Time taken to complete a lap on a circular track Suppose I am running around a circular track of radius $1/10$ mi and I know my speed in miles per hour is given as $v(\theta)=5+3\cos \theta$. I want to know how long it takes me to travel one lap.  What is wrong with the following logic...
The average speed on the interval $0$ to $2\pi$ is $5$ mph as can be seen by integrating the speed function. Notice that the track has length $L=2\pi ^*1/10$ Mi. So isn't finding the time $t$ to travel one lap the same as solving $5=L/t$.  This gives $t=\pi/25$. 
I don't see any errors in my logic.  I know I could solve the problem this way if I was traveling in a straight line. Evidently there is an issue because of the circular track. FYI, the reason I believe I am wrong is that this problem is done many times on different sites and they get $\pi/20$. The way they solve it is much more difficult by looking at the changing theta and integrating. However, I don't see an error I need their logic either. For an example, see
https://www.quora.com/How-do-I-use-calculus-to-figure-out-how-long-it-takes-each-runner-to-run-one-lap-around-a-track-when-the-wind-affects-his-or-her-speed-See-the-image
Thanks!

When you integrated the speed function to get the average, I assume you integrated $v$ with respect to $\theta$. This is the "average of $v$ with respect to $\theta$" in a sense. To use your logic, what you would want to use is the "average with respect to time". 
Think of the values of the speed function $v(\theta)$ as sitting on each point $\theta$ on the circle to which it is associated. During your run, when you get to a point with a higher speed (relative to 5mph), you will speed up and rush to the points where you run at a lower speed. So you will actually spend less time running quickly (>5mph) than you will running slowly (<5mph). This is why the actual answer $t=\pi / 20$ is a slightly longer time than what you got.
Hope this is helpful.
