1. Distance-Time Graphs

Learning outcomes

• I can interpret information presented on a distance-time graph.
• I can determine the speed of an object from the slope of a distance-time graph.
• I can distinguish between constant speed, changing speed, and stationary motion on a graph.
• I can construct a distance-time graph from data.
• I can describe motion using evidence from a distance-time graph.

What Is a Distance-Time Graph?

A distance-time graph is a graph that shows how the distance travelled by an object changes over time.

Instead of simply telling us how far an object travelled, the graph allows us to see how the object moved during its journey.

For example, a distance-time graph can show when an object:

• moves at a constant speed
• moves faster or slower
• stops moving
• changes its speed

Distance-time graphs are useful for studying the motion of cars, runners, cyclists, trains, and many other moving objects.

 

 

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Reading the Axes

Before interpreting any graph, always check the axes.

On a distance-time graph:

• The x-axis (horizontal axis) represents time.
• The y-axis (vertical axis) represents distance.

Time may be measured in seconds, minutes, or hours.

Distance may be measured in metres or kilometres.

For example, a point at (5 s, 20 m) means:

After 5 seconds, the object has travelled 20 metres.

Always check the units shown on the axes before doing any calculations.

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The Slope Tells Us the Speed

The most important feature of a distance-time graph is its slope, also called its gradient.

The slope tells us the speed of the object.

A steep line means that a large distance is covered in a short time.

Therefore:

Steeper line = faster speed

A shallow line means that less distance is covered during the same amount of time.

Therefore:

Shallower line = slower speed

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Calculating Speed from a Distance-Time Graph

We can calculate speed by finding the slope of the graph.

The equation is:

\( speed = \frac{change \ in \ distance}{change \ in \ time} \)

This is sometimes written as:

\( speed = \frac{ \Delta d }{ \Delta t } \)

The symbol Δ means change in.

Worked Example

Imagine an object travels from 10 m at 2 seconds to 40 m at 8 seconds.

First, find the change in distance:

40 - 10 = 30m

Next, find the change in time:

8 - 2 = 6s

Now calculate the speed:

\( speed = \frac{30}{6} = 5 \ m/s \)

The object was travelling at 5 m/s during this section of the graph.

Remember

When calculating the slope, we calculate the change in distance and the change in time.

Do not simply divide any distance value by any time value.

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Recognising Types of Motion

Different shapes on a distance-time graph represent different types of motion.

Constant Speed

A straight sloping line represents constant speed.

This means that the object travels equal distances during equal amounts of time.

For example:

Time (s)	Distance (m)
0	0
1	5
2	10
3	15
4	20

The object travels 5 metres every second, so its speed remains constant.

The steeper the straight line, the greater the constant speed.

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Stationary Motion

A horizontal line means that the object is stationary.

Stationary means not moving.

Time continues to pass, but the distance does not increase. Therefore, the object's speed is:

0 m/s

For example, a car may stop at a traffic light. The time continues to increase, but the car does not travel any additional distance.

Horizontal line = stationary

 

 

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Changing Speed

Sometimes an object's speed does not remain constant.

A curved line on a distance-time graph represents changing speed because the slope of the graph is changing.

Speeding Up

If the curve becomes steeper, the object's speed is increasing.

The object is covering more distance during each unit of time.

Curve getting steeper = speeding up

Slowing Down

If the curve becomes less steep, the object's speed is decreasing.

The object covers less distance during each unit of time.

Curve becoming flatter = slowing down

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Describing Motion from a Graph

A distance-time graph can tell the story of an object's journey.

Imagine a graph showing the following journey:

• From 0 – 10 seconds, the object moves at a constant speed.
• From 10 – 15 seconds, the graph is horizontal.
• From 15 – 20 seconds, the graph rises again with a steeper slope.

We could describe the motion as:

The object moves at a constant speed for the first 10 seconds. It remains stationary from 10 to 15 seconds. It then begins moving again at a faster constant speed.

A good scientific description should include evidence from the graph.

For example, instead of simply saying:

The object stopped.

A stronger answer would be:

The object was stationary between 10 s and 15 s because its distance remained constant at 40 m.

Using values from the graph makes the description more precise.

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Constructing a Distance-Time Graph

Distance-time graphs can also be created from experimental data.

Suppose a student records the following results:

Time (s)	Distance (m)
0	0
2	6
4	12
6	18
8	24

To construct a distance-time graph:

1. Put time on the x-axis.
2. Put distance on the y-axis.
3. Choose a suitable scale for each axis.
4. Label both axes and include the correct units.
5. Plot each pair of values.
6. Connect the points appropriately.
7. Give the graph a clear title.

In this example, the points would form a straight line because the object travels the same distance every 2 seconds.

Its speed is:

$\text{Speed}=\frac{24\text{ m}}{8\text{ s}}$$\boxed{3\text{ m/s}}$

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Real-World Connection

Distance-time graphs can be used to study many real situations.

For example, imagine tracking a student walking from school to a bus stop.

The graph might show:

• a shallow straight line while the student walks slowly
• a horizontal section while the student waits at a crossing
• a steeper line when the student starts running
• another horizontal section when the student reaches the bus stop

By looking only at the graph, we can reconstruct what happened during the journey.

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Common Mistake: Distance-Time vs Speed-Time Graphs

Be careful not to confuse a distance-time graph with a speed-time graph.

On a distance-time graph:

slope = speed

A horizontal line means the object is stationary.

On a speed-time graph, however, a horizontal line usually means the object is moving at a constant speed.

Always check the labels on the axes before interpreting a graph.

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Did You Know?

Modern smartphones and fitness trackers can collect motion data using technologies such as GPS and motion sensors.

Apps can then display this information as graphs, allowing runners and cyclists to analyse different parts of their journey.

The same basic ideas used in classroom distance-time graphs are therefore useful when analysing real motion data.

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Key Vocabulary

Distance-time graph – a graph showing how distance changes with time.

Distance – how far an object has travelled.

Time – how long an event or journey takes.

Slope / Gradient – the steepness of a line on a graph.

Speed – the distance travelled per unit of time.

Constant speed – motion where speed does not change.

Stationary – not moving.

Changing speed – motion in which an object's speed increases or decreases.

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Key Takeaways

• A distance-time graph shows distance against time.
• Time is placed on the x-axis and distance on the y-axis.
• The slope of a distance-time graph represents speed.
• A steeper slope represents a greater speed.
• A straight sloping line represents constant speed.
• A horizontal line represents a stationary object.
• A curve represents changing speed.
• A curve becoming steeper shows an object speeding up.
• A curve becoming flatter shows an object slowing down.
• Speed can be calculated using: ( speed = \frac{change \ in \ distance}{change \ in \ time} \)
• Good descriptions of motion should use specific evidence and values from the graph.

