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Lesson 1 — Scalar and Vector Quantities

Do Now

Write down three physical quantities that can be measured in a physics experiment.

Model Answer Any three from: mass, time, force, temperature, length, speed, distance, energy.

A friend says a car is travelling at "60 mph". What extra piece of information would be useful when following directions on a map?

Model Answer The direction of travel — e.g. "60 mph north". Speed alone doesn't tell you which way the car is going.

Which of the following need a direction to be fully described? Circle your answers: temperature | a push | mass | weight | the length of a room

Model Answer A push and weight need a direction — they are vector quantities. Temperature, mass, and the length of a room are scalars and do not have a direction.

What do we mean when we say a quantity has a magnitude?

Model Answer Magnitude means the size or amount of a quantity — how big or how much it is, expressed as a number with a unit.

Part 1 · Physical Quantities and Scalars

A quantity is something that can be measured. Time, mass and force are all examples.

A scalar quantity has magnitude (size) only — no direction.

Examples of Scalars

Distance
Speed
Energy
Temperature
Mass
Time

It is incorrect to give a scalar quantity a direction — scalars have no direction, so stating one is meaningless.

Questions

Define a scalar quantity.

Model Answer A scalar quantity has magnitude (size) only — no associated direction.

Give three examples of scalar quantities.

Model Answer Any three from: distance, speed, energy, temperature, mass, time.

Molly says "the mass of the parachutist is 75 kg towards the centre of the Earth". What is wrong with this statement? Why is it wrong?

Model Answer Mass is a scalar quantity — it has magnitude only. Giving it a direction ("towards the centre of the Earth") is wrong because scalars do not have direction.

"Energy" is a scalar quantity. Explain why.

Model Answer Energy is measured in joules — a size only. It has no direction, so it is a scalar.

Part 2 · Vector Quantities and Arrow Representation

A vector quantity has both magnitude and direction.

Examples of Vectors

Displacement
Velocity
Force
Acceleration
Weight
Momentum

Vectors are represented by arrows: arrow length = magnitude, arrow direction = direction.

A free body diagram shows all forces acting on a single object as labelled arrows. A scale (vector) diagram draws vectors head-to-tail to find a resultant.

Questions

Define a vector quantity.

Model Answer A vector quantity has both magnitude and a specific direction.

Give three examples of vector quantities.

Model Answer Any three from: displacement, velocity, force, acceleration, weight, momentum.

James says "the force of air resistance on the parachute is 50 N". What else must James include for a complete answer? Explain why this extra information is necessary.

Model Answer James must state the direction (e.g. "upwards" or "opposing motion"). Force is a vector — direction must always be given for a complete description.

Look at the skier diagram. Name the four vector quantities shown.

Model Answer Normal force, air resistance, weight, friction.

Is acceleration a scalar or a vector? Explain.

Model Answer Acceleration is a vector. Velocity is a vector, and the rate of change of a vector must also have a direction.

Q10

Is power a scalar or a vector? Explain.

Model Answer Power is a scalar. Energy is a scalar, and its rate of transfer has no direction.

Part 3 · Exam Question — Scalars and Vectors (8 marks)

The questions below combine ideas from the whole lesson in the style of an AQA exam paper. Mark allocations are shown in brackets.

Exam Questions

Part 1

(a) State the difference between a scalar and a vector, giving one example of each. (4)

Model Answer Scalar: magnitude only, e.g. speed / temperature / mass. (2)
Vector: magnitude and direction, e.g. velocity / force / displacement. (2)

Part 2

(b) A car travels at a speed of 20 m/s and a velocity of 20 m/s north. Explain why these two statements give different information. (2)

Model Answer Speed gives only how fast the car is moving (20 m/s). Velocity gives speed and direction (north), so it provides more information.

Part 3

(c) A book rests on a table. Its weight is 6 N downwards. State the magnitude and direction of the normal contact force, justifying your answer using Newton's First Law. (2)

Model Answer Normal contact force = 6 N upwards. (1) The book is at rest, so by Newton's First Law the resultant force is zero — the normal contact force must equal the weight in magnitude but act in the opposite direction. (1)

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Lesson 2 — Distance and Displacement

Do Now

Define a scalar quantity and give one example.

Model Answer A scalar quantity has magnitude only — no direction. Example: mass / distance / speed / temperature.

Define a vector quantity and give one example.

Model Answer A vector quantity has magnitude and direction. Example: force / velocity / displacement / weight.

Is force a scalar or a vector? Write one sentence to justify your answer.

Model Answer Force is a vector — it has both a size (measured in newtons) and a direction in which it acts.

A sprinter runs 100 m along a straight track from start to finish. Is the distance run the same as the displacement? Explain.

Model Answer Yes — both are 100 m. Because the path is a straight line in one direction (no turning back), the total path length equals the straight-line distance from start to finish.

Part 1 · Definitions

Distance is a scalar — the total length of the path travelled.

Displacement is a vector — the straight-line distance from start to finish, including direction.

Worked Example

An object travels from A to B (6 m up), then B to C (1.5 m right), then C to D (4 m down).

