- #1
FieldDuck
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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.