Putting the tools together
Full kinematics problems mix all the ideas. Differentiate to move from displacement to velocity to acceleration; integrate to move the other way. The particle is at rest when v = 0, its speed is greatest when a = 0, and it changes direction wherever v changes sign.
Key formula /
v = ds/dt, a = dv/dt; s = ∫ v dt, v = ∫ a dt. Total distance over an interval where v changes sign = sum of the magnitudes of the displacements in each part.
Distance versus displacement
When a particle reverses direction, its net displacement can be small even though it has travelled far. To find total distance, find where v = 0, work out the displacement in each stage, then add their magnitudes.
A reliable strategy is to write down s, v and a as they are needed, solve v = 0 to find the rest points and the moments the motion reverses, and only then decide whether the question is asking for displacement or for total distance. Sketching a rough velocity-time picture first helps you see exactly where the motion changes direction before any integration is carried out.
Worked example
A particle has v = 6 − 2t m s⁻¹. It is at rest when v = 0, at t = 3 s. From 0 to 3 s, s = [6t − t²] = 18 − 9 = 9 m. From 3 to 5 s, s = (30 − 25) − 9 = −4 m, a distance of 4 m. Total distance = 9 + 4 = 13 m.
Remember
- Direction change → v = 0.
- Maximum speed/height → a = 0 (or v = 0 for height).
- Total distance adds the magnitudes stage by stage.