Key Idea: Kinematics links position, velocity and acceleration with calculus. Differentiate to go down the chain, integrate to come back up, and keep displacement and distance apart.
| Want | Do |
|---|---|
| v or a from s | Differentiate |
| v or s from a | Integrate, find + C |
| At rest or turning back | Solve v = 0 |
| Distance travelled | ∫|v| dt, split where v = 0 |
Displacement vs distance
- Displacement = ∫ₐᵇ v dt. It is signed: backward motion subtracts.
- Distance = ∫ₐᵇ |v| dt counts every metre and is always ≥ 0.
- By hand: find where the particle stops, split there, integrate each piece, and add the sizes.
Displacement can be small even after a long trip.
v(t) = t² − 6t + 8. When is the cyclist at rest? t = 2 and t = 4.
What is the displacement from t = 0 to t = 5? 20/3 m.
Why is the distance travelled more than 20/3 m? v is negative between t = 2 and t = 4, so the cyclist goes back, and those metres count too.
Given a(t), how do you get v(t)? Integrate, then find C from a known value.