What is Average speed? — Physics

Definition 52.7 Primary & Middle School Physics · Chapter 52 — Motion Graphs and Average Speed

The average speed of a whole journey is the total distance divided by the total time — stops included:

v_{\text{average}} = \frac{d_{\text{total}}}{t_{\text{total}}}.

It is the one steady pace that would have covered the same road in the same overall time — the courier’s $20\,\mathrm{km}$ in one hour: average $20\,\mathrm{km}/\mathrm{h}$ , though the wheels never once turned at that speed.

Examples

Example 52.9 (The trap, sprung)

A cyclist rides $60\,\mathrm{km}$ out at $30\,\mathrm{km}/\mathrm{h}$ , and the same $60\,\mathrm{km}$ home at $20\,\mathrm{km}/\mathrm{h}$ . “Average: $25\,\mathrm{km}/\mathrm{h}$ ”? Check honestly. Out: $t = 60 \div 30 = 2$ hours. Home: $t = 60 \div 20 = 3$ hours. Whole journey: $120\,\mathrm{km}$ in $5$ hours — $v_{\text{average}} = 24\,\mathrm{km}/\mathrm{h}$ . The slow half claimed three hours of the five, and dragged the average below the midpoint. The faster you go on one half, the less time that half even exists.

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Definition 5.6 High School Physics · Chapter 5 — Relative Motion

The average speed of a body, in a given reference frame, is the distance $d$ it travels in that frame divided by the travel time $t$ :

v = \frac{d}{t} .

With $d$ in metres and $t$ in seconds, $v$ is in metres per second ( $\mathrm{m}/\mathrm{s}$ ); everyday life prefers kilometres per hour ( $\mathrm{km}/\mathrm{h}$ ).

Examples

Example 5.8 (Some speeds worth knowing)

A brisk walk: $1.5\,\mathrm{m}/\mathrm{s} \times 3.6 = 5.4\,\mathrm{km}/\mathrm{h}$ . A city car at $54\,\mathrm{km}/\mathrm{h}$ : $54 / 3.6 = 15\,\mathrm{m}/\mathrm{s}$ — about $15\,\mathrm{m}$ covered every second, worth remembering before stepping off a kerb. Sound in air: $340\,\mathrm{m}/\mathrm{s} \times 3.6 = 1224\,\mathrm{km}/\mathrm{h}$ .

Example 5.9 (A hike with a break)

A hiker covers $12\,\mathrm{km}$ in $2\,\mathrm{h}$ of walking, then rests for $1\,\mathrm{h}$ , then covers $6\,\mathrm{km}$ in the final $1.5\,\mathrm{h}$ . Average speed over the whole outing:

v = \frac{d}{t} = \frac{12 + 6}{2 + 1 + 1.5} = \frac{18\,\mathrm{km}}{4.5\,\mathrm{h}} = 4.0\,\mathrm{km}/\mathrm{h}.

The average counts all the elapsed time, breaks included — it is generally not the average of the speeds of the separate legs.

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