Distance = 6 + 1.5 + 4 = 11.5 m
Displacement = 6.5 m to the right (straight-line A→D)

Displacement is always ≤ distance. They are equal only when the path is a straight line in one direction. If an object returns to its starting point, displacement = 0 m.

Questions

Describe the difference between distance and displacement.

Model Answer Distance: total length of the path travelled (scalar, no direction).
Displacement: straight-line distance from start to finish, with direction stated (vector).

Which of distance and displacement is a vector quantity?

Model Answer Displacement is the vector quantity.

True or false: for a moving object, displacement can equal distance but can never be greater. Explain.

Model Answer True. The straight-line distance from start to finish is always ≤ the actual path length. They are equal only when the object travels in a straight line in one direction without turning back.

Draw a diagram for a curved path from A to B labelling distance and displacement.

Model Answer Diagram: a curved arrow labelled "distance" (longer path A→B), and a straight arrow labelled "displacement" (shorter, direct A→B). Direction should be stated for displacement.

Part 2 · Calculating Distance and Displacement

Distance: add up the lengths of every segment of the journey.

Displacement: identify start and end points; draw a straight line between them and state its length and direction.

For 2D paths use Pythagoras: displacement = √(horizontal² + vertical²).

Displacement is negative if the object ends on the opposite side of the start from the defined positive direction.

Questions

A car starts at A (0 km), travels to B (5 km), then to C (9 km), then back to B. Find (a) total distance and (b) displacement.

Model Answer (a) A→B = 5 km, B→C = 4 km, C→B = 4 km. Total distance = 13 km.
(b) Start = 0 km (A), finish = 5 km (B). Displacement = 5 km to the right.

Find (a) the displacement from B to A, and (b) the distance from B to A.

Model Answer (a) Displacement = 5 km to the left (or −5 km).
(b) Distance = 5 km.

Part 3 · Exam Question — Distance and Displacement (7 marks)

The questions below combine ideas from the whole lesson in the style of an AQA exam paper. Mark allocations are shown in brackets.

Exam Questions

Part 1

(a) Using Fig 2.3 (each small square = 2 m): find (i) the distance travelled from A to B and (ii) the displacement from A to B. (2)

Model Answer (i) Count the path squares × 2 m. Distance ≈ 14 m. (1)
(ii) Straight-line A→B = 5 squares × 2 m = 10 m to the right. (1)

Part 2

(b)(i) A sprinter runs one complete lap of a 400 m track and returns to the start. State the total distance. (1)

Model Answer Distance = 400 m.

Part 3

(b)(ii) State the displacement of the sprinter at the end of the lap. (1)

Model Answer Displacement = 0 m.

Part 4

(b)(iii) Explain why the distance and displacement are different in this case. (2)

Model Answer Distance is the total length of the path along the track (400 m). Displacement is the straight-line distance from start to finish — the sprinter returns to the starting point so start and end positions are the same, giving displacement = 0.

Lesson 3 — Distance, Displacement and Speed

Do Now

What is the difference between distance and displacement?

Model Answer Distance is the total length of the path travelled (scalar). Displacement is the straight-line distance from start to finish, including direction (vector).

A car drives 4 km east then 3 km north. What is the total distance driven? What is the magnitude of the displacement from start to finish?

Model Answer Total distance = 4 + 3 = 7 km. Displacement magnitude = √(4² + 3²) = √25 = 5 km (Pythagoras' theorem).

A person walks to school along a winding path (2.1 km) then walks back home along the same path. What is their total distance? What is their displacement when they arrive back home?

Model Answer Total distance = 2.1 + 2.1 = 4.2 km. Displacement = 0 m — they return to their starting point.

Write down the units for: (a) distance (b) time (c) speed.

Model Answer (a) metres, m (b) seconds, s (c) metres per second, m/s

Part 1 · Speed and Velocity — Definitions

Speed is the rate of change of distance. It is a scalar (magnitude only).

speed = distance ÷ time

Unit: m/s. Typical values: walking ≈ 1.5, cycling ≈ 6, car ≈ 30, sound ≈ 330 m/s.

Velocity is the rate of change of displacement. It is a vector (magnitude and direction).

velocity = displacement ÷ time

Velocity must always state direction, e.g. "12 m/s to the right".

Triangle of Equations

v = d / t | d = v × t | t = d / v

Questions

What is the difference between speed and velocity?

Model Answer Speed is scalar (magnitude only). Velocity is a vector (magnitude and direction).

An object travels 25 m to the right in 10 s. Find its velocity.

Model Answer v = 25 / 10 = 2.5 m/s to the right

A hot air balloon travels 125 m upward in 50 s. Find its velocity.

Model Answer v = 125 / 50 = 2.5 m/s upward

A diver travels 640 m downward during a 16 s dive. Find their velocity.

Model Answer v = 640 / 16 = 40 m/s downward

A rocket travels at 150 m/s upward for 30 s. Find its displacement.

Model Answer d = 150 × 30 = 4 500 m upward

An insect crawls for 180 s at 0.25 m/s to the left. Find its displacement.

Model Answer d = 0.25 × 180 = 45 m to the left

A car travels at 24 m/s to the right for 58 s. Find its displacement.

