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Year 12 Maths Standard 2 (2027) Ratios and rates

Distance-Time Graphs

20 practice questions 0 video lessons Theory + worked examples

Learn how to read a distance-time (travel) graph in NSW Year 12 Mathematics Standard 2. You read how far an object is from a starting point over time, and find its speed from the gradient of each straight segment — steeper means faster, and a flat segment means the object is stationary.

You will interpret a whole journey — travelling out, stopping, and returning home — compare the speeds of different segments or travellers, and calculate the average speed as total distance travelled divided by total time. These are core Standard 2 skills for describing real journeys made by cars, trains, cyclists and walkers.

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Theory

A distance-time (travel) graph shows how far an object is from a starting point over time. This Year 12 Standard 2 (NSW) guide shows how to read distances and times, find speed from the gradient of each segment, spot a stationary (flat) part and a return, and work out the average speed for the whole journey.

A distance-time graph (or travel graph) shows how far an object is from a starting point as time passes. Time \(t\) runs along the horizontal axis and distance \(d\) up the vertical axis, so the height of the graph is the distance from the start at that moment.

The gradient (steepness) of each straight segment is the object’s speed — the rise (distance) divided by the run (time). A steeper segment means a faster speed; a horizontal segment means the object is stationary (stopped); and a segment sloping downward means it is returning toward the start.

The average speed for a whole journey is the total distance travelled divided by the total time taken. This is a core Year 12 Mathematics Standard 2 (NSW) skill for reading real journeys made by cars, trains, cyclists and walkers.

A journey out, a rest, then back homeDistance-time graph rising from (0,0) to (2,60), flat to (3,60), then back down to (5,0): travel out, stop, return. t d 1 2 3 4 5 20 40 60
A journey: rising \(=\) travelling out, flat \(=\) stopped, falling \(=\) returning home.
Two cyclists P and Q on the same axesTwo distance-time lines: P straight from (0,0) to (4,80); Q steeper, from (0,0) to (2,80) then flat to (4,80). The steeper line Q is faster. t d 1 2 3 4 20 40 60 80 P Q
Steeper \(=\) faster: cyclist \(Q\) is quicker than \(P\) at first, then rests.

The speed during any straight segment is its gradient — the rise (distance) over the run (time):

\[\text{speed}=\dfrac{\text{rise}}{\text{run}}=\dfrac{\text{distance}}{\text{time}}\]
speed=distancetime

The average speed for the whole journey uses the totals, not the speed of any single segment:

\[\text{average speed}=\dfrac{\text{total distance travelled}}{\text{total time taken}}\]
average speed=total distancetotal time
Flat means stopped. A horizontal segment has gradient \(0\), so the speed is \(0\) — the object is stationary, sitting at a fixed distance from the start. Its width along the time axis is how long the stop lasts.

How to read a distance-time graph

  1. Read a distance or time. Go up from the time on the horizontal axis to the graph, then across to the distance axis (or the reverse).
  2. Find a speed. For a straight segment, speed \(=\dfrac{\text{rise}}{\text{run}}\) \(=\) distance travelled \(\div\) time taken. Steeper means faster.
  3. Spot a stop. A horizontal segment means the object is stationary; its width is how long it is stopped.
  4. Spot a return. A segment sloping down means the object is heading back toward the start (its distance is decreasing).
  5. Average speed. Add up the total distance travelled (count every leg, including the return) and divide by the total time.
Example 1 — Read a distance and a speed
The graph shows a train’s distance \(d\) (km) from the station over time \(t\) (hours). Find its distance after \(1\) hour, and its speed over the first \(2\) hours.
Solution

The first segment rises \(120\) km in \(2\) h, so it climbs at a steady speed.

Example 1 - a train's journeyDistance-time graph rising from (0,0) to (2,120), flat to (3,120), then up to (5,200). t d 1 2 3 4 5 40 80 120 160 200
\(\text{speed}\)\(=\)\(\dfrac{\text{rise}}{\text{run}}=\dfrac{120}{2}\)
\(\text{speed}\)\(=\)\(60\text{ km/h}\)
\(\text{after }1\text{ h}\)\(:\)\(d=60\times 1=60\text{ km}\)
speed=60
Example 2 — How long stationary
The graph shows a delivery van. For how long is the van stationary, and what happens in the last segment?
Solution

The van is stationary where the graph is horizontal (no change in distance).

