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Basic Kinematics problem, why my method is invalid

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data
    A motorist drives along a straight road at a constant speed of 14.4m/s. Just as she passes a parked motorcycle police officer, the officer starts to accelerate at 1.8m/s^2 to overtake her. Assuming the officer maintains this acceleration, determine the time it takes the police officer to reach the motorist (in seconds).



    2. Relevant equations
    vi^2 + 2ad = vf^2
    v=at
    d=vt

    3. The attempt at a solution
    Here is what I did im curious why my method is incorrect.
    (14.4)t=d (speeder)
    vt=d (officer)
    v=at (officer)
    at^2=d (officer)

    1.8t^2 = 14.4 t
    t=0, 8

    Correct answer is 16

    Ok I realized that d = 1/2 vt^2
    this was my mistake, please show mathematically why I can't use these two equations to form d =vt^2

    vt=d
    vf =at (if vi=0)
    (at)t=d
    at^2 =d
     
  2. jcsd
  3. Dec 3, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    vt = d only if velocity is constant over time. In this case it is not since there is acceleration involved. So you must turn to the full kinematic expression:

    ## d(t) = d_o + v_o t + \frac{1}{2} a t^2##
     
  4. Dec 3, 2015 #3
    Ok, thanks, great to know :)
     
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