Model Answer d = 24 × 58 = 1 392 m to the right

A car travelling at 2.75 m/s covers a displacement of 185 m. Find the time.

Model Answer t = 185 / 2.75 = 67.3 s

A rocket travelling at 240 m/s reaches a displacement of 15 000 m. Find the time.

Model Answer t = 15 000 / 240 = 62.5 s

Q10

A snail travels at 0.08 m/s and reaches 0.2 m. Find the time.

Model Answer t = 0.2 / 0.08 = 2.5 s

Part 2 · Further Practice

Apply the same equations to a fresh set of practice values. Watch your units and always state the direction for a vector answer.

Questions

Q11

An object travels 35 m to the right in 40 s. Find its velocity.

Model Answer v = 35 / 40 = 0.875 m/s to the right

Q12

A hot air balloon travels 325 m upward in 59 s. Find its velocity.

Model Answer v = 325 / 59 = 5.51 m/s upward

Q13

A diver travels 645 m downward in 12 s. Find their velocity.

Model Answer v = 645 / 12 = 53.75 m/s downward

Q14

A rocket travels at 350 m/s upward for 20 s. Find its displacement.

Model Answer d = 350 × 20 = 7 000 m upward

Q15

An insect crawls for 380 s at 0.75 m/s to the left. Find its displacement.

Model Answer d = 0.75 × 380 = 285 m to the left

Q16

An object travels at 34 m/s to the right for 48 s. Find its displacement.

Model Answer d = 34 × 48 = 1 632 m to the right

Q17

An object travelling at 3.75 m/s covers 196 m. Find the time.

Model Answer t = 196 / 3.75 = 52.3 s

Q18

A rocket at 170 m/s reaches 46 000 m. Find the time.

Model Answer t = 46 000 / 170 = 270.6 s

Q19

A snail at 0.02 m/s reaches 0.4 m. Find the time.

Model Answer t = 0.4 / 0.02 = 20 s

Part 3 · Exam Question — Velocity and Speed Calculations (9 marks)

The questions below combine ideas from the whole lesson in the style of an AQA exam paper. Mark allocations are shown in brackets.

Exam Questions

Part 1

(a)(i) A cyclist travels 1 200 m north in 3 minutes. Calculate the speed. Give the unit. (3)

Model Answer t = 3 × 60 = 180 s. speed = 1 200 / 180 = 6.67 m/s

Part 2

(a)(ii) State the velocity of the cyclist. (1)

Model Answer Velocity = 6.67 m/s north

Part 3

(b) A car travels at 30 m/s for 2 hours. Calculate the total distance. Give your answer in km. (3)

Model Answer t = 2 × 3 600 = 7 200 s. d = 30 × 7 200 = 216 000 m = 216 km

Part 4

(c)(i) A swimmer completes a 50 m length in 38 s, then swims back 50 m in 40 s. Calculate the average speed. (2)

Model Answer Total distance = 100 m, total time = 78 s. Average speed = 100 / 78 = 1.28 m/s

Part 5

(c)(ii) State the total displacement and explain. (2)

Model Answer Displacement = 0 m. The swimmer returns to the starting point, so start and end positions are the same; displacement = 0.

Lesson 4 — Velocity and Acceleration

Do Now

State the formula for speed and rearrange it to make time the subject.

Model Answer speed = distance / time → t = d / v

An object travels 240 m in 12 s at constant speed. Calculate its speed.

Model Answer v = 240 / 12 = 20 m/s

What is the difference between speed and velocity? Give an example to illustrate the difference.

Model Answer Speed is scalar (magnitude only); velocity is vector (magnitude + direction). Example: "30 m/s" is a speed; "30 m/s north" is a velocity.

A cyclist travels at 6 m/s east for 30 s. Calculate her displacement and state the direction.

Model Answer d = 6 × 30 = 180 m east

Part 1 · Defining Acceleration

Acceleration is the rate of change of velocity. Unit: m/s².

Since velocity is a vector, all of the following count as acceleration:

Speeding up (increasing speed)
Slowing down — also called deceleration (negative acceleration)
Turning a corner (changing direction at constant speed)

An acceleration always requires a resultant (net) force — Newton's Second Law.

Typical values: car ≈ 3, motorbike ≈ 5, free-fall on Earth = 10, space shuttle ≈ 29 (all m/s²).

Questions

Define acceleration and state its unit.

Model Answer Acceleration is the rate of change of velocity. Unit: m/s².

Why does applying the brakes cause the car to accelerate? What changes in velocity?

Model Answer The brakes apply a resultant force opposing the motion. This reduces the car's speed (velocity decreases) — a negative acceleration / deceleration.

Why is a bouncing ball accelerating at the moment it hits the ground?

Model Answer At impact the ball's velocity changes direction (downward → upward). A change in direction is a change in velocity, so there is a very large acceleration over a very short time.

Why is the Moon orbiting the Earth accelerating, even though its speed is roughly constant?

Model Answer The Moon travels in a circle so its direction of motion changes continuously. A change in direction is a change in velocity (velocity is a vector), so the Moon is always accelerating. The centripetal force is provided by gravity.

