Speed and average speed

Speed shows how much path a body covers per unit time, and for non-uniform motion one uses average speed — the ratio of the total path $s$ to the total time of motion $t$:

$$\bar{v} = \frac{s}{t}.$$

The quantity $s$ in this formula is the path, that is, the length of the trajectory (a scalar, $s \ge 0$), not the displacement $\Delta \vec{r}$, which is a vector joining the initial and final points. Therefore average path speed and average displacement speed coincide only for straight-line motion without turns or returns. Instantaneous speed $v$ describes motion at a given instant and equals the average only for uniform motion, where $s = v t$. The main trap: average speed is not the arithmetic mean of the stage speeds, because it is paths and times that must be added, not the speeds themselves. In SI, speed is measured in metres per second, and $1\ \text{m/s} = 3.6\ \text{km/h}$.

A cyclist rides $s = 12$ km along a straight line in $t = 0.5$ h. Find the average speed in km/h and in m/s.

A car covers the first stage $s_1 = 30$ km in $t_1 = 0.5$ h, then waits at a traffic light for $t_2 = 0.25$ h, and covers the last $s_3 = 45$ km in $t_3 = 0.75$ h. Find the average speed over the whole journey. Assume all stages lie on one straight line.

A body covers the first half of the path at speed $v_1 = 30$ km/h and the second half at speed $v_2 = 60$ km/h. Find the average speed over the whole path and explain why it is not $45$ km/h. Obtain a general formula for two halves of equal length.

Measuring the average speed of walking (safe, at home).

1. Mark out a stretch of length $s = 10$ m in a corridor or room (with a tape measure, or with steps of known length checked three times).
2. Walk this stretch at a calm pace three times, timing each run with a phone stopwatch. Record $t_1, t_2, t_3$.
3. For each run compute $\bar{v} = s/t$ and take the mean of the three trials. Convert the result to km/h.
4. Repeat the experiment with a 5 s stop in the middle of the stretch. Verify that the path stayed the same while the average speed decreased.
   Safety rules: measure only indoors or on a closed yard, away from traffic; do not run on stairs or wet floors.
   If you need higher speeds: do not go onto a road — simulate the motion instead, for example in the PhET simulation "Moving Man": set a constant speed, then add a pause in the middle and compare the average slope of the graph with the instantaneous one.

Show that average speed is the slope of the secant on a path-versus-time graph, and that a stop produces a horizontal segment on the graph.

Graph (sketch): axes $t$ (horizontal) and $s$ (vertical). A broken line with three segments: a gentle rise at speed $v_1$, a horizontal plateau of duration $t_2$ (the stop), then a steeper rise at speed $v_3$. A dashed secant is drawn from the start to the end of the broken line, labelled $\bar{v} = s/t$, which is its slope. The slopes of the individual segments are labelled as the instantaneous (stage) speeds, and one can see that the secant slope lies between them.

A runner covers $100\ \text{m}$ along a straight lane in $12.5\ \text{s}$. The tape measure gives the lane length, the stopwatch gives the total time. Dividing: $v_{avg}=\frac{100\ \text{m}}{12.5\ \text{s}}=8.0\ \text{m/s},$ which is $28.8\ \text{km/h}$. The world record of $9.58\ \text{s}$ gives $v_{avg}=\frac{100\ \text{m}}{9.58\ \text{s}}=10.44\ \text{m/s}\approx 37.6\ \text{km/h}.$ Yet at the end of the acceleration the runner moves instantaneously faster — about $12\ \text{m/s}$, i.e. above either of these averages.

The definition of average speed as total path over total time is at work: $v_{avg}=\frac{s}{\Delta t}.$ The path $s=100\ \text{m}$ is measured along the lane in the stadium frame, $\Delta t$ comes from the stopwatch at the start. The instantaneous speed $12\ \text{m/s}$ exceeds the average because the average is spread over the whole interval $\Delta t$, including the slow start.

A car drives along a highway: the first $90\ \text{km}$ at $90\ \text{km/h}$, that is $1.0\ \text{h}$; then another $90\ \text{km}$ at $60\ \text{km/h}$, that is $1.5\ \text{h}$. Total path $s=180\ \text{km}$, total time $\Delta t=2.5\ \text{h}$, hence $v_{avg}=\frac{180\ \text{km}}{2.5\ \text{h}}=72\ \text{km/h}\approx 20\ \text{m/s}.$ Averaging the numbers $90$ and $60$ by mistake gives $75\ \text{km/h}$ — $3\ \text{km/h}$ too high.

Here the average speed is computed from total path and total time, not as the arithmetic mean of two speeds: $v_{avg}=\frac{s_1+s_2}{\Delta t_1+\Delta t_2}=\frac{90+90}{1.0+1.5}=72\ \text{km/h}.$ The path adds up along the highway trajectory in the road frame, the time adds up on the driver's clock. Two legs at different speeds yield one average only by this rule.

After a lightning flash the thunder is heard $3.0\ \text{s}$ later. The speed of sound in air at $20\ ^\circ\text{C}$ is $343\ \text{m/s}$. Then $s=v\,t=343\ \text{m/s}\cdot 3.0\ \text{s}=1030\ \text{m}\approx 1.0\ \text{km}.$ Light from the same flash covers that distance almost instantly: at $c=3.0\cdot 10^{8}\ \text{m/s}$ the time is $3.4\ \mu\text{s}$, so the whole $3.0\ \text{s}$ delay belongs to sound.

The definition of speed $v=s/t$ rearranges into $s=v\,t$; that is exactly the form used to find distance from time. The speed of sound $343\ \text{m/s}$ is the average propagation speed of the wave front in air at $20\ ^\circ\text{C}$, measured in the ground frame (the air is treated as still).