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Kinematics Problem and final velocity

  1. Sep 10, 2004 #1
    Problem: The acceleration of a particle is given by Ax(t) = -2.00m/s^2 + (3.00m/s^2)t. A) Find the initial velocity such that the particle will have the same x-coordinate at t= 4.00s as it had at t= 0. B) What will the velocity be at t= 4.00s?

    Work so far:

    Integrated to get these:

    Vx(t) = -2t + (3/2)t^2
    x(t) = -t^2 + (1/2)t^3

    I am just stuck on what to do. I dont need it worked out as much as I just need a push in the right direction. Thanks.
  2. jcsd
  3. Sep 10, 2004 #2
    Both your equations are wrong. When you integrated the acceleration, you forgot about the constant of integration or the initial velocity.
  4. Sep 10, 2004 #3
    So then they are:

    Vx(t) = Vi + [-2t + (3/2)t^2]
    x(t) = Xi + Vi(t) + [-t^2 + (1/2)t^3]

    With Vi meaning initial velocity and Xi meaning initial position.

    I just solved it, thanks for the correction.
    Last edited: Sep 10, 2004
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