---
title: "Motion Graphs and Average Speed"
book: "Primary & Middle School Physics"
subject: physics
language: en
chapter: 52
exercises: 12
source: https://one-course.com/books/physics/1/en/chapter/52-motion-graphs-and-average-speed
---

# Chapter 52 — Motion Graphs and Average Speed

Two years ago, [speed](#def-g7-motion-average-speed-formula) meant “kilometres covered in one hour,” counted on your fingers. Now, armed with letters and graphs, the idea grows teeth: a formula that computes in three directions, a picture that shows a whole journey at a glance — and a famous trap about averages that catches adults daily.

## 52.1 The formula

**Definition 52.1 (Speed, by formula).**

For a journey (or stretch of one) covered at a steady pace, the *speed* $v$ is the distance $d$ divided by the travel time $t$:

$$
v = \frac{d}{t}.
$$

With $d$ in kilometres and $t$ in [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock), $v$ speaks *kilometres per hour* ($\mathrm{km}/\mathrm{h}$); with [metres](https://one-course.com/books/physics/1/en/chapter/7-measuring-length#def-g2-measuring-length-units) and seconds, *[metres](https://one-course.com/books/physics/1/en/chapter/7-measuring-length#def-g2-measuring-length-units) per second* ($\mathrm{m}/\mathrm{s}$) — the scientist’s favorite. Like [density](https://one-course.com/books/physics/1/en/chapter/46-volume-mass-density#def-g7-volume-mass-density-formula)’s, this formula computes in all three directions: $v = d/t$, $d = v \times t$, $t = d/v$.

**Example 52.2 (The recipes at work).**

A train covers $240\,\mathrm{km}$ in $2$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock): $v = 240 \div 2 =
120\,\mathrm{km}/\mathrm{h}$. How far at that pace in $5$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock)? $d = 120
\times 5 = 600\,\mathrm{km}$. How long for $300\,\mathrm{km}$? $t = 300
\div 120 = 2.5$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock) — two [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock) and thirty minutes (mind the rebel: $0.5$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock) is $30$ minutes, never “$50$”).

**Example 52.3 (Two units, one speed).**

A sprinter runs $100\,\mathrm{m}$ in $10\,\mathrm{s}$: $v = 10\,\mathrm{m}/\mathrm{s}$. How fast is that in road [units](https://one-course.com/books/physics/1/en/chapter/37-measurement-in-science-units-and-instruments#def-g6-measurement-in-science-measuring)? In one hour — $3600$ seconds — the sprinter would cover $10 \times 3600 = 36\,000\,\mathrm{m} =
36\,\mathrm{km}$: so $10\,\mathrm{m}/\mathrm{s} = 36\,\mathrm{km}/\mathrm{h}$. The general exchange rate: each $\mathrm{m}/\mathrm{s}$ is worth $3.6\,\mathrm{km}/\mathrm{h}$ — because an hour holds $3600$ seconds and a kilometre only $1000$ [metres](https://one-course.com/books/physics/1/en/chapter/7-measuring-length#def-g2-measuring-length-units). [Sound](https://one-course.com/books/physics/1/en/chapter/19-sound-around-us#def-g3-sound-around-us-sound)’s $340\,\mathrm{m}/\mathrm{s}$, in road [units](https://one-course.com/books/physics/1/en/chapter/37-measurement-in-science-units-and-instruments#def-g6-measurement-in-science-measuring): over $1200\,\mathrm{km}/\mathrm{h}$.

## 52.2 The journey as a picture

**Proposition 52.4 (The graph laws).**

Plot distance covered against time, and motion writes its autobiography:

1. [uniform motion](https://one-course.com/books/physics/1/en/chapter/40-describing-motion-trajectory-and-speed#def-g6-describing-motion-uniform) draws a *straight* line — equal distances in equal times, step after step;
2. the *steeper* the line, the faster the motion: steepness is [speed](#def-g7-motion-average-speed-formula) made visible;
3. a *flat* stretch is a stop: time passes, distance stands.

