Distance counts every metre you travelled. Displacement only cares how far you are from the start, in which direction. Once you read each graph with that in mind, the lines stop looking contradictory.
This page uses one original journey to show both ideas, and the graph-model and residual explorer and rate and energy graph interpreter let you practise reading gradient and area on other graphs.
Why do the two get mixed up?
In daily speech, “how far did you go” could mean either one. In Physics, the words are separate quantities with separate graphs, and questions test whether you can tell which graph you are looking at.
A quick rule: distance never decreases, displacement can. Distance is a scalar, so it has size only. Displacement is a vector, so it has size and direction.
Worked example: a walk to the shop
Aina walks 60 m east from home to a shop in 40 s. She stays there for 20 s. She then walks 60 m back home in 60 s.
Total time: 40 + 20 + 60 = 120 s.
Distance travelled: 60 + 60 = 120 m.
Final displacement: she ends at home, so 0 m.
Average speed: 120 m ÷ 120 s = 1.0 m/s.
Average velocity: 0 m ÷ 120 s = 0 m/s.
Now the graphs.
On the distance-time graph the line rises for 40 s, stays flat for 20 s, then rises again for 60 s. It never goes down. On the displacement-time graph the line rises to 60 m, stays flat, then falls back to 0 m.
The same flat section in both graphs means the same thing: she is stationary, because the gradient is zero.
Reading a velocity-time graph with a reversal
If your syllabus treats velocity as signed, a velocity-time graph can go below the axis. Suppose a trolley moves at +3 m/s for 4.0 s, then at −2 m/s for 2.0 s.
Forward part: 3 × 4.0 = +12 m.
Backward part: −2 × 2.0 = −4 m.
Displacement: 12 − 4 = 8 m from the start.
Distance: 12 + 4 = 16 m, because distance adds both parts as positive lengths.
The mistake to watch for
A student looks at Aina’s displacement-time graph, sees the line falling in the last section and writes “she is slowing down”.
The line is falling because she is moving back towards the start, not because her speed is dropping. Her return is slower (60 ÷ 60 = 1.0 m/s against 60 ÷ 40 = 1.5 m/s on the way there), but the size of the gradient shows that, not the fact that the line slopes down.
The fix: read the axis label, then ask two questions. Is the gradient steep or shallow, which gives the size of the speed? Is it positive or negative, which gives the direction?
Check yourself
1. An athlete runs one 400 m lap of a track in 50 s and finishes where she started. Give her average speed and average velocity.
Show answer
Average speed = 400 ÷ 50 = 8.0 m/s. Displacement is 0 m, so average velocity is 0 m/s.
2. A ball is thrown straight up 5.0 m and caught at the launch point. State the distance and displacement.
Show answer
Distance = 5.0 + 5.0 = 10 m. Displacement = 0 m, since it ends where it started.
3. A velocity-time graph shows +5 m/s for 2.0 s, then −5 m/s for 3.0 s. Find the displacement and the distance.
Show answer
First part: 5 × 2.0 = +10 m. Second part: −5 × 3.0 = −15 m. Displacement = 10 − 15 = −5 m, meaning 5 m behind the start. Distance = 10 + 15 = 25 m.
Where this leads next
Practise the separation in distinguishing displacement from total distance, then compare distance-time and speed-time graphs. The motion and graphs module gathers the rest, and the Physics terminology guide lists other pairs that cause the same trouble.
If graph questions still feel uncertain after you practise, a teacher can go through your own marked graphs in online one-to-one Physics tuition.