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Reading Kinematics Graphs

Key idea: Learn to read displacement–time and velocity–time graphs using gradient and area to find velocity, acceleration, and displacement (O Level Physics).

  • SEC G3 Physics 2027
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Learning objectives

  • State what speed means
  • State what velocity means, including its direction
  • Calculate average speed from total distance and total time
  • Calculate acceleration as change in velocity divided by time taken
  • State what uniform acceleration means
  • Interpret examples of non-uniform acceleration
  • Plot and interpret displacement–time and velocity–time graphs in one dimension
  • Deduce rest and uniform or non-uniform velocity from a displacement–time graph
  • Deduce rest, uniform velocity and uniform or non-uniform acceleration from a velocity–time graph
  • Use signed area under a velocity–time graph to determine displacement
  • Recall constant free-fall acceleration near Earth as approximately 10 m/s²

1. Definition

A kinematics graph shows how a motion quantity changes with time (for motion in one dimension).

Syllabus focus (6091)

You should be able to plot and interpret:

  • a displacement–time graph
  • a velocity–time graph and use the area under a velocity–time graph to find displacement (for uniform velocity or uniform acceleration).
Prerequisites

Interactive Concept Check: Kinematics Graph Explorer

Use this explorer to connect all three graph views quickly:

  • slope of s-t at a time gives v
  • slope of v-t at a time gives a
  • signed area under v-t gives displacement

Concept Explorer: Kinematics Graph Relationships

Switch graph-shape presets, move the time marker, and connect slope/area ideas across position-time, velocity-time, and acceleration-time graphs.

BetaO LevelA LevelMotionBest for: O Level and A Level kinematics revision
  • Gradient Interpretation
  • Signed Area
  • Direction From Velocity
  • Graph Translation

Open the full interactive simulation on its own page

Use the standalone simulation page for the live controls, SVG scene, run modes, and scoring flow.

Open Interactive Simulation

The lesson stays lightweight and links out to the dedicated simulation page.

Explore the graphs directly in the Kinematics Graph Explorer.

2. Key Ideas

  • Always check the axes and units first:
    • time, t (s) is on the horizontal axis
    • displacement, s (m) or velocity, v (m s⁻¹) is on the vertical axis
  • Displacement–time graph (s–t):
    • gradient = velocity: v = (Δ s)/(Δ t)
  • Velocity–time graph (v–t):
    • gradient = acceleration: a = (Δ v)/(Δ t)
    • signed area = displacement: Δ s = A_signed
  • Units check:
    • gradient of s–t is m/s (velocity)
    • gradient of v–t is m/s² (acceleration)
    • area under v–t is m (displacement)
Relationships for distance–time and speed–time graphsTwo graphs share a time axis. The distance–time graph becomes steeper as speed increases. The speed–time graph rises linearly; its gradient gives acceleration and its area gives distance travelled.One journey, two Science graph viewsThe axes decide what gradient and area mean.
Read downwards using gradients: displacement–time to velocity–time to acceleration–time. Read upwards using signed areas: acceleration–time to change in velocity, then velocity–time to change in displacement.

The figure uses one constant-acceleration example. Other motions produce different shapes, but the gradient and signed-area relationships remain the same.

3. Detailed Explanations

A. Displacement–time graph (s–t)

Use the gradient to tell what the velocity is doing.

ShapeWhat it means
horizontal lineat rest (velocity = 0)
straight line (constant gradient)uniform velocity
curve (changing gradient)non-uniform velocity (velocity is changing)

Direction matters:

  • positive gradient → positive velocity (moving in the chosen positive direction)
  • negative gradient → negative velocity (moving in the opposite direction)

Average velocity between two points:

v = (Δ s)/(Δ t) = (s₂-s₁)/(t₂-t₁)

Reading an s–t graph (schematic)

A schematic displacement–time graph: flat section (rest), positive gradient (positive velocity), then negative gradient (moving back).

Scroll across the graph to read all labels.

A schematic displacement–time graph: flat section (rest), positive gradient (positive velocity), then negative gradient (moving back).A schematic displacement–time graph: flat section (rest), positive gradient (positive velocity), then negative gradient (moving back).
Horizontal section → at rest. Positive gradient → positive velocity. Negative gradient → moving back (negative velocity).
Open full-size graph
View figure data
Values for Reading an s–t graph (schematic)
Time (s)s–t
00
20
56
72
Curved graphs (non-uniform motion)

For a curve, the gradient changes. At O Level, you are usually asked for the average velocity over a time interval using two points.

B. Velocity–time graph (v–t)

Use the value of v (above/below the axis) to tell direction, and the gradient to tell acceleration.

ShapeWhat it means
horizontal line at v = 0at rest
horizontal line above/below 0uniform velocity (constant v)
straight sloping lineuniform acceleration (constant a)
curvenon-uniform acceleration (acceleration is changing)

Gradient (acceleration):

a = (Δ v)/(Δ t)

Area (displacement):

  • area above the time-axis → positive displacement
  • area below the time-axis → negative displacement

For full area methods + more practice, see: Area Under a Velocity–Time Graph.

