Travel Graphs
A travel graph, or distance–time graph, shows how far someone is from a starting point as time passes. The gradient of each section is the speed, a flat section means stopped, and a curved section means the speed is changing. In NSW Year 10 Advanced maths you read and draw travel graphs and use them to compare journeys (NESA outcome MA5‑RAT‑P‑02).
You will find speeds in km/h and m/s, total distances, rest times and average speeds, and draw graphs from tables and descriptions. Journeys include a bus route, a train, a swimmer and a marble on a ramp. You will find where and when two travellers meet, such as a walker and a cyclist or a bus and a car. Harder questions form an equation for a round trip and show why average speed is not the mean of two speeds. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.
Every question with a fully worked solution.
- Distance Time Graphs Watch
Theory
A travel graph (distance-time graph) shows how far something is from a starting point as time passes. Its gradient is the speed: steeper is faster, flat is stopped, and a falling line is a return trip. This Year 10 Advanced maths page covers reading and drawing travel graphs, average speed, unit changes, curved graphs, and two travellers on one graph.
A distance-time graph has time on the horizontal axis and distance from a fixed point on the vertical axis. Each point tells you where the traveller is at that time.
The gradient of a straight section is \(\dfrac{\text{rise}}{\text{run}} = \dfrac{\text{distance}}{\text{time}}\), which is the speed. A horizontal section means the traveller is stationary. A section sloping down means the distance from the start is shrinking: the traveller is heading back.
A curved section means the speed is changing: getting steeper means speeding up, flattening out means slowing down.
Average speed is the total distance travelled divided by the total time taken, stops included. It is usually not the mean of the separate speeds.
It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
For any section of a journey:
| Change | Rule |
|---|---|
| minutes to hours | divide by \(60\): \(45\) min \(= 0.75\) h, \(20\) min \(= \dfrac{1}{3}\) h |
| m/s to km/h | multiply by \(3.6\) |
| km/h to m/s | divide by \(3.6\) |
How to read a travel graph
- Check the scales on both axes: what one grid square is worth, and whether time is in minutes, hours or clock times.
- Split the graph into straight sections. Read the distance change and time taken for each.
- Convert the time to hours (or seconds) to match the units of speed asked for.
- Divide distance by time for each section; use totals for average speed.
Problem types
- Matching a graph to a story: constant speed is a straight line, a stop is horizontal, faster is steeper, returning slopes down to \(0\).
- Drawing a graph: work out each section's time with time \(=\) distance \(\div\) speed, plot the end of each section, and join with straight lines. A table of values is plotted point by point.
- Fastest section: the steepest one. Compare gradients, not lengths of line.
- Curved graphs: describe the speed as increasing or decreasing; for an average speed, use the gradient of the chord between the two points.
- Two travellers: draw both on one graph. Where the lines cross they are at the same place at the same time. Algebra confirms it: write each distance in terms of the same \(t\), allowing for a later start (\(t - 0.5\) for half an hour later) or a stop, and solve. Travellers moving towards each other use \(d\) and \(D - d\).
- Gap between travellers: at a given time, find both distances and subtract.
- Unknown speed: add the times \(\dfrac{d}{v}\) for each part and set equal to the total time. Clearing fractions gives a quadratic; reject a negative speed, or one that makes another speed negative (Example 4).
- Required speed for a target average: write the average speed as total distance over total time with the unknown speed in it, then solve.
(i) her speed on the way out
(ii) her speed on the way home
(iii) her average speed for the whole trip, in km/h.
Read each section: out \(16\) km in \(48\) min, stopped \(48\) to \(60\) min, home \(16\) km in \(40\) min. Change minutes to hours first.
| \(\text{(i) speed}\) | \(=\) | \(16 \div \dfrac{48}{60} = 16 \div 0.8 = \mathbf{20}\text{ km/h}\) |
| \(\text{(ii) speed}\) | \(=\) | \(16 \div \dfrac{40}{60} = 16 \times \dfrac{3}{2} = \mathbf{24}\text{ km/h}\) |
(iii) Average speed uses the total distance (out and back) and the total time, including the stop.
| \(\text{average speed}\) | \(=\) | \(\dfrac{32}{100 \div 60}\) |
| \(\) | \(=\) | \(32 \times \dfrac{60}{100} = \mathbf{19.2}\text{ km/h}\) |
(i) Describe the motion.
(ii) Find the average speed between \(t = 2\) and \(t = 6\) in m/s and in km/h.
(i) The graph gets steeper, so the gradient (the speed) is increasing: the go-kart is speeding up.
(ii) Average speed is the gradient of the chord from \((2,\ 6)\) to \((6,\ 54)\):
| \(\text{average speed}\) | \(=\) | \(\dfrac{54 - 6}{6 - 2}\) |
| \(\) | \(=\) | \(\dfrac{48}{4} = \mathbf{12}\text{ m/s}\) |
| \(12 \times 3.6\) | \(=\) | \(\mathbf{43.2}\text{ km/h}\) |
\(1\) m/s \(= 3600\) m per hour \(= 3.6\) km/h, so multiply by \(3.6\).
Let \(t\) be the time in hours after 9:00 am. The car starts \(40\) min \(= \dfrac{2}{3}\) h later, so it has been driving for \(t - \dfrac{2}{3}\) hours.
| \(\text{cyclist: } d\) | \(=\) | \(18t\) |
| \(\text{car: } d\) | \(=\) | \(54\left(t - \dfrac{2}{3}\right) = 54t - 36\) |
They meet when the distances are equal:
| \(18t\) | \(=\) | \(54t - 36\) |
| \(36t\) | \(=\) | \(36\) |
| \(t\) | \(=\) | \(1\) |
The car catches the cyclist at 10:00 am, \(18 \times 1 = \mathbf{18}\) km from the town. On the graph this is where the two lines cross.
Time \(=\) distance \(\div\) speed. Add the two times:
| \(\dfrac{24}{v} + \dfrac{24}{v + 4}\) | \(=\) | \(5\) |
| \(24(v + 4) + 24v\) | \(=\) | \(5v(v + 4)\) |
| \(48v + 96\) | \(=\) | \(5v^2 + 20v\) |
| \(5v^2 - 28v - 96\) | \(=\) | \(0\) |
| \((v - 8)(5v + 12)\) | \(=\) | \(0\) |
So \(v = 8\) or \(v = -\dfrac{12}{5}\). A speed must be positive, so \(v = \mathbf{8}\) km/h.
| \(\text{out}\) | \(=\) | \(24 \div 8 = \mathbf{3}\text{ h}\) |
| \(\text{back}\) | \(=\) | \(24 \div 12 = \mathbf{2}\text{ h}\) |
Check: \(3 + 2 = 5\) h. Average speed \(= 48 \div 5 = 9.6\) km/h, not the mean of \(8\) and \(12\), because more time is spent at the slower speed.
Common pitfalls
Frequently asked questions
How do you find speed from a distance-time graph?
Find the gradient of the section: divide the change in distance by the change in time. Convert the time to hours for km/h or to seconds for m/s.
What does a horizontal line on a travel graph mean?
The distance is not changing, so the object is stopped. The length of the flat section is how long it stopped for.
How do you calculate average speed?
Divide the total distance travelled by the total time taken, including any stops. Do not just average the different speeds.
How do you convert m/s to km/h?
Multiply by 3.6, because 1 metre per second is 3600 metres per hour, which is 3.6 kilometres per hour. Divide by 3.6 to go back.
How do you find when two people meet on a travel graph?
Read the time and distance where the two lines cross. To check with algebra, write each person's distance as an expression in the same time variable and solve the equation formed by setting them equal.