What is the relationship between resultant force and acceleration?

Model Answer A resultant force causes acceleration. F = m × a: acceleration is proportional to force and inversely proportional to mass.

Part 2 · The Acceleration Formula

a = (v − u) / t

where u = initial velocity, v = final velocity, t = time taken. Rearranged:

Δv = a × t | t = Δv / a

If an object slows down, the change in velocity is negative, giving a negative acceleration (deceleration).

Questions

Object accelerates from 5 to 10 m/s in 20 s. Find a.

Model Answer Δv = 5. a = 5 / 20 = 0.25 m/s²

Object accelerates from 0.2 to 12.5 m/s in 5 s. Find a.

Model Answer Δv = 12.3. a = 12.3 / 5 = 2.46 m/s²

Object goes from 20 to 8 m/s in 3 s. Find a.

Model Answer Δv = −12. a = −12 / 3 = −4 m/s²

Object decelerates from 100 to 75 m/s in 5 s. Find the deceleration.

Model Answer Δv = −25. a = −25 / 5 = −5 m/s²

Q10

From 45 to 13 m/s in 12 s. Find the deceleration.

Model Answer Δv = −32. a = −32 / 12 = −2.67 m/s²

Q11

From 48 to 3 m/s in 30 s. Find the deceleration.

Model Answer Δv = −45. a = −45 / 30 = −1.5 m/s²

Q12

Object from rest to 14 m/s in 3.5 s. Find a.

Model Answer a = 14 / 3.5 = 4 m/s²

Q13

Object from rest to 105 m/s in 55 s. Find a.

Model Answer a = 105 / 55 = 1.91 m/s²

Q14

From 35 m/s to rest in 70 s. Find the deceleration.

Model Answer a = −35 / 70 = −0.5 m/s²

Q15

From 12 m/s to rest in 2.5 s. Find the deceleration.

Model Answer a = −12 / 2.5 = −4.8 m/s²

Q16

From 58 m/s to rest in 4 s. Find the deceleration.

Model Answer a = −58 / 4 = −14.5 m/s²

Q17

Deceleration 2 m/s², from 10 to 4 m/s. Find t.

Model Answer Δv = −6. t = 6 / 2 = 3 s

Q18

Acceleration 4.3 m/s², from 9 to 85 m/s. Find t.

Model Answer Δv = 76. t = 76 / 4.3 = 17.7 s

Q19

Acceleration 0.45 m/s², from rest to 3 m/s. Find t.

Model Answer t = 3 / 0.45 = 6.67 s

Q20

a = 6 m/s², 15 s. Find Δv.

Model Answer Δv = 6 × 15 = 90 m/s

Q21

a = −18 m/s², 76 s. Find Δv.

Model Answer Δv = −18 × 76 = −1 368 m/s

Q22

a = 0.9 m/s², 200 s. Find Δv.

Model Answer Δv = 0.9 × 200 = 180 m/s

Part 3 · Exam Question — Acceleration (8 marks)

The questions below combine ideas from the whole lesson in the style of an AQA exam paper. Mark allocations are shown in brackets.

Exam Questions

Part 1

(a)(i) A train accelerates from rest to 55 m/s in 220 s. Calculate the acceleration. (2)

Model Answer a = (55 − 0) / 220 = 0.25 m/s²

Part 2

(a)(ii) Describe what is meant by deceleration. (1)

Model Answer Deceleration is negative acceleration — a decrease in velocity / speed.

Part 3

(b)(i) A skydiver jumps with v = 0. After 6 s they reach 60 m/s downward. Find the acceleration. (2)

Model Answer a = (60 − 0) / 6 = 10 m/s² downward

Part 4

(b)(ii) The skydiver opens a parachute and slows from 60 to 8 m/s in 10 s. Find the deceleration. (2)

Model Answer a = (8 − 60) / 10 = −52 / 10 = −5.2 m/s² (deceleration of 5.2 m/s²)

Part 5

(b)(iii) The skydiver later reaches terminal velocity. Explain what this means and why it occurs. (3)

Model Answer Terminal velocity is the maximum constant velocity reached when the driving force (weight, downward) equals the resistive force (air resistance/drag, upward), giving a zero resultant force. With no resultant force there is no acceleration, so velocity remains constant.

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Lesson 5 — Distance-Time Graphs

Do Now

State the formula for acceleration and write down its unit.

Model Answer a = (v − u) / t or a = Δv / t. Unit: m/s²

A motorbike accelerates from 0 to 24 m/s in 8 s. Calculate the acceleration.

Model Answer a = (24 − 0) / 8 = 3 m/s²

On any graph, what does the gradient (steepness of the line) tell us?

Model Answer The gradient tells us the rate of change of the y-variable with respect to the x-variable.

Describe in words the motion of an object with zero acceleration. Give two different examples of an object that could have zero acceleration.

Model Answer Zero acceleration means the object is moving at constant velocity (constant speed in a straight line) or is stationary. Examples: a car cruising at steady speed on a motorway; a book sitting still on a table.

Part 1 · Key Rules for Distance-Time Graphs

A distance-time graph has time on the x-axis and distance on the y-axis.

