approach to the problem.
glid02
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Here's the question:
A particles position is defined by the equation x=y^2/6 and it moves at a constant velocity in the y-direction of 3 in/s. Find the velocity and acceleration when x=6 in.

I know v=sqrt(vx^2+vy^2) and a is found the same way, but I have absolutely no idea how to find the velocity of in the x-direction through the equation.

The only way I can think to do it is find the change in the x position over a very small interval of time and go from there, but I know there's an easier way.

If anyone could give me a starting point it'd be great.

Thanks a lot.
 
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well if Vy is constant, y"=0 leaving only the x directions to worry about fpr acceleration.
so can't you just differentiate with respect to x and plug in the x value given? For velocity add the constant as you propose, vector summation.
 
Last edited:
yeah, I'm half retarded. I needed to make y a function of t and go from there. Thanks.
 
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