ForcesAQA GCSE Physics (8463)

Forces and Motion

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

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Master the topic

Worked examples7 Key formulae3 Definitions8 Common mistakes6

Step-by-step worked examples

01.Acceleration of a car
Foundation
Question
A car speeds up from 8 m/s to 22 m/s in 7 s. Find its acceleration.
Solution
1. 1
  Formula.
  $a = (v - u)/t$
2. 2
  Substitute.
  $a = (22 - 8)/7 = 14/7 = 2\,\text{m/s}^2$
Answer
2 m/s².
02.Final speed using SUVAT
Higher
Question
A motorcyclist accelerates from 12 m/s at 4 m/s² over 100 m. Find final speed.
Solution
1. 1
  Formula.
  $v^2 = u^2 + 2as$
2. 2
  Substitute.
  $v^2 = 12^2 + 2 \times 4 \times 100 = 144 + 800 = 944$
3. 3
  Square root.
  $v = \sqrt{944} \approx 30.7\,\text{m/s}$
Answer
~30.7 m/s.
03.Apply F = ma
Foundation
Question
A 5 kg ball experiences a net force of 20 N. Find its acceleration.
Solution
1. 1
  Rearrange.
  $a = F/m$
2. 2
  Substitute.
  $a = 20/5 = 4\,\text{m/s}^2$
Answer
4 m/s².
04.Net force with friction
Higher
Question
A 500 kg car has a driving force of 1200 N and total friction of 700 N. Find its acceleration.
Solution
1. 1
  Net force.
  $F_{net} = 1200 - 700 = 500\,\text{N}$
2. 2
  Acceleration.
  $a = F/m = 500/500 = 1\,\text{m/s}^2$
Answer
1 m/s².
05.Newton's third law example
Foundation
Question
A swimmer pushes water backwards with their hands. Identify the action-reaction pair and the direction of resulting motion.
Solution
1. 1
  Action.
  $\text{Swimmer pushes water BACKWARDS.}$
2. 2
  Reaction.
  $\text{Water pushes swimmer FORWARDS (equal and opposite).}$
3. 3
  Result.
  $\text{Net force forward on swimmer → forward acceleration.}$
Answer
Swimmer pushes water back; water pushes swimmer forward. Net forward force accelerates the swimmer forward.
06.Calculate stopping distance
Higher
Question
A car travels at 25 m/s. Reaction time is 0.8 s; deceleration during braking is 5 m/s². Find (a) thinking distance, (b) braking distance, (c) total stopping distance.
Solution
1. 1
  Thinking distance.
  $s_t = v \times t = 25 \times 0.8 = 20\,\text{m}$
2. 2
  Braking distance via v² − u² = 2as.
  $0 - 25^2 = 2 \times -5 \times s \Rightarrow s = 625/10 = 62.5\,\text{m}$
3. 3
  Total.
  $20 + 62.5 = 82.5\,\text{m}$
Answer
Thinking 20 m; braking 62.5 m; total 82.5 m.
07.Brake heating
Higher
Question
A 1500 kg car decelerates from 30 m/s to 0. How much energy is transferred to the thermal store of its brakes?
Solution
1. 1
  Initial KE.
  $E_k = \tfrac{1}{2} \times 1500 \times 30^2 = 675{,}000\,\text{J}$
2. 2
  Final KE = 0; difference goes to brakes.
  $\Delta E = 675\,\text{kJ}$
Answer
675 kJ → thermal store of brake discs (assuming 100 % conversion).

Key formulae

Acceleration
$a = (v - u)/t$
- $a$
  — Acceleration (m/s²)
- $v$
  — Final velocity (m/s)
- $u$
  — Initial velocity (m/s)
- $t$
  — Time (s)
When to use
Whenever velocity changes uniformly over time. Recall (not on equation sheet).
v² − u² = 2as
$v^2 - u^2 = 2 a s$
- $v$
  — Final velocity (m/s)
- $u$
  — Initial velocity (m/s)
- $a$
  — Acceleration (m/s²)
- $s$
  — Distance (m)
When to use
When time is unknown but distance is given. On AQA equation sheet.
Newton's second law
$F = m a$
- $F$
  — Resultant force (N)
- $m$
  — Mass (kg)
- $a$
  — Acceleration (m/s²)
When to use
Whenever a net force causes acceleration. Recall (not on equation sheet).

