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Time and velocity

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data
    A speeder passes a parked police car at a constant speed of 27.6 m/s. At the instant, the police car starts from rest with a uniform acceleration of 2.27 m/s^2. How much t passes before the speeder is overtaken by the police car.


    I tried solving for t using t=Vf-Vi /a and i get 12.15sec, which is wrong what am i doing wrong?
     
  2. jcsd
  3. Oct 23, 2008 #2

    LowlyPion

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    Observe that to catch him they must both be at the same X at the same Time.

    What is the equation for the PoPo's displacement with respect to time?
    And for the reckless miscreant?

    When they are equal, it's book'em Danno.
     
  4. Oct 23, 2008 #3
    OMMMMG I love u I GOT IT!!!!!
    x=1/2a(t)^2 + Vi(t)

    Cops= 1/2(2.27)(12.15)+0(12.15) = 13.79025
    Speeder= 1/2(0)12.15+27.6(12.15) = 335.34
    335.34/13.79025 = 24.32s
     
    Last edited: Oct 23, 2008
  5. Oct 23, 2008 #4

    LowlyPion

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    ??

    I was thinking of something considerably more direct.

    1/2*a*t2 = V * t

    t = 2*V/a = 2*27.6/2.27 = 24.32 s
     
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