Reading a v–t graph: gradient and signed area (schematic)

A schematic velocity–time graph where the gradient gives acceleration and the area above/below the axis gives positive/negative displacement.

Scroll across the graph to read all labels.

A schematic velocity–time graph where the gradient gives acceleration and the area above/below the axis gives positive/negative displacement.A schematic velocity–time graph where the gradient gives acceleration and the area above/below the axis gives positive/negative displacement.
Gradient gives acceleration. The signed area above/below the time-axis gives displacement (negative area means motion in the opposite direction).
Open full-size graph
View figure data
Values for Reading a v–t graph: gradient and signed area (schematic)
Time (s)v–t
00
28
58
70
9-4

C. Finding a gradient (draw a large triangle)

When the graph section is a straight line, you can find the gradient reliably using a large gradient triangle.

Large gradient triangle on a velocity–time graphA straight velocity–time line passes through one second and two metres per second, and five seconds and ten metres per second. A dashed triangle shows a velocity change of eight metres per second over four seconds, giving acceleration two metres per second squared.Gradient = change in velocity ÷ time takentime / svelocity/ m s⁻¹(1, 2)(5, 10)Δt = 4 sΔv = 8 m s⁻¹a = Δv / Δta = 8 / 4 = 2.0 m s⁻²
Scroll diagram horizontally to read all labels.
Choose two well-separated points on the line. Here, gradient = Δv/Δt = 8/4 = 2.0 m s⁻².

Exam reminders:

  • Pick two well-separated points on the straight line (not necessarily the original plotted points).
  • Show your working as: gradient = Δy/Δx and include the unit.
  • Use the correct meaning:
    • s–t gradient → velocity (m s⁻¹)
    • v–t gradient → acceleration (m s⁻²)

D. How to describe motion from a graph (exam workflow)

  1. Split the graph into time intervals where the shape changes.
  2. For a displacement–time graph, choose from at rest, uniform velocity or changing velocity.
  3. For a velocity–time graph, choose from at rest, uniform velocity, uniform acceleration or changing acceleration.
  4. If asked for a gradient, show Δ y/Δ x with units.
  5. If asked for an area, split it into simple shapes and add the signed areas.

4. Common Mistakes

A. Mixing up gradient and y-value

  • On an s–t graph, the gradient is velocity (not the y-value).
  • On a v–t graph, the gradient is acceleration (not the y-value).

B. Ignoring sign (direction)

  • A negative gradient on an s–t graph means motion in the negative direction.
  • A negative v on a v–t graph means motion in the negative direction.

C. Wrong units

  • Gradient of s–t: m/s
  • Gradient of v–t: m/s²
  • Area under v–t: m

D. Using “distance” when the graph is displacement

  • Displacement can decrease; distance travelled cannot.

5. Exam Tips

  • Start every question by writing: “gradient = Δ y/Δ x”.
  • Use two well-separated points on a straight section (less percentage error).
  • Always show the unit at the end (it is an easy mark).
  • For curved sections, describe what is happening (velocity/acceleration changing) unless the question specifically asks for an average value over an interval.
  • Use the mark-scheme words: “at rest”, “uniform velocity”, “uniform acceleration”, “non-uniform acceleration”.
  • For v–t area questions that cross the time-axis, split into above-axis and below-axis sections before combining signed displacement.

6. Worked Examples

Modelled example 1

Reading an s–t graph (rest + velocity)

Core

Problem

A displacement–time graph is horizontal at 5 m from 0–2 s, rises from 5 m to 13 m from 2–6 s, then falls to 9 m from 6–8 s. Identify when the object is at rest and find each interval’s velocity.

Study the worked solution

  1. Identify the horizontal interval

    Method

    Read zero gradient as zero velocity.

    Reason

    Velocity is the gradient of displacement against time.

    Working

    From 0–2 s, v = 0, so the object is at rest.

  2. Find the positive velocity

    Method

    Calculate the gradient from 2–6 s.

    Reason

    Displacement increases uniformly over this interval.

    Working

    v = (13-5)/(6-2) = 2.0 m s⁻¹

  3. Find the negative velocity

    Method

    Calculate the gradient from 6–8 s.

    Reason

    Falling displacement gives a negative gradient and reverse motion.

    Working

    v = (9-13)/(8-6) = -2.0 m s⁻¹

Guided practice 2

Reading a v–t graph (acceleration)

About 4 min

Problem

An object’s velocity rises uniformly from 4.0 m s⁻¹ to 14.0 m s⁻¹ in 5.0 s. Find its acceleration.

Calculate the graph gradient

Unit: m s^-2

Hints

Hint 1: use the graph relationship
Acceleration is the velocity–time gradient.
Hint 2: calculate the changes
Use Δ v = 10.0 m s⁻¹ and Δ t = 5.0 s.
View solution step by step

  1. Calculate the gradient

    Method

    Divide velocity change by time change.

    Reason

    A straight velocity–time line has constant acceleration equal to its gradient.

    Working

    a = (14.0-4.0)/5.0 = 2.0 m s⁻²

Common misconception 3

Displacement vs distance from a v–t graph

Find and correct the mistake

Learner response

Velocity is + 4.0 m s⁻¹ from 0–3 s, then -2.0 m s⁻¹ from 3–7 s. A student adds the area magnitudes and claims both displacement and distance are 20 m. Locate the error and find both quantities.

