Displacement is the integral of velocity over time, and it measures net change in position. Total distance needs the integral split wherever the particle stops and turns around, because reversing cancels the earlier movement.
This lesson is part of SPM Additional Mathematics integration. The motion side is covered in kinematics of linear motion.
Worked example: zero displacement, long distance
A particle moves in a straight line with velocity v = 6t − 3t² m/s for 0 ≤ t ≤ 3. Find its displacement and its total distance.
Step 1: displacement. ∫ (6t − 3t²) dt from 0 to 3 = [3t² − t³] from 0 to 3 = (27 − 27) − 0 = 0 m.
The particle ends where it started. That does not mean it stayed still.
Step 2: find the turning point. v = 3t(2 − t) = 0 at t = 0 and t = 2. So the velocity is positive for 0 < t < 2 and negative for 2 < t ≤ 3.
Step 3: integrate each section. F(t) = 3t² − t³, with F(0) = 0, F(2) = 12 − 8 = 4 and F(3) = 0.
- 0 to 2: F(2) − F(0) = 4, moving forward.
- 2 to 3: F(3) − F(2) = −4, moving back.
Step 4: add the sizes. Total distance = 4 + 4 = 8 m.
The mistake that costs marks
The slip is using the displacement as the distance. Here one integral from 0 to 3 gives 0, and the working is neat, but the question asked for distance.
| Step | Wrong | Right |
|---|---|---|
| Interval | One integral, 0 to 3 | Split at t = 2 |
| Values | 0 | +4 and −4 |
| Distance | 0 m | 4 + 4 = 8 m |
The same idea appears as area in finding area between a curve and an axis. Distance is the area between the velocity graph and the time axis, with all parts counted as positive.
Finding a position from an initial condition
When the question gives a position at a particular time, use it to find the constant. If s = 2 when t = 1 and v = 3t² − 4t, then s = t³ − 2t² + c, so 1 − 2 + c = 2 and c = 3.
The position at any time is now s = t³ − 2t² + 3. A definite integral between two times never needs this constant.
A method you can reuse
- Write what the question asks for: displacement, distance or position.
- Solve v = 0 and note the times inside the interval.
- Integrate v, then evaluate over each section.
- For distance, take the positive size of each section and add. For displacement, add the signed values.
- State units and whether the answer is a displacement, a distance or a position.
Check yourself
A particle has velocity v = 2t − 8 m/s for 0 ≤ t ≤ 6. Find the displacement and the total distance.
Answer
F(t) = t² − 8t. Then v = 0 at t = 4.
F(0) = 0, F(4) = 16 − 32 = −16 and F(6) = 36 − 48 = −12.
Displacement = F(6) − F(0) = −12 m.
Distance = |F(4) − F(0)| + |F(6) − F(4)| = 16 + |−12 − (−16)| = 16 + 4 = 20 m.
What to study next
Go deeper on the motion side with distinguishing total distance from displacement. Then try the mixed integration practice set.
For a teacher to go through motion questions with you, see online one-to-one Additional Mathematics tuition. The mistake log helps you track repeated slips.