Displacement is the net change in position, and total distance is the whole length of path travelled. They differ whenever the particle turns round, so distance needs the journey split at each turning point.
This lesson is part of kinematics of linear motion in SPM Additional Mathematics. It builds on finding displacement from a velocity function.
Worked example: one journey, two answers
A particle moves in a straight line with velocity v = t² − 6t + 8 m/s for 0 ≤ t ≤ 5. Find its displacement and its total distance in the 5 seconds.
Step 1: integrate. F(t) = t³/3 − 3t² + 8t.
Step 2: displacement. F(5) = 125/3 − 75 + 40 = 125/3 − 35 = 20/3, and F(0) = 0. Displacement = 20/3 m.
Step 3: turning points. v = (t − 2)(t − 4), so v = 0 at t = 2 and t = 4. Both are inside the interval.
| Interval | Sign of v | F at end | Change |
|---|---|---|---|
| 0 to 2 | positive | F(2) = 20/3 | +20/3 |
| 2 to 4 | negative | F(4) = 16/3 | −4/3 |
| 4 to 5 | positive | F(5) = 20/3 | +4/3 |
Here F(2) = 8/3 − 12 + 16 = 20/3, and F(4) = 64/3 − 48 + 32 = 16/3.
Step 4: add the sizes. Distance = 20/3 + 4/3 + 4/3 = 28/3 = 9⅓ m.
The displacement is 6⅔ m, but the particle travelled 9⅓ m. The difference is twice the distance it went backwards, 2 × 4/3.
The mistake that costs marks
The slip is to report F(5) − F(0) as the distance. The positive and negative parts partly cancel, so the number is smaller than the path length.
| Step | Wrong | Right |
|---|---|---|
| Interval | One integral, 0 to 5 | Split at t = 2 and t = 4 |
| Values | 20/3 | +20/3, −4/3, +4/3 |
| Distance | 20/3 m | 28/3 m |
A second slip is to split at the wrong times, for example solving a = 0 instead of v = 0. Direction changes where the velocity is zero, not the acceleration.
A method you can reuse
- Integrate v to get F(t), the position up to a constant.
- Solve v = 0 and keep roots inside the interval.
- Evaluate F at the start, each root and the end.
- Take the positive size of each difference and add them for distance.
- Add the signed differences for displacement.
If you want a picture of the journey, draw the velocity graph and shade the regions above and below the time axis. Distance is the total shaded area.
Check yourself
A particle has velocity v = 6 − 2t m/s for 0 ≤ t ≤ 5. Find the displacement and the total distance.
Answer
F(t) = 6t − t². Then v = 0 at t = 3, inside the interval.
F(0) = 0, F(3) = 18 − 9 = 9 and F(5) = 30 − 25 = 5.
Displacement = 5 − 0 = 5 m.
Distance = |9 − 0| + |5 − 9| = 9 + 4 = 13 m.
Check: the particle goes 9 m forward, then returns 4 m, so it is 5 m from its start.
What to study next
Understand what happens at each turning point in interpreting direction changes and instantaneous rest. Then try the kinematics practice set.
The area idea behind this lesson is in using integration for displacement and distance. For a teacher to go through your working, see online one-to-one Additional Mathematics tuition.