Practice with flashcards

Card 1 of 80 known · 0 to review

Key definitions and keywords

DisplacementExaminer keyword: The straight-line distance from start to finish, in a specific direction. Vector.
VelocityExaminer keyword: Rate of change of displacement; speed with direction. Vector, m/s.
AccelerationExaminer keyword: Rate of change of velocity. Vector, m/s².
InertiaExaminer keyword: The tendency of an object to remain in its current state of motion (at rest or constant velocity).
Inertial mass (HT): A measure of how hard it is to change an object's velocity. Defined as the ratio of force to acceleration: m = F/a.
Stopping distanceExaminer keyword: Total distance from when the driver sees a hazard to coming to rest. Thinking distance + braking distance.
Thinking distanceExaminer keyword: Distance travelled during the driver's reaction time, before brakes are applied.
Braking distanceExaminer keyword: Distance travelled while the brakes are applied, until the vehicle stops.

Common mistakes and misconceptions

Mistake
Treating speed and velocity as interchangeable.
Why it happens
Both are 'how fast'.
How to avoid it
Speed = magnitude only. Velocity has direction (e.g. '5 m/s north').
Mistake
Reporting acceleration in mph/s.
Why it happens
Mixing units.
How to avoid it
Use m/s² in physics answers.
Mistake
Saying the two forces in Newton's 3rd law act on the same object.
Why it happens
Confusing balanced forces with action-reaction.
How to avoid it
Action-reaction forces always act on DIFFERENT objects (one pushes, one is pushed).
Mistake
Calling inertia a force.
Why it happens
Common misconception.
How to avoid it
Inertia is a PROPERTY of mass — the tendency to resist changes. It is not a force.
Mistake
Saying braking distance depends only on the driver.
Why it happens
Confusing thinking with braking.
How to avoid it
Driver affects THINKING distance. Braking is set by speed, mass, tyres, brakes, road.
Mistake
Saying doubled speed doubles braking distance.
Why it happens
Linear thinking.
How to avoid it
Braking distance ∝ v², so doubled speed = QUADRUPLED braking distance.

ForcesAQA GCSE Physics (8463)

Forces and Motion

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

Previous Next

Master the topic

Worked examples7 Key formulae3 Definitions8 Common mistakes6

Step-by-step worked examples

01.Acceleration of a car
Foundation
Question
A car speeds up from 8 m/s to 22 m/s in 7 s. Find its acceleration.
Solution
1. 1
  Formula.
  $a = (v - u)/t$
2. 2
  Substitute.
  $a = (22 - 8)/7 = 14/7 = 2\,\text{m/s}^2$
Answer
2 m/s².
02.Final speed using SUVAT
Higher
Question
A motorcyclist accelerates from 12 m/s at 4 m/s² over 100 m. Find final speed.
Solution
1. 1
  Formula.
  $v^2 = u^2 + 2as$
2. 2
  Substitute.
  $v^2 = 12^2 + 2 \times 4 \times 100 = 144 + 800 = 944$
3. 3
  Square root.
  $v = \sqrt{944} \approx 30.7\,\text{m/s}$
Answer
~30.7 m/s.
03.Apply F = ma
Foundation
Question
A 5 kg ball experiences a net force of 20 N. Find its acceleration.
Solution
1. 1
  Rearrange.
  $a = F/m$
2. 2
  Substitute.
  $a = 20/5 = 4\,\text{m/s}^2$
Answer
4 m/s².
04.Net force with friction
Higher
Question
A 500 kg car has a driving force of 1200 N and total friction of 700 N. Find its acceleration.
Solution
1. 1
  Net force.
  $F_{net} = 1200 - 700 = 500\,\text{N}$
2. 2
  Acceleration.
  $a = F/m = 500/500 = 1\,\text{m/s}^2$
Answer
1 m/s².
05.Newton's third law example
Foundation
Question
A swimmer pushes water backwards with their hands. Identify the action-reaction pair and the direction of resulting motion.
Solution
1. 1
  Action.
  $\text{Swimmer pushes water BACKWARDS.}$
2. 2
  Reaction.
  $\text{Water pushes swimmer FORWARDS (equal and opposite).}$
3. 3
  Result.
  $\text{Net force forward on swimmer → forward acceleration.}$
Answer
Swimmer pushes water back; water pushes swimmer forward. Net forward force accelerates the swimmer forward.
06.Calculate stopping distance
Higher
Question
A car travels at 25 m/s. Reaction time is 0.8 s; deceleration during braking is 5 m/s². Find (a) thinking distance, (b) braking distance, (c) total stopping distance.
Solution
1. 1
  Thinking distance.
  $s_t = v \times t = 25 \times 0.8 = 20\,\text{m}$
2. 2
  Braking distance via v² − u² = 2as.
  $0 - 25^2 = 2 \times -5 \times s \Rightarrow s = 625/10 = 62.5\,\text{m}$
3. 3
  Total.
  $20 + 62.5 = 82.5\,\text{m}$
Answer
Thinking 20 m; braking 62.5 m; total 82.5 m.
07.Brake heating
Higher
Question
A 1500 kg car decelerates from 30 m/s to 0. How much energy is transferred to the thermal store of its brakes?
Solution
1. 1
  Initial KE.
  $E_k = \tfrac{1}{2} \times 1500 \times 30^2 = 675{,}000\,\text{J}$
2. 2
  Final KE = 0; difference goes to brakes.
  $\Delta E = 675\,\text{kJ}$
Answer
675 kJ → thermal store of brake discs (assuming 100 % conversion).