2. Velocity-Time Graphs

Learning outcomes

• I can interpret information presented on a velocity-time graph.
• I can determine acceleration from the slope of a velocity-time graph.
• I can identify periods of constant velocity, acceleration, and deceleration.
• I can distinguish between positive and negative velocity on a graph.
• I can construct and analyze velocity-time graphs.

Velocity-Time Graphs

A velocity-time graph shows how an object's velocity changes over time.

Velocity-time graphs allow us to determine much more than how fast an object is moving. We can use them to identify:

• velocity at a particular time
• direction of motion
• constant velocity
• acceleration
• deceleration
• changes in direction

The slope (gradient) of a velocity-time graph represents the object's acceleration.

 

 

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Reading a Velocity-Time Graph

A velocity-time graph has two axes.

Horizontal Axis

The horizontal or x-axis represents:

Time (s)

Vertical Axis

The vertical or y-axis represents:

Velocity (m/s)

Unlike a speed-time graph, the vertical axis can contain both positive and negative values.

This is because velocity includes direction.

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Velocity Includes Direction

Velocity describes both:

• how fast an object is moving
• the direction in which it is moving

We usually choose one direction to be positive.

For example:

Right = positive

Left = negative

An object travelling at:

+8 m/s

could therefore be moving right at 8 m/s.

An object travelling at:

−8 m/s

would be moving at the same speed but in the opposite direction.

The negative sign does not mean the object is moving slowly.

It tells us about its direction.

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Positive and Negative Velocity

On a velocity-time graph:

Above the time axis → Positive velocity

Below the time axis → Negative velocity

On the time axis → Zero velocity

 

 

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For example:

Velocity.	Meaning
+10 m/s	Moving in the positive direction
+4 m/s	Moving in the positive direction
0 m/s	Stationary at that instant
−4 m/s	Moving in the negative direction
−10 m/s	Moving in the negative direction

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Constant Velocity

A horizontal line on a velocity-time graph represents constant velocity.

For example, suppose a car travels at:

12 m/s for 5 seconds

Its velocity does not change.

Therefore:

Acceleration = 0 m/s²

The velocity-time graph would show a horizontal line at +12 m/s.

Constant velocity

An object travels at a constant velocity of 12 m/s.

 

A horizontal line therefore means:

Constant velocity → Zero acceleration

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Acceleration and Slope

The slope of a velocity-time graph represents acceleration.

The equation is:

\( a = \frac{ \Delta v }{ \Delta t } = \frac{v_f - v_i}{t} \)

where:

• a = acceleration in m/s²
• v_f = final velocity in m/s
• v_i = initial velocity in m/s
• t = time in seconds

A steeper slope represents a larger magnitude of acceleration.

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Calculating Acceleration from the Graph

Suppose an object's velocity increases from:

4 m/s to 16 m/s

during:

6 seconds

Step 1 – Find the change in velocity

Δv = 16 - 4 = 12 m/s

Step 2 – Divide by the time

\( a = \frac{12}{6} = 2 m/s^2 \)

The object accelerates at 2 m/s².

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Positive Acceleration

Consider an object whose velocity changes as follows:

Time (s)	Velocity (m/s)
0	0
1	2
2	4
3	6
4	8
5	10

 

Positive acceleration

Velocity increases by 2 m/s every second.

 

The line slopes upward.

Acceleration:

\( a = \frac{10 - 0}{5} = 2m/s^2 \)

The object has positive acceleration.

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Negative Acceleration

Now suppose the velocity changes from:

20 m/s → 0 m/s

over 5 seconds.

\( a = \frac{0 - 20}{5} = -4m/s^2 \)

The negative acceleration means the velocity is changing in the negative direction.

In this particular example, the object has positive velocity but is slowing down, so we can also say it is decelerating.

 

 

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Deceleration

Deceleration means that an object's speed is decreasing.

For an object moving in the positive direction, deceleration appears as a line sloping downward toward zero.

For example:

15 m/s → 10 m/s → 5 m/s → 0 m/s

The object becomes progressively slower until it stops.

However, we need to be careful:

Negative acceleration does not always mean deceleration.

An object moving in the negative direction can have negative acceleration and actually speed up.

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Acceleration vs Deceleration

The easiest way to determine whether an object is speeding up or slowing down is to look at the magnitude of its velocity.

Moving Away from Zero

Speed is increasing.

Object is speeding up.

Moving Toward Zero

Speed is decreasing.

Object is slowing down.

For example:

−2 m/s → −4 m/s → −6 m/s

The values are becoming more negative, but the object's speed is increasing:

2 m/s → 4 m/s → 6 m/s

So the object is speeding up in the negative direction.

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Crossing the Time Axis

One of the most important features of a velocity-time graph occurs when the line crosses:

v = 0

At that instant, the object has zero velocity.

If the velocity then changes sign, the object has changed direction.

For example:

+6 m/s → +3 m/s → 0 m/s → −3 m/s → −6 m/s

The object:

1. moves in the positive direction
2. slows down
3. momentarily stops
4. reverses direction
5. speeds up in the negative direction

 

 

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This is a major difference between speed-time graphs and velocity-time graphs.

A speed-time graph cannot have negative speed.

A velocity-time graph can have negative velocity.

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Interpreting Different Sections

A velocity-time graph may contain several different sections.

Imagine the following journey:

Section A

Velocity increases from:

0 → +10 m/s

The object is accelerating in the positive direction.

Section B

Velocity remains:

+10 m/s

The object travels at constant positive velocity.

Section C

Velocity decreases:

+10 → 0 m/s

The object decelerates to a stop.

Section D

Velocity becomes:

0 → −5 m/s

The object changes direction and speeds up in the negative direction.

Section E

Velocity remains:

−5 m/s

The object moves at constant velocity in the negative direction.

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A Complete Journey

Consider this example:

Time (s)	Velocity (m/s)
0	0
2	6
4	6
6	0
8	−4
10	−4

 

A complete velocity-time journey

The object accelerates, moves at constant velocity, slows to a stop, reverses direction, and then travels at constant negative velocity.

 

We can analyse each section.

0–2 seconds

Velocity:

0 → +6 m/s

The object accelerates.

\( a = \frac{6 - 0}{2} = 3m/s^2 \)

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2–4 seconds

Velocity remains:

+6 m/s

The object moves at constant positive velocity.

a = 0 m/s²

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4–6 seconds

Velocity:

+6 → 0 m/s

\( a = \frac{0 - 6}{2} = -3m/s^2 \)

The object slows to a stop.

