This morning I went for a bike ride. The weather forecast said that it would be breezy but there was little chance of rain. Here in the midwest, our weather and prevailing winds generally arrive out of the west. I learned long ago that when there is wind, always ride into it on the way out and you will be rewarded on the way home.

In the past, I have made comments to the effect that my 20 mile ride takes longer on windy days like today, only to be met with skepticism. The comments often go something like “well, whatever you lose on the way out you gain on the way back, so it all comes out even.”

Students often found the idea of averaging speeds counter-intuitive. They did not distinguish between situations where two speeds were undertaken for equal distances or equal amounts of time.

**Warning: If you are going to use these examples with students, fix the significant digits.**

The classic problem goes something like this:

If I drive halfway to the state capitol at 50 mph, how fast will my sister have to drive the rest of the way so that we average 75 mph? Most students will wonder briefly why they are getting such an easy problem and quickly write 100 mph. Some will have trouble with the concept even after it is explained and demonstrated with real numbers.

Imagine that the trip length is 150 miles. Making up an example is another thing that students often fail to try. In this example, I ask them “do you think the trip length actually affects the answer?” I point out that if it did, then it would have needed to be supplied in the problem or there would be no way to calculate the answer. Since the problem did not specify the length of the trip, it must not matter and we are free to make up an example to simplify the problem conceptually. In the end, if we want to check to see if our assumption that trip length did not matter was valid, we could make up a second example with a different trip length and see if we got the same answer. Of course, you could use x and 2x for the distances and in the end the x’s will disappear, but for some students numbers are easier.

This problem actually appears in our 7th grade algebra book. I keep trying to explain to the math teachers that it is conceptually more complicated than they think because students do not always have a firm grasp of the motion concept and will therefore have a difficult time applying the algebra. Here is a table of values that are either supplied in the problem or made up for my first example.

Using the formula for average speed the other values can be calculated. Rather than moving at 100 mph (as some students predict), my sister must actually drive 150 mph if we are to average 75. The lesson is that speeds can only be averaged if they two drivers drive for equal amounts of TIME, not if they drive equal DISTANCES.

To check the original assumption that trip length did not matter, a second example with a trip length of 300 miles yields an identical answer, 150 mph.

Back to the bike ride. How does wind affect my speed. Assuming I give a constant effort, the wind will decrease my speed on the way out when I am riding into the wind and increase my speed on the way back when I am riding with the wind. In the simplest example, the speed with be increased and decreased an equal amount.

I usually ride 12 mph and my normal ride is 20 miles. This means when there is no wind, I complete the ride in 100 minutes (1.67 hours).

10 miles / 12 mph = 0.833 hours out

10 miles / 12 mph = 0.833 hours back

** total time = 1.67 hours**

If I assume that the wind can increase and decrease my speed by 3 mph then:

10 miles / 9 mph = 1.11 hours out

10 miles / 15 mph = 0.67 hours back

** total time = 1.78 hours**

The total time of the ride with wind is longer than the time riding without the wind. This is because the speed were changed for equal DISTANCES, not equal amounts of TIME, therefore cannot be simply averaged. The wind did not “make up” on the way back what it took on the way out in terms of time.

If I assume that the wind was even more severe, robbing and returning 6 mph, the calculations are even more extreme.

10 miles / 6 mph = 1.67 hours out (actually the original total!!!!)

10 miles / 18 mph = 0.55 hours back

** total time = 2.23 hours**

This is a full 0.56 hours (34 minutes) (33%) increase in time over the ride calculated with no wind. No wonder I hate wind. Even the numbers fail to convince some students. One student suggested that if I rode the other direction first, perhaps I would benefit and complete the ride in less time rather than more. Oh, physics. You are a harsh mistress.

Of course, this is not really how wind affects speed. Instead, wind contributes to the total resistance that I encounter when I ride. And that resistance is a function of the speed. That discussion would be much more complex but still fun.

As always, your polite and helpful comments are welcome.