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CIE IGCSE Physics: Co- Your notes

ordinated Sciences (Double Award)
Motion
Contents
Speed & Velocity
Acceleration
Distance-Time Graphs
Speed-Time Graphs
Calculating Acceleration from Speed-Time Graphs
Freefall

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Speed & Velocity

Your notes
Speed
The speed of an object is defined as
Distance travelled per unit time
Speed is a scalar quantity
This is because it only contains a magnitude (without a direction)
For objects that are moving at a constant speed, the equation for calculating speed is:
s
v=
t
Where:
v = speed, measured in metres per second (m/s)
s = distance travelled, measured in metres (m)
t = time, measured in seconds (s)

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Average speed
Your notes
The speed of an object can vary throughout its journey
Therefore, it is often more useful to know an object's average speed
Examples of average speeds

A hiker might have an average speed of 2.0 m/s, whereas a particularly excited bumble bee can have
average speeds of up to 4.5 m/s

The equation for calculating the average speed of an object is:

distance travelled
Average speed =
time taken
Average speed considers the total distance travelled and the total time taken

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Your notes

Formula triangle for average speed, distance moved and time taken

How to use formula triangles

Formula triangles are really useful for knowing how to rearrange physics equations
To use them:
1. Cover up the quantity to be calculated, this is known as the 'subject' of the equation
2. Look at the position of the other two quantities
If they are on the same line, this means they are multiplied
If one quantity is above the other, this means they are divided - make sure to keep the order of
which is on the top and bottom of the fraction!

In the example below, to calculate average speed, cover-up the variable speed so that only distance
and time are left
The equation is revealed as:
distance
speed =
time

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Your notes

To use a formula triangle, simply cover up the quantity you wish calculate and the structure of the
equation is revealed

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Worked example
Your notes
Planes fly at typical average speeds of around 250 m/s.
Calculate the distance travelled by a plane moving at this average speed for 2 hours.
Answer:
Step 1: List the known quantities
Average speed = 250 m/s
Time taken = 2 hours
Step 2: Write the relevant equation

distance travelled
average speed =
time taken
Step 3: Rearrange to make distance moved the subject

distance travelled = average speed × time taken

Step 4: Convert any units
The time given in the question is not in standard units
Convert 2 hours into seconds:
2 hours = 2 × 60 × 60
2 hours = 7200 s
Step 5: Substitute the values for average speed and time taken

distance travelled = 250 × 7200

distance travelled = 1 800 000 m

Examiner Tip
Rearranging equations is an important skill in Physics. You can use the equation triangles to help you
practice, but it is better not to rely on them because they do not work for all equations you may need to
rearrange in the exam.

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Velocity
Extended tier only Your notes
Velocity is a vector quantity with magnitude and direction
Velocity is defined as:
Speed in a given direction
The direction of a velocity can be given in words
For example, 20 m/s east
Or the direction of velocity can be given using a positive or negative value
For example, −20 m/s
A positive direction is typically in the direction of the initial motion, to the right, or upward
A negative velocity is typically in the opposite direction to the initial velocity, to the left, or downward
Comparing speed and velocity

The cars in the diagram above have the same speed (a scalar quantity) but different velocities (a vector
quantity). Fear not, they are in different lanes!

Examiner Tip
The positive and negative values of velocity can be assigned to any direction as long as the negative
velocity is in the opposite direction to the positive value. You can decide which direction you assign to
be positive as long as you are consistent throughout a question.

The equation for velocity is very similar to the equation for speed:
s
v=
t
Where:
v = velocity in metres per second (m/s)
s = displacement, measured in metres (m)
t = time, measured in seconds (s)
Velocity is a vector quantity, so it uses displacement, s, which is another vector quantity

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Acceleration
Your notes
Acceleration
An object can change its velocity in several ways:
speeding up
slowing down
changing direction
Any change in an object's velocity is an acceleration
When an object speeds up, it is accelerating
When an object slows down, it is decelerating

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Calculating acceleration
Extended tier only Your notes
Acceleration describes how the velocity of an object changes over time
Acceleration is defined as:
The rate of change of velocity
In other words, acceleration is the change in velocity per unit time
The acceleration of an object is often changing throughout an object's journey

