#### FieldDuck

1. 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.

2. Homework Equations

3. 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.

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#### 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.

#### FieldDuck

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.