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Cars persecution

  1. Sep 23, 2008 #1
    1. The problem statement, all variables and given/known data
    A thief steals a car in front of two cops, A and B.
    The thief's initial velocity is 25m/s, cop A starts his car with 3m/s2, and cop B starts 10 seconds after cop A, but with 5m/s2. After 200m from his initial point, the thief notices the cops and start accelerating at 1m/s2

    Which cop catches the thief? At what time does he catch it?

    2. Relevant equations
    d= Vot+1/2at2
    I think that's the only required formula


    3. The attempt at a solution
    I tried to set an equality between the distance of the thief and the distance of cop A, but I get stuck in the math to solve for t.

    (25m/s)(t)+(1/2)(m/s2)+200m=(1.5m/s2)(t+8)2
    t is the time since the thief started accelerating, and that happens 8s after A started.
    after doing math here I get completely stuck with the algebra.
     
  2. jcsd
  3. Sep 24, 2008 #2
    I tried a new equation:
    (25m/s)t+(1/2)(m/s2)t2-200m = (1.5m/s2)t2
    I got that t = 10.73seconds, but I don't know if its correct because the left side of the equation which is the thief's distance seems that it was always accelerating but started 200m behind cop A, so would anyone tell me what equation I could use?
     
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