Therefore, is it often useful to know the average acceleration

∆v
a=
∆t
Where:
a = acceleration in metres per second squared (m/s2)
∆ v = change in velocity in metres per second (m/s)
∆ t = time taken in seconds (s)
Formula triangle for acceleration, change in velocity and change in time

To use a formula triangle, simply cover up the quantity you wish calculate and the structure of the
equation is revealed
Information on how to use a formula triangle can be found in Speed & velocity

Change in velocity
The change in velocity is the difference between the initial and final velocity:

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∆v = v − u
Where: Your notes
∆ v = change in velocity in metres per second (m/s)
v = final velocity in metres per second (m/s)
u = initial velocity in metres per second (m/s)

If an object speeds up, its acceleration is positive

If an object slows down, its acceleration is negative
Acceleration is positive if it is in the same direction as the motion of the object
Acceleration of different objects

A rocket speeding up (accelerating) and a car slowing down (decelerating)

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Worked example
Your notes
A Japanese bullet train decelerates at a constant rate in a straight line. The velocity of the train
decreases from 50 m/s to 42 m/s in 30 seconds.
(a) Calculate the change in velocity of the train.
(b) Calculate the deceleration of the train, and explain how your answer shows the train is slowing
down.
Answer:
Part (a)
Step 1: List the known quantities

Initial velocity, u = 50 m/s

Final velocity, v = 42 m/s

Step 2: Write the equation for change in velocity

∆v = v − u
Step 3: Substitute values for final and initial velocity

∆ v = 42 − 50

∆ v = − 8 m/s
Part (b)
Step 1: List the known quantities

Change in velocity, ∆ v = − 8 m/s

Time taken, ∆ t = 30 s

Step 2: Write the equation for acceleration

∆v
a=
∆t
Step 3: Substitute the values for change in velocity and time

−8
a=
30
a = − 0 . 27 m/s2

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Step 4: Interpret the value for deceleration

The answer is negative, which indicates the train is slowing down Your notes

Examiner Tip
Remember, the units for acceleration are metres per second squared, m/s2. In other words,
acceleration measures how much the velocity (m/s) changes every second, so the units are metres per
second per second (m/s/s).

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Distance-Time Graphs
Your notes
Distance-time graphs
A distance-time graph is used to describe the motion of an object and calculate its speed

Distance-time graph of an object moving at a constant speed

The graph shows a moving object moving further away from its origin at a constant speed

Constant speed on a distance-time graph

If an object is moving at a constant speed, the distance-time graph will be a straight line
If the constant speed is zero, the line will be horizontal
If the constant speed is non-zero, the line will have a gradient
If an object has a speed of zero, the object is stationary
The distance moved by the object over time is zero
The gradient of a distance-time graph represents the magnitude of the object's velocity, or its speed
A steeper slope, or a higher gradient, represents a greater speed

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A shallower slope, or a lower gradient, represents a slower speed

Different speeds on a distance-time graph Your notes

Both of these objects are moving at a constant speed, because the lines are straight. The steeper slope
represents the faster speed and the shallower line represents the slower speed.

Changing speed on a distance-time graph

Often, the speed of an object is not constant
If the speed of an object is changing, the object is accelerating
If an object is accelerating, the distance-time graph will be a curved line
A curve on a distance-time graph is a changing gradient
If the gradient increases over time, the speed is increasing over time
If the gradient decreases over time, the speed is decreasing over time
Speed of an object increasing and decreasing on a distance-time graph

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Your notes

Changing speeds are represented by changing slopes, or gradients. The red line shows a decreasing
gradient and represents an object slowing down, or decelerating. The green line shows an increasing
gradient and represents an object speeding up, or accelerating.

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Using distance-time graphs

The speed of a moving object can be calculated from the gradient of the line on a distance-time Your notes
graph:
∆y
speed = gradient =
∆x

The speed of an object can be found by calculating the gradient of a distance-time graph
∆ y is the change in y (distance) values
∆ x is the change in x (time) values

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Worked example
Your notes
A distance-time graph is drawn below for part of a train journey. The train is travelling at a constant
speed.

Calculate the speed of the train.

