Forces and motion

What you'll learn

How to describe motion using distance, displacement, speed and velocity.
How to calculate speed and acceleration, including from distance–time and velocity–time graphs.
How Newton’s three laws explain changes in motion.
Why stopping distance depends on reaction time, speed, road conditions and braking.

Distance and displacement

When you describe motion, start by asking: “How far did it move?” and “In what direction did it end up?”

Definition

Scalars and vectors

A scalar quantity has magnitude only, meaning size only. A vector quantity has magnitude and direction.

Distance is how far an object moves along its actual path. It has no direction, so distance is a scalar.

Displacement is the straight-line change in position from the start point to the finish point, including the direction. Displacement is a vector.

This diagram shows why distance and displacement can be different, even for the same journey.

Diagram comparing distance as total path length and displacement as straight-line change in position with direction

Example

Finding distance and displacement

A student walks 40 m east, then 15 m west.

The distance is the total path length: $m40\text{ m} + 15\text{ m} = 55\text{ m}$ .
Treat east as positive, so the final position is $m40\text{ m} - 15\text{ m} = 25\text{ m}$ from the start.
The displacement is 25 m east, because displacement needs both size and direction.

Common Mistake

Forgetting direction

“25 m” is not a full displacement answer. You need something like 25 m east or 25 m to the right.

Speed and average speed

Speed is how fast an object is moving. It does not include direction, so speed is a scalar.

For an object moving at constant speed:

s = v t

where $s$ is distance travelled in metres, $v$ is speed in metres per second, and $t$ is time in seconds.

For motion that is not uniform, use:

time\text{average speed} = \frac{\text{total distance}}{\text{total time}}

Typical speeds you should know are: walking about 1.5 m/s, running about 3 m/s, cycling about 6 m/s, and sound in air about 330 m/s. Transport values vary, but a car is often around 25 m/s, a train around 50 m/s, and an aeroplane around 250 m/s.

Example

Calculating average speed

A cyclist travels 900 m in 3 minutes.

Convert the time to seconds: 3 minutes is $\times 60 = 180\text{ s}$ .
Substitute into the speed equation: $\frac{900}{180}$ .
Calculate the average speed: $5\text{ m/s}$ , which is a sensible cycling speed.

Tip

Unit conversion shortcut

To convert from km/h to m/s, divide by 3.6. To convert from m/s to km/h, multiply by 3.6.

Velocity

Velocity is speed in a given direction. It is a vector.

For example, 10 m/s north and 10 m/s south are the same speed, but different velocities, because the directions are different.

If you are on Higher Tier, you also need this idea: an object moving in a circle at constant speed is still changing velocity, because its direction is constantly changing.

Distance–time graphs

A distance–time graph shows how the distance travelled changes with time. The gradient tells you the speed:

time\text{speed} = \frac{\text{change in distance}}{\text{change in time}}

A steeper line means a greater speed. A horizontal line means the object is stationary. A curve means the speed is changing. On Higher Tier, the speed at one instant can be found by drawing a tangent to the curve and finding its gradient.

Distance-time and velocity-time graph summary showing gradients, acceleration and area under the graph

Example

Finding speed from a distance–time graph

A straight section of a graph goes from 5 m at 2 s to 29 m at 8 s.

Find the change in distance: $m29\text{ m} - 5\text{ m} = 24\text{ m}$ .
Find the change in time: $s8\text{ s} - 2\text{ s} = 6\text{ s}$ .
Find the gradient: $\frac{24}{6} = 4\text{ m/s}$ .

Common Mistake

Using one point instead of a gradient

For a graph, do not just read one coordinate and call it the speed. Speed comes from the gradient, so you need a change in distance divided by a change in time.

Acceleration

Acceleration is the rate of change of velocity. It tells you how quickly velocity changes.

Definition

Acceleration

Acceleration is calculated using:

\frac{\Delta v}{t}

where $a$ is acceleration in m/s², $Δv\Delta v$ is change in velocity in m/s, and $t$ is time in seconds.

An object that slows down is decelerating. Near the Earth’s surface, an object falling freely under gravity accelerates at about 9.8 m/s², if air resistance is ignored.

Example

Calculating acceleration

A car increases its velocity from 5 m/s to 25 m/s in 10 s.

