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Kinematics car chase question

  1. Mar 19, 2017 #1
    1. The problem statement, all variables and given/known data
    We have that a police car is travelling at ##26.4## m/s, and that at a time, ##t=0##, a car speeds by at ##37.5## m/s. After ##1## s, the police car accelerates constantly at ##2## m/s2. We are asked to find the time, ##t##, where the police car catches the speeder.

    2. Relevant equations
    ##v_f=v_i+at##, and ##s=v_0t+\frac{1}{2}at^2##. (Standard kinematics equations.)

    3. The attempt at a solution
    Granted that the police car is not accelerating for the first second, I chose to simply begin analyzing the motion at the ##1## s mark. In the first second, the police car moves, ##26.4## m, while the speeder moves, ##37.5## m. We have then, in this frame, that the speeder's displacement after ##1## s is given by, ##(37.5-26.4)+37.5t##. That is, I simply state that the initial position of the speeder is given by the displacement between the two vehicles after ##1## s. The police car's displacement is ##26.4t+t^2##. We simply rearrange then to solve the quadratic for ##t##, such that ##t=12.2## s. My book says the answer is ##13## s. I'm not sure if this is simply due to a rounding error, or if there's something wrong with my logic.
     
  2. jcsd
  3. Mar 19, 2017 #2
    Use: X-Xo = VoT + 1/2AT^2
    For the police car:
    X-Xo = 1/2(2)(T-1)^2 + (26.4)T
    For the car:
    X-Xo = (37.5)T

    Set them equal to each other & solve for T :)
    The trick is realizing that the police car has two components for distance...
     
  4. Mar 19, 2017 #3
    I got 12.02 s as an answer.

    Edit: I thought your method looked good.

    Edit2: Oops. We both forgot to add back in the 1 second, so the answer looks like it should be 13.02 s.
     
  5. Mar 19, 2017 #4
    That will work too, but for the police car I think it should be:
    x-xo = (0.5)(2)(t-1)2 + 26.4(t-1)
    Edit: I'm sorry. If you are using xo = 0, then your equation is correct as it is written. That is probably what you intended.
     
    Last edited: Mar 19, 2017
  6. Mar 19, 2017 #5
    Well that explains it. I knew I was missing something. Thanks.
     
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