Average velocity, half and half

Problem

A car covers the first half of a trip at $50{.}0\;\text{km/h}$ and the second half at $75{.}0\;\text{km/h}$. What is the average velocity over the whole trip? Explain why the answer is not the arithmetic mean of the two velocities.

Solution

Correct definition. The average velocity is the total distance divided by the total time, not the average of the two velocities. The difference arises because the car spends more time on the slow leg than on the fast one, so the low velocity “weighs” more.

Setup. Let $d$ be the total length of the trip: each half measures $d/2$. The times of the two legs are

$t_1 = \frac{d/2}{v_1} = \frac{d/2}{50{.}0}, \qquad t_2 = \frac{d/2}{v_2} = \frac{d/2}{75{.}0}$

Average velocity.

$\bar v = \frac{d}{t_1 + t_2} = \frac{d}{\dfrac{d/2}{50{.}0} + \dfrac{d/2}{75{.}0}} = \frac{1}{\dfrac{1}{2\cdot 50{.}0} + \dfrac{1}{2\cdot 75{.}0}}$

The distance $d$ cancels: the result does not depend on the length of the trip. We recognise the harmonic mean of the two velocities:

$\bar v = \frac{2\, v_1 v_2}{v_1 + v_2} = \frac{2\cdot 50{.}0\cdot 75{.}0}{50{.}0 + 75{.}0} = \frac{7500}{125} = 60{.}0\;\text{km/h}$

$\ev{\bar v = 60{.}0\;\mathrm{km/h}}$

Why not the arithmetic mean. The arithmetic mean $(50{.}0 + 75{.}0)/2 = 62{.}5\;\text{km/h}$ would be correct only if the car travelled for equal time intervals at the two velocities. Here instead the legs have equal length, and at $50{.}0\;\text{km/h}$ it takes more time than at $75{.}0\;\text{km/h}$: the slow leg occupies a larger fraction of the trip, so the true average velocity ($60{.}0\;\text{km/h}$) is shifted towards the low value and turns out lower than the arithmetic mean.

Collegamenti

Argomenti: Kinematics Concetti: Average velocity