Displacement is the net change in position. Distance travelled is the total length of the path. They differ whenever the particle reverses direction, because backward motion cancels forward motion in displacement but still adds to distance.
The method comes straight from splitting an interval when signed areas change, applied to velocity. You also need calculating displacement from velocity.
How do the two quantities differ?
Picture a bus that drives 4 km east, then 4 km west back to the depot. Its displacement is 0 km because it ends where it started. Its distance travelled is 8 km.
In calculus, displacement is ∫ v dt, which lets forward and backward parts cancel. Distance is ∫ |v| dt, which treats every part as positive. In practice you find it by splitting where v = 0.
How to find distance travelled
- Solve v = 0 to find when the particle is at rest.
- Keep the times inside the interval. These are the split points.
- Integrate v over each section to get the displacement for that section.
- Take the positive value of each section.
- Add them. That total is the distance.
For displacement alone, integrate once over the whole interval.
Worked example
A particle moves with velocity v = 3t² − 12t + 9 m/s for 0 ≤ t ≤ 4. Find (a) the displacement and (b) the distance travelled.
Step 1, factorise: v = 3(t² − 4t + 3) = 3(t − 1)(t − 3). So v = 0 at t = 1 and t = 3, both inside the interval.
Step 2, integrate: the displacement from the start at time t is s = t³ − 6t² + 9t, since we measure from t = 0.
s(0) = 0, s(1) = 1 − 6 + 9 = 4, s(3) = 27 − 54 + 27 = 0, s(4) = 64 − 96 + 36 = 4.
Part (a), displacement: s(4) − s(0) = 4 − 0 = 4 m.
Part (b), distance by section:
- 0 to 1: s goes from 0 to 4, so 4 m forward.
- 1 to 3: s goes from 4 to 0, so 4 m backward.
- 3 to 4: s goes from 0 to 4, so 4 m forward.
Distance = 4 + 4 + 4 = 12 m.
Check: the sign of v confirms the directions. At t = 0, v = 9 (forward). At t = 2, v = 12 − 24 + 9 = −3 (backward). At t = 4, v = 48 − 48 + 9 = 9 (forward).
The mistake to watch for
A common slip is to give the displacement when the question asks for the distance.
Mistaken answer: the distance is s(4) − s(0) = 4 m.
The student integrated once over the whole interval. The forward and backward parts cancelled partly, so 4 m is the net change, not the length of the path.
The correction is to list the moments v = 0 first. If any of them lies strictly inside the interval and v changes sign there, the two answers will differ. Always underline the word in the question: distance, displacement, position or final position.
Check yourself
Try these on paper, then open each answer.
1. A particle has velocity v = 8 − 2t m/s for 0 ≤ t ≤ 6. Find the distance travelled.
Show answer
v = 0 at t = 4. The displacement function is s = 8t − t².
s(0) = 0, s(4) = 32 − 16 = 16, s(6) = 48 − 36 = 12.
From 0 to 4: 16 m forward. From 4 to 6: 4 m backward.
Distance = 16 + 4 = 20 m (the displacement is 12 m)
2. A particle has velocity v = t² − 4 m/s for 0 ≤ t ≤ 3. Find the displacement and the distance travelled.
Show answer
v = 0 at t = 2. The displacement function is s = t³/3 − 4t.
s(0) = 0, s(2) = 8/3 − 8 = −16/3, s(3) = 9 − 12 = −3.
Displacement = s(3) − s(0) = −3 m.
From 0 to 2: 16/3 m backward. From 2 to 3: −3 − (−16/3) = 7/3 m forward.
Distance = 16/3 + 7/3 = 23/3 m (7⅔ m)
3. A particle has velocity v = 2t + 1 m/s for 0 ≤ t ≤ 3. Without splitting anything, explain why the distance equals the displacement, and find it.
Show answer
For t ≥ 0, 2t + 1 is always positive, so the particle never reverses.
s = t² + t. s(3) − s(0) = 12 − 0.
Distance = displacement = 12 m
Where this leads next
Next, connecting position, velocity and acceleration links these integrals with differentiation in one chain. You can rehearse the signed arithmetic with the non-calculator working trainer and check your factorising with the quadratic structure explorer.
Questions that test distance and displacement together reward a steady routine more than speed. Our teachers can help you build that routine in online one-to-one Additional Mathematics tuition.