Calculate the change in velocity: $m/s25\text{ m/s} - 5\text{ m/s} = 20\text{ m/s}$ .
Substitute into the equation: $\frac{20}{10}$ .
Calculate the acceleration: $2\text{ m/s}^2$ .

Velocity–time graphs

On a velocity–time graph, the gradient gives acceleration:

\frac{\Delta v}{\Delta t}

For Higher Tier, the area under a velocity–time graph gives the distance travelled, as long as the velocity is above the time axis. If the graph has awkward shapes, you can estimate the area by counting squares.

Example

Using a velocity–time graph

A vehicle’s velocity increases uniformly from 0 m/s to 12 m/s in 6 s.

The acceleration is the gradient: $\frac{12 - 0}{6 - 0} = 2\text{ m/s}^2$ .
For Higher Tier, the distance is the triangular area under the graph: $m\frac{1}{2} \times 6 \times 12 = 36\text{ m}$ .
The graph shows uniform acceleration because the velocity–time line is straight.

Uniform acceleration equation

For uniform acceleration, you can use:

v^2 - u^2 = 2as

where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration, and $s$ is distance.

This equation is useful when time is not given.

Example

Using the uniform acceleration equation

A skateboarder starts from rest and accelerates uniformly at 2 m/s² over 25 m. Find the final velocity.

Identify the values: $u = 0$ , $a = 2$ , and $s = 25$ .
Substitute into the equation: $v2−02=2×2×25=100v^2 - 0^2 = 2 \times 2 \times 25 = 100$ .
Square root both sides: $10\text{ m/s}$ .

Terminal velocity

A fluid is a liquid or gas. When an object falls through a fluid, gravity acts downwards and drag acts upwards.

At first, gravity is bigger than drag, so the object accelerates. As speed increases, drag increases. Eventually, drag balances weight, so the resultant force is zero and the object moves at a constant terminal velocity.

Key Idea

Terminal velocity

Terminal velocity happens when the resultant force on a falling object is zero, so acceleration is zero and the object continues at constant velocity.

Newton’s First Law

Key Idea

Newton’s First Law

If the resultant force on an object is zero, a stationary object stays stationary, and a moving object continues at the same velocity.

The resultant force is the overall force after all forces have been combined, taking direction into account.

A vehicle travelling at a steady speed has balanced forces: the driving force forwards equals the resistive forces backwards. Its velocity only changes if there is a resultant force.

If you are on Higher Tier, the tendency of objects to keep their state of rest or uniform motion is called inertia.

Newton’s Second Law

Newton’s Second Law says acceleration is proportional to resultant force and inversely proportional to mass. In symbols, for constant mass, $\propto F$ .

The equation is:

F = ma

where $F$ is resultant force in newtons, $m$ is mass in kilograms, and $a$ is acceleration in m/s².

Example

Calculating resultant force

A 900 kg car accelerates at 1.6 m/s².

Choose Newton’s Second Law: $F = ma$ .
Substitute the values: $\times 1.6$ .
Calculate the resultant force: $1440\text{ N}$ .

For Higher Tier, inertial mass means how difficult it is to change an object’s velocity. It can be found from:

\frac{F}{a}

Required practical: force, mass and acceleration

You need to understand how to investigate $F = ma$ .

Use a trolley on a runway with a way of measuring acceleration, such as light gates, a motion sensor, or a data logger. To test force, keep the mass constant and vary the pulling force. To test mass, keep the force constant and add masses to the trolley.

Repeat readings, calculate means, and plot graphs. Acceleration should increase when force increases, and decrease when mass increases.

Newton’s Third Law

Key Idea

Newton’s Third Law

Whenever two objects interact, the forces they exert on each other are equal and opposite.

These forces act on different objects, so they do not cancel each other out.

Example

Applying Newton’s Third Law

A person pushes backwards on the ground to start running forwards.

The person exerts a backwards force on the ground.
The ground exerts an equal and opposite forwards force on the person.
The forwards force from the ground accelerates the person forwards.

Common Mistake

Mixing up balanced forces and force pairs

The weight of a book and the support force from the table can be balanced, but they are not a Newton’s Third Law pair because both act on the same object: the book.

Stopping distance

Stopping distance is the total distance a vehicle travels from the moment the driver notices a hazard to the moment the vehicle stops.

distance\text{stopping distance} = \text{thinking distance} + \text{braking distance}

Thinking distance happens during the driver’s reaction time. Braking distance happens while the brakes slow the vehicle down.

