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Final Exam Review: Police Car

  1. May 6, 2012 #1
    1. While driving on the highway at a constant speed v0, significantly above the speed limit,
    you pass in front of a parked police car without noticing it. After a few seconds, during which time you have moved a distance d, the police car starts chasing you with a constant acceleration a0. Give your answers in terms of v0, d, and a0 and check the units of your answers. Assuming that you keep moving at a constant speed, and that the highway can be considered straight:

    [a] How long will it take for the police car to reach you?

    [b.] How far has the police car traveled when it reaches you?

    [c] What is the speed of the police car when it reaches you?



    2. Relevant equations

    For the Police:

    Vp = ∫a0p(dt) → a0pt + v0p

    Xp = ∫Vp(dt) → 0.5a0pt2 + v0pt + X0p


    3. The attempt at a solution

    [a] The police car will reach me when our positions are the same, ∴ XM = XP

    XM is d ∴ d = 0.5a0pt2 + v0pt + X0p

    However, the v0pt is = 0 because the cop car is initially at rest. X0p is also at rest because the placement of the cop car is at the origin.

    ∴ d = 0.5a0pt2

    Solving for t gives: t = (2d/a0p)0.5. You can discard the negative value for t.

    [b.] For part B, the distance the cop car has travelled can be calculated simply by plugging in the value of t into 0.5a0pt2 + v0pt + X0p

    Once again, the v0pt is = 0 because the cop car is initially at rest. X0p is also at rest because the placement of the cop car is at the origin.

    ∴XP = 0.5a0p(2d/a0p)0.5)2

    which gives d as an output. I am not quite sure if this is the correct answer or not as I have no solutions manual to compare my answer to.

    [c.] For part C, the speed of the police car when it reaches me should be equivalent to my speed, which is V0

    therefore, VP = a0pt

    = a0p(2d/a0p)0.5)2
     
    Last edited: May 6, 2012
  2. jcsd
  3. May 6, 2012 #2
    For part A: You have not considered the total distance covered by the speeding car. Cop does not start until speeder has a distance, d, on him. But what about the distance speeder covers while the cop accelerates?
     
  4. May 6, 2012 #3
    I'm not quite sure how to incorporate that into this. I'm sorry
     
  5. May 6, 2012 #4
    OK, no problem.

    "Assuming that you keep moving at a constant speed,...."

    The distance you will have traveled during that time is Vyou * t, where the t is the same time the police car accelerates until he catches you. That adds additional distance to the equation.
     
  6. May 6, 2012 #5
    Oh! I forgot to integrate my V0, which should have given me XM = V0Mt + X0M.

    Which should have been:

    V0Mt + d = 0.5a0pt2 + v0pt + X0p

    At which point, solving for t should be solved using a quadratic equation?
     
  7. May 6, 2012 #6
    And when this quadratic is solved:

    t = (2V0M + (4V0m2 + 8da0p)0.5)/2a0P.

    The negative solution for time t can be discarded. This also alters my answers to b and c
     
  8. May 6, 2012 #7
    You could solve by 'completing the square' as it was called when I was in school. Or you could use the quadratic formula which is derived by completing the square.
     
  9. May 6, 2012 #8
    is the answer I provided above correct?
     
  10. May 6, 2012 #9
    Yes but I would simplify it a bit by putting the 2 inside the radical and removing it from the first term.
     
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