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6–8 seconds

Velocity:

0 → −4 m/s

\( a = \frac{-4 - 0}{2} = -2m/s^2 \)

The object accelerates in the negative direction.

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8–10 seconds

Velocity remains:

−4 m/s

The object moves at a constant velocity in the negative direction.

a = 0 m/s²

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Understanding the Slope

The slope tells us how rapidly velocity is changing.

Graph Shape	Meaning
Horizontal line	Constant velocity
Upward slope	Positive acceleration
Downward slope.	Negative acceleration
Steep slope	Large acceleration magnitude
Gentle slope	Small acceleration magnitude

Remember:

Slope = acceleration

This is one of the most important relationships when interpreting a velocity-time graph.

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Constructing a Velocity-Time Graph

Suppose we are given the following description:

A cyclist starts from rest and accelerates uniformly to 8 m/s in 4 seconds. The cyclist travels at 8 m/s for another 3 seconds before slowing uniformly to rest over 2 seconds.

First, create a table.

Time	Velocity
0 s	0 m/s
4 s	8 m/s
7 s	8 m/s
9 s	0 m/s

Then:

1. put time on the x-axis
2. put velocity on the y-axis
3. choose an appropriate scale
4. plot the points
5. connect the points with straight lines

 

Cyclist's velocity-time graph

The cyclist accelerates, travels at constant velocity, and then decelerates to rest.

 

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Analysing the Cyclist's Graph

0–4 seconds

The cyclist accelerates:

\( a = \frac{8 - 0}{4} = 2m/s^2 \)

4–7 seconds

Velocity remains constant at: 8m/s

Therefore:

a = 0 m/s²

7–9 seconds

The cyclist slows:

\( a = \frac{0 - 8}{2} = -4m/s^2 \)

Notice that the final section is steeper than the first.

Therefore, the magnitude of the cyclist's deceleration is greater than the magnitude of the initial acceleration.

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Velocity-Time Graphs and Real Motion

Velocity-time graphs can describe many real situations.

For example, a car approaching traffic lights might:

Accelerate → Constant velocity → Decelerate → Stop

An elevator might:

Accelerate upward → Constant upward velocity → Decelerate → Stop

Then later:

Accelerate downward → Constant negative velocity → Decelerate → Stop

 

 

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A Good Strategy for Reading Any Velocity-Time Graph

When you see a velocity-time graph, work through it systematically.

First: Look at whether the graph is above or below zero.

This tells you the direction of motion.

Second: Look at the slope.

This tells you the acceleration.

Third: Look for horizontal sections.

These represent constant velocity.

Fourth: Look for places where the graph reaches or crosses zero.

These may represent the object stopping or changing direction.

Finally: Calculate slopes when numerical acceleration values are required.

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Common Misconception

A downward-sloping line does not always mean the object is slowing down.

Consider:

0 → −5 → −10 m/s

The graph slopes downward, so acceleration is negative.

But the object's speed changes:

0 → 5 → 10 m/s

The object is actually speeding up in the negative direction.

Therefore:

Negative acceleration ≠ always slowing down

Instead, compare the direction of velocity and acceleration.

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Did You Know?

Velocity-time graphs can also tell us an object's displacement.

The displacement during a time interval is equal to the signed area between the graph and the time axis.

Areas above the axis represent displacement in the positive direction.

Areas below the axis represent displacement in the negative direction.

This makes velocity-time graphs especially powerful: the slope tells us acceleration, while the area tells us displacement.

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Key Terms

Velocity-time graph – A graph showing how velocity changes with time.

Velocity – Speed in a specified direction.

Acceleration – Rate of change of velocity.

Deceleration – A decrease in speed.

Slope – The steepness of a graph; on a velocity-time graph it represents acceleration.

Constant velocity – Motion with unchanged velocity.

Positive velocity – Motion in the chosen positive direction.

Negative velocity – Motion in the direction opposite to the chosen positive direction.

Zero velocity – The object is stationary at that instant.

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Key Takeaways

• A velocity-time graph shows how velocity changes with time.
• Time is plotted on the x-axis and velocity on the y-axis.
• The slope of a velocity-time graph represents acceleration.
• Acceleration can be calculated using \( a = \frac{ \Delta v }{ \Delta t } = \frac{v_f - v_i}{t} \).
• A horizontal line represents constant velocity and zero acceleration.
• An upward slope represents positive acceleration.
• A downward slope represents negative acceleration.
• Velocity above the time axis is positive.
• Velocity below the time axis is negative.
• Negative velocity means movement in the opposite direction, not negative speed.
• If the graph crosses the time axis, the object may be changing direction.
• Deceleration means speed is decreasing, so negative acceleration does not always mean deceleration.
• Velocity-time graphs can be constructed from motion descriptions or numerical data.
• The slope gives acceleration, while the signed area under the graph gives displacement.

3. Area Under a Graph

Learning outcomes

• I can explain the significance of the area under a velocity-time graph.
• I can calculate displacement from the area under a velocity-time graph.
• I can calculate areas using rectangles and triangles.
• I can distinguish between slope and area when interpreting graphs.
• I can solve motion problems using graphical methods.

Area Under a Velocity-Time Graph

A velocity-time graph tells us how an object's velocity changes over time.

There are two especially important features to interpret:

• the slope tells us the object's acceleration
• the area between the graph and the time axis tells us the object's displacement

This means a single velocity-time graph can provide information about both how the velocity changes and how far the object moves from its starting position.

 

 

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Why Does Area Represent Displacement?

Start with the equation for constant velocity: \( v = \frac{ \Delta x }{ \Delta t } \)

Rearranging: Δx = vΔt

where:

• Δx = displacement
• v = velocity
• Δt = time interval

On a velocity-time graph:

• velocity is the height
• time is the width

Therefore: Area = (velocity)(time)

So: Area under a velocity-time graph = displacement

The units confirm this: \( (\frac{m}{s})(s) = m \)

The result is measured in metres, which is the unit of displacement.

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Constant Velocity: Rectangle Area

Suppose a car travels at a constant velocity of 8 m/s for 5s

The velocity-time graph forms a rectangle.

 

 

The area of a rectangle is: A = (base)(height)

Therefore: A = (5)(8) = 40m

The car's displacement is 40m.

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Accelerating Motion: Triangle Area

Suppose an object starts from rest and accelerates uniformly to 12 m/s over 4s. 

The graph forms a triangle.

