5 · IB Maths AA
Displacement and distance travelled
Watch a particle move along a line while the area under its velocity–time graph builds up: signed area gives displacement, total area gives distance travelled. Drag across the graph to scrub through time.
area above t-axis (counts +)area below t-axis (counts −)
s(t) = ∫₀ᵗ v dt (signed area)
distance = ∫₀ᵗ |v| dt (total area)
x(t) = x₀ + s(t), x₀ = 0 m
Time t
Velocity v(t)
Position x(t)
Displacement, 0 to t
Distance, 0 to t
Displacement, 0 to T
Distance, 0 to T
Direction changes at
Try this
- Quadratic with p = 1 s, q = 3 s, T = 4 s: confirm the displacement is 1.33 m but the distance is 4.00 m. Why is the gap exactly twice the red area?
- Linear with u = 4 m s⁻¹, a = −2 m s⁻², T = 4 s: watch the particle turn at t = 2 s and arrive back at O, giving displacement 0 m but distance 8.00 m.
- Quadratic with p = q: v touches zero but never changes sign. Check that distance equals displacement and that no direction change is listed.
Common error: finding distance by integrating v straight across a time where v = 0 lets the positive and negative areas cancel; integrate |v| instead, or split the interval at each root and add the magnitudes.