Bends tell of change: curving upward, speeding up; flattening, slowing down.

![A courier’s morning told by one line: brisk riding, a fifteen-minute stop, then a gentler pace. The graph is the journey’s autobiography.](https://one-course.com/images/onecourse/chapters/physics-1/g7-motion-average-speed/fig-1ad2931e8298.svg)

*A courier’s morning told by one line: brisk riding, a fifteen-minute stop, then a gentler pace. The graph is the journey’s autobiography.*

**Method 52.5 (Reading a distance–time graph).**

1. check the axes and their [units](https://one-course.com/books/physics/1/en/chapter/37-measurement-in-science-units-and-instruments#def-g6-measurement-in-science-measuring) first — minutes or [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock) , [metres](https://one-course.com/books/physics/1/en/chapter/7-measuring-length#def-g2-measuring-length-units) or kilometres;
2. split the line at its bends into stretches; label each: straight-and-steep, straight-and-gentle, flat;
3. for any straight stretch, read off its rise and its run — distance gained, time taken — and divide: the stretch’s [speed](#def-g7-motion-average-speed-formula) ;
4. for the whole story, read total distance at the final time — and remember the flat stretches are part of the total time.

**Example 52.6 (The courier, decoded).**

Apply the method to the figure. First stretch: $10\,\mathrm{km}$ in $20$ minutes — a third of an hour — so $v = 10 \div
\tfrac{1}{3} = 30\,\mathrm{km}/\mathrm{h}$. Second: flat from minute $20$ to $35$ — a delivery stop. Third: $10\,\mathrm{km}$ in $25$ minutes, a shade under $24\,\mathrm{km}/\mathrm{h}$. Total: $20\,\mathrm{km}$ in one hour — which hands us the chapter’s next idea on a plate.

## 52.3 Average speed — and its famous trap

**Definition 52.7 (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](#def-g7-motion-average-speed-formula).

**Proposition 52.8 (The trap).**

The [average speed](#def-g7-motion-average-speed-average) of a journey is *not*, in general, the midpoint of its [speeds](#def-g7-motion-average-speed-formula). Slow stretches eat more time than fast ones, so they weigh more heavily in the average. Only the full recipe — total distance over total time — is trustworthy; averaging the [speed](#def-g7-motion-average-speed-formula) numbers themselves is the most seductive wrong move in the chapter.

**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](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock). Home: $t = 60 \div 20 = 3$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock). Whole journey: $120\,\mathrm{km}$ in $5$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock) — $v_{\text{average}} = 24\,\mathrm{km}/\mathrm{h}$. The slow half claimed three [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock) of the five, and dragged the average below the midpoint. The faster you go on one half, the less time that half even exists.

**Remark 52.10 (What the speedometer knows).**

[Average speed](#def-g7-motion-average-speed-average) describes a whole journey; the speedometer needle answers a different question — *how fast right now*. On the graph, “right now” lives in the line’s steepness *at a single point* — easy to see on a straight stretch, subtle where the line curves. Making “steepness at a point” precise is one of mathematics’ greatest inventions, and it waits for you at the end of high school. Until then: straight stretches get numbers, curves get stories.

## 52.4 Exercises

**Exercise 52.1 ★.**

Write the [speed](#def-g7-motion-average-speed-formula) formula and its two everyday [units](https://one-course.com/books/physics/1/en/chapter/37-measurement-in-science-units-and-instruments#def-g6-measurement-in-science-measuring). Which recipe finds a distance? A time?

**Solution of Exercise 52.1.**

$v = d/t$; [units](https://one-course.com/books/physics/1/en/chapter/37-measurement-in-science-units-and-instruments#def-g6-measurement-in-science-measuring) $\mathrm{km}/\mathrm{h}$ and $\mathrm{m}/\mathrm{s}$. Distance: $d = v \times t$; time: $t = d/v$.