The area of a triangle is: \( A = \frac{1}{2}(base)(height) \)

Therefore: \( A = \frac{1}{2}(4)(12) = 24 m \)

So the displacement is 24 m.

 

 

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Why a Triangle Appears During Constant Acceleration

If an object starts from rest and accelerates at a constant rate, its velocity increases steadily.

For example:

Time	Velocity
0 s	0 m/s
1 s	3 m/s
2 s	6 m/s
3 s	9 m/s
4 s	12 m/s

Because the velocity changes at a constant rate, the velocity-time graph is a straight sloping line.

The area beneath this line forms a triangle.

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Combining Shapes

More complicated velocity-time graphs can often be divided into simple shapes.

Common shapes include:

• rectangles
• triangles
• trapezoids

Calculate each area separately and then add them together.

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Worked Example: Accelerate, Then Constant Velocity

A cyclist:

1. accelerates from 0 to 10 m/s during the first 4 seconds
2. then travels at 10 m/s for another 6 seconds

The graph can be divided into:

• one triangle
• one rectangle

Triangle

\( A = \frac{1}{2}(4)(10) = 20m \)

Rectangle

A = 6(10) = 60m

Total Displacement

20 + 60 = 80

Displacement = 80 m

 

 

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A Complete Journey

Suppose a car has the following motion:

• accelerates from 0 to 12 m/s in 4 s
• travels at 12 m/s for 5 s
• slows uniformly to rest over 3 s

The graph contains:

• Triangle A
• Rectangle B
• Triangle C

Triangle A

\( A = \frac{1}{2}(4)(12) = 24 m\)

Rectangle B

A = 5(12) = 60 m

Triangle C

\( A = \frac{1}{2}(3)(12) = 18m \)

Total

24 + 60 + 18 = 102

Total displacement = 102 m

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Using a Trapezoid

Sometimes an object begins with a non-zero velocity and then accelerates.

For example:

Initial velocity: 4 m/s

Final velocity: 10 m/s

Time: 3 s

The area under the graph forms a trapezoid.

The trapezoid area can be calculated using: \( A = \frac{1}{2}(a + b)h \)

For a velocity-time graph: \( A = \frac{1}{2}(v_i + v_f)t \)

Therefore: \( A = \frac{1}{2}(4 + 10)(3) = \frac{1}{2}(14)(3) = 21 m \)

The displacement is 21 m.

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Another Way to Handle a Trapezoid

Instead of using the trapezoid formula, we can divide it into:

• a rectangle
• a triangle

Rectangle

A = 4(3) = 12 m

Triangle

Difference in velocity: 10 - 4 = 6 m/s

Area: \( A = \frac{1}{2}(3)(6) = 9 m \)

Total: 12 + 9 = 21 m

The answer is the same.

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Positive Area

If the graph is above the time axis, the velocity is positive.

Therefore, the area represents positive displacement.

For example: +6 m/s for 5 s

gives: Δx = 6(5) = + 30 m

The object moves 30 m in the positive direction.

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Negative Area

If the graph is below the time axis, the velocity is negative.

The area therefore represents negative displacement.

For example: -4 m/s for 5 s

gives: Δx = (-4)(5) = -20 m

The negative sign means the displacement is in the negative direction.

 

 

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When the Graph Crosses the Axis

A velocity-time graph can contain both positive and negative areas.

Suppose an object:

• moves forward
• stops
• reverses direction
• moves backward

The area above the time axis represents displacement in the positive direction.

The area below the axis represents displacement in the negative direction.

To calculate the final displacement:

Net displacement = positive area + negative area

Remember that the area below the axis has a negative sign.

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Worked Example: Changing Direction

Suppose the positive area is: +50 m

and the negative area is: -20 m

The net displacement is: 50 + (-20) = 30 m

The object finishes 30 m in the positive direction from its starting point.

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Displacement vs Distance

This is an important distinction.

Displacement includes direction.

Distance is the total amount of ground travelled.

Suppose an object moves: +50m then -20 m

Displacement

50 - 20 = 30 m

Distance

50 + 20 = 70 m

Therefore:

Distance = 30 m

but:

Distance = 70 m

On a velocity-time graph:

Displacement = signed area

For total distance, add the magnitudes of all areas.

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Slope and Area Are Different

One of the most important skills is distinguishing between the slope and area of a velocity-time graph.

Slope

The slope tells us: Acceleration because:

\( a = \frac{ \Delta v }{ \Delta t } \)

Area

The area tells us: Displacement

because: Δx = vΔt

 

 

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A useful memory rule is:

Slope → Acceleration

Area → Displacement

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Don't Confuse the Two

Consider a horizontal line on a velocity-time graph.

Its slope is: 0

Therefore: a = 0

But there may still be a large area beneath the graph.

So the object can have:

Zero acceleration

while still having:

Large displacement

For example, travelling at a constant 20 m/s for 10 seconds:

Acceleration: 0 m/s²

Displacement: 20(10) = 200 m

This is why slope and area must be interpreted separately.

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Solving Motion Problems Graphically

Consider this journey:

A train:

• accelerates from 0 to 20 m/s in 10 s
• travels at 20 m/s for 30 s
• slows to rest in 10 s

We can solve the displacement entirely from the graph.

Acceleration Section

Triangle:

\( A = \frac{1}{2}(10)(20) = 100 m \)

Constant Velocity Section

Rectangle:

A = (30)(20) = 600 m

Deceleration Section

Triangle:

\( A = \frac{1}{2}(10)(20) = 100 m \)

Total Displacement

100 + 600 + 100 = 800 m

No displacement equation was needed beyond calculating the areas of geometric shapes.

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Graphical Methods Can Give Multiple Answers

A velocity-time graph can often tell us several things about the same motion.

For example:

Slope

Provides:

Acceleration

Area

Provides:

Displacement

Height

Provides:

Velocity

Position above or below axis

Provides:

Direction of motion

Crossing the axis

May indicate:

A change in direction

Velocity-time graphs therefore contain a large amount of information.

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Constructing and Analysing a Graph

Suppose a cyclist:

• starts from rest
• accelerates uniformly to 6 m/s in 3 s
• travels at 6 m/s for 4 s
• slows uniformly to rest in 2 s

First identify the points:

Time	Velocity
0 s	0 m/s
3 s	6 m/s
7 s	6 m/s
9 s	0 m/s

After constructing the graph, calculate the areas.

First Triangle

\( A = \frac{1}{2}(3)(6) = 9m \)

Rectangle

A = (4)(6) = 24 m

Final Triangle

\( A = \frac{1}{2}(2)(6) = 6 m \)

Total:

9 + 24 + 6 = 39 m

The cyclist's displacement is 39 m.

