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Police car's acceleration physics problem

  1. Sep 6, 2004 #1
    A police car traveling a constant 95 km/h is passed by a speeder traveling 140 km/h. Precisely 1.00 s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 2.00 m/s2, how much time passes after the speeder passes before the police car overtakes the speeder (assumed moving at constant speed)?

    I converted 95 km/h to 40 m/s, and 95 to 26 m/s, then I use the equation

    x = x + v*t + 1/2 * a *t^2
    40 m/s * t + 40 m = 0 + 26 m/s * t +1/2*2.0*t^2

    t=15, but that's wrong....any ideas?
  2. jcsd
  3. Sep 6, 2004 #2


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    ... where a is the acceleration with appropriate units, of course! :-)
  4. Sep 6, 2004 #3
    I still get t=15 with that equation. :confused:
  5. Sep 6, 2004 #4


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    You did? I got 6.3 minutes if I did my arithmetic right - seems too long though! Check to see if you converted the units on your acceleration correctly.
  6. Sep 6, 2004 #5
    well, since the final answer is in seconds, I converted 140 km/h and 95km/h to m/s. So

    95km/h * 1h/60min * 1min/60s * 1000m/1km= 26.389 m/s and
    140km/h * 1h/60min * 1min/60s * 1000m/1km= 38.889 m/s.

    so 26.389*t+1/2*2.0*t^2=38.889(t+1)
    and t=15 s.

  7. Sep 6, 2004 #6


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    Oh! The t+1 in my original equation was incorrect. I was mixing seconds with hours. When I correct for that I get something closer to what you have (15.9s).

    How do you know your answer is wrong?
  8. Sep 6, 2004 #7
    The hw is online and when I entered 15s and submitted, it said it was incorrect.
  9. Sep 6, 2004 #8


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    You are getting distance, not acceleration, from that formula. The original formula Tide gave is correct. You need only solve for time. Keep trying.
    Last edited: Sep 6, 2004
  10. Sep 6, 2004 #9
    Chronos, can you clarify what you said?
  11. Sep 6, 2004 #10

    Doc Al

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    This isn't quite right for several reasons.

    Measuring everything from the moment the speeder passes the police car (at t = 0):
    Distance police car travels: [itex]x_p = v_p t + \frac{1}{2}a(t-1)^2[/itex], note that acceleration only begins at t = 1 second, [itex]v_p[/itex] is the initial speed of the police car in m/s.
    Distance speeder travels: [itex]x_s = v_s t[/itex], [itex]v_s[/itex] is the speed of the speeder in m/s.

    So, when the police car overtakes the speeder:
    [tex]v_p t + \frac{1}{2}a(t-1)^2 = v_s t[/tex]
    All times measured in seconds. Now plug in your values and solve for t.
  12. Sep 6, 2004 #11


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    Thanks, Al, I stand corrected!
  13. Sep 6, 2004 #12
    I got 14.4 seconds, is that correct?
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