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Kinematics Graphs (A Level)

Key idea: Learn how to interpret distance–time, displacement–time, velocity–time and acceleration–time graphs using gradients, areas and tangents (A Level Physics).

  • GCE A-Level H2 Physics 2027
On this page

Learning objectives

  • Interpret position, displacement, velocity and acceleration using equations and graphs.
  • Derive the uniformly accelerated motion equations from the definitions of velocity and acceleration.
  • Derive and apply uniformly accelerated motion equations with a stated sign convention.

1. Definitions (Must Know)

A kinematics graph shows how a motion quantity changes with time.

Two graph skills that show up again and again:

  • Gradient (slope): gradient = Δy / Δx (units: “y-unit per x-unit”)
  • Area under a graph: add up areas of simple shapes (rectangles/triangles/trapeziums). Units: “y-unit × x-unit”.

A. Instantaneous value (tangent idea)

For a curved displacement–time graph, instantaneous velocity is found from the gradient of the tangent at that point. For a curved velocity–time graph, the tangent gradient gives instantaneous acceleration. The graph’s vertical coordinate itself is the instantaneous value of the quantity named on its vertical axis.

Tangent method (displacement–time example)

A curve s=t^2 with a tangent at t=3 s. The gradient of the tangent gives instantaneous velocity.

Scroll across the graph to read all labels.

A curve s=t^2 with a tangent at t=3 s. The gradient of the tangent gives instantaneous velocity.A curve s=t^2 with a tangent at t=3 s. The gradient of the tangent gives instantaneous velocity.
For s = t², the tangent at t = 3 s has gradient ds/dt = 6 m s⁻¹.
Open full-size graph
View figure data
Values and uncertainty for Tangent method (displacement–time example)
SeriesTime (s)Time uncertaintyDisplacement (m)Displacement uncertainty
Curve: s = t²00
Curve: s = t²0.50.25
Curve: s = t²11
Curve: s = t²1.52.25
Curve: s = t²24
Curve: s = t²2.56.25
Curve: s = t²39
Curve: s = t²3.512.25
Curve: s = t²416
Curve: s = t²4.520.25
Curve: s = t²525
Tangent at t = 3 s23
Tangent at t = 3 s415
Point at t = 3 s39

2. Key Ideas (What Earns Marks)

GraphGradient tells youArea tells you
Distance–timespeed (m s⁻¹)(not usually used)
Displacement–timevelocity (m s⁻¹)(not usually used)
Velocity–timeacceleration (m s⁻²)displacement, s (m, signed)
Acceleration–time(not usually used)change in velocity, Δ v (m s⁻¹)
  • Sign matters for displacement and velocity. Decide your positive direction first.
Exam pitfall: gradient vs area

On a velocity-time graph, gradient gives acceleration while area gives displacement. Swapping these two ideas usually costs all method marks in graph questions.

Velocity and acceleration direction determine whether speed changesFour panels show an object moving right or left. Solid velocity arrows and dashed acceleration arrows point in the same direction for speeding up and opposite directions for slowing down.Moving right, speeding upvelocity +acceleration +same direction → speed increasesMoving right, slowing downvelocity +acceleration −opposite directions → speed decreasesMoving left, speeding upvelocity −acceleration −same direction → speed increasesMoving left, slowing downvelocity −acceleration +opposite directions → speed decreases
Scroll diagram horizontally to read all labels.
Speed increases when velocity and acceleration point in the same direction. Speed decreases when they point in opposite directions; a negative acceleration does not automatically mean slowing down.
Negative acceleration is not always deceleration

A negative gradient on a velocity–time graph means negative acceleration. The object slows down only when velocity and acceleration have opposite signs; if both are negative, its speed increases in the negative direction.

Use this explorer to stress-test your graph intuition across presets and marker positions:

  • slope of s-t gives v
  • slope of v-t gives a
  • 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 motion graphs in the Kinematics Graph Explorer.

