# Walking back and forth.

1. Sep 20, 2011

### Quark Itself

Problem : Two hikers, Call them Carl and David started to walk at 12pm (noon) along the same path but in opposite directions. Let's assume Carl walked from Point A to point B and David did the opposite, walked from Point B to Point A. Both of them have different constant speed. They pass each other at a point at 1500 hours. Carl arrived at Point B , 2.5 hours before David arrived at point A. When did David arrive at point A?

Attempt: Carl Case 1) Xf = Xi + ViT assuming that his initial point Xi = 0 makes David's Xf = 0 as well.
Xf = 0 + 3Vi. this Xf is the meeting point, therefore they have been walking for 3 hours, hence the 3Vi.
Then I got stuck..

Thanks in advance !

2. Sep 20, 2011

### daveb

Since they meet at 3:00, their elapsed times are equal, though the distance traveled are different. When the walk is over their distances are equal though the elapsed times are different.

3. Sep 20, 2011

### Quark Itself

Yeah, I got that part and everything. It is very clear, but from there how do you proceed?

4. Sep 20, 2011

### Ray Vickson

Set it up carefully and explicitly. For example, let A be at x = 0 and B be at x = L. Let Carl's speed be c and David's speed be d. Then Carl's position xc(t) is xc(t) = c*t for t <= L/c and David's position is xd(t) = L-d*t for t <= L/d. Now you have xc(3)=xd(3), and you have one other relationship regarding the endpoints. If you write things properly you will see that you do (believe it or not) have enough information to complete the question.

RGV

Last edited: Sep 20, 2011
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