Time taken to complete a lap on a circular track

Question

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!

Answer 1

Your logic only works if the radius of the circle is one. You are talking about distances around a track and mixing them with angles on the circle.

Answer 2

Let $s$ equal distance traveled along the circumference of the track, which implies $v=ds/dt$, which in turn implies $1/v=dt/ds$, which finally gets us to the relationship

$$dt=\frac{ds}{v(\theta)}$$

We also know that (in radian measure) $s=r\theta$. Differentiating with respect to $\theta$ will give us $ds=r\,d\theta$, and we will introduce this into our previous relationship, yielding

$$dt=\frac{r\,d\theta}{v(\theta)}$$

One lap, enduring from $t_0$ to $t_f$, is correlated with $\theta$ varying from $0$ to $2\pi$. Integrate over these bounds:

$$\int_{t_0}^{t_f}dt = \int_{0}^{2\pi}\frac{r\,d\theta}{v(\theta)}$$

Adding all the incremental changes in time $dt$ over the course of one lap gets you the time for one lap $T$, which is still equal to the right-hand-side integral:

$$T=\int_0^{2\pi} \frac{r\,d\theta}{5+3\cos\theta}$$

Since we’re considering a circular track, $r$ is constant and can come outside the integral. I will leave the calculations to you.

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I wish I could help you understand why your calculations are slightly off, but I don’t actually quite understand what you did with your “average speed” calculation. If you post your solution in detail, I would love to try to help you on that front.

Answer 3

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.