Velocity–Time Graphs: Use Gradient for Acceleration and Area for Displacement

A velocity-time graph combines two calculations that should stay separate. Its gradient gives acceleration; its signed area gives displacement. If the graph crosses below the time axis, the object moves in the negative direction, and total distance must be found by adding area magnitudes rather than cancelling them.

These worked examples of velocity–time graphs use three original, fully specified one-dimensional journeys. Every graph has a coordinate description so you can complete the questions without relying on colour or a drawing. Straight segments represent constant acceleration; sharp joins are idealisations, not claims that a real object changes acceleration instantaneously.

Read velocity before calculating anything

Choose a positive direction, called forward here. A positive velocity means forward motion; a negative velocity means reverse motion. Speed is the magnitude of velocity, so −4 m/s has speed 4 m/s. A negative velocity is not a negative speed.

A horizontal segment above or below zero represents constant velocity and zero acceleration. A horizontal segment on zero represents rest. A line sloping downward has negative acceleration, but the object might be slowing down or speeding up: compare the signs of velocity and acceleration.

For a straight segment, acceleration = change in velocity ÷ elapsed time. Units are (m/s)/s = m/s². For area, velocity multiplied by time gives (m/s) × s = m. That unit check helps expose a solution that accidentally uses gradient when asked for displacement.

Journey A: the complete graph

The object accelerates forward from 0 to 4 s, moves steadily until 7 s, slows to rest by 9 s, rests until 11 s, then moves in reverse and reaches −4 m/s at 13 s. That reverse velocity continues until 16 s. The rest interval has duration 2 s, not 9 s or 11 s.

Worked example 1: acceleration on every segment

Calculate gradients using the endpoints of each straight segment. Keep the subtraction order consistent: final minus initial for both velocity and time.

From 7 to 9 s, velocity is positive while acceleration is negative, so speed decreases. From 11 to 13 s, velocity becomes increasingly negative and acceleration is negative, so speed increases in reverse. Calling every negative acceleration “slowing down” would misdescribe the latter interval.

The overall average acceleration is (−4 − 0)/16 = −0.25 m/s². That is a valid whole-journey average, but it is not the acceleration of every segment. It also cannot replace the segment values in a constant-acceleration equation covering the entire journey.

Worked example 2: signed area and displacement

Split the area between the graph and time axis into simple shapes. Area above zero is positive; area below zero is negative. Use interval widths rather than the time coordinate at the right edge.

• 0–4 s: triangle, ½ × 4 × 8 = +16 m.
• 4–7 s: rectangle, (7 − 4) × 8 = +24 m.
• 7–9 s: triangle, ½ × 2 × 8 = +8 m.
• 9–11 s: zero velocity, so displacement is 0 m.
• 11–13 s: triangle below zero, −½ × 2 × 4 = −4 m.
• 13–16 s: rectangle below zero, −3 × 4 = −12 m.

Total displacement = 16 + 24 + 8 + 0 − 4 − 12 = +32 m. The object ends 32 m forward of its starting point. It had reached 48 m forward by 9 s, stayed there during the rest interval, and then returned 16 m.

Displacement does not identify an absolute position without an initial position. If the starting coordinate were 100 m, the final coordinate would be 132 m, while displacement would remain +32 m. The area describes a change of position.

Worked example 3: distance and two kinds of average

Total distance counts all movement, irrespective of direction. Add the magnitudes: 16 + 24 + 8 + 0 + 4 + 12 = 64 m. The reverse journey adds distance even though it subtracts from displacement.

Average speed = total distance/total time = 64/16 = 4.00 m/s. Average velocity = total displacement/total time = 32/16 = +2.00 m/s. Both calculations include the two seconds spent at rest because the question concerns the full 16-second journey.

The simple mean of initial and final velocity, [0 + (−4)]/2 = −2 m/s, is not the journey’s average velocity. That endpoint rule applies to a single constant-acceleration interval; Journey A includes several different gradients and constant-velocity periods.

As a quick check, total distance must be at least the magnitude of displacement. Here 64 ≥ 32. If a solution gives distance 32 m and displacement 64 m for this graph, the two procedures have probably been swapped.

Worked example 4: crossing zero within a segment

Complete text alternative: Journey B is one straight line from (0 s, +6 m/s) to (6 s, −6 m/s). It crosses zero velocity at 3 s. Positive direction is forward. Acceleration is constant throughout the interval.

Acceleration = (−6 − 6)/6 = −2 m/s². To locate the crossing, use v = u + at: 0 = 6 − 2t, giving t = 3 s. The positive triangle has area ½ × 3 × 6 = +9 m; the negative triangle contributes −9 m.

Displacement is 0 m but distance is 18 m. The object returns to its starting position after moving 9 m forward and 9 m back. It was not stationary for six seconds. At the single crossing instant its velocity is zero, while its acceleration remains −2 m/s².

Check the displacement with s = ut + ½at²: s = 6 × 6 + ½ × (−2) × 6² = 36 − 36 = 0 m. This equation works across Journey B because its acceleration is constant. As another segment check, Journey A’s first four seconds give s = 0 × 4 + ½ × 2 × 4² = 16 m, agreeing with its triangle.

Diagnose the common gradient-area mix-up

A learner divides 8 by 4 and writes “displacement = 2 m” for Journey A’s first segment. The numerical operation found the gradient, so the correct quantity is acceleration 2 m/s². Displacement is the triangular area, 16 m.

Another learner uses 8 × 4 = 32 m for that segment. This treats the final velocity as though it applied throughout. Velocity increases from zero, so the triangular area is half the corresponding rectangle. A third learner treats the downward-sloping section as negative displacement even while it remains above zero; slope direction determines acceleration, whereas vertical sign determines displacement sign.

Three practice tasks with solved answers

Complete text alternative: Journey C joins (0 s, 2 m/s), (3 s, 8 m/s), (5 s, 8 m/s) and (9 s, 0 m/s) with straight segments. All velocities are non-negative.

Practice 1: two gradients

Find Journey C’s acceleration from 0 to 3 s and from 5 to 9 s.

Answer: First acceleration = (8 − 2)/3 = +2 m/s². Final acceleration = (0 − 8)/(9 − 5) = −2 m/s². The middle horizontal segment has zero acceleration.

Practice 2: trapezium, rectangle and triangle

Find Journey C’s displacement and average speed over all nine seconds.

Answer: First area = ½(2 + 8) × 3 = 15 m. Middle area = 8 × 2 = 16 m. Final area = ½ × 4 × 8 = 16 m. Total is 47 m. With no reverse motion, distance is also 47 m, giving average speed 47/9 ≈ 5.22 m/s.

Practice 3: constant reverse motion

A separate supplied segment has velocity −3 m/s throughout a four-second interval. Find acceleration, displacement and distance.

Answer: Acceleration is 0 m/s² because velocity does not change. Displacement = −3 × 4 = −12 m. Distance = 12 m. A negative horizontal line means steady reverse motion, not rest.

Sources and further reading

These references support the scientific background. Examples, datasets, diagrams and practice answers on this page are original. Reference links are for background reading, not instructions to perform a practical activity.

• OpenStax Physics: acceleration equations and graphs
• OpenStax College Physics 2e: graphical motion analysis

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