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A Useful Problem-Solving Method

When asked to find displacement from a velocity-time graph:

Step 1: Identify the time interval.

Step 2: Divide the area into simple shapes.

Step 3: Calculate each area.

Step 4: Give areas below the time axis a negative sign.

Step 5: Add the areas.

Step 6: Give the answer in metres.

For distance rather than displacement, add the absolute values of the areas.

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Common Misconceptions

The slope does not give displacement.

Slope gives acceleration.

The area does not give acceleration.

Area gives displacement.

Areas below the axis are not automatically ignored.

They represent negative displacement.

Distance and displacement are not always equal.

If an object changes direction, distance is usually greater than the magnitude of displacement.

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Did You Know?

The area-under-the-graph idea is an early example of a much more powerful mathematical concept called integration.

In calculus, displacement can be found from a velocity function by integrating velocity over time.

For straight-line sections, however, there is no need for calculus. We can simply calculate the areas of familiar geometric shapes such as rectangles, triangles, and trapezoids.

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Key Terms

Area under a graph – The region between a graph line and the horizontal axis.

Displacement – Change in position, including direction.

Distance – Total length of the path travelled.

Velocity-time graph – A graph showing velocity against time.

Slope – The steepness of a graph; on a velocity-time graph it represents acceleration.

Rectangle – A shape with area A = bh.

Triangle – A shape with area A = \( \frac{1}{2}bh \).

Trapezoid – A four-sided shape with one pair of parallel sides.

Net displacement – The overall change in position after positive and negative displacements are combined.

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Key Takeaways

• The area under a velocity-time graph represents displacement.
• This works because vt gives units of metres.
• A rectangle has area: A = bh
• A triangle has area: A = \( \frac{1}{2}bh \)
• More complicated graphs can be divided into rectangles, triangles, and trapezoids.
• Area above the time axis represents positive displacement.
• Area below the time axis represents negative displacement.
• Net displacement is found by adding positive and negative areas.
• Total distance is found by adding the magnitudes of all areas.
• The slope of a velocity-time graph gives acceleration.
• The area under a velocity-time graph gives displacement.
• Velocity-time graphs can be used to solve motion problems graphically, often without needing more advanced equations.

 

 

 

 

4. Comparing Multiple Motions

Why compare motion graphs?

Displaying the motions of two or more objects on the same axes makes it easier to compare their behaviour.

We can investigate questions such as:

• Which object is moving faster?
• Which object is accelerating more rapidly?
• When does one object catch up with another?
• When do two objects have the same velocity?
• Is the distance between the objects increasing or decreasing?

The meaning of a graph depends on its axes. A higher line, a steeper slope and an intersection each mean different things on different types of motion graph.

Before interpreting a graph, identify the quantities and units shown.

Understanding the different graph types

When comparing positions, the objects must use the same origin, positive direction and time reference.

A cumulative distance–time graph does not necessarily tell us where an object is. Two people could each travel 100 m along different routes and finish in different places.

For deciding whether objects meet, a position–time graph is usually the clearest choice.

Comparing speeds using position–time graphs

The gradient of a position–time graph gives velocity:

v = Δx / Δt

For a straight line, velocity is constant. For a curved graph, the gradient of a tangent gives velocity at a particular instant.

To compare speed, compare the magnitudes of the gradients.

• A steeper line represents greater speed.
• A horizontal line represents an object at rest.
• A positive gradient represents motion in the positive direction.
• A negative gradient represents motion in the negative direction.

An object with a steep negative gradient can be moving faster than one with a shallow positive gradient.

For example:

• Object A has velocity +3 m/s.
• Object B has velocity −5 m/s.

Object B is moving faster because its speed is 5 m/s, compared with A’s speed of 3 m/s.

The height of a position–time graph tells us position, not speed.

Worked example: catching up with another object

Two objects move along the same straight track:

• Object A starts at the origin and moves at 4 m/s.
• Object B starts 6 m ahead and moves in the same direction at 2 m/s.

Their position equations are:

xₐ = 4t

xᵦ = 6 + 2t

Object B begins ahead, but A’s position–time line is steeper. A is moving faster and closes the gap.

To find when they reach the same position, set their positions equal:

4t = 6 + 2t
2t = 6
t = 3 s

Substitute into either equation:

x = 4(3) = 12 m

They have the same position after 3 seconds, at 12 metres from the origin.

The left graph shows A and B reaching the same position. The right graph shows a separate example in which C and D reach the same velocity. These intersections represent different events.

Interpreting intersections on position–time graphs

When two position–time graphs intersect, the objects have the same position at the same time.

Depending on the situation, this may represent:

• One runner catching another.
• Two objects passing in opposite directions.
• A moving object reaching a stationary object.

An intersection does not mean their velocities are equal. Compare their slopes to determine their velocities.

In the example above, at t = 3 s:

• Both objects are at x = 12 m.
• A still moves at 4 m/s.
• B still moves at 2 m/s.

A catches and passes B.

The graph alone does not establish that a physical collision occurs. Objects may be in separate lanes, or the model may treat them as points moving along the same coordinate axis.

Comparing separation on position–time graphs

At a chosen time, the vertical gap between two position–time graphs gives their separation along the position axis:

Separation = |xᵦ − xₐ|

For A and B at t = 1 s:

Separation = |8 − 4| = 4 m

At t = 2 s:

Separation = |10 − 8| = 2 m

At t = 3 s:

Separation = 0 m

The gap decreases until A catches B.

Useful patterns include:

• Parallel straight lines: Same constant velocity and constant separation.
• Decreasing vertical gap: Objects are getting closer.
• Increasing vertical gap: Objects are getting farther apart.
• Intersecting lines: Objects have the same position at that instant.

Comparing velocities using velocity–time graphs

On a velocity–time graph, read velocity directly from the vertical axis.

At the same time:

• Equal vertical coordinates mean equal velocities.
• Values with opposite signs indicate opposite directions.
• The value with the larger magnitude represents greater speed.

For example, an object at −8 m/s is moving faster than one at +5 m/s.

Its graph lies lower, but its speed is greater.

On a velocity–time graph, compare vertical coordinates to compare velocities and compare slopes to compare accelerations.

Comparing accelerations using slopes

The gradient of a velocity–time graph gives acceleration:

a = Δv / Δt

For straight-line sections:

• A steeper upward slope means a greater positive acceleration.
• A horizontal line means zero acceleration.
• A downward slope means negative acceleration.
• A greater gradient magnitude means a greater acceleration magnitude.

