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Police car and speeder when do they meet?

  1. Jan 15, 2007 #1
    The problem statement, all variables and given/known data
    An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 140 km/h. Precisely 1.00s after the speeder passes, the policeman steps on the accelerator; if the police car's acceleration is 2.00m/s2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?

    v0p = 95 km/h = 26.389 m/s
    v0s = 140 km/h = 38.889 m/s
    ap = 2.00 m/s
    t1 = 1.00 s
    vs = v0s

    2. Relevant equations
    x = x0 + v0t + 0.5 * a * t2

    3. The attempt at a solution
    Let t2 = the time it takes them to meet after the initial 1.00s.
    After 1.00s, the xp = 26.389m and xs = 38.889m. Taking these two quantities as our initial positions, x0p and x0s, we have:
    xp = x0p + v0pt2 + 0.5 * ap * t22
    xs = x0s + v0st2
    Setting these two equal, I get a quadratic equation and find t2 = 13.43 s or -0.9307 s. The total time would then be 14.43s...

    But wait. Shouldn't the negative answer for t2 be -1.00s?? Because 1.00s in the past, the cars met... So I must be doing something wrong...
     
  2. jcsd
  3. Jan 15, 2007 #2

    Hootenanny

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    Firstly, nice presentation :approve: Secondly, don't forget from 0.00s to 1.00s the police car wasn't accelerating, so the symmetry is slightly misleading.
     
  4. Jan 15, 2007 #3
    What do you mean "the symmetry is slightly misleading"?

    Is my answer correct or wrong?
     
  5. Jan 15, 2007 #4

    Hootenanny

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    Your method looks correct, but I ain't checking the math for you. The symmetry is misleading because t = -0.9307 would have been the time they would have met if the police car would have been accelerating from t = 0.

    Does that make sense?
     
  6. Jan 15, 2007 #5
    Oh I see...
    I forgot about that.

    thanks :approve:
     
  7. Jan 15, 2007 #6

    Hootenanny

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    Pleasure, once again great presentation.
     
  8. Jul 29, 2011 #7
    i cannt understand it
     
  9. Jul 29, 2011 #8
    The position of the police officier is a piece-wise function.
     
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