Displacement s, velocity v and acceleration a form a chain. Differentiate to move from s to v to a, and integrate to move backwards from a to v to s.
This lesson is the start of kinematics of linear motion in SPM Additional Mathematics. It relies on differentiating powers of t, and integration for the backward direction.
What does each quantity tell you?
The three quantities answer three different questions about a particle moving on a line, measured from a fixed point O.
| Quantity | Meaning | Link |
|---|---|---|
| s (displacement) | Where it is relative to O | the starting quantity |
| v (velocity) | How fast and in which direction | v = ds/dt |
| a (acceleration) | How the velocity is changing | a = dv/dt |
Going backwards: v = ∫a dt and s = ∫v dt. Each integration adds a constant that comes from a given condition.
Worked example: find v and a at a given time
A particle moves in a straight line so that its displacement from O is s = t³ − 6t² + 9t metres after t seconds. Find s, v and a when t = 2.
Differentiate once. v = ds/dt = 3t² − 12t + 9.
Differentiate again. a = dv/dt = 6t − 12.
Substitute t = 2.
- s = 8 − 24 + 18 = 2 m.
- v = 12 − 24 + 9 = −3 m/s.
- a = 12 − 12 = 0 m/s².
The particle is 2 m from O, moving in the negative direction at 3 m/s, and at this instant the velocity is not changing.
The mistake that costs marks
The slip is to read a = 0 as “the particle is stationary”. It is tempting because zero sounds like stillness.
| Reading at t = 2 | Wrong | Right |
|---|---|---|
| a = 0 | The particle is at rest | The velocity is momentarily constant |
| v = −3 | The particle is slowing down | It moves in the negative direction |
| At rest when | a = 0 | v = 0, at t = 1 and t = 3 |
Here v = 3(t − 1)(t − 3), so the particle is at rest at t = 1 and t = 3. At t = 2 it is moving fastest in the negative direction, which is where the acceleration passes through zero.
Going backwards with conditions
If you know a, integrate twice and use conditions to fix each constant. Suppose a = 2t and the particle starts from rest at O (v = 0 and s = 0 when t = 0).
v = t² + c₁, and v = 0 at t = 0 gives c₁ = 0, so v = t².
s = t³/3 + c₂, and s = 0 at t = 0 gives c₂ = 0, so s = t³/3.
Practise both directions with your own functions using the word problem structure worksheet.
Check yourself
A particle has displacement s = 2t³ − 9t² + 12t metres. Find s, v and a when t = 3, and find when a = 0.
Answer
v = 6t² − 18t + 12 and a = 12t − 18.
At t = 3: s = 54 − 81 + 36 = 9 m, v = 54 − 54 + 12 = 12 m/s and a = 36 − 18 = 18 m/s².
For a = 0: 12t − 18 = 0, so t = 1.5 s.
Check: at t = 1.5, v = 13.5 − 27 + 12 = −1.5 m/s, so the particle is not at rest even though a = 0.
What to study next
The next step integrates a velocity function. Continue with finding displacement from a velocity function, then test the chapter with the kinematics 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 record which link slips.