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Acceleration Based on Position

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data
    speeder passes a police car at 34 m/s. Police car from rest accelerates with a uniform acceleration of 2.2 m/s^2. How much time till the police car overtakes the speeder?

    2. Relevant equations
    V=Vi + at
    X=Xi +Vit + 1/2at^2
    a= 2.2m/s^2 V=34m/s Vi=0

    3. The attempt at a solution
    not sure what equation to use, I think the second one just because it's a question passed on position. Vi=0 so the equation then would be t^2 = 2(X-Xi)/a, but where do I get X and Xi from?
  2. jcsd
  3. Sep 8, 2010 #2
    You have two bodies - police car and speeder car. So it's good idea to use two X's, one for police car position Xp and other for speeder Xs, best choice for origin would seem to be at the point where speeder passes police car. So write equation for Xp and Xs with values given in problem statement and solve for t.
  4. Sep 10, 2010 #3
    bump, I still have no clue how to even start this question. Do I need to set up two equations, one for the car and one for the cop, then set them equal, I'm just confused.
  5. Sep 10, 2010 #4
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