Worked example

Over the same 4-second interval:

• Object P changes velocity from 2 m/s to 10 m/s.
• Object Q changes velocity from 8 m/s to 12 m/s.

For P:

aₚ = (10 − 2) / 4
aₚ = 2 m/s²

For Q:

aᵩ = (12 − 8) / 4
aᵩ = 1 m/s²

P has the greater acceleration, even though Q has the greater velocity throughout this interval.

Moving faster and accelerating faster are different ideas.

For curved velocity–time graphs, compare tangent gradients at the time of interest. A gradient calculated between two separated points gives average acceleration over that interval.

Same velocity does not mean same position

Consider the right-hand graph above:

• Object C starts from rest and accelerates uniformly at 2 m/s².
• Object D moves at a constant velocity of 6 m/s.

Their velocity equations are:

v꜀ = 2t

vᴅ = 6

They have equal velocities when:

2t = 6
t = 3 s

At that instant, both move at 6 m/s.

However, the intersection does not show that they are side by side. To compare their positions, we need their starting positions and their displacements.

Using areas to compare displacement

The signed area between a velocity–time graph and the time axis gives displacement:

• Area above the axis contributes positive displacement.
• Area below the axis contributes negative displacement.

To find position:

Final position = initial position + displacement

Suppose C and D start at the same position at t = 0.

During the first 3 seconds, C’s displacement is the triangular area under its graph:

Displacement of C = ½ × 3 × 6
Displacement of C = 9 m

D’s displacement is the rectangular area:

Displacement of D = 3 × 6
Displacement of D = 18 m

Therefore, at t = 3 s:

• They have the same velocity.
• D is 9 m ahead of C.

The area between their velocity graphs over this interval represents D’s extra displacement.

Determining when an accelerating object catches up

Continue the same example, with C and D starting together.

At time t, C’s velocity is 2t. Its displacement is the triangular area:

Displacement of C = ½ × t × 2t
Displacement of C = t²

D’s displacement is:

Displacement of D = 6t

Set their displacements equal:

t² = 6t
t(t − 6) = 0

The solutions are:

• t = 0 s: Their shared starting position.
• t = 6 s: C catches D again.

At t = 6 s:

Position of C = 6² = 36 m

Position of D = 6 × 6 = 36 m

Their velocities are then:

• C: 2 × 6 = 12 m/s
• D: 6 m/s

They have equal positions but different velocities.

From 0 to 3 seconds, D pulls farther ahead. After 3 seconds, C is faster and begins closing the gap. C catches D at 6 seconds.

Objects moving towards each other

Opposite directions can be represented using positive and negative velocities.

Worked example

Two cyclists begin 100 m apart on a straight path.

• Cyclist A starts at x = 0 and travels at +6 m/s.
• Cyclist B starts at x = 100 m and travels at −4 m/s.

Their position equations are:

xₐ = 6t

xᵦ = 100 − 4t

Set their positions equal:

6t = 100 − 4t
10t = 100
t = 10 s

Their meeting position is:

x = 6(10) = 60 m

On a position–time graph, A’s line slopes upwards and B’s line slopes downwards. The lines intersect at (10 s, 60 m).

Their separation decreases at 10 m/s because they move towards each other at 6 m/s and 4 m/s.

Relative velocity

Relative velocity describes how one object’s position changes compared with another object.

For motion along one axis:

Velocity of A relative to B = vₐ − vᵦ

For the earlier catching-up example:

vₐ − vᵦ = 4 − 2 = 2 m/s

A closes the initial 6 m gap at 2 m/s:

Time to catch up = 6 / 2 = 3 s

For the cyclists moving towards each other:

vₐ − vᵦ = 6 − (−4) = 10 m/s

Signs matter. Subtracting a negative velocity gives the combined closing rate in this situation.

Making reliable graphical comparisons

Before drawing a conclusion:

1. Identify the graph type. Position, distance, velocity and speed are different quantities.
2. Check scales and units. Visual steepness cannot be compared directly across graphs with different scales.
3. Compare at the same time. Read vertically above the chosen time.
4. Use the appropriate feature. Interpret heights, slopes, intersections and areas according to the axes.
5. Check starting conditions. Equal displacements imply equal final positions only if starting positions are equal.
6. State your conclusion with evidence. Include values, units and the graphical feature used.

For example:

“Object A is moving faster because its position–time gradient is 4 m/s, compared with 2 m/s for B.”

This is more precise than saying, “A’s graph is higher.”

Common misconceptions

• “The higher line always shows the faster object.” On a position–time graph, height shows position.
• “A steeper velocity–time graph means greater velocity.” It means greater acceleration magnitude.
• “Intersecting graphs always mean the objects meet.” Only a common position at a common time establishes this in a shared position model.
• “Equal velocity means equal acceleration.” Two velocity graphs can intersect with different slopes.
• “Equal areas mean equal positions.” Initial positions must also be considered.
• “Negative velocity means slowing down.” It indicates direction; compare velocity with acceleration to determine whether speed decreases.
• “A distance–time intersection means the same location.” Equal cumulative distances do not necessarily imply equal positions.

Did you know?

Two objects can have the same velocity at an instant when their separation is greatest.

In the C and D example, D’s lead grows until t = 3 s. At that moment, their velocities are equal. Afterward, C is faster, so the gap begins to shrink.

Key terms

• Position–time graph: A graph showing position relative to an origin as time changes.
• Velocity–time graph: A graph showing velocity as time changes.
• Gradient: Change in the vertical quantity divided by change in the horizontal quantity.
• Intersection: A point shared by two graphs.
• Separation: The distance between two objects at a particular time.
• Displacement: Change in position, including direction.
• Relative velocity: The velocity of one object measured relative to another.
• Closing speed: The rate at which the separation between approaching objects decreases.
• Tangent: A straight line matching a curve’s instantaneous slope at a point.
• Initial position: An object’s position at the chosen starting time.

Key takeaways

• Position–time gradients give velocity; their magnitudes give speed.
• Velocity–time gradients give acceleration.
• Position–time intersections show equal positions.
• Velocity–time intersections show equal velocities.
• Signed area under a velocity–time graph gives displacement.
• Use starting positions and displacements together to determine whether objects meet.
• Always check the axes, scales, directions and time reference before comparing motions.

5. Graphical Problem Solving

Why use graphs to solve motion problems?

Motion graphs show how quantities such as position, velocity and acceleration change over time. They can reveal information that may be difficult to recognize from an equation or written description.