**Exercise 52.2 ★.**

A ferry covers $45\,\mathrm{km}$ in $3$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock); a hare runs $100\,\mathrm{m}$ in $8\,\mathrm{s}$. Compute both [speeds](#def-g7-motion-average-speed-formula), each in its natural [unit](https://one-course.com/books/physics/1/en/chapter/37-measurement-in-science-units-and-instruments#def-g6-measurement-in-science-measuring).

**Solution of Exercise 52.2.**

Ferry: $45 \div 3 = 15\,\mathrm{km}/\mathrm{h}$. Hare: $100 \div 8 =
12.5\,\mathrm{m}/\mathrm{s}$.

**Exercise 52.3 ★.**

Convert, with the exchange rate: $5\,\mathrm{m}/\mathrm{s}$ to $\mathrm{km}/\mathrm{h}$; $72\,\mathrm{km}/\mathrm{h}$ to $\mathrm{m}/\mathrm{s}$.

**Solution of Exercise 52.3.**

$5 \times 3.6 = 18\,\mathrm{km}/\mathrm{h}$; $72 \div 3.6 = 20\,\mathrm{m}/\mathrm{s}$.

**Exercise 52.4 ★.**

State the three graph laws. What does a bend that flattens gradually tell?

**Solution of Exercise 52.4.**

Straight line: [uniform motion](https://one-course.com/books/physics/1/en/chapter/40-describing-motion-trajectory-and-speed#def-g6-describing-motion-uniform); steeper: faster; flat: stopped. A gradually flattening bend tells of slowing down.

**Exercise 52.5 ★.**

On the courier’s graph: between which minutes is the pace gentlest (but not zero)? How can you tell without computing?

**Solution of Exercise 52.5.**

From minute $35$ to $60$ — the third stretch: it is the least steep of the rising stretches, read directly from its gentler slant.

**Exercise 52.6 ★.**

A walker’s graph shows a straight line through the points ($30$ min, $2\,\mathrm{km}$) and ($60$ min, $4\,\mathrm{km}$). Uniform or varied? [Speed](#def-g7-motion-average-speed-formula) in $\mathrm{km}/\mathrm{h}$?

**Solution of Exercise 52.6.**

Uniform — one straight line. It gains $2\,\mathrm{km}$ each half hour: $v = 4\,\mathrm{km}/\mathrm{h}$.

**Exercise 52.7 ★.**

Define [average speed](#def-g7-motion-average-speed-average). A hike: $12\,\mathrm{km}$ in $4$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock) including a one-hour picnic. [Average speed](#def-g7-motion-average-speed-average) — and [average speed](#def-g7-motion-average-speed-average) *while walking*?

**Solution of Exercise 52.7.**

Total distance over total time, stops included. Whole hike: $12 \div 4 = 3\,\mathrm{km}/\mathrm{h}$. Walking only: $12 \div 3 =
4\,\mathrm{km}/\mathrm{h}$.

**Exercise 52.8 ★★.**

Why is a slow stretch “heavier” in an average than a fast one of equal length? Answer with time, not with formulas.

**Solution of Exercise 52.8.**

Over equal distances, the slow stretch simply lasts longer — more of the journey’s clock is spent living at the slow pace, so the slow pace speaks with more of the journey’s voice.

**Exercise 52.9 ★★.**

Replay the trap: $30\,\mathrm{km}$ out at $15\,\mathrm{km}/\mathrm{h}$, $30\,\mathrm{km}$ back at $30\,\mathrm{km}/\mathrm{h}$. Predicted midpoint, honest average — and which half of the journey owned most of the clock?

**Solution of Exercise 52.9.**

Midpoint guess: $22.5\,\mathrm{km}/\mathrm{h}$. Honestly: out $30 \div 15 = 2$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock), back $30 \div 30 = 1$ hour; total $60\,\mathrm{km}$ in $3$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock): $20\,\mathrm{km}/\mathrm{h}$. The slow half owned two of the three [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock).

**Exercise 52.10 ★★.**

Sketch (or describe precisely) the graph of this trip: uniform $40\,\mathrm{km}/\mathrm{h}$ for half an hour; stopped for a quarter hour; uniform $60\,\mathrm{km}/\mathrm{h}$ for a quarter hour. Then compute the trip’s [average speed](#def-g7-motion-average-speed-average).

