Why Are Distance and Displacement Different? A Time-Graph Simulation
1D Kinematics: Distance vs. Displacement
- Finish back at the start after rolling 8 m: Δx = 0, d = 8 m
- Roll 6 m without ever turning back: d = |Δx| = 6 m
- Finish 3 m left of the start after rolling 9 m: Δx = −3 m, d = 9 m
Displacement is a change of position: Δx = x − x0. It is a vector, so its sign gives the direction: + right of the start, − left of it. Only the start and the finish count, never the path between them.
Distance is the total ground covered. Every meter rolled adds to d, whichever way the ball goes, so the teal d line never dips while the orange Δx line can. The strip under the pitch stacks one layer per pass. Any turn back makes d bigger than |Δx|: d ≥ |Δx| always.
So average velocity and average speed differ. Average velocity is Δx/Δt and average speed is d/Δt, with Δt from the first kick to the last stop; out and back, the first is zero and the second is not. Each kick is a brief push the sim does not model; the grass then slows the ball to a stop one kick length on.
Can Distance Ever Be Smaller Than Displacement?
In short: Displacement is the change in position, Δx = x − x₀, and it can be positive, negative or zero. Distance is the total ground covered, so it only ever grows. That is why distance is never smaller than the size of the displacement, d ≥ |Δx|. The two are equal only when the object never turns back.
Common doubts
Can displacement be zero when distance is not? Yes. Roll the ball 4 m right and then 4 m back: Δx = 0, but d = 8 m.
Does a negative displacement mean the ball moved less? No. The sign only gives the direction (left of the start). Compare the size |Δx| with the distance.
Why do average velocity and average speed differ? Average velocity is Δx/Δt and uses displacement. Average speed is d/Δt and uses distance. After a round trip the first is zero and the second is not.
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