Read motion graphs without guessing

A position graph can tell you where something is without telling you how far it has travelled. This worked lesson follows one small, hypothetical journey, checks the answer in two ways, then removes an assumption to show what the measurements cannot establish.

By Yosef Gerson, also publishing as Yossi Gerson. For curious readers learning introductory physics. You need subtraction, division and the ability to read a graph; no calculus is required.

1. State what the graph represents

Imagine a cart moving on one straight track. Choose its starting point as x = 0 metres and call movement to the right positive. In this idealized model it moves at a constant velocity between each pair of marked times: (0 s, 0 m), (2 s, 4 m), (4 s, 4 m), (7 s, -2 m).

The horizontal axis is time. The vertical axis is position along the track, not height above the ground and not accumulated distance. The cart does not climb a hill when the line rises. It is farther to the right.

The straight segments are part of the supplied model, not a discovery made from four readings. The abrupt starts and stops simplify the arithmetic. Real carts change velocity over some time; this graph is not a construction or control specification.

2. Read movement from slope

From 0 to 2 seconds, position changes by +4 m in 2 s, so velocity is +2 m/s. From 2 to 4 seconds, the position stays at 4 m, so velocity is zero. From 4 to 7 seconds, position changes by -6 m in 3 s, so velocity is -2 m/s.

The cart is stationary during the highest flat part. Later, when its position passes through zero at 6 s, it is still moving left at 2 m/s. A zero position does not mean a zero velocity. The sign tells you direction relative to the chosen axis; it does not say the cart is slowing down.

Within either sloping segment, the speed is 2 m/s. The two velocities have opposite signs but equal magnitudes. The magnitude of velocity is speed.

Idealized position-time graph: from 0 seconds at 0 metres to 2 seconds at 4 metres; stationary until 4 seconds; then reaches -2 metres at 7 seconds. Segment velocities are +2, 0 and -2 metres per second.

3. Add the whole path before dividing

Distance travelled is 4 m outward plus 6 m back, for 10 m. Displacement is final position minus initial position: -2 m - 0 m = -2 m. The minus sign means the endpoint is 2 m to the left of the start.

Use all 7 seconds, including the pause. Average speed = 10 m / 7 s, approximately 1.43 m/s. Average velocity = -2 m / 7 s, approximately -0.286 m/s. These answers describe different questions. Neither says the cart maintained that value throughout the journey.

A tempting shortcut is to average the three segment velocities: (2 + 0 - 2) / 3 = 0. That gives each segment equal weight even though the durations are 2, 2 and 3 seconds. The time-weighted calculation gives (2 × 2 + 0 × 2 - 2 × 3) / 7 = -2/7 m/s, matching displacement divided by elapsed time.

For a different limitation of averages, What an eight-hour average leaves out shows how two weeks with the same mean can have very different daily schedules.

One-dimensional path begins at x equals zero, travels four metres right, pauses, then travels six metres left to x equals minus two. Total distance is ten metres and displacement is minus two metres.

4. Cross-check with a velocity graph

For each constant-velocity interval, multiply velocity by its duration. The first rectangle contributes +4 m, the pause contributes 0 m and the last contributes -6 m. Their signed sum is -2 m, the displacement.

To recover distance, add the magnitudes: 4 + 0 + 6 = 10 m. The negative rectangle does not subtract distance already travelled. This rectangle method applies directly to the constant-velocity segments shown here. More complicated velocity curves need a suitable area calculation.

This check is useful because it catches a common mismatch: using signed displacements in a speed calculation, or discarding signs in an average-velocity calculation. Keep the quantity, unit and time interval beside each answer.

The same rate-and-time idea also explains electricity use. Why a Small Device Can Use More Electricity Than a Powerful One compares a briefly used kettle with a continuously running router to show why watts alone do not determine monthly energy use.

Velocity is +2 metres per second for the first two seconds, zero for the next two, then -2 for three seconds. Signed areas total -2 metres of displacement; area magnitudes total 10 metres of distance. Abrupt changes are idealized.

5. Four samples cannot establish every move

Now change the evidence. Suppose only the four black dots were measured, and nobody told you that the cart moved uniformly between them. The flat connecting line from 2 s to 4 s is then one possible model, not proof that the cart stopped.

A different model keeps the other segments unchanged but uses straight segments through x = 6 m at 3 s, returning to x = 4 m at 4 s. It passes through exactly the same four measured positions. That extra excursion adds 4 m, bringing distance to 14 m and average speed to 2 m/s. Final displacement and average velocity remain -2 m and -2/7 m/s.

The four positions determine the net displacement and elapsed time. They do not uniquely determine distance travelled, peak speed or what happened between samples. Ten metres is the minimum distance compatible with these readings on this straight track; hidden reversals can add distance. More observations or justified motion assumptions are needed to narrow the possibilities.

Four black measured positions are compatible with the teal ten-metre model or a fourteen-metre model with the dashed detour: from four to six and back to four metres between two and four seconds, with the other segments unchanged. Both models use straight segments. The measurements alone do not identify distance travelled.

6. Try the checks before reading the answers

A. In the original model, at t = 5 s, what are position and velocity? Does a positive position require positive velocity?

B. Keep the same two moving segments but remove the two-second pause. What changes: distance, displacement, average speed or average velocity? Treat the instant direction change as the same kind of idealization used above.

C. Someone says, “The two readings at 4 m prove the cart was stopped between them.” What extra assumption have they made?

Answers. A: position is +2 m and velocity is -2 m/s. It is right of the origin while moving left. B: distance stays 10 m and displacement stays -2 m. Total time becomes 5 s, so average speed is 2 m/s and average velocity is -0.4 m/s. C: they assumed no unobserved motion between the readings. Equal endpoints establish zero displacement over that interval, not zero distance.

Read further

The numerical journey, graphics and sample-data exercise here are original teaching examples. For standard definitions and further practice, see OpenStax University Physics Volume 1, sections 3.1 and 3.2, and OpenStax Physics, section 2.3. No source illustrations are reproduced.

OpenStax 3.1

OpenStax 3.2

OpenStax Physics 2.3

This lesson is also available as a visual webpage on Adobe Express: Read motion graphs without guessing.