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Velocity and Acceleration

  1. Sep 14, 2004 #1
    Velocity and Acceleration Please Help!

    My problem reads:

    A particle is observed to move with the coordinates x(t)=(1.5m/s)t + (-0.5 m/s^2)t^2 and y(t) = 6m + (-3m/s)t + (1.5 m/s^2)t^2. What are the particle's position, velocity, and acceleration? At what time(s) are the velocity's horizontal and vertical components equal?

    I got the derivative of x(t) and y(t) to get v(x) and v(y). Where do I go from there :confused:
  2. jcsd
  3. Sep 14, 2004 #2
    do get the acceleration take the derivative of V(x) and V(y).

    For the what time thing, set them equal, solve for "t"?
  4. Sep 14, 2004 #3
    For the second question, solve for x(t) = y(t).
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