**Solution of Exercise 52.10.**

Graph: straight to ($30\,\mathrm{min}$, $20\,\mathrm{km}$); flat to $45$ min; straight and steeper to ($60\,\mathrm{min}$, $35\,\mathrm{km}$). Average: $35\,\mathrm{km}$ in one hour — $35\,\mathrm{km}/\mathrm{h}$.

**Exercise 52.11 ★★.**

Storm-counting, upgraded: thunder arrives $6\,\mathrm{s}$ after the flash. With [sound](https://one-course.com/books/physics/1/en/chapter/19-sound-around-us#def-g3-sound-around-us-sound) at $340\,\mathrm{m}/\mathrm{s}$, use $d = v \times t$ for the storm’s distance — and check the old “divide by three for kilometres” rule against your formula.

**Solution of Exercise 52.11.**

$d = 340 \times 6 = 2040\,\mathrm{m}$ — about $2\,\mathrm{km}$. The old rule: $6 \div 3 = 2$ kilometres — agreeing, because $340\,\mathrm{m}/\mathrm{s}$ means very nearly a kilometre every three seconds.

**Exercise 52.12 ★★★.**

An old chestnut, worth every minute: a driver covers the first half *of the distance* of a trip at $30\,\mathrm{km}/\mathrm{h}$ and wants an overall average of $60\,\mathrm{km}/\mathrm{h}$. Show — with the total-distance-over-total-time recipe on a $60\,\mathrm{km}$ trip — that the second half would have to be covered in zero time: the wish is impossible, not merely difficult.

**Solution of Exercise 52.12.**

First half: $30\,\mathrm{km}$ at $30\,\mathrm{km}/\mathrm{h}$ costs exactly $1$ hour. An overall $60\,\mathrm{km}/\mathrm{h}$ over $60\,\mathrm{km}$ allows a total of exactly $1$ hour — already spent to the last second. The remaining $30\,\mathrm{km}$ would need to take $0$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock): no [speed](#def-g7-motion-average-speed-formula), however heroic, makes the average; the budget is gone.

## 52.5 Problem: The Courier’s Friday

**Problem 52.1.**

Weekend problem — one bicycle courier, one city Friday; the dispatcher’s graph, the trap in the bonus sheet

The dispatcher records courier Lena’s Friday on a distance–time graph and settles her pay from it. You audit the sheet.

**Part I — The morning, from the graph.** The graph shows: a straight climb from ($0$ min, $0\,\mathrm{km}$) to ($30$ min, $9\,\mathrm{km}$); flat until $45$ min; straight from there to ($75$ min, $15\,\mathrm{km}$); flat until $90$ min.

1. Tell the morning’s story in words: stretches, stops, and what the stops presumably were.
2. [Speed](#def-g7-motion-average-speed-formula) on the first stretch, in $\mathrm{km}/\mathrm{h}$ ?
3. [Speed](#def-g7-motion-average-speed-formula) on the second riding stretch?
4. Lena’s *average* [speed](#def-g7-motion-average-speed-formula) over the whole $90$ -minute morning, stops included?

**Part II — The afternoon, from the formula.**

5. Afternoon leg one: $12\,\mathrm{km}$ at a steady $24\,\mathrm{km}/\mathrm{h}$ . How long did it take, in minutes?
6. Leg two: a delivery uptown, $20\,\mathrm{minutes}$ at $18\,\mathrm{km}/\mathrm{h}$ . How many kilometres?
7. Leg three: the long ride to the depot, $10\,\mathrm{km}$ , done in $25$ minutes. Steady-pace [speed](#def-g7-motion-average-speed-formula) in $\mathrm{km}/\mathrm{h}$ ?
8. Total afternoon: add the distances and the times (legs only, no breaks): what [average speed](#def-g7-motion-average-speed-average) did the wheels keep while rolling?