Keep signs for displacement only

Unit: m
Unit: m

View solution step by step

  1. Find the two signed areas

    Method

    Use rectangles above and below the axis.

    Reason

    Velocity–time area carries the sign of velocity.

    Working

    Δ s₁ = (4.0)(3.0) = 12 m, Δ s₂ = (-2.0)(4.0) = -8.0 m

  2. Distinguish displacement and distance

    Method

    Combine signed areas for displacement and magnitudes for distance.

    Reason

    Displacement retains direction; distance accumulates path length.

    Working

    Δ s = 12-8 = 4.0 m, d = 12 + 8 = 20 m

Examiner practice 4

Displacement from a v–t triangle (area)

3 marks

Examination question

A velocity–time graph rises uniformly from 0 to 12 m s⁻¹ in 6.0 s. Find the displacement. [3 marks]

Name the graph quantity and show the area

View solution step by step

  1. Identify signed area as displacement

    1 mark

    Method

    Use the area under the velocity–time line.

    Reason

    Velocity multiplied by time has units of displacement.

    Working

    Δ s = Aᵥ₋ₜ

  2. Use the triangle area

    1 mark

    Method

    Apply half base times height.

    Reason

    The uniform rise from zero encloses a triangle.

    Working

    Δ s = (1/2)(6.0)(12)

  3. State the displacement

    1 mark

    Reason

    The whole triangle lies above the time axis.

    Working

    Δ s = 36 m

Challenge 5

Direction change from a v–t graph

Minimal support

Direction-change transfer

Velocity is + 6.0 m s⁻¹ from 0–4 s, then falls uniformly to -2.0 m s⁻¹ at 8 s. Find the acceleration during the sloping section and the time when the object changes direction.

Combine gradient with the zero crossing

Unit: m s^-2
Unit: s

Hints

Hint 1: find the sloping gradient
Calculate (-2.0-6.0)/(8.0-4.0).
Hint 2: locate the zero crossing
Starting at t = 4.0 s, find how long + 6.0 m s⁻¹ takes to reach zero.
View solution step by step

  1. Find the acceleration

    Method

    Calculate the velocity–time gradient.

    Reason

    The line is straight from 4–8 s, so acceleration is uniform.

    Working

    a = (-2.0-6.0)/(8.0-4.0) = -2.0 m s⁻²

  2. Find the time to reach zero velocity

    Method

    Use the uniform rate of velocity decrease.

    Reason

    The object changes direction where the line crosses v = 0.

    Working

    Δ t = (0-6.0)/(-2.0) = 3.0 s, t = 4.0 + 3.0 = 7.0 s

7. Mind Stretchers

Mind stretcher 1: Distance vs displacement on a graphExtension

A graph of “distance travelled” against time slopes downwards in one section.

Is this possible? What is the graph more likely showing?

Show Answer

No. Distance travelled cannot decrease (it is the total path length).

If the graph goes down, it is more likely a displacement–time (position–time) graph, because displacement can decrease when the object moves in the opposite direction.

Mind stretcher 2: What does a steeper s–t curve mean?Extension

On a displacement–time graph, the curve becomes steeper as time increases (the gradient increases).

What does this tell you about the velocity?

Show Answer

The gradient of an s–t graph is velocity. If the gradient increases, the velocity is increasing.

So the object is speeding up in the positive direction.

Mind stretcher 3: Can acceleration be zero when velocity is not zero?Extension

On a v–t graph, the line is horizontal at v = +5.0 m s⁻¹.

Is the acceleration zero? Is the object moving?

Show Answer

Yes, acceleration is zero because the gradient is zero.

The object is still moving because its velocity is + 5.0 m s⁻¹ (not zero).

Mind stretcher 4: Describe the motion (quick practice)Extension

For each description, state what the object is doing using mark-scheme words like at rest, uniform velocity, uniform acceleration, speeding up, slowing down, or changes direction.

  1. s–t graph is a horizontal line.
  2. s–t graph is a straight line with a constant positive gradient.
  3. s–t graph is a curve that gets steeper with time.
  4. v–t graph is a horizontal line above the time-axis.
  5. v–t graph is a straight line sloping upwards and stays above the time-axis.
  6. v–t graph is a straight line sloping downwards but stays above the time-axis.
  7. v–t graph is a straight line that crosses the time-axis from positive to negative v.
  8. v–t graph is a horizontal line on the time-axis (v = 0).
Show Answers
  1. At rest (velocity = 0).
  2. Uniform velocity in the positive direction.
  3. Velocity is increasing (speeding up) in the positive direction (gradient increasing).
  4. Uniform velocity (constant v) in the positive direction.
  5. Uniform acceleration (constant positive acceleration): the object is speeding up in the positive direction.
  6. Uniform negative acceleration: the object is slowing down while still moving in the positive direction.
  7. Uniform negative acceleration with a direction change: the object slows to v = 0, then moves in the negative direction (velocity changes sign).
  8. At rest (velocity = 0).

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027