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Speed and acceleration - running for the bus

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data

    A bus goes from the bus stop with the acceleration 0.75 m/s^2. A girl runs after the bus. She has the speed 8.0 m/s, and she was 17.5 m behind the bus as it started.

    a) How long time does it take the girl to catch up with the bus?


    2. Relevant equations

    THe equations for constant acceleration [itex]v=v_0 +at[/itex] and [itex]s=v_0 t + \frac 12 at^2[/itex]

    3. The attempt at a solution

    I only came as far as to figure out which variables I have for each object:

    BUS: [itex]v_0=0 m/s[/itex]since the bus has no speed at the stop, a=0.75m/s^2

    GIRL: [itex]v_0=v=8.0 m/s[/itex] Velocity at start is equal to velocity later when the speed is konstant, a=0m/s^2 No acceleration when an object has constant speed. [itex]s_0=-17.5 m[/itex] Starting point 17.5 m behind the bus

    However I can't figure out how to use these variables and formulas to get the time it takes her to catch up with the bus....

    Please help!
     
  2. jcsd
  3. Sep 7, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Write an equation for the position with respect to time for each of them. They meet when their distances are equal, so equate the two and solve for the time variable.
     
  4. Sep 7, 2011 #3
    I just found that was the way I could solve it :smile: Many thanks for your help, anyway.
     
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