Stopping distance diagram showing thinking distance, braking distance, and factors that increase each

Reaction times vary from person to person, but typical values are about 0.2 s to 0.9 s. Tiredness, drugs, alcohol and distractions increase reaction time, so they increase thinking distance.

You can measure reaction time using a ruler-drop test or a computer reaction timer. Good experiments use repeats, calculate a mean, and control variables such as the same hand, same ruler position, and same instructions.

Example

Calculating stopping distance

A car travels at 20 m/s. The driver’s reaction time is 0.7 s, and the braking distance is 25 m.

Calculate thinking distance: $20 \times 0.7 = 14\text{ m}$ .
Add the braking distance: $m14\text{ m} + 25\text{ m} = 39\text{ m}$ .
The stopping distance is 39 m.

Braking distance and safety

Braking distance is increased by adverse road conditions, such as wet or icy roads, and poor vehicle condition, especially worn tyres or faulty brakes.

When brakes are applied, friction does work. This reduces the vehicle’s kinetic energy and increases the temperature of the brakes. Greater speed means more kinetic energy, so a larger braking force is needed to stop in the same distance.

Large decelerations can be dangerous because they may cause loss of control and can make brakes overheat.

Example

Estimating braking force

A 1200 kg car slows from 30 m/s to rest in 6 s.

Calculate acceleration: $\frac{0 - 30}{6} = -5\text{ m/s}^2$ .
Use Newton’s Second Law: $\times -5$ .
The braking force is $N-6000\text{ N}$ , meaning 6000 N backwards.

Exam technique

In the exam

Always use SI units: metres, seconds, kilograms, newtons, and metres per second.
For graph questions, write gradient as change in vertical axis divided by change in horizontal axis.
For vector answers, include direction: for example, “12 m east” or “4 m/s upwards”.
For braking questions, separate factors that affect thinking distance from factors that affect braking distance.

Self review

Check yourself

A walker travels 80 m north then 30 m south. What are the distance and displacement?
What does the gradient of a distance–time graph tell you?
Why does an icy road increase braking distance but not the driver’s reaction time?

What you'll learn

How to describe motion using distance, displacement, speed and velocity.
How to calculate speed and acceleration, including from distance–time and velocity–time graphs.
How Newton’s three laws explain changes in motion.
Why stopping distance depends on reaction time, speed, road conditions and braking.

Distance and displacement

When you describe motion, start by asking: “How far did it move?” and “In what direction did it end up?”

Definition

Scalars and vectors

A scalar quantity has magnitude only, meaning size only. A vector quantity has magnitude and direction.

Distance is how far an object moves along its actual path. It has no direction, so distance is a scalar.

Displacement is the straight-line change in position from the start point to the finish point, including the direction. Displacement is a vector.

This diagram shows why distance and displacement can be different, even for the same journey.

Diagram comparing distance as total path length and displacement as straight-line change in position with direction

Example

Finding distance and displacement

A student walks 40 m east, then 15 m west.

The distance is the total path length: $m40\text{ m} + 15\text{ m} = 55\text{ m}$ .
Treat east as positive, so the final position is $m40\text{ m} - 15\text{ m} = 25\text{ m}$ from the start.
The displacement is 25 m east, because displacement needs both size and direction.

Common Mistake

Forgetting direction

“25 m” is not a full displacement answer. You need something like 25 m east or 25 m to the right.

Speed and average speed

Speed is how fast an object is moving. It does not include direction, so speed is a scalar.

For an object moving at constant speed:

s = v t

where $s$ is distance travelled in metres, $v$ is speed in metres per second, and $t$ is time in seconds.

For motion that is not uniform, use:

time\text{average speed} = \frac{\text{total distance}}{\text{total time}}

Example

Calculating average speed

A cyclist travels 900 m in 3 minutes.

Convert the time to seconds: 3 minutes is $\times 60 = 180\text{ s}$ .
Substitute into the speed equation: $\frac{900}{180}$ .
Calculate the average speed: $5\text{ m/s}$ , which is a sensible cycling speed.

Tip

Unit conversion shortcut

To convert from km/h to m/s, divide by 3.6. To convert from m/s to km/h, multiply by 3.6.

Velocity

Velocity is speed in a given direction. It is a vector.

For example, 10 m/s north and 10 m/s south are the same speed, but different velocities, because the directions are different.

