# Simple problem I thought

• physmurf
In summary, the speaker assigned a problem to their students without realizing it would be too difficult for high school students. They got stuck on finding the third equation for the average speed over the entire trip, but eventually realized that one of the unknowns cancels out and the average speed is 3.75 m/s. The speaker also admits to not knowing how to do the problem themselves and expresses relief that it has a well-defined answer.

#### physmurf

I assigned a couple of problem to my students today without realizing that these would be too difficult for high school students. However, after starting the problem, I got stumped. I know I am missing something, but what is it? The problem reads as follows:

5. A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s. What is
a. her average speed over the entire trip and
b. her average velocity over the entire trip?

Part b is simple, but 'a' is where I got stuck. You know delta x is the same, and you know that t1 and t2 will be different as well.

So, delta x = v1*t1, and x=v2*t2.

However this leaves us with three unknowns and two equations.

What am I missing for my third equation?

Thanks!

x= distance traveled from A to B
v_1=5 m/s
v_2=3 m/s

x=5t_1
x=3t_2

t_1=x/5
t_2=x/3

avg. speed = distance/total time= 2x/(t_1 +t_2)=(2x)/(x/3+x/5)=3.75 m/s

one of your unknowns simply cancels out.

since the average speed will be the same no matter what the distance, it is unimportant what the distance between a and b are.

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Thanks,

I knew it was something like that.

Okay, so we are all relieved that the problem has a well-defined answer!

Are you telling us that you assigned this problem to your students without knowing how to do it yourself?

I agree that this problem is not too difficult for high school students- but I'm wondering about you!