Read position, displacement and distance

Begin with a journey, not a formula

A cart travels along a straight line. At time zero it is at 0 m. Three seconds later it is at +6 m. At six seconds it is at +3 m. It has gone forward and then partly returned. Before reading a calculation, sketch where the cart was at these three times and decide what you mean by “how far it went.”

A position locates the cart relative to a chosen origin and positive direction. A displacement is its signed change in position. Distance traveled adds the lengths of all parts of the journey, regardless of direction. These distinctions and average velocity are introduced in OpenStax position, displacement and average velocity. The numerical journey here is an original synthetic teaching example.

Read the axes before the line

The horizontal graph axis is time in seconds; the vertical axis is position in metres. The graph is not a drawing of a hilly track. A line rising on this graph means position increases with time along our straight physical track. The solid curve always describes the ground frame. The dashed curve describes the selected observer frame; they overlap at the default settings.

Keep the observer velocity at 0 m/s. The cart’s first segment rises by 6 m in 3 s, so its slope is +2 m/s. The second falls by 3 m in 3 s, so its slope is −1 m/s. Negative velocity means motion toward decreasing coordinate values, not a negative amount of motion. Negative position means a location on the negative side of the origin; it does not by itself tell you the direction of motion.

The ledger supplies exact synthetic positions at each integer second. Straight segments between those points are part of the model definition. With real sampled observations, the connecting line would be an interpretation: a cart could move between samples in a way that the samples miss. Additional observations or a justified motion model would be needed to exclude that possibility.

Keep the endpoint change separate from total travel

Write xᵢ for initial position, x_f for final position, and Δx for displacement; the symbol Δ means “change in.” Then Δx = x_f − xᵢ. Read this as “final position minus initial position.” For the full journey, Δx = 3 − 0 = +3 m.

The first segment contributes 6 m of distance and the second contributes 3 m. Total distance is 9 m, although displacement is +3 m. The difference is not an error: the two quantities answer different questions. Distance cannot be negative. In one dimension it is at least as large as the magnitude of displacement, with equality when the motion never reverses direction in the chosen frame.

Let Δt be elapsed time, here 6 s. Average velocity = Δx/Δt = 3/6 = +0.5 m/s. Average speed = total distance/Δt = 9/6 = 1.5 m/s. A speedometer-like record and an endpoint-only calculation therefore need not give the same number. The arithmetic mean of segment speeds works here only because the two segment durations are equal. If durations differ, use total distance divided by total time instead.

Move the origin without moving the observer

Change “Observer origin at time zero” from 0 m to +4 m while keeping observer velocity zero. The observer assigns the cart positions −4 m, +2 m and −1 m at the three key times. Subtract 4 m from every ground position to obtain these coordinates.

Your action: compare captures before and after this origin shift. Calculate final minus initial in each. The shifted displacement is −1 − (−4) = +3 m. Distance, slopes and both averages remain unchanged. A fixed origin shift changes coordinate labels, but cancels when taking a position difference. This makes a useful check on a calculation: if you change a fixed origin and obtain a new displacement, inspect your subtraction.

The graph’s position limits remain −20 to +20 m for all settings. That shared scale makes changes comparable instead of magnifying every curve to fill the plot. At the default frame the trace occupies a small part of the scale; that does not mean the cart has stopped.

Know what the sharp corner leaves out

At 3 s, the ideal line switches abruptly from slope +2 to slope −1 m/s. Its left and right slopes differ, so the model does not assign one instantaneous velocity at that corner. A real cart changes velocity over some nonzero interval. This lesson does not model the force, acceleration history or mechanism causing the reversal. Describing a path is called kinematics; explaining its dynamical cause requires further physics.

Transfer: In a different six-second journey, an object starts at +2 m, reaches +8 m after two seconds, and ends at −1 m after another four seconds, with straight position–time segments. Find displacement, distance, average velocity and average speed. Explain why averaging the two segment speed magnitudes without weighting would give a wrong answer.

Check the motion transfer

Displacement is −1−2 = −3 m. Distance is 6+9 = 15 m. Average velocity is −3/6 = −0.5 m/s; average speed is 15/6 = 2.5 m/s. Segment speeds are 3 and 2.25 m/s but last two and four seconds respectively; their unweighted mean 2.625 m/s is not the journey average speed.