Graphs can help us determine:

• Where an object is located.
• How fast it is moving.
• Which direction it is moving.
• Whether it is speeding up or slowing down.
• How far it travels.
• When its motion changes.
• Whether two objects meet.

To solve a graphical problem successfully, begin by identifying the quantities and units shown on the axes.

The main types of motion graphs

Position–time graphs

A position–time graph shows an object’s position relative to a chosen origin.

The vertical coordinate gives the object’s position.

The gradient gives velocity:

Velocity = change in position ÷ change in time

v = Δx / Δt

A positive gradient represents motion in the positive direction. A negative gradient represents motion in the negative direction.

Distance–time graphs

A distance–time graph usually shows the total distance travelled.

Its gradient gives speed:

Speed = change in distance ÷ change in time

The total distance normally cannot decrease, so a cumulative distance–time graph should not slope downwards.

Velocity–time graphs

A velocity–time graph shows the object’s velocity at each moment.

Its gradient gives acceleration:

Acceleration = change in velocity ÷ change in time

a = Δv / Δt

The signed area between the graph and the time axis gives displacement.

Speed–time graphs

A speed–time graph shows how quickly an object moves without representing direction.

Its gradient gives the rate of change of speed, and the area under the graph gives distance travelled.

Acceleration–time graphs

An acceleration–time graph shows how acceleration changes.

The signed area under the graph gives the change in velocity:

Change in velocity = area under an acceleration–time graph

This relationship allows us to move from an acceleration graph to a velocity graph.

Connecting the graph types

The same motion can be represented using several different graphs.

These relationships form a chain:

Position → gradient → velocity → gradient → acceleration

Working in the opposite direction involves accumulation:

Acceleration → signed area → change in velocity → signed area → displacement

An area alone gives a change. An initial value is needed to find the final quantity.

For example:

Final velocity = initial velocity + change in velocity

Final position = initial position + displacement

The three graphs show the same 12-second journey. The vertical dashed lines divide it into three stages.

Interpreting the example journey

The object begins from rest at a position of 0 m.

Its motion has three stages:

• From 0 to 4 seconds, it accelerates uniformly.
• From 4 to 8 seconds, it moves at constant velocity.
• From 8 to 12 seconds, it decelerates uniformly to rest.

Each graph describes these stages in a different way.

Stage 1: accelerating from rest

From 0 to 4 seconds, the acceleration is:

a = +2 m/s²

The area under the acceleration–time graph gives the change in velocity:

Δv = 2 × 4
Δv = 8 m/s

Because the initial velocity is zero:

v = 0 + 8
v = 8 m/s

On the velocity–time graph, velocity rises in a straight line from 0 to 8 m/s.

The area under this part of the velocity–time graph is a triangle:

Displacement = ½ × base × height
Displacement = ½ × 4 × 8
Displacement = 16 m

The position therefore increases from 0 m to 16 m.

On the position–time graph, the curve becomes progressively steeper. This shows that velocity is increasing.

Stage 2: moving at constant velocity

From 4 to 8 seconds, the velocity remains at:

v = 8 m/s

The horizontal velocity–time line has zero gradient, so:

a = 0 m/s²

The displacement is the rectangular area under the velocity–time graph:

Displacement = base × height
Displacement = 4 × 8
Displacement = 32 m

The object began this stage at 16 m:

Final position = 16 + 32
Final position = 48 m

On the position–time graph, this stage is a straight line. Its constant gradient represents constant velocity.

A velocity of 8 m/s does not require an acceleration of 8 m/s². An object can maintain a non-zero velocity while its acceleration is zero.

Stage 3: slowing to rest

From 8 to 12 seconds, velocity falls from 8 m/s to 0 m/s.

The acceleration is the gradient of the velocity–time graph:

a = (0 − 8) / (12 − 8)
a = −8 / 4
a = −2 m/s²

The negative acceleration acts opposite to the positive velocity, so the object slows down.

The displacement is the triangular area:

Displacement = ½ × 4 × 8
Displacement = 16 m

The object began this stage at 48 m:

Final position = 48 + 16
Final position = 64 m

The position–time graph gradually becomes less steep and ends horizontally. The horizontal tangent at 12 seconds shows that the final velocity is zero.

Finding total distance and displacement

All velocities in the example are positive. The object never reverses direction.

The total displacement equals the complete area under the velocity–time graph:

Displacement = 16 + 32 + 16
Displacement = 64 m

Because the object moves in only one direction:

Total distance = 64 m

If part of a velocity–time graph lies below the time axis, total distance and displacement must be handled differently.

• Displacement uses signed areas.
• Distance uses the magnitudes of all areas.

Finding average speed and average velocity

Average speed is:

Average speed = total distance ÷ total time

Average speed = 64 / 12
Average speed = 5.33 m/s

Average velocity is:

Average velocity = displacement ÷ total time

Average velocity = 64 / 12
Average velocity = 5.33 m/s in the positive direction

They are equal in this example because the object never changes direction.

If the object reversed direction, total distance would be greater than the magnitude of displacement.

Extracting values from gradients

A gradient measures the rate at which the vertical quantity changes relative to the horizontal quantity.

Gradient of a straight line

Choose two clear points that lie far apart:

Gradient = (y₂ − y₁) / (x₂ − x₁)

Using widely separated points generally reduces the effect of small reading errors.

Gradient of a curve

A curved graph has a changing gradient.

To estimate an instantaneous value:

1. Locate the required point.
2. Draw a tangent that touches the curve at that point.
3. Choose two widely separated points on the tangent.
4. Calculate the tangent’s gradient.

The chosen points should lie on the tangent. They do not have to lie on the original curve.

Worked example: velocity from a position graph

Between 4 and 8 seconds in the example:

v = (48 − 16) / (8 − 4)
v = 32 / 4
v = 8 m/s

This agrees with the velocity–time graph.

Extracting values from areas

The units of an area come from multiplying the units on the two axes.

For a velocity–time graph:

(m/s) × s = m

Therefore, the area represents displacement.

For an acceleration–time graph:

(m/s²) × s = m/s

Therefore, the area represents change in velocity.

Common area formulas

Rectangle

Area = base × height

Triangle

Area = ½ × base × perpendicular height

Trapezium

Area = ½ × (sum of parallel sides) × perpendicular distance

For a curved graph, the area may be estimated using strips or calculated using integration.

Worked example: motion involving a change of direction

A velocity–time graph shows:

• Velocity increases uniformly from 0 to 6 m/s during the first 3 seconds.
• Velocity decreases uniformly from 6 m/s at 3 seconds to −2 m/s at 7 seconds.

