Integrals of Motion
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
Integrating goes the other way: Δv = ∫a dt and Δx = ∫v dt. On a graph, each integral is the area under the curve. To get v(t) or x(t) itself, add the starting value (the constant of integration).
1. Area under the curve
2. Displacement vs. distance
The integral of v gives displacement: areas below the axis count as negative.
For distance traveled, find where v changes sign, integrate each piece separately, and add the absolute values.
An indefinite integral has a “+ C”. Find C from a known value, such as x = 3 m at t = 1 s. It doesn’t have to be at t = 0.
3. Example, then try it
A particle starts at x = 0 with v = 3 m/s, and its acceleration is a(t) = 4 − 2t (m/s²). Find v and x at t = 3 s.
v(t) = 3 + 4t − t², so v(3) = 3 + 12 − 9 = 6 m/s.
x(t) = 3t + 2t² − t³/3, so x(3) = 9 + 18 − 9 = 18 m.
A particle has v(t) = 3 − t (m/s) and starts at x = 1 m. Find its displacement and distance traveled from t = 0 to 5 s, and its position at t = 5 s.
Displacement: ∫05(3 − t) dt = [3t − t²/2] = 15 − 12.5 = 2.5 m.
Where v = 0: t = 3 s. From 0 to 3: ∫ = 9 − 4.5 = 4.5 m. From 3 to 5: ∫ = 2.5 − 4.5 = −2.0 m.
Distance: 4.5 + 2.0 = 6.5 m.
Position: x(5) = 1 + 2.5 = 3.5 m.
Answer: Δx = 2.5 m, distance = 6.5 m, x(5) = 3.5 m.
4. Common mistakes
What does the area under a velocity–time graph between t₁ and t₂ represent?
v = dx/dt, so ∫v dt from t₁ to t₂ is Δx, the displacement.
Ready to practice?
5 questions on Integrals of Motion, each with a hint and an explanation for every wrong answer.