The Average of Average Speeds

I was giving a series of questions to complete from a friend. They were from a job application test. The questions were not very sophisticated, but the last one was as follows (verbatim):

A cyclist has to complete 2 laps of a track at an average of 60 Mph to beat the World record. If he completes the first lap at 30 Mph, what must his average speed be on the second lap for the record to be broken?

A naïve approach would be to say that if the first lap was at 30 mph, and the second lap at 90 mph, then the overall average speed is the average of these two: 60 mph.

However, this is wrong. The intutive reason is that when the lap is completed with a higher average speed, the lap took less time to complete and so contributes to the average less.

Let us be explicit. Suppose that each lap is of distance D. Let us say that the first lap was completed with an average speed of v1, which took time t1. Likewise, let us say the second lap was completed with an average speed of v2, which took time t2.

Straight away we have the following relationships:

v_1 = \frac{D}{t_1}\qquad v_2 = \frac{D}{t_2}

Let us call the overall average speed u. Then u is:

u = \frac{\mbox{Overall distance}}{\mbox{Overall time}} = \frac{2D}{t_1+t_2} = \frac{2D}{\left(\frac{D}{v_1} + \frac{D}{v_2}\right)} = \frac{2v_1 v_2}{v_1 + v_2}

Suppose, as in the question, that our first lap was at 30 mph. We therefore have that u = \frac{60 \times v_2}{30 + v_2}. Suppose we wish u = 60. Then we have that 1 = \frac{v_2}{30+v_2}. Note that there is no value of v2 that makes this true!

Therefore the cyclist has blown his chance! He cannot possibly beat the world record!

Another way to see the conclusion is as follows. The world record time is distance/avg. speed = 2D/60 = D/30. The first lap was took a time of distance/avg. speed = D/30. Therefore all the time is taken, and we reach the same conclusion as before.

Given the lack of sophistication in the previous questions, it seemed this was not the answer they were looking for. It is certainly a lot harder than the others, especially given how it is a trick question! I thought it safer just to give the naïve answer, especially considering that it might be considered suspicious if the person I was doing it on the behalf of got it right.


2 Responses to The Average of Average Speeds

  1. Daniel Krügler says:

    Just a suggested typo fix: “overall speed” in the denominator of the definition of u should be “overall time”

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