Key Idea: Kinematics applies calculus to motion in a straight line: differentiate down to find velocity and acceleration, integrate up to recover them. It appears on both papers, often as a multi-part 'find when…' question.
🚗 The s ↔ v ↔ a chain
- displacement (position) at time t
- velocity — the derivative of displacement
- acceleration — the derivative of velocity
| You have… | Differentiate ↓ to get | Integrate ↑ to get |
|---|---|---|
| Displacement s | Velocity v = ds/dt | — |
| Velocity v | Acceleration a = dv/dt | Displacement s = ∫v dt (+ C) |
| Acceleration a | — | Velocity v = ∫a dt (+ C) |
Every integration up the chain adds a + C that an initial condition (a value at t = 0) pins down. The key moments: at rest (or changes direction) when v = 0; max/min velocity when a = 0; displacement = ∫v dt (signed) but total distance = ∫|v| dt (split where v = 0, add the magnitudes).
✏️ IB-style worked examples
IB-style question — velocity and acceleration by differentiating
A particle moves with displacement s = t³ − 5t² + 3t metres (t ≥ 0). Find its velocity and acceleration at t = 2.
Step by step:
Differentiate s for velocity, then again for acceleration.
Substitute t = 2 into each.
At t = 2: velocity −5 m/s, acceleration 2 m/s².
IB-style question — integrate up with an initial condition
A particle has acceleration a = 6t − 4 m/s² and velocity 5 m/s at t = 0. Find an expression for v(t).
Step by step:
Integrate a to get v — don't forget the + C.
Use v(0) = 5 to find C.
v(t) = 3t² − 4t + 5 m/s.
IB-style question — at rest, and maximum velocity
Particle A has velocity v = 2t² − 10t + 12. Find when it is at rest. Particle B has velocity v = 9 + 6t − 3t²; find its maximum velocity.
Step by step:
At rest: set v = 0 and factor.
Max velocity: set a = 0, i.e. dv/dt = 0, then read off v.
A is at rest at t = 2 s and t = 3 s; B's maximum velocity is 12 m/s (at t = 1 s).
IB-style question — displacement vs total distance
A particle has velocity v = t² − 4t + 3 m/s. Find the displacement and the total distance travelled from t = 0 to t = 3.
Step by step:
Displacement = ∫v dt over [0, 3].
Distance: v = 0 at t = 1 and t = 3. Integrate each leg and add magnitudes.
Displacement = 0 m; total distance travelled = 8/3 ≈ 2.67 m.
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Important: When you integrate up, you must add + C and use an initial condition to find it — a definite-integral shortcut skips this. And for 'how far did it travel', use ∫|v| (distance), not ∫v (displacement) — they differ whenever the particle changes direction.
Tap each card to reveal the answer.
How do you get velocity from displacement? Differentiate: v = ds/dt — and a = dv/dt = d²s/dt².
How do you get displacement from velocity? Integrate: s = ∫v dt (+ C) — find C from an initial condition.
What does 'the particle is at rest' mean? v = 0 — solve the velocity equation for t (it may also change direction here).
When is the velocity a maximum? When a = 0 — velocity is extreme where its derivative (acceleration) is zero.
Displacement vs distance over [0, 4]? Displacement = ∫v dt; distance = ∫|v| dt — split where v = 0 and add magnitudes.
a = 4 m/s² constant, v(0) = 3. Find v(t). v = 4t + 3 — integrate a, then C = 3 from v(0) = 3.
Exam tips
- Differentiate DOWN the chain (s → v → a); integrate UP (a → v → s) and add + C.
- Find C from an initial condition — the value of v or s at t = 0.
- 'At rest' or 'changes direction' → solve v = 0. 'Max/min velocity' → solve a = 0.
- Displacement = ∫v dt (signed); total distance = ∫|v| dt (split where v = 0, add magnitudes).
- On Paper 2 use fnInt: ∫|v| for distance, ∫v for displacement — read which one is asked.