Average speed of the round trip

The times for the trip to and from the airport are $t_1 = \frac{\SI{40}{mi}}{\SI{40}{mph}} = \SI{1}{h};\ \ \ \ t_2 = \frac{\SI{40}{mi}}{\SI{60}{mph}} = \tfrac23 \SI{\ }{h}.$ The average speed is therefore $\langle v \rangle = \frac{d_1 + d_2}{t_1 + t_2} = \frac{\SI{40}{mi} + \SI{40}{mi}}{\SI{1}{h} + \tfrac23\SI{\ }{h}} = \frac{\SI{80}{mi}}{\tfrac53\SI{\ }{h}} = \boxed{\SI{48}{mph}}.$

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So why is it not simply $\tfrac12(40 + 60) = 50$ mph?

Basically because Bob spends more time traveling at 40 mph than at 60 mph. Therefore the slower speed, 40 mph, contributes more to the average than the higher speed of 60 mph. The average speed is therefore less than the mean of these two speeds.

Formally, if the two speeds are $v_1$ and $v_2$ , and the one-way distance is $d$ , then $\langle v \rangle = \frac{d + d}{t_1 + t_2} = \frac{2d}{\frac d{v_1} + \frac d{v_2}} = \frac{2}{\frac 1{v_1} + \frac 1{v_2}}.$ This may be written as $\langle v \rangle = \frac{2v_1v_2}{v_1 + v_2}\ \ \ \ \text{or}\ \ \ \ \frac 1{\langle v\rangle} = \frac12\left(\frac 1{v_1} + \frac 1{v_2}\right).$ The second expression shows that in this case, the inverses of the speed (or "slowness") $1/v$ does work with a simple mean. On the way there, the slowness was 1.5 minute per mile; on the way back, 1.0 minute per mile; the average slowness was therefore 1.25 minute per mile.

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The average defined as $\langle v_h \rangle = \frac1 {n}\left(\frac 1{v_1} + \dots + \frac 1{v_n}\right)$ is known as the harmonic mean. In general, the harmonic mean of positive values is less than their "regular", arithmetic mean.

48 mile per hour.

The total time the round trip is 40 / 40 + 40 / 60 = 1 $\frac{2}{3}$ hours.

The total distance is 40 + 40 = 80 miles.

The average speed is 80 / 1 $\frac{2}{3}$ = 48 miles per hour.