- #1

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## Homework Statement

A person walks in a straight line from point A to point B at a constant speed of 5.00m/s and walks back from point b to point a, with a constant speed at 3.00 m/s. What is the average speed of the entire trip.

## Homework Equations

## The Attempt at a Solution

First, I know how to solve this problem.

[tex] 5.00m/s = \frac {d}{t_1} [/tex]

and

[tex] -3.00 m/s = -\frac{d}{t_2}[/tex]

Therefore

[tex] V_avg = \frac{Total Distance}{Total Time}= \frac{d+d}{\frac{d}{5.00m/s}+\frac{d}{3.00m.s}}[/tex]

Some simple math and we end up with answer of 3.75 m/s

And that is the answer in the book. The problem is I only got this after I realized my first attempt was wrong and I couldn't figure out why. I tracked down the problem to the simple fact that I didn't write -3.00m/s. I don't particularly understand why it would be -3.00m/s

The way I thought about this initial was that average speed, the direction doesn't matter since they are not vectors, so when I wrote [tex] 3.00=\frac{d}{t_2} [/tex] it made sense to me. So I guess my question is, why is it negative and not just positive? Maybe i'm just so tired tonight.