Reading Motion Graphs

You’ll be able to

Interpret the slope of a position–time graph as velocity
Interpret the slope of a velocity–time graph as acceleration
Find displacement from the area under a velocity–time graph

On an x–t graph, slope is velocity

A position–time (x–t) graph plots where an object is against time. Its slope, rise over run, is Δx / Δt — which is exactly velocity. A steep line means fast motion; a flat, horizontal line means the position is not changing, so the object is at rest. A downward slope means negative velocity.

On a v–t graph, slope is acceleration and area is displacement

A velocity–time (v–t) graph plots velocity against time. Now the slope is Δv / Δt — the acceleration. A horizontal line means constant velocity (zero acceleration). The area between the line and the time axis equals the displacement, because velocity × time is a distance. Area below the axis counts as negative displacement.

Slope and area

slope of x–t = v · slope of v–t = a · area under v–t = Δx

Two different graphs, three different meanings — keep straight which graph you are reading.

Worked example

On a v–t graph an object’s velocity rises in a straight line from 4 m/s to 10 m/s over 3 s. Find its acceleration and its displacement.

1.Acceleration is the slope: a = Δv / Δt = (10 − 4) / 3 = 6 / 3 = 2 m/s².
2.Displacement is the area under the line — a trapezoid with parallel sides 4 and 10 and width 3.
3.Area of a trapezoid = ½(base₁ + base₂) × height = ½(4 + 10)(3).
4.Evaluate: ½ × 14 × 3 = 21 m.

Answer: a = 2 m/s² and Δx = 21 m

Watch out

A point high up on a v–t graph does not mean the object is far away — it means it is moving fast. Height is velocity, not position. Never read an x–t graph and a v–t graph the same way.

Checkpoint

On a position–time graph, a horizontal (flat) line indicates that the object is:

Checkpoint

An object moves at a constant 6 m/s for 10 s. What displacement does the area under its velocity–time graph give?

Checkpoint

A velocity–time graph is a straight line rising from 0 to 8 m/s over 4 s. What is the acceleration?

On the exam

Free-response graph questions love the chain "slope of x–t → v, slope of v–t → a, area under v–t → Δx". Memorize those three links and you can convert between any pair of graphs.

Answer the 3 checkpoints as you read.