## 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.

## The Attempt at a Solution

First, I know how to solve this problem.

$$5.00m/s = \frac {d}{t_1}$$
and

$$-3.00 m/s = -\frac{d}{t_2}$$

Therefore
$$V_avg = \frac{Total Distance}{Total Time}= \frac{d+d}{\frac{d}{5.00m/s}+\frac{d}{3.00m.s}}$$

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 $$3.00=\frac{d}{t_2}$$ 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.

Related Introductory Physics Homework Help News on Phys.org
jedishrfu
Mentor
Since it says avg speed, you don't use minus 3m/s just 3 m/s

Distance from a to b is the same as distance from b to a

(2 * d) / ( t1 + t2 ) = avg speed

Just sub in for the t values and simplify.

I think the problem is stated wrong in your book as the minus 3 m/s in the answer implies avg velocity and velocity is direction dependent.

You know what, I think you just confirmed that my brain was fried because as I reworked it, it became obvious -_-. Now I feel a bit like an idiot. I'll take this as a sign I need sleep.