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Basic Kinematics Question

  1. Aug 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A speeder passes a hiding speed cop at 55m/s. The speed cop puts down his coffee and starts his engine (which takes 5 sec). He then accelerates from rest at 5m/s^2 until he catches the speeder.
    How far does the cop travel before he catches up and how fast is the cop going?


    2. Relevant equations

    V=x/t --> x=vt
    Vf^2 = Vi^2 + 2ax


    3. The attempt at a solution

    I know that the final distances can be set to equal each other but I am confused on what to put for time. If i can solve for time, I can use the time multiplied by 55m/s to find how far they traveled. Also, if the cop takes 5 seconds to start moving, the speeder is already 275m away

    Velocity of the cop
    Vf^2 = Vi^2 +2ax
    Vf^2 = 0^2 + 2(5)(275-0)
    Vf = 52.44 m/s

    Time of cop
    v=x/t
    52.44 = 275m / t
    t = 5.24 sec

    Xspeeder=Xcop
    Vspeeder x Tspeeder = Vcop x Tcop
    55 m/s x Tspeeder + 275m = 52.44 m/s x 5.24 sec
    But when I solve for Tspeeder I get a negative number and time can't be negative.
    Any help would be appreciated!
    Thank you
     
  2. jcsd
  3. Aug 30, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    The distance moved by the speeder in ( t + 5 ) seconds is equal to the distance traveled by cop in t seconds.
    Equate them and solve the quadratic to get the time.
     
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