# Homework Help: Started out simple enough

1. Sep 9, 2007

### josephpatricks

Hi everybody. I haven't been in school a few years, and its second week into physics. I have what the book describes as the answer, and through several methods seem to come .5m away from the stated answer...I don't know if they just simplify it or not.

1. The problem statement, all variables and given/known data
A dog runs back and forth between its two owners, who are walking toward one another. The dog starts running when the owners are 10m apart. If the dog runs with a speed of 3.0 m/s, and the owners walk with a speed of 1.3 m/s, how far has the dog traveled when the owners meet?

2. Relevant equations
limit of change in x / change in t

3. The attempt at a solution
From as far as I can see, the owners are covering 2.6 m/s on a 10ft gap, while the dog's traveling of 3.0m/s is being hastened as a limit from 10 to 0. I don't know how to use latex yet, but I took the limit of (3/2.6)x t, where t is 10. the answer I receive is 11.5m or 150/13, but the book puts the answer as 11m. I am frustrated by now and just want to bs some attribution to the dog not starting at the side of an owner, accounting for the .5 m, but I offer this to anyone who could give me a better explanation.

2. Sep 9, 2007

### learningphysics

I'm also getting 11.5m . It seems like you didn't use limits, but just regular distance = speed * time, which is fine.

The time when the owners meet is t=(10/2.6). So the distance the dog travelled in that time is 3*t which is 11.5m.

Does the problem ask you to use limits?

3. Sep 10, 2007

### josephpatricks

no it doesn't. i can understand the distance part, i.e. the dog is allowed travel of 3 m/s during the time frame the owners travel (10/2.6). I went over the problem with a kid at the tutoring site at our school, we all get the same answer. I'll ask the teacher, he might have a solutions manual. If there is a good reason, I'll update. thanks!