Speed = gradient of the line = change in distance ÷ change in time.

Steeper gradient → higher speed
Horizontal (flat) line → object is stationary (gradient = 0)
Downward slope → object moving back towards the starting point
Curved line → speed is changing (acceleration or deceleration)

Example: an object that travels 40 m in 5 s has speed = 40 / 5 = 8 m/s.

Questions

What does the gradient of a distance-time graph represent?

Model Answer The speed of the object.

What does a horizontal line on a distance-time graph tell us?

Model Answer The object is stationary (speed = 0).

Calculate the speed for: (a) 60 m in 5 s, (b) 100 m in 5 s, (c) 20 m in 5 s.

Model Answer (a) 60 / 5 = 12 m/s
(b) 100 / 5 = 20 m/s
(c) 20 / 5 = 4 m/s

Calculate the speed for: (a) 10 m in 10 s, (b) 100 m in 10 s, (c) 40 m in 10 s.

Model Answer (a) 10 / 10 = 1 m/s
(b) 100 / 10 = 10 m/s
(c) 40 / 10 = 4 m/s

Part 2 · Reading Speed from a Graph

To read speed from a distance-time graph: choose two clear points on the line, then:

speed = (d₂ − d₁) / (t₂ − t₁)

If the line does not start at the origin, use the change in distance, not the absolute value on the y-axis.

Questions

Using Fig 5.3 above, find the speed of each object (a–f).

Model Answer a) gradient = 3/4 = 0.75 m/s
b) gradient = 3/4 = 0.75 m/s (same gradient, different starting distance)
c) gradient = 4/2 = 2 m/s
d) gradient = 8/2 = 4 m/s
e) gradient = 8/8 = 1 m/s
f) gradient = 12/8 = 1.5 m/s

Using Fig 5.4 above, find the speed of each object (a–c).

Model Answer a) gradient = 20/8 = 2.5 m/s
b) gradient = 6/4 = 1.5 m/s
c) gradient = 240/12 = 20 m/s

Part 3 · Curved Graphs

A curved distance-time graph means speed is changing — the object is accelerating.

Curve becoming steeper → positive acceleration (speeding up)
Curve becoming flatter → negative acceleration (slowing down)

To find instantaneous speed at a given time: draw a tangent to the curve at that point, then calculate the gradient of the tangent.

Questions

Q7–10

For each sketch graph, state (i) positive or negative acceleration, and (ii) is the object moving forwards or backwards?

Model Answer Graph 7 — curve becomes flatter: negative acceleration / deceleration; moving forwards.
Graph 8 — curve becomes flatter: deceleration; moving forwards.
Graph 9 — distance decreasing: deceleration; moving backwards.
Graph 10 — curve becomes steeper: positive acceleration; moving forwards.

Q11

How do we find the instantaneous speed of an object from a curved distance-time graph?

Model Answer Draw a straight tangent to the curve at the required point. The gradient of the tangent gives the speed: speed = Δd / Δt.

Q12

Describe the motion shown by a straight line with a negative gradient on a distance-time graph. What does the gradient tell us?

Model Answer The object is moving back towards the starting point at constant speed. The magnitude of the gradient gives the speed; the negative sign indicates the object is moving back towards the starting point.

Exam Question — Distance-Time Graphs (5 marks)

The questions below combine ideas from the whole lesson in the style of an AQA exam paper. Mark allocations are shown in brackets.

Exam Questions

Part 1

(a) A straight line on a distance-time graph slopes downward. Describe the motion. (2)

Model Answer The object is moving back towards the starting point at constant speed.

Part 2

(b) A curved line becomes progressively flatter. Describe the motion. (2)

Model Answer The object is decelerating (slowing down) while moving forwards.

Part 3

(c) An object travels 80 m in 10 s, stays still for 5 s, then returns to the start in 8 s. Sketch the d-t graph and find the speed in each phase. (5)

Model Answer Phase 1: speed = 80 / 10 = 8 m/s.
Phase 2: speed = 0 m/s (stationary).
Phase 3: speed = 80 / 8 = 10 m/s (returning, line slopes back down to 0).

📊

Lesson 6 — Velocity-Time Graphs

Do Now

On a distance-time graph, what does the gradient of the line represent?

Model Answer The speed of the object.

An object travels 80 m in 10 s at constant speed. Calculate its speed and describe what its distance-time graph looks like.

Model Answer speed = 80 / 10 = 8 m/s. The d-t graph is a straight line with a constant positive gradient, starting at the origin.

On a distance-time graph, what does a horizontal (flat) line tell us about the object's motion?

Model Answer The object is stationary — distance is not changing, so speed = 0.

A car travels at 15 m/s for 20 s then stops. Calculate the total distance travelled.

Model Answer d = 15 × 20 = 300 m. The d-t graph is a straight line rising to 300 m at t = 20 s, then horizontal.

Part 1 · Key Rules — Don't Confuse with Distance-Time Graphs!

A velocity-time graph has time on the x-axis and velocity on the y-axis.

Velocity-time graphs look similar to distance-time graphs but tell us completely different things. Always check the y-axis label first.

