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1D Kinematics of 2 Objects

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A car traveling at a constant speed of 108 km/hr passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets in a chase after the car with a constant acceleration of 3.0 m/s2. How far does the trooper travel before he overtakes the speeding car?

    2. Relevant equations
    [tex]\Delta x = v_0 t + \frac{1}{2} a t^2[/tex]

    [tex]108 \frac {km}{hr} = 30 \frac {m}{sec}[/tex]


    3. The attempt at a solution
    Well, since they're both moving along the same axis, I set the equations equal to each other.

    [tex]v_0 t = \frac{1}{2} a t^2[/tex]

    [tex]v_0 = \frac{1}{2} a t[/tex]

    [tex] \frac{2 v_0}{a} = t[/tex]

    and I got t = 20. Then,

    [tex]\Delta x = \frac{1}{2} a t^2[/tex]

    [tex]\Delta x = \frac{1}{2} (30 \frac{m}{s}) (20sec)^2[/tex]

    and I got 600 m but that was wrong.
     
    Last edited: Sep 3, 2009
  2. jcsd
  3. Sep 3, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Zhalfirin8! Welcome to PF! :smile:
    erm :redface: … what happened to that one second? :wink:
     
  4. Sep 3, 2009 #3
    Try equating both displacements, and don't forget that the cop will have a delayed time (i.e t-1s)

    EDIT: tiny-tim was faster, sorry for the repetitive post.
     
  5. Sep 3, 2009 #4
    What would that change? The cop would still catch up to the guy in 20 seconds right?

    So the 2nd equation would be:

    [tex]\Delta x = \frac{1}{2} (30 \frac{m}{s}) (21sec)^2[/tex]

    ?
     
  6. Sep 4, 2009 #5

    tiny-tim

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    (just got up :zzz: …)
    It would change t (in one of the equations only) to either t + 1 or t - 1. :wink:

    (and then you have two equations which use the same t, so you can solve them jointly)

    Have a go! :smile:
     
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