Answer:
Step 1: Draw a large gradient triangle on the graph
The image below shows a large gradient triangle drawn with dashed lines
∆ y and ∆ x are labelled, using the units as stated on each axes

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Your notes

Step 2: Convert units for distance and time into standard units

The distance travelled, s = 8 km × 1000 m = 8000 m

The time taken, t = 6 min × 60 s = 360 s

Step 3: State that speed is equal to the gradient of a distance-time graph

The gradient of a distance-time graph is equal to the speed of a moving object:
∆y s
gradient = v = =
∆x t
Step 4: Substitute values to calculate the speed

8000
v=
360
v = 22. 2 m/s

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Worked example
Your notes
A student decides to take a stroll to the park. They find a bench in a quiet spot, take a seat, and read a
book on black holes. After some time reading, the student realises they lost track of time and runs
home.
A distance-time graph for the trip is drawn below.

(a) How long does the student spend reading the book?
(b) Which section of the graph represents the student running home?
(c) What is the total distance travelled by the student?

Answer:
Part (a)
The student spends 40 minutes reading his book
The flat section of the line (section B) represents an object which is stationary, so section B
represents the student sitting on the bench reading
This section lasts for 40 minutes

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Your notes

Part (b)
Section C represents the student running home
The slope of the line in section C is steeper than the slope in section A
This means the student was moving at a faster speed (running) in section C
Part (c)
The total distance travelled by the student is 0.6 km
The total distance travelled by an object is given by the final point on the line; in this case, the line
ends at 0.6 km on the distance axis

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Your notes

Examiner Tip
When calculating a gradient, use the entire line where possible. Examiners tend to award credit if they
see a large gradient triangle used, so you need to actually draw the lines directly on the graph itself!
Remember to check the units on each axis. These may not always be in standard units; in our example,
the unit of distance was km and the unit of time was minutes. Double-check which units to use in your
answer.
You can read more about the use of graphs in exams in the article Graph skills for GCSE Physics

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Speed-Time Graphs
Your notes
Speed-time graphs
A speed-time graph is used to describe the speed of an object and calculate its acceleration

Constant acceleration on a speed-time graph

If an object is moving at a constant acceleration, the speed-time graph will be a straight line
If the constant acceleration is zero, the line will be horizontal
If the constant speed is non-zero, the line will have a gradient
If an object has an acceleration of zero, the object is travelling at a constant velocity
Its velocity is not changing over time
If the constant speed is zero, then the object is stationary
Motion on a speed-time graph

This image shows how to interpret the slope of a speed-time graph

The gradient of a speed-time graph represents the object's acceleration

A steeper slope, or a higher gradient, represents a greater acceleration
A shallower slope, or a lower gradient, represents a slower acceleration

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If the gradient is positive, the line slopes upward

A positive gradient represents an increasing speed, or acceleration
Your notes
If the gradient is negative, the line slopes downward
A negative gradient represents a decreasing speed, or deceleration

Speeding up and slowing down on a speed-time graph

Both of these objects are moving at a constant acceleration, because the lines are straight. The positive
gradient represents an increasing speed or positive acceleration. The negative gradient represents a
decreasing speed or negative acceleration.

Examiner Tip
For CIE IGCSE Physics, you may be asked plot a graph of your own, or to interpret information given to
you in a graph. You can read more about graph skills in the article Graph skills in GCSE Physics

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Calculating distance from speed-time graphs

Your notes
Speed-time graphs can also be used to determine the distance travelled by an object

The area under a speed-time graph

The area under a speed-time graph represents the distance travelled

The area under a speed-time graph represents the distance travelled

If the area of a section of the speed-time graph forms a triangle, the area can be calculated using:
1
AT = bh
2
If the area of a section of the speed-time graph forms a rectangle, the area can be determined using:

A R = bh
Where:
A T = area of a triangle
A R = area of a rectangle

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b = base
h = height Your notes
The total distance travelled can be determined by finding the total area under the speed-time graph
The distance travelled for part of the journey can be determined by finding the area under the graph
for a specific time interval
Area under a speed-time graph split into sections

The area under a speed-time graph can split into triangular and rectangular sections

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Worked example
Your notes
The speed-time graph below shows a car journey that lasts for 160 seconds.

Calculate the total distance travelled by the car on this journey.