3. Detailed Explanations

A. Distance–time graphs

  • Distance never decreases.
  • Horizontal line → object is stationary.
  • Steeper gradient → higher speed.
  • Curve → speed is changing (non-uniform speed).

B. Displacement–time graphs

  • Displacement can be positive, negative, or zero (depends on your chosen origin and direction).
  • Displacement can increase, decrease, or stay constant.
  • Straight line → uniform velocity.
  • Negative gradient → motion in the opposite direction.
  • For a curve, the instantaneous velocity is the gradient of the tangent (A Level idea).

C. Velocity–time graphs

  • Horizontal line → constant velocity.
  • Straight line with positive gradient → uniform acceleration.
  • Straight line with negative gradient → uniform negative acceleration. Whether the object speeds up or slows down depends on the velocity sign.
  • Curve → acceleration is changing.
  • Area under the graph → displacement (signed area).
  • Total distance travelled → add areas using absolute values (treat areas below the axis as positive).

D. Acceleration–time graphs

  • Area under the graph → change in velocity.
  • A horizontal line above zero means constant positive acceleration; below zero means constant negative acceleration.

E. Relationship between the graphs (summary)

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.
For the same motion, differentiation moves from displacement to velocity to acceleration; signed area moves back through changes in velocity and displacement.

This schematic shows constant positive acceleration. It is not a universal set of shapes: the relationships between gradient and signed area are the transferable rule.

F. Deriving the SUVAT equations (constant acceleration)

If acceleration is constant, the velocity–time graph is a straight line, so:

  • gradient gives acceleration: a = (v-u)/t ⇒ v = u + at
  • area gives displacement: s = v_avgt = (1/2)(u + v)t

Substitute v = u + at into the average-velocity result:

s = ut + (1/2)at²

Eliminate t between v = u + at and s = (1/2)(u + v)t:

v² = u² + 2as

These equations apply only to straight-line motion with constant acceleration. Define the positive direction before substituting signed values for u, v, a and s.

4. Common Mistakes

  • Mixing up distance vs displacement (sign matters for displacement).
  • Forgetting that area below the time-axis is negative displacement (but still adds to distance).
  • Finding a gradient using two points that are too close together (large percentage error).
  • Calling every negative gradient “deceleration”. Compare the signs of velocity and acceleration before deciding whether speed decreases.

5. Exam Tips

  • When asked for an instantaneous value on a curve, draw a tangent and use two well-separated points on the tangent.
  • Always show units: gradient of v–t is m s⁻², and area under v–t is m.

6. Worked Examples

Modelled example 1

Acceleration–time graph (area idea)

Core

Problem

The acceleration–time graph shows an object that starts from rest. At which labelled point is its speed largest?
Acceleration–time graph with four labelled pointsAcceleration starts at zero, rises linearly to positive point 1, falls through zero at point 2 to negative point 3, then rises to zero at point 4.timeacceleration1234O
Study the worked solution

  1. Translate graph area

    Method

    The signed area accumulated from the start gives the object’s velocity.

    Reason

    Area under an acceleration–time graph is change in velocity, and the initial velocity is zero.

    Working

    v(t) = 0 + signed area from 0 to t

  2. Compare the labelled points

    Method

    Point 2 has the largest speed.

    Reason

    At point 2, the magnitude of the accumulated signed area—and therefore the velocity magnitude—is greatest.

    Working

    largest |v| ↔ largest accumulated |area|

Guided practice 2

Average speed vs average velocity (out-and-back)

About 5 min

Problem

A person walks from A to B at 5.00 m s⁻¹ and returns from B to A at 3.00 m s⁻¹. Find the average speed and average velocity for the whole trip.

Try this before viewing the solution

Unit: m s^-1
Unit: m s^-1

Hints

Hint 1: use an arbitrary one-way distance
Let the A-to-B distance be d; it cancels after you add the two travel times.
View solution step by step

  1. Add the travel times

    Method

    The total time is 8d/15.

