A body moves along a straight path. Over successive segments…

A body moves along a straight path. Over successive segments of equal length $s$, it keeps constant speeds $v_1, v_2, v_3, \ldots, v_n$. What is the average speed of the body?

Explanation of Method

$n$ segments of equal length.

Step 1: Time on each segment

Every segment has the same length $s$, so the time spent on the $i$-th segment is

\[ t_i = \frac{s}{v_i}, \qquad i = 1, 2, \ldots, n. \]

$s$ $t_1$ $v_1$ $s$ $t_2$ $v_2$ $s$ $t_3$ $v_3$ $s$ $t_n$ $v_n$ $\cdots$ $\cdots$ $\text{distance}$ $\text{time}$ $t_i = \dfrac{s}{v_i}$

Equal distances (top) take unequal times (bottom): the slower the segment, the longer it lasts. Here $v_2$ is the slowest and $v_3$ the fastest.

Step 2: Total distance and total time

\[ s_{\text{total}} = n\,s, \]

\[ t_{\text{total}} = \frac{s}{v_1} + \frac{s}{v_2} + \cdots + \frac{s}{v_n} = s\left(\frac{1}{v_1} + \cdots + \frac{1}{v_n}\right). \]

Step 3: Average speed

\[ v_{\text{avg}} = \frac{s_{\text{total}}}{t_{\text{total}}} = \frac{n\,s}{s\left(\dfrac{1}{v_1} + \cdots + \dfrac{1}{v_n}\right)}. \]

The segment length $s$ cancels: \[v_{\text{avg}} = \frac{n}{\dfrac{1}{v_1} + \dfrac{1}{v_2} + \cdots + \dfrac{1}{v_n}} = \frac{n}{\displaystyle\sum_{i=1}^{n} \frac{1}{v_i}}\]

Special cases

$ n = 2: \quad v_{\text{avg}} = \dfrac{2}{\dfrac{1}{v_1} + \dfrac{1}{v_2}} = \dfrac{2 v_1 v_2}{v_1 + v_2} \quad $ (the formula from Translational Kinematics 002)

$ n = 3: \quad v_{\text{avg}} = \dfrac{3}{\dfrac{1}{v_1} + \dfrac{1}{v_2} + \dfrac{1}{v_3}} $

Result

\[\boxed{\,v_{\text{avg}} = \frac{n}{\dfrac{1}{v_1} + \dfrac{1}{v_2} + \cdots + \dfrac{1}{v_n}} = \frac{n}{\displaystyle\sum_{i=1}^{n} \frac{1}{v_i}}\,}\]

Remark

Conclusion. For segments of equal length, the average speed is the harmonic mean of the segment speeds. The harmonic mean is never larger than the ordinary (arithmetic) mean, because slow segments take longer and so count for more.