Displacement is the definite integral of velocity. If a particle has velocity v(t), its change in position from t = a to t = b is the integral of v from a to b. The sign matters: a positive displacement means the particle ended up in the positive direction from where it started.
This lesson applies the area idea from finding area between a curve and an axis to a velocity-time graph. The area under the graph is the displacement, and the same splitting rule from the previous lesson applies when velocity changes sign.
How does the area link to motion?
Distance is speed × time when speed is constant. A velocity-time graph draws velocity on the vertical axis, so a thin strip of the graph has height v and width dt, and its area is velocity × time, which is a small displacement.
Adding all the strips gives displacement = ∫ v dt. Integrating without limits and adding a constant gives position instead: s = ∫ v dt + c.
How to set out a displacement question
- Write the velocity function and the time interval.
- Integrate term by term to get F(t).
- Substitute the limits as F(upper) − F(lower) for displacement.
- For position, use the given information, such as s = 5 when t = 0, to find the constant.
- State the answer with units and a sign if direction matters.
Worked example
A particle moves in a straight line with velocity v = 2t + 3 m/s for t ≥ 0. At t = 0 it is 5 m from the origin O, on the positive side.
Find (a) the displacement from t = 0 to t = 4, and (b) its position at t = 4.
Part (a), integrate: F(t) = t² + 3t.
F(4) = 16 + 12 = 28 and F(0) = 0.
Displacement = 28 m.
Part (b), position: s = t² + 3t + c. At t = 0, s = 5, so c = 5.
At t = 4: s = 16 + 12 + 5 = 33 m from O.
Check: the velocity is 3 m/s at the start and 11 m/s at the end, so the average is 7 m/s. Over 4 seconds that gives 7 × 4 = 28 m. The position is that displacement added to the starting 5 m, which gives 33 m.
The mistake to watch for
A common slip is to multiply a single velocity by the time, as if velocity were constant.
Mistaken answer: at t = 4, v = 11, so displacement = 11 × 4 = 44 m.
The student used s = vt with the final velocity. But the velocity changes from 3 to 11 during the four seconds.
The correction is to integrate. The formula s = vt only works when velocity is constant, and here it is not. Whenever velocity is given as a formula in t, integrate.
Check yourself
Try these on paper, then open each answer.
1. A particle has velocity v = 6t² m/s. Find its displacement from t = 1 to t = 3.
Show answer
F(t) = 2t³. F(3) = 54 and F(1) = 2.
54 − 2 = 52 m
2. A particle has velocity v = 8 − 2t m/s. Find its displacement from t = 0 to t = 6.
Show answer
F(t) = 8t − t². F(6) = 48 − 36 = 12 and F(0) = 0.
Displacement = 12 m
Check: the velocity is zero at t = 4. The particle moves forward 16 m in the first 4 seconds, then back 4 m, so the net is 12 m.
3. A particle has velocity v = 4t − 3 m/s and is 2 m from O when t = 1. Find its position when t = 3.
Show answer
s = 2t² − 3t + c. At t = 1: 2 − 3 + c = 2, so c = 3.
At t = 3: s = 18 − 9 + 3 = 12 m from O.
Check by displacement: from 1 to 3 it is (18 − 9) − (2 − 3) = 10, and 2 + 10 = 12.
Where this leads next
The next lesson looks at distance travelled versus displacement, which matters when the particle turns round. After that, connecting position, velocity and acceleration puts the integrals and derivatives together in one chain.
If motion questions feel like a different subject from calculus, that gap is worth closing early. Our teachers can link the two using your own worked solutions in online one-to-one Additional Mathematics tuition.