Horizontal line → constant velocity (zero acceleration). The object may still be moving!
A line at v = 0 → object is stationary
Positive gradient (sloping up) → accelerating
Negative gradient (sloping down) → decelerating
Gradient = acceleration
Area under the graph = distance travelled

Questions

Q1–4

1–4. State what each of the following graphs shows (acceleration / constant velocity / deceleration / stationary).

Model Answer Graph 1: Deceleration — velocity decreasing (negative gradient).
Graph 2: Deceleration — velocity decreasing.
Graph 3: Constant velocity — horizontal line (zero gradient).
Graph 4: Acceleration — velocity increasing (positive gradient).

Part 2 · Finding Acceleration from the Gradient

a = (v − u) / t

Always use the change in velocity (final − initial), not just the final velocity value.

v from 0 → 2 m/s in 10 s → a = 2 / 10 = 0.2 m/s²
v from 2 → 8 m/s in 10 s → a = 6 / 10 = 0.6 m/s²
v from 6 → 0 m/s in 10 s → a = −6 / 10 = −0.6 m/s² (deceleration)

Questions

Object 1: v from 10 → 30 m/s in 8 s. Find a.

Model Answer Δv = 20. a = 20 / 8 = 2.5 m/s²

Object 2: v from 10 → 20 m/s in 8 s. Find a.

Model Answer Δv = 10. a = 10 / 8 = 1.25 m/s²

Object 3: v from 30 → 15 m/s in 8 s. Find a.

Model Answer Δv = −15. a = −15 / 8 = −1.875 m/s²

Object 4: v from 20 → 0 m/s in 4 s. Find a.

Model Answer Δv = −20. a = −20 / 4 = −5 m/s²

Object 5: v from 10 → 25 m/s in 6 s. Find a.

Model Answer Δv = 15. a = 15 / 6 = 2.5 m/s²

Q10

Object 6: v from 30 → 0 m/s in 6 s. Find a.

Model Answer Δv = −30. a = −30 / 6 = −5 m/s²

Q11

Object 7: v from 10 → 30 m/s in 4 s. Find a.

Model Answer Δv = 20. a = 20 / 4 = 5 m/s²

Q12

Object 8: v from 30 → 10 m/s in 8 s. Find a.

Model Answer Δv = −20. a = −20 / 8 = −2.5 m/s²

Part 3 · Area Under the Graph = Distance

Rectangle: area = base × height = time × velocity
Triangle (v starts or ends at zero): area = ½ × base × height
Trapezium: area = ½ × (v₁ + v₂) × t
Complex shapes: split into rectangles and triangles, then add the areas

Exam Question — Velocity-Time Graphs (14 marks)

The questions below combine ideas from the whole lesson in the style of an AQA exam paper. Mark allocations are shown in brackets.

Exam Questions

Part 1

(a)(i) A car accelerates from rest to 20 m/s in 10 s, runs at 20 m/s for 30 s, then decelerates to rest in 5 s. Sketch the v-t graph. (2)

Model Answer Straight line from (0, 0) to (10, 20); horizontal at v = 20 from t = 10 to t = 40; straight line from (40, 20) down to (45, 0).

Part 2

(a)(ii) Calculate the acceleration during the first 10 s. (2)

Model Answer a = (20 − 0) / 10 = 2 m/s²

Part 3

(a)(iii) Calculate the total distance travelled. (3)

Model Answer Area = triangle + rectangle + triangle
= (½ × 10 × 20) + (30 × 20) + (½ × 5 × 20)
= 100 + 600 + 50 = 750 m

Part 4

(b)(i) An object accelerates uniformly from rest to 30 m/s in 8 s. Find a. (2)

Model Answer a = 30 / 8 = 3.75 m/s²

Part 5

(b)(ii) Calculate the distance travelled during the 8 s. (2)

Model Answer Area = ½ × 8 × 30 = 120 m

Part 6

Model Answer Initially the graph has a steep positive gradient (large acceleration). As speed increases, air resistance increases, so the gradient decreases. When air resistance equals weight, the resultant force is zero and the gradient becomes zero — the line becomes horizontal and terminal velocity is reached.

v²

Lesson 7 — v² = u² + 2as

Do Now

Write the formula a = Δv / t and rearrange it to make Δv the subject.

Model Answer a = Δv / t → Δv = a × t

A car accelerates from 10 m/s to 34 m/s in 6 s. Calculate the acceleration.

Model Answer a = (34 − 10) / 6 = 24 / 6 = 4 m/s²

On a velocity-time graph, what does the area under the line represent?

Model Answer The distance travelled by the object.

A cyclist decelerates uniformly from 12 m/s to rest. The deceleration is 3 m/s². Calculate how long the cyclist takes to stop.

Model Answer t = Δv / a = 12 / 3 = 4 s

Part 1 · The Equation and Identifying Variables

v² = u² + 2as

This equation links four quantities:

v = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration (m/s²)
s = displacement (m)

This formula is provided on the AQA formula sheet. Use it when the time t is not known.

Special cases: object starting from rest → u = 0; object slowing to a stop → v = 0.