Answer:
Step 1: Recall that the area under a velocity-time graph represents the distance travelled
In order to calculate the total distance travelled, the total area underneath the line must be
determined
Step 2: Identify each enclosed area
In this example, there are five enclosed areas under the line
These can be labelled as areas 1, 2, 3, 4 and 5, as shown in the image below:

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Your notes

Step 3: Calculate the area of each enclosed shape under the line
Area 1 = area of a triangle
1
A1 = bh
2
1
A1 = × 40 × 17. 5
2
A 1 = 350 m
Area 2 = area of a rectangle
A 2 = bh

A 2 = 30 × 17. 5

A 2 = 525 m
Area 3 = area of a triangle
1
A3 = bh
2

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1
A3 = × 20 × 7 . 5
2 Your notes
A 3 = 75 m
Area 4 = area of a rectangle
A 4 = bh

A 4 = 20 × 17. 5

A 4 = 350 m
Area 5 = area of a triangle
1
A5 = bh
2
1
A5 = × 70 × 25
2
A 5 = 875 m
Step 4: Calculate the total distance travelled by finding the total area under the line
Add up each of the five areas enclosed:
total distance = A + 1
A2 + A3 + A4 + A5

total distance = 350 + 525 + 75 + 350 + 875

total distance = 2175 m

Examiner Tip
Some areas will need to be split into a triangle and a rectangle to determine the area for a specific time
interval, like areas 3 & 4 in the worked example above.
If you are asked to find the distance travelled for a specific time interval, then you just need to find the
area of the section above that time interval.
For example, the distance travelled between 70 s and 90 s is the sum of Area 3 + Area 4

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Calculating Acceleration from Speed-Time Graphs

Your notes
Interpreting speed-time graphs
Extended tier only
When interpreting speed-time graphs, the shape of the graph can show:
constant acceleration
changing acceleration
The gradient of a speed-time graph shows the acceleration of a moving object
The gradient is positive if the object is accelerating (speed increases with time)
The gradient is negative if the object is decelerating (speed decreases with time)
Interpreting constant positive acceleration on a speed-time graph
When the acceleration is constant and non-zero:
the graph is a straight line
velocity is increasing at a constant rate, i.e. speed changes by the same amount in equal intervals
of time
Constant positive acceleration on a speed-time graph

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Your notes

A speed-time graph for an object with constant positive acceleration. Its speed increases by 5 m/s
every 20 s, showing that the rate at which the speed increases is constant.
Interpreting increasing positive acceleration on a speed-time graph
When the acceleration is increasing:
the graph is a curve
velocity is increasing at an increasing rate, i.e. the speed changes by the same amount in
increasingly shorter time intervals
Increasing positive acceleration on a speed-time graph

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Your notes

A speed-time graph for an object with changing positive acceleration. The time taken for the speed to
increase by 5 m/s decreases over time, showing that acceleration is increasing.
Interpreting decreasing positive acceleration on a speed-time graph
When the acceleration is decreasing:
the graph is a curve
velocity is increasing at a decreasing rate, i.e. the speed changes by the same amount in
increasingly longer time intervals
Decreasing positive acceleration on a speed-time graph

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Your notes

A speed-time graph for an object with changing positive acceleration. The time taken for the speed to
increase by 5 m/s increases over time, showing that acceleration is decreasing.
Interpreting decreasing negative acceleration on a speed-time graph
When the deceleration is decreasing:
the graph is a curve
the velocity is decreasing at a decreasing rate, i.e. the speed changes by the same amount in
increasingly longer intervals of time
Decreasing negative acceleration on a speed-time graph

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Your notes

A speed-time graph for an object with changing negative acceleration. The time taken for the speed to
decrease by 5 m/s increases over time, showing that deceleration is decreasing.

Examiner Tip
Interpreting graphs can be difficult, and students often struggle with this key skill
In CIE IGCSE Physics, interpreting graphs is a required skill
For your exam, you are also expected to calculate the gradient of a graph and the area under a
graph
Finding the area under a graph is covered in the revision note Speed-time graphs

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Calculating acceleration from speed-time graphs

Extended tier only Your notes
The acceleration of an object can be calculated from the gradient of a speed-time graph

