Average Speed Over Two Laps
A car races two laps of a track, each lap the same distance. It completes the first lap at 100 mph and the second lap at 200 mph. What is the average speed over the full two-lap race?
Hints
- Average speed is total distance divided by total time -- not the average of the two speeds.
- Let each lap have length $d$; compute the time for each lap and add them.
- When two legs cover equal distances at speeds $v_1$ and $v_2$, the average speed is the harmonic mean: $2v_1 v_2 / (v_1 + v_2)$.
Worked Solution
How to Think About It: Most people's first instinct is to average the speeds: $(100 + 200)/2 = 150$ mph. That is wrong. Average speed is total distance divided by total time -- and the two laps do not take equal time, so the speeds are not equally weighted. The car spends twice as long crawling through the first lap at 100 mph as it does blazing through the second at 200 mph. The slow lap dominates the total time, pulling the average below 150.
Quick Estimate: Let each lap be 100 miles. Time for lap 1: $100/100 = 1$ hour. Time for lap 2: $100/200 = 0.5$ hours. Total: 200 miles in 1.5 hours. Average speed: $200/1.5 \approx 133$ mph. This is the harmonic mean of 100 and 200.
Approach: Average speed = total distance / total time; this gives the harmonic mean, not the arithmetic mean, when legs have equal distance.
Formal Solution:
Let each lap have length $d$.
- Time for lap 1: $t_1 = d/100$
- Time for lap 2: $t_2 = d/200$
Average speed: $$\bar{v} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{t_1 + t_2} = \frac{2d}{\frac{d}{100} + \frac{d}{200}} = \frac{2d}{\frac{2d + d}{200}} = \frac{2d \cdot 200}{3d} = \frac{400}{3}.$$
This is the harmonic mean of the two speeds: $$\bar{v} = \frac{2 v_1 v_2}{v_1 + v_2} = \frac{2 \times 100 \times 200}{100 + 200} = \frac{40000}{300} = \frac{400}{3} \approx 133.33 \text{ mph}.$$
Answer: $\dfrac{400}{3} \approx 133.33$ mph.
Intuition
This is the canonical example of why the harmonic mean, not the arithmetic mean, is the right average for rates over equal distances. The arithmetic mean weights the two speeds equally; the harmonic mean weights them by the time spent at each speed. Since the car spends more time at the lower speed (the slow lap takes twice as long), the time-weighted average is pulled below the midpoint.
The general principle: when you are averaging rates over equal amounts of the denominator quantity (distance, in this case), use the harmonic mean. When you are averaging over equal amounts of the numerator quantity (time), use the arithmetic mean. Confusing these is a common mistake in finance too -- for example, averaging growth rates over equal time periods (arithmetic mean of returns) versus averaging rates over equal investment amounts (harmonic mean, relevant for dollar-cost averaging). Getting the right type of average requires asking: what is held constant across the legs being averaged?