    Reason

    Times add, while speeds cannot be averaged directly because the two equal distances take unequal times.

    Working

    t = d/5.00 + d/3.00 = 8d/15

  2. Calculate average speed

    Method

    3.75 m s⁻¹.

    Reason

    Average speed is total distance 2d divided by total time.

    Working

    v bar _speed = 2d/(8d/15) = 3.75 m s⁻¹

  3. Calculate average velocity

    Method

    0 m s⁻¹.

    Reason

    Returning to A makes the net displacement zero.

    Working

    v bar = Δ s/Δ t = 0

Guided practice 3

Hare and tortoise (photo finish)

About 5 min

Problem

A tortoise runs at 0.200 m s⁻¹ over a 1.00 km course. A hare runs 0.800 km at 8.00 m s⁻¹, then rests. How far from the finish should the tortoise be when the hare restarts so that they reach the finish together to photo-finish precision?

Try this before viewing the solution

Unit: m

Hints

Hint 1: match remaining times
When the hare restarts, both competitors must have the same time remaining.
View solution step by step

  1. Find the hare's remaining distance

    Method

    The hare has 200 m left.

    Reason

    It has completed 0.800 km of the 1.00 km course.

    Working

    1.00-0.800 = 0.200 km = 200 m

  2. Find the hare's remaining time

    Method

    25.0 s.

    Reason

    At constant speed, time is distance divided by speed.

    Working

    t = 200/8.00 = 25.0 s

  3. Match the tortoise distance

    Method

    The tortoise should be 5.00 m from the finish.

    Reason

    It covers this distance during the hare’s remaining 25.0 s.

    Working

    d = (0.200)(25.0) = 5.00 m

Common misconception 4

Displacement vs distance from a velocity–time graph

Find and correct the mistake

Learner claim

An object travels at + 4.0 m s⁻¹ for 3.0 s and then at -2.0 m s⁻¹ for 5.0 s. A learner reports both displacement and distance as 2.0 m by adding signed areas. Diagnose the method and find both quantities.

Try this before viewing the solution

Unit: m
Unit: m

View solution step by step

  1. Find the two signed areas

    Method

    The areas are + 12 m and -10 m.

    Reason

    Each constant-velocity interval forms a rectangle; the negative velocity gives negative displacement.

    Working

    (4.0)(3.0) = +12 m, (-2.0)(5.0) = -10 m

  2. Calculate displacement

    Method

    Δ s = +2.0 m.

    Reason

    Displacement retains the signs of the two areas.

    Working

    Δ s = 12-10 = 2.0 m

  3. Calculate total distance

    Method

    22 m.

    Reason

    Distance counts path length, so both area magnitudes are positive contributions.

    Working

    d = |12| + |-10| = 22 m

Examiner practice 5

Quadratic in time (two possible times)

5 marks

Examination question

A particle has initial velocity 20 m s⁻¹ and constant acceleration -2.0 m s⁻². Find both times when its displacement from the start is 30 m, and explain why two positive times occur. [5 marks]

Try this before viewing the solution

Unit: s
Unit: s

View solution step by step

  1. Form the displacement equation

    2 marks

    Method

    t²-20t + 30 = 0.

    Reason

    Constant acceleration permits s = ut + (1/2)at² with signed values.

    Working

    30 = 20t + (1/2)(-2.0)t² ⇒ t²-20t + 30 = 0

  2. Solve the quadratic

    2 marks

    Method

    t = 1.63 s or 18.4 s.

    Reason

    The quadratic formula gives two positive roots.

    Working

    t = (20± square root of (400-120))/2 = 1.63 s or 18.4 s

  3. Interpret both times

    1 mark

    Method

    The particle passes 30 m once outward and once after turning back.

    Reason

    Its negative acceleration reduces the positive velocity to zero and then reverses the motion.

