For a particle moving along a line with displacement x at time t: velocity v = [frac dx/dt] and acceleration a = [frac dv/dt] = [frac d[sup 2]x/dt[sup 2]]. Speed is |v|.

The particle is at rest when v = 0, moving in the positive direction when v > 0, and speeding up when v and a have the same sign (slowing down when they have opposite signs).

Worked example. x = t³ − 6t² + 9t. Then v = 3t² − 12t + 9 = 3(t − 1)(t − 3), so it is at rest at t = 1 and t = 3, and a = 6t − 12.

Displacement is the change in position, x(end) − x(start); it can be negative. Distance covered counts the full length of the path, so you must find where the particle changes direction (where v = 0) and add the distances for each stage.

Worked example. x = t² − 4t for 0 ≤ t ≤ 5. v = 2t − 4 = 0 at t = 2. x(0) = 0, x(2) = −4, x(5) = 5. Displacement = 5 m; distance = 4 + 9 = 13 m.

Reverse the process: v = ∫ a dt and x = ∫ v dt. Each integration adds a constant, found from an initial condition such as the starting velocity or position.

∫ from t₁ to t₂ of v dt is the change in displacement. For the distance covered, split the integral where v changes sign and add the absolute values.

Worked example. a = 6t with v(0) = 2 and x(0) = 1. Then v = 3t² + 2 and x = t³ + 2t + 1.