# Physics Distance Question

1. Sep 9, 2007

### Larrytsai

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. Sep 9, 2007

### bob1182006

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.

$$d_{Mary}=d_0+vt$$
$$d_{John}=d_0+vt$$

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