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Physics Distance Question

  1. Sep 9, 2007 #1
    Mary left home at noon running at 7m/s towards school. Her brother John left 35 seconds later running at 8m/s. How much time passed and how far were they from the house when John caught up to Mary.



    2. Relevant equations
    My solution i tried was to the x-y graph. My notes saids im supposed to use these formulas
    d-d0=vt , d=vt

    3. The attempt at a solution
    I tried graphing as an university friend/student has told me but it just doesnt work out because when I get to john i have no clue to where i start on the graph to graph johns distance. All i know right now is that in 35 seconds, when john leaves Mary is at 245 metres.
     
    Last edited: Sep 9, 2007
  2. jcsd
  3. Sep 9, 2007 #2
    ok those equations are correct the thing is you need to have 2 equations for the distance.
    One will represent Mary, another will represent John.

    [tex]d_{Mary}=d_0+vt[/tex]
    [tex]d_{John}=d_0+vt[/tex]

    since velocity is constant a=0 and we loose that a/2*t^2 part.

    So when should time start for these equations?
    t=0 Mary leaves the house? or
    t=0 John leaves the house, Mary has a 35s lead and is how many m away from the house already?
     
    Last edited: Sep 9, 2007
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