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Velocity and acceleration from a position equation

  1. Feb 22, 2007 #1
    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.
  2. jcsd
  3. Feb 22, 2007 #2
    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: Feb 22, 2007
  4. Feb 22, 2007 #3
    yeah, I'm half retarded. I needed to make y a function of t and go from there. Thanks.
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