Finding the gradient when acceleration is constant

When acceleration is constant, the speed-time graph will be a straight line
The gradient of a straight line can be found using:
∆y
gradient =
∆x
Where:
∆ y = change in y (speed) values
∆ x = change in x (time) values
Therefore, the gradient is equal to:
∆v
a=
∆t
Where:
a = acceleration, measured in metres per second squared (m/s2)
∆ v = change in speed, measured in metres per second (m/s)
∆ t = change in time, measured in seconds (s)

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The gradient of a speed-time graph for constant acceleration can be found using Δy divided by Δx
Your notes
Finding the gradient when acceleration is changing
When acceleration is changing, the speed-time graph will be a curve
The gradient of a point on a curve can be found by drawing a tangent to the curve
A tangent to the curve

The value of the gradient at a single point on a curve can be determined by finding the gradient of the
tangent to that point

The tangent provides a gradient that is representative of the gradient at a specific point on the curve
The gradient of the tangent can be found using:
∆y
gradient =
∆x
The value of the gradient at specific point on the curve represents the acceleration of the object at
that moment
This is called instantaneous acceleration

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Worked example
Your notes
A cyclist is training for a cycling tournament.
The speed-time graph below shows the cyclist's motion as they cycle along a flat, straight road.

(a) In which section (A, B, C, D, or E) of the speed-time graph is the cyclist's acceleration the largest?
(b) Calculate the cyclist's acceleration between 5 and 10 seconds.

Answer:
Part (a)
Step 1: Recall that the slope of a speed-time graph represents the magnitude of acceleration
The slope of a speed-time graph indicates the magnitude of acceleration
Therefore, the only sections of the graph where the cyclist is accelerating are sections B and D
Sections A, C, and E are flat; in other words, the cyclist is moving at a constant velocity (therefore,
not accelerating)

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Step 2: Identify the section with the steepest slope

Section D of the graph has the steepest slope Your notes
Hence, the largest acceleration is shown in section D

Part (b)
Step 1: Recall that the gradient of a speed-time graph gives the acceleration
Calculating the gradient of a slope on a speed-time graph gives the acceleration for that time
period
Step 2: Draw a large gradient triangle at the appropriate section of the graph
A gradient triangle is drawn for the time period between 5 and 10 seconds

Step 3: Calculate the size of the gradient and state this as the acceleration
The acceleration is given by the gradient, which can be calculated using:
∆y
a=
∆x

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5
a=
5 Your notes
a = 1 m/s2
Therefore, the cyclist accelerated at 1 m/s2 between 5 and 10 seconds

Examiner Tip
Use the entire slope, where possible, to calculate the gradient. Examiners tend to award credit if they
see a large gradient triangle used.
Remember to actually draw the lines directly on the graph itself, particularly when the question asks
you to use the graph to calculate the acceleration.

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Worked example
Your notes
A skydiver jumps from a plane and reaches terminal velocity after 15 seconds. A speed-time graph of
their motion is shown below.
Use the graph to find the acceleration at 5 seconds.

Answer:
Step 1: Draw a tangent to the curve at the point where t = 5 s

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Your notes

Step 2: Calculate the gradient of the tangent

∆y
a=
∆x

58 − 20
a=
9 . 75 − 0 . 75
38
a=
9.0
a = 4 . 2 m/s2 (2 s . f . )

Examiner Tip
The CIE IGCSE Co-ordinated Sciences specification includes knowing how to calculate the gradient
of a tangent to the curve in the maths skills section. This means that you could be asked to
demonstrate this skill in any topic. The skills in this revision note are applicable to any type of graph. For
more information on interpreting graphs in Physics, see the article Graph skills in GCSE Physics

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Freefall
Your notes
Acceleration of free fall
Extended tier only
In the absence of air resistance, all objects fall with the same acceleration regardless of their mass
This is called the acceleration of freefall
This is also sometimes called acceleration due to gravity

acceleration of freefall = g = 9 . 8 m/s2

In the absence of air resistance, Galileo discovered that all objects (near Earth's surface) fall with an
acceleration of about 9.8 m/s2
This means that for every second an object falls, its velocity will increase by 9.8 m/s

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Examiner Tip
Your notes
The symbol also stands for the gravitational field strength, and can be used to calculate the
force of weight acting an object using its mass:

Where:
= the force of weight acting on an object, measured in newtons (N)
= mass of object, measured in kilograms (kg)
= gravitational field strength, measured in newtons per kilogram (N/kg)

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