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Year 8 Maths Time and Travel

Travel Graphs

20 practice questions 1 video lesson Theory + worked examples

A travel graph, or distance–time graph, shows how far someone is from a starting point as time passes, with numbers on both axes. In NSW Year 8 maths, students read these graphs and calculate the speed of each straight section (NESA outcome MA4‑RAT‑C‑01).

You will read distances and times, explain what flat and downward sections mean, and find speeds in km/h or m/s from each part of a journey. You will find the average speed with stops included, compare two travellers on one graph to see where they meet or how far apart they are, choose the graph that matches a description, and finish a trip that is only partly drawn. Rail-trail rides, whale watching and a drive to Sydney are examples. It is part of Year 8 Time and Travel.

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Theory

A travel graph has time across and distance up. Each straight section has a speed \(=\) distance \(\div\) time: a car that goes \(30\) km in the first \(40\) minutes is doing \(30 \div \dfrac{2}{3} = 45\) km/h. This Year 8 maths page shows how to read and draw travel graphs.

Work out the scales first. If \(6\) squares cover \(30\) minutes, one square is \(5\) minutes. Do the same for the distance axis.

Reading a point: go up from the time to the line, then across. A point between grid lines is read as part of a square.

The shape: up is moving away, flat is stopped, down is coming back.

Speed of a section \(=\) distance \(\div\) time for that section; steeper means faster. From minutes to km/h, divide by the minutes and multiply by \(60\). A downward section still has a positive speed.

Average speed \(=\) total distance \(\div\) total time, counting stops. On a return trip count the distance out and back.

Two travellers: where the lines cross, they are at the same place at the same time; the gap straight up is how far apart they are. Coming towards each other, they close the gap at the sum of their speeds.

Drawing one: plot the (time, distance) point at the end of each stage and join the points with straight lines.

A travel graph in four sections Distance from home against time of day from 08:00 to 10:00. A: up to 6 km at 08:30. B: flat to 08:45. C: up to 18 km at 09:15. D: back down to 3 km at 10:00. 0 3 6 9 12 15 18 21 08:00 08:30 09:00 09:30 10:00 A B C D Time of day Distance (km)
One square across is \(15\) min, one square up is \(3\) km. A: \(6\) km in \(30\) min, \(12\) km/h. B: stopped. C: \(12\) km in \(30\) min, \(24\) km/h (steepest). D: back \(15\) km in \(45\) min, \(20\) km/h. Average: \(33 \div 2 = 16.5\) km/h.

Speed from a section with the time in minutes:

SectionWorkingSpeed
6 km in 30 min\(6 \div 30 \times 60\)12 km/h
12 km in 30 min\(12 \div 30 \times 60\)24 km/h
15 km in 45 min\(15 \div 45 \times 60\)20 km/h
One square is not always one unit. Check both scales before reading anything.
Coming back is not a negative speed. \(20\) km home in \(25\) min is \(20 \div 25 \times 60 = 48\) km/h.

Finding the speed of a section

  1. Read the distance at the start and end of the section and subtract.
  2. Read the time taken.
  3. Divide distance by time, changing minutes to hours for km/h.
Example 1 — Two speeds
The graph shows a car journey.
(i) Find the speed in the first \(40\) minutes.
(ii) Find the speed in the last \(40\) minutes.
A car journey Distance in km against time in minutes: up to 30 km at 40 minutes, flat until 60 minutes, then up to 100 km at 100 minutes. 0 20 40 60 80 100 0 20 40 60 80 100 30 km Time (minutes) Distance (km)
Solution
\(30 \div \dfrac{2}{3}\)\(=\)\(45\)
\((100 - 30) \div \dfrac{2}{3}\)\(=\)\(105\)
45 km / h 105 km / h

(i) \(45\) km/h (\(40\) min \(= \dfrac{2}{3}\) h)
(ii) \(105\) km/h

Example 2 — Reading between grid lines
A car leaves at 09:00 and reaches \(30\) km at 09:30, stops until 09:45, then reaches \(60\) km at 10:30. (One square across is \(15\) min, one square up is \(10\) km.)
(i) How far had it gone at 09:15?
(ii) How long did it stop?
(iii) Find its speed on the last section.
Solution
\(60 - 30\)\(=\)\(30\)
\(30 \div 45 \times 60\)\(=\)\(40\)
15 km 15 min 40 km / h

(i) Halfway between \(10\) and \(20\): \(15\) km
(ii) \(15\) min
(iii) \(40\) km/h

Example 3 — Drawing a travel graph
A ferry sails \(21\) km to an island in \(45\) min, waits \(30\) min, then sails back in \(45\) min.
(i) Draw the travel graph.
(ii) Find the average speed for the whole trip.
Solution

Plot \((0, 0)\), \((45, 21)\), \((75, 21)\), \((120, 0)\) and join them.

A ferry trip Distance from the wharf: up to 21 km at 45 minutes, flat until 75 minutes, back to 0 at 120 minutes. 0 3 6 9 12 15 18 21 24 0 30 60 90 120 Time (minutes) Distance (km)
\(21 + 21\)\(=\)\(42\)
\(45 + 30 + 45\)\(=\)\(120 \text{ min}\)
\(42 \div 2\)\(=\)\(21\)
21 km / h

(ii) \(21\) km/h. Each sail is \(28\) km/h; the wait pulls the average down.

Example 4 — Metres and seconds
An athlete jogs, then sprints to the finish.
(i) Find her speed on each section in m/s.
(ii) Write the faster speed in km/h.
Jog then sprint Distance in metres against time in seconds: 100 m at 40 s, then 400 m at 100 s. 0 100 200 300 400 0 20 40 60 80 100 Time (s) Distance (m)
Solution
\(100 \div 40\)\(=\)\(2.5\)
\(300 \div 60\)\(=\)\(5\)
\(5 \times 3.6\)\(=\)\(18\)
2.5 m / s and 5 m / s 18 km / h

(i) Jog \(2.5\) m/s; sprint \(300\) m in \(60\) s \(= 5\) m/s
(ii) \(18\) km/h

Example 5 — Two travellers
Zoe walks from the pier and Hugo rides towards her from a lighthouse \(12\) km away.
(i) How far apart are they at 09:30?
(ii) When and where do they pass?
Two travellers Distance from the pier against time from 09:00: Zoe walks from 0 km to 8 km by 11:00; Hugo rides from 12 km back to 0 km by 10:30. The lines cross at 10:00, 4 km from the pier. 0 2 4 6 8 10 12 09:00 09:30 10:00 10:30 11:00 Zoe Hugo Time of day From the pier (km)
Solution

At 09:30 Zoe is \(2\) km from the pier and Hugo is \(8\) km.

\(8 - 2\)\(=\)\(6\)
6 km 1000,4 km from the pier

(i) \(6\) km
(ii) Where the lines cross: 10:00, \(4\) km from the pier

Common pitfalls

Assuming one square is one unit. Read the scale on each axis.
Using clock times as the time taken. From 09:30 to 10:15 is \(45\) min.
Leaving stops out of the average. Average speed = total distance ÷ total time.
Counting only one way on a return trip. Add the distance out and the distance back.

Frequently asked questions

What does a flat line on a travel graph mean?

The traveller is stopped: time passes but the distance does not change.

How do I find speed from a travel graph?

Divide the distance covered in a section by the time it took.

Which section is fastest?

The steepest one.

What does it mean when two lines cross?

The two travellers are at the same place at the same time.

How do I find the average speed for a whole trip?

Divide the total distance travelled by the total time, including stops.