Questions

Section A · Identify u, v, a, s. "Find the final velocity of an object whose initial velocity is 9 m/s, acceleration 7.8 m/s², over a displacement of 80 m."

Model Answer u = 9, a = 7.8, s = 80, v = ?
v² = 81 + 1 248 = 1 329 → v = 36.5 m/s

Identify u, v, a, s. "a = 7 m/s². After 500 m the object reaches 280 m/s. Find the initial velocity."

Model Answer v = 280, a = 7, s = 500, u = ?
78 400 = u² + 7 000 → u² = 71 400 → u = 267.2 m/s

Identify u, v, a, s. "a = 0.8 m/s², over 2 500 m, what u is needed to reach 560 m/s?"

Model Answer v = 560, a = 0.8, s = 2 500, u = ?
313 600 = u² + 4 000 → u² = 309 600 → u = 556.4 m/s

Section B · v = 10, u = 3, a = 2.5. Find s.

Model Answer 100 = 9 + 5s → 5s = 91 → s = 18.2 m

u = 4, a = 3.8, s = 31. Find v.

Model Answer v² = 16 + 235.6 = 251.6 → v = 15.86 m/s

v = 40, a = 4.1, s = 100. Find u.

Model Answer 1 600 = u² + 820 → u² = 780 → u = 27.9 m/s

Part 2 · Rearranging — Finding v, u, a and s

To find v: v = √(u² + 2as)
To find u: u = √(v² − 2as)
To find a: a = (v² − u²) / (2s)
To find s: s = (v² − u²) / (2a)

Questions

Section C · u = 2, a = 30, s = 45. Find v.

Model Answer v² = 4 + 2 700 = 2 704 → v = 52 m/s

u = 40, a = 6.8, s = 230. Find v.

Model Answer v² = 1 600 + 3 128 = 4 728 → v = 68.8 m/s

u = 650, a = 1 300, s = 490. Find v.

Model Answer v² = 422 500 + 1 274 000 = 1 696 500 → v = 1 302.5 m/s

Q10

u = 3, a = 4.8, s = 24. Find v.

Model Answer v² = 9 + 230.4 = 239.4 → v = 15.5 m/s

Q11

u = 25, a = 0.5, s = 200. Find v.

Model Answer v² = 625 + 200 = 825 → v = 28.7 m/s

Q12

u = 0.2, a = 1.9, s = 140. Find v.

Model Answer v² = 0.04 + 532 = 532.04 → v = 23.1 m/s

Q13

Section D · v = 135.7, a = 22.1, s = 100. Find u.

Model Answer u² = 135.7² − 2(22.1)(100) = 18 414 − 4 420 = 13 994 → u = 118.3 m/s

Q14

v = 655, a = 25, s = 239. Find u.

Model Answer u² = 655² − 2(25)(239) = 429 025 − 11 950 = 417 075 → u = 645.8 m/s

Q15

v = 0.64, a = 0.21, s = 0.31. Find u.

Model Answer u² = 0.64² − 2(0.21)(0.31) = 0.4096 − 0.1302 = 0.2794 → u = 0.529 m/s

Q16

a = 3.1, s = 700, v = 207. Find u.

Model Answer u² = 207² − 2(3.1)(700) = 42 849 − 4 340 = 38 509 → u = 196.2 m/s

Q17

a = 2.1, s = 900, v = 607. Find u.

Model Answer u² = 607² − 2(2.1)(900) = 368 449 − 3 780 = 364 669 → u = 603.9 m/s

Q18

a = 1.8, s = 1 600, v = 560. Find u.

Model Answer u² = 560² − 2(1.8)(1 600) = 313 600 − 5 760 = 307 840 → u = 554.8 m/s

Part 3 · Finding a and s; Deceleration

For deceleration, v < u so v² − u² is negative and a comes out negative.

Special case (u = 0): v² = 2as, so a = v²/(2s) and s = v²/(2a).

Special case (v = 0): 0 = u² + 2as, so s = −u² / (2a). Note a is negative here, so s comes out positive.

Questions

Q19

Section E · v = 560, u = 352, s = 28. Find a.

Model Answer (560² − 352²) / (2 × 28) = 189 696 / 56 = 3 387 m/s²

Q20

v = 28.3, u = 15.1, s = 0.7. Find a.

Model Answer (28.3² − 15.1²) / (2 × 0.7) = 572.9 / 1.4 = 409 m/s²

Q21

v = 1 560, u = 1 550, s = 145. Find a.

Model Answer (1 560² − 1 550²) / (2 × 145) = 31 100 / 290 = 107.2 m/s²

Q22

From 12 → 63 m/s over 125 m. Find a.

Model Answer (63² − 12²) / (2 × 125) = 3 825 / 250 = 15.3 m/s²

Q23

Section F · u = 2, v = 3, a = 0.25. Find s.

Model Answer (9 − 4) / (2 × 0.25) = 5 / 0.5 = 10 m

Q24

u = 6, v = 30, a = 4. Find s.

Model Answer (900 − 36) / (2 × 4) = 864 / 8 = 108 m

Q25

u = 20, v = 300, a = 18. Find s.

Model Answer (90 000 − 400) / (2 × 18) = 89 600 / 36 = 2 489 m

Q26

Section G · v = 20, u = 30, s = 5. Find a.