Example 2 - a delivery van's stopDistance-time graph up to (2,40), flat to (5,40), then up to (6,70). t d 1 2 3 4 5 6 10 20 30 40 50 60 70
\(\text{stop}\)\(=\)\(5-2=3\text{ h}\)
\(\text{last leg}\)\(:\)\(40\to 70\text{ km}=30\text{ km in }1\text{ h}\)
\(\text{speed}\)\(=\)\(30\text{ km/h}\)
stop=3

The van is stationary for \(3\) h, then travels a further \(30\) km at \(30\) km/h.

Example 3 — Which segment is fastest
The journey is made of three straight segments. During which segment is the object travelling fastest?
Solution

The steepest segment has the greatest speed — compare each gradient.

Example 3 - three segmentsDistance-time graph through (0,0),(2,40),(3,100),(5,120); the middle segment is steepest. t d 1 2 3 4 5 20 40 60 80 100 120
\(0\text{--}2\text{ h}\)\(:\)\(\dfrac{40}{2}=20\text{ km/h}\)
\(2\text{--}3\text{ h}\)\(:\)\(\dfrac{100-40}{1}=60\text{ km/h}\)
\(3\text{--}5\text{ h}\)\(:\)\(\dfrac{120-100}{2}=10\text{ km/h}\)
60 km/h

Fastest between \(2\) h and \(3\) h, at \(60\) km/h.

Example 4 — Out and back: average speed
A cyclist rides to a lookout, rests, then rides home. Find the total distance travelled and the average speed for the whole trip.
Solution

Add every leg — the return counts as extra distance, it does not cancel.

Example 4 - a cyclist there and backDistance-time graph up to (2,60), flat to (3,60), then back down to (6,0). t d 1 2 3 4 5 6 20 40 60
\(\text{total distance}\)\(=\)\(60+60=120\text{ km}\)
\(\text{total time}\)\(=\)\(6\text{ h}\)
\(\text{average speed}\)\(=\)\(\dfrac{120}{6}=20\text{ km/h}\)
average speed=20

The trip is \(120\) km in total, at an average of \(20\) km/h.

Common pitfalls

Flat is not “home”. A horizontal segment means the object is stationary at a fixed distance from the start — usually not zero. It has only gone home if the line is flat along the \(d=0\) axis.
Down means returning, not slowing. A segment sloping downward shows the distance from the start falling, so the object is heading back. Speed is the steepness of the line, not its direction.
Average speed uses totals. Divide the total distance travelled by the total time — not the average of the segment speeds. On a there-and-back trip, add both legs.

Frequently asked questions

How do you find speed from a distance-time graph?

Speed is the gradient of the segment: the rise (distance travelled) divided by the run (time taken). For example, a segment that rises 120 km over 2 hours has a speed of 120 divided by 2, which is 60 km per hour.

What does a flat (horizontal) line mean on a travel graph?

A horizontal segment means the object is stationary, or stopped. Its distance from the start is not changing, so its speed is zero. The width of the flat part along the time axis tells you how long the object is stopped.

What does a line sloping downward mean?

A downward-sloping segment means the object is returning toward the start, because its distance from the start is decreasing. It does not mean the object is slowing down; speed is shown by how steep the line is, not by its direction.

How do you find the average speed for a whole journey?

Add up the total distance travelled during the whole journey and divide by the total time taken. Do not average the separate segment speeds. On a there-and-back trip, count both the outward and the return distance.

Does a steeper line mean a faster speed?

Yes. The steeper the segment, the greater the gradient, so the faster the object is moving. A gentle slope is slow, a steep slope is fast, and a flat line is stopped.

What is the difference between distance travelled and distance from the start?

The height of the graph is the distance from the start. The distance travelled is how far the object has actually moved, adding every leg. On an out-and-back trip the object can end 0 km from the start yet have travelled a large total distance.