I'm going through my old College Algebra book to shake off some of the dust, and I came across this example:

"To qualify for the Labor Day 400 auto race, each driver must complete two laps of the track at an average speed of 100 mi/h. Due to some problems at the start, Naomi averages only 50 mi/h on her first lap. How fast must she go on her second lap to qualify for the race?"

I first tried to solve it on my own, and I saw "average" so did this:

$(50 + s)/2 = 100$


which is, as it turns out, incorrect. The book arrived at the conclusion that there is no speed she could hit that would guarantee her a qualification. I understand how they arrived at the conclusion, but I'll provide their process for reference:

$50 mi/h = \frac{L mi}{t_1 h},$ where $L$ is the length of the track in miles and $t_1$ is the time it took her to complete it in hours.

$(S)mi/h = \frac{L mi}{t_2 h},$ $S$ being the speed she needs to average at for the second lap and $L$ being the length again.

$100mi/h = \frac{2L mi}{(t_1 + t_2) h}.$

This comes out to $t_2=0$, so there is no way for her to qualify.

I understand why this works, using units and all, so I'm good there. My question is why my original method isn't sufficient? There's obviously some conceptual understanding that I'm missing, and I'm hoping to close that gap. What about this problem should tip me off to the unit method and not the formula for average? Any help would be appreciated!


Your first "arithmetic mean" of speed is when the object moves at each speed for the same time $t$. Then the average speed is

$$\frac{50t + st}{2t} = \frac{50+s}{2}$$

But for the average speed when the object moves at each speed for the same distance $L$, the average would be a harmonic mean:

$$\frac{2L}{\dfrac L{50} + \dfrac L{s}} = \frac{2}{\dfrac 1{50} + \dfrac1{s}}$$

For your question, the two speeds happen each for a lap, so their harmonic mean should be used.

  • $\begingroup$ Thank you, this makes sense now! $\endgroup$
    – Lo12
    Dec 11 '20 at 1:19

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