Displacement, velocity, and acceleration are linked by calculus. Differentiate to step down the chain s → v → a; integrate to climb back up a → v → s. Differentiation handles rates and “at rest”; integration recovers motion from a rate and, with limits, measures displacement and distance.
📘 What you need to know
Down by differentiating: v = dsdt and a = dvdt = d2sdt2.
Up by integrating: v = ∫ a dt and s = ∫ v dt.
“At rest” ⟺ v = 0; solve v(t) = 0 for the times.
Constant of integration: use an initial/boundary condition to find c.
Displacement between two times: ∫t1t2v dt (signed).
Differentiating moves s → v → a; integrating moves a → v → s. Acceleration is the second derivative of displacement.
The kinematics calculus linksv = dsdt, a = dvdt = d2sdt2, s = ∫ v dt, v = ∫ a dt✓ these relationships are in the formula booklet
🧠 “Differentiate down, integrate up”
Going down the chain s → v → a you differentiate; going upa → v → s you integrate. Each integration brings a “+ c” you must pin down with a condition.
Displacement vs distance with integrals
Displacement∫v dtSigned. Areas below the time axis subtract — the net change in position.
Distance∫ |v| dtAll areas add. Split where v = 0, or integrate the modulus on a GDC.
Key point: displacement and distance agree only if the particle never reverses. If v changes sign in the interval, you must use |v| (or split at the turning times) for distance.
🧭 Recipe — motion from a rate
Differentiate for v or a; integrate for v or s.
Find “+ c“ using an initial (“initially”, t = 0) or boundary condition.
“At rest”: solve v(t) = 0.
Displacement over [t1, t2]: ∫v dt.
Distance: ∫ |v| dt — use the GDC’s modulus graph.
Worked examples
Examples 1–2 use displacement s(t) = 2t3 − 27t2 + 84t; examples 3–5 use velocity v(t) = 8t3 − 12t2 − 2t with the particle starting at the origin.
GDC evaluates the definite integral and the modulus graph directly on Paper 2.
⚠ Common mistakes
Using v for distance when the particle reverses — distance needs |v|.
Dropping “+ c“ when integrating, or forgetting to evaluate it.
Mixing up the directions — differentiating when you should integrate.
Setting a = 0 for “at rest” instead of v = 0.
Confusing the second derivative: a = s″, not s′.
That wraps up Kinematics. The unit began with the language of motion — displacement, velocity, and acceleration as signed quantities, with speed and distance as their unsigned shadows — and read them off a velocity–time graph. Here calculus made the links exact: differentiate down the s → v → a chain, integrate up it, pin down the constant with an initial condition, and use definite integrals for displacement (v) and distance (|v|). The gradient-and-area intuition from the graph and the exact calculus are two views of the same motion.
Need help with Kinematics?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