    Working

    v = 20-2t = 0 ⇒ tₜᵤᵣₙ = 10 s, between the two roots.

Challenge 6

Negative acceleration while speeding up

Minimal support

Independent transfer

Take right as positive. A particle’s velocity changes uniformly from -2.0 m s⁻¹ to -8.0 m s⁻¹ in 3.0 s. Find its acceleration and state whether it speeds up or slows down.

Try this before viewing the solution

Unit: m s^-2
Speed change

Hints

Hint 1: compare signs after calculating
Negative acceleration means leftward acceleration; decide whether that reinforces or opposes the negative velocity.
View solution step by step

  1. Calculate signed acceleration

    Method

    a = -2.0 m s⁻².

    Reason

    The velocity becomes more negative by 6.0 m s⁻¹ over 3.0 s.

    Working

    a = (-8.0-(-2.0))/3.0 = -2.0 m s⁻²

  2. Compare velocity and acceleration

    Method

    The particle speeds up in the negative direction.

    Reason

    Velocity and acceleration have the same sign, so the velocity magnitude grows.

    Working

    |v|:2.0 → 8.0 m s⁻¹

7. Mind Stretchers

Mind stretcher 1: Sketch the matching graph (shape only)Extension

A car’s velocity–time graph is:

  • starts at v = 0
  • increases linearly for 4 s
  • then stays constant for 3 s
  • then decreases linearly back to 0 over 2 s

Sketch the displacement–time graph (shape only) and label where the gradient is largest.

Show Answer

The displacement–time graph must always increase (since velocity is never negative).

  • From 0–4 s, velocity increases, so the displacement–time graph becomes steeper (increasing gradient, curve).
  • From 4–7 s, velocity is constant, so displacement–time is a straight line (constant gradient).
  • From 7–9 s, velocity decreases to 0, so the displacement–time graph still increases but flattens (decreasing gradient) and becomes horizontal at 9 s.

The gradient is largest at the start of the constant-velocity section (at 4 s), since that is where v is largest.

Mind stretcher 2: Non-uniform acceleration (tangent method)Extension

On a curved velocity–time graph, explain how you would find the acceleration at a particular time.

Show Answer

Draw a tangent to the curve at that time, then choose two well-separated points on the tangent and calculate its gradient: a = (Δ v)/(Δ t)

Mind stretcher 3: Optional (Enrichment)Extension

A. Calculus notation (if you are studying it)

If you use calculus language:

  • velocity is the gradient of the displacement–time graph: v = ds/dt
  • acceleration is the gradient of the velocity–time graph: a = dv/dt

B. Average velocity from a position–time equation

If you are given an equation for position/displacement, e.g. s(t):

  • Average velocity from t₁ to t₂: v bar = (s(t₂)-s(t₁))/(t₂-t₁)
  • If you are using calculus, the instantaneous velocity is the derivative: v = ds/dt

Example: s = 3t² + 2t (SI units).

Show Answer

Average velocity from t = 1.0 s to t = 3.0 s:

v bar = (s(3)-s(1))/(3-1) = ((3 · 9 + 2 · 3)-(3 · 1 + 2 · 1))/2 = (33-5)/2 = 14 m s⁻¹

A particle moves according to x = 10 t², where x is in metres and t is in seconds.

Find the average velocity:

  1. from t = 2.00 s to t = 3.00 s
  2. from t = 2.00 s to t = 2.10 s
Show Answer

Average velocity is: v_avg = (Δ x)/(Δ t) = (x₂ - x₁)/(t₂ - t₁)

x(3.00) = 10(3.00)² = 90 m x(2.00) = 10(2.00)² = 40 m v_avg = (90 - 40)/(3.00 - 2.00) = 50 m s⁻¹

x(2.10) = 10(2.10)² = 44.1 m x(2.00) = 40 m v_avg = (44.1 - 40)/(2.10 - 2.00) = 41 m s⁻¹

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027