Reversing the derivatives
Because differentiation takes s to v to a, integration takes a back to v and v back to s. Each integration introduces a constant, which is found from a known value (usually the value at t = 0).
Key formula /
v = ∫ a dt + c and s = ∫ v dt + c. Displacement between times t₁ and t₂ is the definite integral ∫ v dt from t₁ to t₂. Distance uses the absolute value where v changes sign.
Definite integrals for displacement
To find the displacement over an interval, evaluate the definite integral of v. If the velocity keeps one sign over the interval, this equals the distance travelled; if v changes sign, split the interval and add the magnitudes.
Never omit the constant of integration. Without it the velocity or displacement is known only up to a fixed shift, and the initial condition is exactly what pins that shift down. When instead a definite integral is used to find the change in displacement between two given times, the constant cancels out, so no initial value is needed for that part of the working.
Worked example
A particle has a = 6t and v = 5 m s⁻¹ at t = 0. Integrate: v = 3t² + c; using v(0) = 5 gives c = 5, so v = 3t² + 5. At t = 2, v = 12 + 5 = 17 m s⁻¹. For s = ∫ v dt from 0 to 2 with v = 6 − 2t: [6t − t²] = 12 − 4 = 8 m.
Remember
- Integrate a → v → s.
- Use initial conditions to find each constant.
- Displacement can be negative; distance cannot.