If you are on Higher Tier, you also need this idea: an object moving in a circle at constant speed is still changing velocity, because its direction is constantly changing.

Distance–time graphs

A distance–time graph shows how the distance travelled changes with time. The gradient tells you the speed:

time\text{speed} = \frac{\text{change in distance}}{\text{change in time}}

Distance-time and velocity-time graph summary showing gradients, acceleration and area under the graph

Example

Finding speed from a distance–time graph

A straight section of a graph goes from 5 m at 2 s to 29 m at 8 s.

Find the change in distance: $m29\text{ m} - 5\text{ m} = 24\text{ m}$ .
Find the change in time: $s8\text{ s} - 2\text{ s} = 6\text{ s}$ .
Find the gradient: $\frac{24}{6} = 4\text{ m/s}$ .

Common Mistake

Using one point instead of a gradient

For a graph, do not just read one coordinate and call it the speed. Speed comes from the gradient, so you need a change in distance divided by a change in time.

Acceleration

Acceleration is the rate of change of velocity. It tells you how quickly velocity changes.

Definition

Acceleration

Acceleration is calculated using:

\frac{\Delta v}{t}

where $a$ is acceleration in m/s², $Δv\Delta v$ is change in velocity in m/s, and $t$ is time in seconds.

An object that slows down is decelerating. Near the Earth’s surface, an object falling freely under gravity accelerates at about 9.8 m/s², if air resistance is ignored.

Example

Calculating acceleration

A car increases its velocity from 5 m/s to 25 m/s in 10 s.

Calculate the change in velocity: $m/s25\text{ m/s} - 5\text{ m/s} = 20\text{ m/s}$ .
Substitute into the equation: $\frac{20}{10}$ .
Calculate the acceleration: $2\text{ m/s}^2$ .

Velocity–time graphs

On a velocity–time graph, the gradient gives acceleration:

\frac{\Delta v}{\Delta t}

Example

Using a velocity–time graph

A vehicle’s velocity increases uniformly from 0 m/s to 12 m/s in 6 s.

The acceleration is the gradient: $\frac{12 - 0}{6 - 0} = 2\text{ m/s}^2$ .
For Higher Tier, the distance is the triangular area under the graph: $m\frac{1}{2} \times 6 \times 12 = 36\text{ m}$ .
The graph shows uniform acceleration because the velocity–time line is straight.

Uniform acceleration equation

For uniform acceleration, you can use:

v^2 - u^2 = 2as

where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration, and $s$ is distance.

This equation is useful when time is not given.

Example

Using the uniform acceleration equation

A skateboarder starts from rest and accelerates uniformly at 2 m/s² over 25 m. Find the final velocity.

Identify the values: $u = 0$ , $a = 2$ , and $s = 25$ .
Substitute into the equation: $v2−02=2×2×25=100v^2 - 0^2 = 2 \times 2 \times 25 = 100$ .
Square root both sides: $10\text{ m/s}$ .

Terminal velocity

A fluid is a liquid or gas. When an object falls through a fluid, gravity acts downwards and drag acts upwards.

Key Idea

Terminal velocity

Terminal velocity happens when the resultant force on a falling object is zero, so acceleration is zero and the object continues at constant velocity.

Newton’s First Law

Key Idea

Newton’s First Law

If the resultant force on an object is zero, a stationary object stays stationary, and a moving object continues at the same velocity.

The resultant force is the overall force after all forces have been combined, taking direction into account.

A vehicle travelling at a steady speed has balanced forces: the driving force forwards equals the resistive forces backwards. Its velocity only changes if there is a resultant force.

If you are on Higher Tier, the tendency of objects to keep their state of rest or uniform motion is called inertia.

Newton’s Second Law

Newton’s Second Law says acceleration is proportional to resultant force and inversely proportional to mass. In symbols, for constant mass, $\propto F$ .

The equation is:

F = ma

where $F$ is resultant force in newtons, $m$ is mass in kilograms, and $a$ is acceleration in m/s².

Example

Calculating resultant force

A 900 kg car accelerates at 1.6 m/s².

Choose Newton’s Second Law: $F = ma$ .
Substitute the values: $\times 1.6$ .
Calculate the resultant force: $1440\text{ N}$ .

For Higher Tier, inertial mass means how difficult it is to change an object’s velocity. It can be found from:

\frac{F}{a}

Required practical: force, mass and acceleration

You need to understand how to investigate $F = ma$ .