Find the acceleration from 3 to 7 seconds

a = [−2 − 6] / (7 − 3)
a = −8 / 4
a = −2 m/s²

Find when the object changes direction

The object changes direction when velocity crosses zero.

Starting at 6 m/s and decreasing at 2 m/s each second:

Time needed to reach zero = 6 / 2
Time needed = 3 s

This occurs 3 seconds after t = 3 s:

Direction changes at t = 6 s.

Find the displacement

From 0 to 3 seconds:

Area = ½ × 3 × 6
Area = 9 m

From 3 to 6 seconds:

Area = ½ × 3 × 6
Area = 9 m

From 6 to 7 seconds, the graph lies below the axis:

Signed area = −(½ × 1 × 2)
Signed area = −1 m

Total displacement:

9 + 9 − 1 = 17 m

Find the total distance

Use the magnitude of each area:

9 + 9 + 1 = 19 m

The displacement and distance differ because the object reverses direction.

Combining information from two graphs

Sometimes one graph does not contain enough information to answer a question.

Suppose an acceleration–time graph shows an acceleration of 3 m/s² for 5 seconds. The area gives:

Δv = 3 × 5
Δv = 15 m/s

This does not give the final velocity unless the initial velocity is known.

If a velocity–time graph or written statement shows an initial velocity of 4 m/s:

Final velocity = 4 + 15
Final velocity = 19 m/s

Similarly, the area under a velocity–time graph gives displacement. To find final position, add the initial position.

Comparing multiple objects

When two objects appear on the same graph, an intersection has a meaning determined by the vertical axis.

On a position–time graph

An intersection means the objects have the same position at the same time. They may meet or pass one another.

Their velocities may still be different. Compare the gradients at the intersection.

On a velocity–time graph

An intersection means the objects have the same velocity at the same time.

It does not show that they have the same position.

To determine whether they meet, compare their starting positions and accumulated displacements.

On an acceleration–time graph

An intersection means the objects have the same acceleration at that time.

It does not establish equal velocity or position.

Real-world example: analyzing a vehicle journey

A delivery van starts from a traffic light, travels along a straight road and stops at another light.

A velocity–time graph shows:

• 0–5 s: velocity rises uniformly from 0 to 10 m/s.
• 5–20 s: velocity remains at 10 m/s.
• 20–25 s: velocity falls uniformly to 0 m/s.

Acceleration during the first stage

a = (10 − 0) / 5
a = 2 m/s²

Acceleration during braking

a = (0 − 10) / 5
a = −2 m/s²

Total distance

First triangle:

½ × 5 × 10 = 25 m

Rectangle:

15 × 10 = 150 m

Final triangle:

½ × 5 × 10 = 25 m

Total distance:

25 + 150 + 25 = 200 m

Average speed

Average speed = 200 / 25
Average speed = 8 m/s

A strong written conclusion would be:

“The van travelled 200 m. This is the total area under the speed–time graph: two triangular areas of 25 m and a rectangular area of 150 m. Dividing by the 25-second journey time gives an average speed of 8 m/s.”

Explaining answers with graphical evidence

A complete graphical explanation should include:

• The relevant graph feature.
• The values read from the graph.
• The calculation or comparison.
• The correct units.
• A conclusion connected to the motion.

For example:

“The object accelerated at 2 m/s² from 0 to 4 seconds because the velocity–time graph rose from 0 to 8 m/s. Its gradient was (8 − 0)/(4 − 0) = 2 m/s².”

Avoid vague statements such as:

“The line goes up, so the object is faster.”

A rising line has different meanings on different graph types.

Checking whether an answer is reasonable

Use several checks before accepting an answer.

Check the units

• Position–time gradient: m/s.
• Velocity–time gradient: m/s².
• Velocity–time area: m.
• Acceleration–time area: m/s.

Check the sign

• Positive velocity: motion in the chosen positive direction.
• Negative velocity: motion in the opposite direction.
• Negative acceleration: acceleration in the negative direction.

A negative value does not automatically mean the object is slowing down.

Compare connected graphs

• Increasing velocity should match non-zero acceleration.
• Constant velocity should match zero acceleration.
• A stationary object should have a horizontal position–time graph.
• A position–time graph should become steeper as speed increases.

Check the scale

Do not assume each grid square represents one unit. Read labels carefully.

Check the physical context

A mathematical graph may continue beyond the period that makes sense physically. For example, a vehicle’s speed should not become negative unless it actually reverses direction.

Common misconceptions

• “The area under a position–time graph gives distance.” The gradient of a position–time graph gives velocity.
• “The gradient of a velocity–time graph gives velocity.” It gives acceleration.
• “A horizontal velocity–time line means the object is stationary.” It means constant velocity; the object is stationary only if that velocity is zero.
• “A line below the time axis means negative distance.” On a velocity graph, it represents motion in the negative direction.
• “Zero velocity means zero acceleration.” An object can have zero velocity at an instant while still accelerating.
• “An intersection always means two objects meet.” That interpretation applies to compatible position–time graphs.
• “Area gives the final value directly.” Area often gives a change, which must be combined with an initial value.
• “A steeper-looking graph always represents a greater rate.” Scales and units must be considered.

Did you know?

A position–time graph can be curved even when the acceleration is constant.

When acceleration is constant, velocity changes linearly with time. Since position accumulates that changing velocity, its graph becomes quadratic.

This is why the position–time graph in the main example curves during acceleration and deceleration, while the velocity–time graph consists of straight lines.

Key terms

• Position–time graph: A graph showing position relative to an origin as time changes.
• Velocity–time graph: A graph showing velocity as time changes.
• Acceleration–time graph: A graph showing acceleration as time changes.
• Gradient: Change in the vertical quantity divided by change in the horizontal quantity.
• Tangent: A line used to estimate a curve’s instantaneous gradient.
• Signed area: Area counted as positive above the time axis and negative below it.
• Displacement: Change in position, including direction.
• Distance: Total length of the path travelled.
• Instantaneous velocity: Velocity at a particular moment.
• Average velocity: Total displacement divided by total time.
• Change of direction: The event that occurs when velocity passes through zero and changes sign.

Key takeaways

• Begin by identifying the graph type, axes, scales and units.
• Position–time gradient gives velocity.
• Velocity–time gradient gives acceleration.
• Velocity–time area gives displacement.
• Acceleration–time area gives change in velocity.
• Add initial values when an area gives only a change.
• Separate positive and negative areas when calculating displacement or distance.
• Combine evidence from related graphs to check your interpretation.
• Support conclusions with graph values, calculations and units.