**Part III — The bonus sheet’s trap.** The bonus rule: “[average speed](#def-g7-motion-average-speed-average) over a full tour above $20\,\mathrm{km}/\mathrm{h}$ earns the fast-rider bonus.”

9. Lena’s full Friday: $37\,\mathrm{km}$ in $3$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock) (all stops included). Does she earn the bonus?
10. Her colleague Max claims it: “I rode half my tour at $30\,\mathrm{km}/\mathrm{h}$ and half at $15\,\mathrm{km}/\mathrm{h}$ — that averages $22.5$ !” The dispatcher checks: both halves were $12\,\mathrm{km}$ . Compute Max’s true average and rule on his bonus.
11. Explain to Max, in time-language, where his $22.5$ went wrong.
12. Lena proposes a fairer bonus rule for couriers — one that does not punish delivery stops. Suggest one (the graph knows how), and say which [speed](#def-g7-motion-average-speed-formula) it [measures](https://one-course.com/books/physics/1/en/chapter/37-measurement-in-science-units-and-instruments#def-g6-measurement-in-science-measuring) .

**Part IV — The dispatcher’s lesson.**

13. Friday review: write the dispatcher’s three sentences — what a straight stretch, a flat stretch, and a steeper stretch each mean on the graph — and the one warning about averaging [speeds](#def-g7-motion-average-speed-formula) .

**Solution of Problem 52.1.**

**1.** A brisk ride ($30$ min), a $15$-minute stop (delivery), a gentler ride ($30$ min), and a second $15$-minute stop — two runs, two calls. **2.** $9\,\mathrm{km}$ in half an hour: $18\,\mathrm{km}/\mathrm{h}$. **3.** $15 - 9 = 6\,\mathrm{km}$ in half an hour: $12\,\mathrm{km}/\mathrm{h}$. **4.** $15\,\mathrm{km}$ in $1.5$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock): $10\,\mathrm{km}/\mathrm{h}$. **5.** $t = 12 \div 24 = 0.5$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock): $30$ minutes. **6.** $20$ minutes is a third of an hour: $d = 18 \times
\tfrac{1}{3} = 6\,\mathrm{km}$. **7.** $25$ minutes is $\tfrac{25}{60}$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock); $v = 10
\div \tfrac{25}{60} = 24\,\mathrm{km}/\mathrm{h}$. **8.** Distances: $12 + 6 + 10 = 28\,\mathrm{km}$; times: $30 + 20 + 25 = 75$ minutes $= 1.25$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock); rolling average: $28 \div 1.25 = 22.4\,\mathrm{km}/\mathrm{h}$. **9.** $37 \div 3 \approx 12.3\,\mathrm{km}/\mathrm{h}$ — far below $20$: no bonus (her stops are honest work, but the rule counts them). **10.** Max: fast half $12 \div 30 = 0.4$ h; slow half $12 \div 15 = 0.8$ h; total $24\,\mathrm{km}$ in $1.2$ h: exactly $20\,\mathrm{km}/\mathrm{h}$ — not *above* $20$: no bonus. **11.** His $22.5$ averaged the two *numbers*, as if each pace owned half the clock. In truth the slow half owned $0.8$ of his $1.2$ [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock) — two thirds of the journey’s voice — and dragged the true average down to $20$. **12.** For example: pay the bonus on *rolling* [average speed](#def-g7-motion-average-speed-average) — total distance divided by riding time only, the flat stretches of the graph excluded. It [measures](https://one-course.com/books/physics/1/en/chapter/37-measurement-in-science-units-and-instruments#def-g6-measurement-in-science-measuring) the pace the wheels actually keep, rewarding fast riding rather than skipped deliveries. **13.** “A straight stretch is a steady pace; a flat stretch is a stop with the clock still running; a steeper stretch is a faster pace — steepness is [speed](#def-g7-motion-average-speed-formula). And never [average speed](#def-g7-motion-average-speed-average) numbers: average the journey — total distance over total time — or the slow [hours](https://one-course.com/books/physics/1/en/chapter/8-measuring-time#ex-g2-measuring-time-clock) will make a fool of the sheet.”