Repeat readings, calculate means, and plot graphs. Acceleration should increase when force increases, and decrease when mass increases.

Newton’s Third Law

Key Idea

Newton’s Third Law

Whenever two objects interact, the forces they exert on each other are equal and opposite.

These forces act on different objects, so they do not cancel each other out.

Example

Applying Newton’s Third Law

A person pushes backwards on the ground to start running forwards.

The person exerts a backwards force on the ground.
The ground exerts an equal and opposite forwards force on the person.
The forwards force from the ground accelerates the person forwards.

Common Mistake

Mixing up balanced forces and force pairs

The weight of a book and the support force from the table can be balanced, but they are not a Newton’s Third Law pair because both act on the same object: the book.

Stopping distance

Stopping distance is the total distance a vehicle travels from the moment the driver notices a hazard to the moment the vehicle stops.

distance\text{stopping distance} = \text{thinking distance} + \text{braking distance}

Thinking distance happens during the driver’s reaction time. Braking distance happens while the brakes slow the vehicle down.

Stopping distance diagram showing thinking distance, braking distance, and factors that increase each

Reaction times vary from person to person, but typical values are about 0.2 s to 0.9 s. Tiredness, drugs, alcohol and distractions increase reaction time, so they increase thinking distance.

Example

Calculating stopping distance

A car travels at 20 m/s. The driver’s reaction time is 0.7 s, and the braking distance is 25 m.

Calculate thinking distance: $20 \times 0.7 = 14\text{ m}$ .
Add the braking distance: $m14\text{ m} + 25\text{ m} = 39\text{ m}$ .
The stopping distance is 39 m.

Braking distance and safety

Braking distance is increased by adverse road conditions, such as wet or icy roads, and poor vehicle condition, especially worn tyres or faulty brakes.

Large decelerations can be dangerous because they may cause loss of control and can make brakes overheat.

Example

Estimating braking force

A 1200 kg car slows from 30 m/s to rest in 6 s.

Calculate acceleration: $\frac{0 - 30}{6} = -5\text{ m/s}^2$ .
Use Newton’s Second Law: $\times -5$ .
The braking force is $N-6000\text{ N}$ , meaning 6000 N backwards.

Exam technique

In the exam

Always use SI units: metres, seconds, kilograms, newtons, and metres per second.
For graph questions, write gradient as change in vertical axis divided by change in horizontal axis.
For vector answers, include direction: for example, “12 m east” or “4 m/s upwards”.
For braking questions, separate factors that affect thinking distance from factors that affect braking distance.

Self review

Check yourself

A walker travels 80 m north then 30 m south. What are the distance and displacement?
What does the gradient of a distance–time graph tell you?
Why does an icy road increase braking distance but not the driver’s reaction time?

What you'll learn

Distance and displacement

Speed and average speed

Velocity

Distance–time graphs

Acceleration

Velocity–time graphs

Uniform acceleration equation

Terminal velocity

Newton’s First Law

Newton’s Second Law

Required practical: force, mass and acceleration

Newton’s Third Law

Stopping distance

Braking distance and safety

Review Forces and motion by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Forces and motion

What you'll learn

Distance and displacement

Speed and average speed

Velocity

Distance–time graphs

Acceleration

Velocity–time graphs

Uniform acceleration equation

Terminal velocity

Newton’s First Law

Newton’s Second Law

Required practical: force, mass and acceleration

Newton’s Third Law

Stopping distance

Braking distance and safety

Review Forces and motion by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Forces and motion

What you'll learn

Distance and displacement

Speed and average speed

Velocity

Distance–time graphs

Acceleration

Velocity–time graphs

Uniform acceleration equation

Terminal velocity

Newton’s First Law

Newton’s Second Law

Required practical: force, mass and acceleration

Newton’s Third Law

Stopping distance

Braking distance and safety

Review Forces and motion by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Forces and motion Revision Guide

Forces and motion

What you'll learn

Distance and displacement

Speed and average speed

Velocity

Distance–time graphs

Acceleration

Velocity–time graphs

Uniform acceleration equation

Terminal velocity

Newton’s First Law

Newton’s Second Law

Required practical: force, mass and acceleration

Newton’s Third Law

Stopping distance

Braking distance and safety

Review Forces and motion by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Forces and motion Revision Guide