A body moves along a straight path. In successive equal time…

A body moves along a straight path. In successive equal time intervals of duration $\Delta t$, it keeps constant speeds $v_1, v_2, v_3, \ldots, v_n$. What is the average speed of the body?

Explanation of Method

$n$ equal time intervals.

Step 1: Distance in each interval

Every interval lasts the same time $\Delta t$, so the distance covered in the $i$-th interval is

\[ s_i = v_i\,\Delta t, \qquad i = 1, 2, \ldots, n. \]

$\Delta t$ $s_1$ $v_1$ $\Delta t$ $s_2$ $v_2$ $\Delta t$ $s_3$ $v_3$ $\Delta t$ $s_n$ $v_n$ $\cdots$ $\cdots$ $\text{time}$ $\text{distance}$ $s_i = v_i\,\Delta t$

Equal time intervals (top) cover unequal distances (bottom): the faster the interval, the farther the body goes. Each speed counts for the same amount of time.

Step 2: Total distance and total time

\[ s_{\text{total}} = v_1\Delta t + v_2\Delta t + \cdots + v_n\Delta t = \Delta t\,(v_1 + \cdots + v_n), \]

\[ t_{\text{total}} = n\,\Delta t. \]

Step 3: Average speed

\[ v_{\text{avg}} = \frac{s_{\text{total}}}{t_{\text{total}}} = \frac{\Delta t\,(v_1 + v_2 + \cdots + v_n)}{n\,\Delta t}. \]

The interval length $\Delta t$ cancels: \[v_{\text{avg}} = \frac{v_1 + v_2 + \cdots + v_n}{n} = \frac{1}{n}\sum_{i=1}^{n} v_i\]

Special cases

$ n = 2: \quad v_{\text{avg}} = \dfrac{v_1 + v_2}{2} \quad $ (the formula from Translational Kinematics 001)

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

Result

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

Remark

Conclusion. For equal time intervals, the average speed is the arithmetic mean of the speeds.

The trip is split into equal…	Average speed	Type of mean
time intervals $\Delta t$	$\dfrac{v_1 + \cdots + v_n}{n}$	arithmetic
distances $s$	$\dfrac{n}{\frac{1}{v_1} + \cdots + \frac{1}{v_n}}$	harmonic