Model Answer (400 − 900) / (2 × 5) = −500 / 10 = −50 m/s² (deceleration)

Q27

u = 4, v = 3, a = −0.25. Find s.

Model Answer (9 − 16) / (2 × −0.25) = −7 / −0.5 = 14 m

Q28

From rest to 5 m/s with a = 0.9. Find s.

Model Answer u = 0: s = v² / (2a) = 25 / 1.8 = 13.9 m

Q29

From rest, a = 5, s = 15. Find v.

Model Answer u = 0: v² = 2(5)(15) = 150 → v = 12.25 m/s

Q30

From rest reaches 165 m/s in 30 m. Find a.

Model Answer u = 0: a = 165² / (2 × 30) = 27 225 / 60 = 453.75 m/s²

Q31

Slows to rest with a = −20 m/s² over 60 m. Find u.

Model Answer v = 0: u² = −2(−20)(60) = 2 400 → u = 49.0 m/s

Q32

Car at 35 m/s comes to rest in 10.2 m. Find a.

Model Answer v = 0: a = −35² / (2 × 10.2) = −1 225 / 20.4 = −60.1 m/s²

Q33

Stone enters ground at 1.3 m/s, a = −135 m/s². Find penetration depth.

Model Answer v = 0: s = −1.3² / (2 × −135) = 1.69 / 270 = 0.00626 m (≈ 6.3 mm)

Exam Question — v² = u² + 2as (10 marks)

The questions below combine ideas from the whole lesson in the style of an AQA exam paper. Mark allocations are shown in brackets.

Exam Questions

Part 1

(a) State the names, symbols and units of the four quantities in v² = u² + 2as. (4)

Model Answer v = final velocity (m/s), u = initial velocity (m/s), a = acceleration (m/s²), s = displacement (m). One mark each.

Part 2

(b)(i) A car at 30 m/s brakes to rest over 45 m. Calculate the deceleration. (3)

Model Answer v = 0, u = 30, s = 45. 0 = 900 + 2a(45) → 90a = −900 → a = −10 m/s²

Part 3

(b)(ii) State one assumption made in your calculation. (1)

Model Answer The deceleration is uniform (constant) throughout.

Part 4

(c)(i) A ball is dropped from rest under gravity (g = 10 m/s²). Find v after falling 20 m. (3)

Model Answer u = 0, a = 10, s = 20. v² = 0 + 2(10)(20) = 400 → v = 20 m/s

Part 5

(c)(ii) Find v after falling 80 m. (2)

Model Answer v² = 2(10)(80) = 1 600 → v = 40 m/s

FORCES

Lesson 1 — Scalar and Vector Quantities

Do NowExpand All

Part 1 · Physical Quantities and Scalars

Examples of Scalars

QuestionsExpand All

Part 2 · Vector Quantities and Arrow Representation

Examples of Vectors

QuestionsExpand All

Part 3 · Exam Question — Scalars and Vectors (8 marks)

Exam QuestionsExpand All

Lesson 2 — Distance and Displacement

Do NowExpand All

Part 1 · Definitions

Worked Example

QuestionsExpand All

Part 2 · Calculating Distance and Displacement

QuestionsExpand All

Part 3 · Exam Question — Distance and Displacement (7 marks)

Exam QuestionsExpand All

Lesson 3 — Distance, Displacement and Speed

Do NowExpand All

Part 1 · Speed and Velocity — Definitions

Triangle of Equations

QuestionsExpand All

Part 2 · Further Practice

QuestionsExpand All

Part 3 · Exam Question — Velocity and Speed Calculations (9 marks)

Exam QuestionsExpand All

Lesson 4 — Velocity and Acceleration

Do NowExpand All

Part 1 · Defining Acceleration

QuestionsExpand All

Part 2 · The Acceleration Formula

QuestionsExpand All

Part 3 · Exam Question — Acceleration (8 marks)

Exam QuestionsExpand All

Lesson 5 — Distance-Time Graphs

Do NowExpand All

Part 1 · Key Rules for Distance-Time Graphs

QuestionsExpand All

Part 2 · Reading Speed from a Graph

QuestionsExpand All

Part 3 · Curved Graphs

QuestionsExpand All

Exam Question — Distance-Time Graphs (5 marks)

Exam QuestionsExpand All

Lesson 6 — Velocity-Time Graphs

Do NowExpand All

Part 1 · Key Rules — Don't Confuse with Distance-Time Graphs!

QuestionsExpand All

Part 2 · Finding Acceleration from the Gradient

QuestionsExpand All

Part 3 · Area Under the Graph = Distance

Exam Question — Velocity-Time Graphs (14 marks)

Exam QuestionsExpand All

Lesson 7 — v² = u² + 2as

Do NowExpand All

Part 1 · The Equation and Identifying Variables

QuestionsExpand All

Part 2 · Rearranging — Finding v, u, a and s

QuestionsExpand All

Part 3 · Finding a and s; Deceleration

QuestionsExpand All

Exam Question — v² = u² + 2as (10 marks)

Exam QuestionsExpand All

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