When velocity is given as a function of time, integrating it gives the position s, with a constant that comes from a known position. A definite integral of the velocity gives the displacement between two times instead.
This lesson is part of kinematics of linear motion in SPM Additional Mathematics. It builds on relating displacement, velocity and acceleration.
Worked example: position and displacement
A particle moves in a straight line with velocity v = 3t² − 4t m/s. When t = 1, its displacement from O is 2 m. Find (a) its displacement from O when t = 3, and (b) its displacement between t = 1 and t = 3.
Find s as a function of t. s = ∫ (3t² − 4t) dt = t³ − 2t² + c.
Use t = 1, s = 2: 1 − 2 + c = 2, so c = 3. Then s = t³ − 2t² + 3.
(a) Position at t = 3. s = 27 − 18 + 3 = 12 m from O.
(b) Displacement from t = 1 to t = 3. Use the definite integral, which needs no constant:
∫ (3t² − 4t) dt from 1 to 3 = [t³ − 2t²] from 1 to 3 = (27 − 18) − (1 − 2) = 9 + 1 = 10 m.
Reconciling the two answers
The position at t = 3 is 12 m and the position at t = 1 was 2 m. The change is 12 − 2 = 10 m, which matches part (b).
This is the check you can run every time: displacement over an interval equals the final position minus the initial position.
The mistake that costs marks
The slip is to give the position when the question asks for displacement, or the reverse. Both use the same integration, so the working looks right.
| Wording | Meaning | Answer here |
|---|---|---|
| “displacement from O when t = 3” | Position at t = 3 | 12 m |
| “displacement between t = 1 and t = 3” | Change in position | 10 m |
| “total distance” | Different skill, needs splitting | Not found by either |
Underline the times and the reference point in the question before you start. The phrase “from O” asks for position, and “between” or “during” asks for a change.
Using a definite integral directly
If only the displacement between two times is needed, skip finding s altogether. Write the definite integral, integrate, and subtract.
If the question gives no condition for the constant, a definite integral is probably what it wants.
Check yourself
A particle has velocity v = 4t − 2 m/s. At t = 0 its displacement from O is 5 m. Find its displacement from O at t = 3, and the displacement during the interval t = 1 to t = 3.
Answer
s = 2t² − 2t + c. Using t = 0, s = 5 gives c = 5, so s = 2t² − 2t + 5.
At t = 3: s = 18 − 6 + 5 = 17 m from O.
For the interval: ∫ (4t − 2) dt from 1 to 3 = [2t² − 2t] from 1 to 3 = (18 − 6) − (2 − 2) = 12 m.
Check: s(1) = 2 − 2 + 5 = 5, and 17 − 5 = 12.
What to study next
A displacement of zero can hide a long journey. Continue with distinguishing total distance from displacement, then try the kinematics practice set.
The integration side is covered in evaluating definite integrals. For a teacher to go through your working, see online one-to-one Additional Mathematics tuition.