Key formulae

Acceleration
$a = (v - u)/t$
- $a$
  — Acceleration (m/s²)
- $v$
  — Final velocity (m/s)
- $u$
  — Initial velocity (m/s)
- $t$
  — Time (s)
When to use
Whenever velocity changes uniformly over time. Recall (not on equation sheet).
v² − u² = 2as
$v^2 - u^2 = 2 a s$
- $v$
  — Final velocity (m/s)
- $u$
  — Initial velocity (m/s)
- $a$
  — Acceleration (m/s²)
- $s$
  — Distance (m)
When to use
When time is unknown but distance is given. On AQA equation sheet.
Newton's second law
$F = m a$
- $F$
  — Resultant force (N)
- $m$
  — Mass (kg)
- $a$
  — Acceleration (m/s²)
When to use
Whenever a net force causes acceleration. Recall (not on equation sheet).

Practice with flashcards

Card 1 of 80 known · 0 to review

Key definitions and keywords

DisplacementExaminer keyword: The straight-line distance from start to finish, in a specific direction. Vector.
VelocityExaminer keyword: Rate of change of displacement; speed with direction. Vector, m/s.
AccelerationExaminer keyword: Rate of change of velocity. Vector, m/s².
InertiaExaminer keyword: The tendency of an object to remain in its current state of motion (at rest or constant velocity).
Inertial mass (HT): A measure of how hard it is to change an object's velocity. Defined as the ratio of force to acceleration: m = F/a.
Stopping distanceExaminer keyword: Total distance from when the driver sees a hazard to coming to rest. Thinking distance + braking distance.
Thinking distanceExaminer keyword: Distance travelled during the driver's reaction time, before brakes are applied.
Braking distanceExaminer keyword: Distance travelled while the brakes are applied, until the vehicle stops.

Common mistakes and misconceptions

Mistake
Treating speed and velocity as interchangeable.
Why it happens
Both are 'how fast'.
How to avoid it
Speed = magnitude only. Velocity has direction (e.g. '5 m/s north').
Mistake
Reporting acceleration in mph/s.
Why it happens
Mixing units.
How to avoid it
Use m/s² in physics answers.
Mistake
Saying the two forces in Newton's 3rd law act on the same object.
Why it happens
Confusing balanced forces with action-reaction.
How to avoid it
Action-reaction forces always act on DIFFERENT objects (one pushes, one is pushed).
Mistake
Calling inertia a force.
Why it happens
Common misconception.
How to avoid it
Inertia is a PROPERTY of mass — the tendency to resist changes. It is not a force.
Mistake
Saying braking distance depends only on the driver.
Why it happens
Confusing thinking with braking.
How to avoid it
Driver affects THINKING distance. Braking is set by speed, mass, tyres, brakes, road.
Mistake
Saying doubled speed doubles braking distance.
Why it happens
Linear thinking.
How to avoid it
Braking distance ∝ v², so doubled speed = QUADRUPLED braking distance.

Forces and Motion

Master the topic

Step-by-step worked examples

01.Acceleration of a car

02.Final speed using SUVAT

03.Apply F = ma

04.Net force with friction

05.Newton's third law example

06.Calculate stopping distance

07.Brake heating

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions

Forces and Motion

Master the topic

Step-by-step worked examples

01.Acceleration of a car

02.Final speed using SUVAT

03.Apply F = ma

04.Net force with friction

05.Newton's third law example

06.Calculate stopping distance

07.Brake heating

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions