Overview
What you’ll learn
- Sort scalars from vectors and calculate weight with W = mg.
- Combine forces along a line and predict the motion from the resultant.
- Read speed, acceleration and distance straight off motion graphs.
- Apply all three of Newton's laws, including F = ma calculations.
- Write up the acceleration and force–extension required practicals.
- Explain stopping distances, the speed-squared rule and terminal velocity.
Specification checklist (AQA Trilogy 8464 · Edexcel 1SC0 · OCR Gateway J250)
Statements are paraphrased from the shared DfE combined science subject content and grouped by tier. Foundation content is studied by every student; Higher statements are examined on Higher-tier papers only. This chapter covers only what Combined Science (double award) examines — separate-science extras live on the GCSE Biology, Chemistry and Physics hubs. QuizQuip is not affiliated with AQA, Pearson Edexcel or OCR.
Foundation — every student
- 27.1Distinguish scalars (speed, distance, mass, time, energy) from vectors (velocity, displacement, force, weight, acceleration); recall that weight is a force measured in newtons, calculated as W = m g, acting through the centre of mass.
- 27.2Find the resultant of forces along a straight line and draw free-body diagrams; explain that zero resultant force means rest or constant velocity, and that a non-zero resultant force changes the velocity.
- 27.3Interpret distance–time and velocity–time graphs: speed is the gradient of a d–t graph, acceleration is the gradient of a v–t graph, and distance travelled is the area under a v–t graph.
- 27.4Calculate acceleration with a = (v − u) ÷ t, including decelerations as negative accelerations.
- 27.5Apply Newton's first law (no resultant force → no change in velocity), Newton's second law as F = m a, and Newton's third law (pairs of equal, opposite forces acting on different objects).
- 27.6Carry out the acceleration required practical: vary the applied force on a trolley of constant mass (falling weights over a pulley), using light gates or ticker timers, with a tilted runway compensating friction; the graph of acceleration against force is a straight line through the origin.
- 27.7Carry out the force–extension required practical: adding loads to a spring gives a straight-line graph of force against extension in the linear region, F = k e with spring constant k in N/m; beyond the limit of proportionality the graph curves and the spring may deform.
- 27.8Describe stopping distance as thinking distance plus braking distance: thinking distance grows with speed and reaction time (tiredness, alcohol, distractions); braking distance grows with speed, vehicle condition (brakes, tyres), road condition and mass.
- 27.9Explain why braking distance grows with the square of the speed: the work done by the brakes equals the kinetic energy (½ m v²), so doubling the speed quadruples the braking distance; very large decelerations overheat brakes and endanger passengers.
- 27.10Explain terminal velocity qualitatively: as speed grows, drag grows until it balances the weight, the resultant force falls to zero and the object moves at a steady speed (e.g. a skydiver before the parachute opens).
Key points
- 1Scalars (speed, mass, time) have size only; vectors (velocity, force, acceleration) have size and direction. Weight W = m g acts through the centre of mass.
- 2Forces along a line add or subtract to a resultant. Zero resultant → rest or constant velocity; non-zero → the velocity changes.
- 3Motion graphs: speed = gradient of d–t; acceleration = gradient of v–t; distance = area under v–t.
- 4Newton 1: no resultant force, no change in velocity. Newton 2: F = m a. Newton 3: pairs act on different objects, equal and opposite.
- 5Acceleration practical: vary force (falling weights) at constant trolley mass, tilt the runway to cancel friction; a against F is a straight line through the origin.
- 6Force–extension practical: F = k e in the linear region (k in N/m); past the limit of proportionality the line curves.
- 7Stopping distance = thinking + braking. Thinking grows with reaction time (tiredness, alcohol, phones); braking grows with speed, mass, worn tyres and slick roads.
- 8Braking distance ∝ speed² (the brakes must absorb ½ m v²): double the speed → four times the distance. A skydiver reaches terminal velocity when drag balances weight.
Notes · 1 of 4
Forces, weight and resultants
Pushes, pulls, and what happens when they disagree.
A force (newtons, N) is a push or pull, and it is a vector — direction matters as much as size. Scalars such as speed, mass and time carry no direction. Weight is the force of gravity on an object: W = m g, with g ≈ 10 N/kg on Earth. An 850 kg car weighs 850 × 10 = 8 500 N — its mass never changes, but its weight would shrink on the Moon. Weight acts as if concentrated at the centre of mass.
Several forces along one line collapse into a single resultant: 4 500 N of engine thrust against 1 200 N of drag gives 4 500 − 1 200 = 3 300 N forwards. Newton's first law reads the verdict: zero resultant force means no change in velocity — a plane cruising level at steady speed has drag exactly matching thrust, not zero forces. A non-zero resultant always changes the velocity: speeding up, slowing down or turning.
Worked example — a tug of war with numbers
Q: A boat's engine drives 2 800 N forwards; water drag is 900 N backwards. Find the resultant and state the motion. A: Resultant = 2 800 − 900 = 1 900 N forwards. A non-zero resultant means the boat accelerates forwards. Only when drag grows to match the thrust does it settle at a steady cruising speed.
Notes · 2 of 4
Motion graphs and acceleration
Gradient for speed and acceleration; area for distance.
On a distance–time graph the gradient is the speed: a horizontal line is rest, a steeper line is faster. On a velocity–time graph the gradient is the acceleration, a horizontal line is constant velocity, and the area under the line is the distance travelled — count rectangles and triangles. A cyclist covering 240 m in 48 s cruises at 240 ÷ 48 = 5.0 m/s.
Acceleration is the rate of change of velocity: a = (v − u) ÷ t. A cart surging from 0 to 18 m/s in 6.0 s accelerates at 18 ÷ 6.0 = 3.0 m/s²; a negative result is a deceleration. On a v–t graph, a curved line whose gradient steadily falls shows acceleration shrinking — often as drag builds.
Worked example — area under a v–t graph
Q: A train speeds up from rest to 12 m/s in 4.0 s, cruises at 12 m/s for 6.0 s, then brakes to rest in 4.0 s. Find the total distance. A: Triangle (0–4 s): ½ × 4.0 × 12 = 24 m. Rectangle (4–10 s): 12 × 6.0 = 72 m. Triangle (10–14 s): ½ × 4.0 × 12 = 24 m. Total = 24 + 72 + 24 = 120 m.
Notes · 3 of 4
Newton's laws and the two practicals
F = ma, equal-and-opposite pairs, and two experiments to write up.
Newton's second law quantifies the push: F = m a. A 950 kg car accelerating at 2.5 m/s² needs a resultant of 950 × 2.5 = 2 375 N. Newton's third law pairs are subtle: the tyre pushes the road backwards while the road pushes the tyre forwards — equal sizes, opposite directions, same type of force, acting on different objects. That road-on-tyre push is what actually accelerates the car.
Required practical — acceleration. A trolley runs along a track with a string over a pulley to a hanging weight that provides the force. Vary the force by changing the hanging weights while keeping the trolley plus load mass constant; measure acceleration with light gates or a ticker timer. Tilt the runway slightly first so gravity along the slope cancels friction. The graph of acceleration against force is a straight line through the origin: acceleration is proportional to force at fixed mass (double F, double a).
Required practical — force and extension. Hang a spring from a clamp, add loads, and measure the extension each time. In the linear region the force–extension graph is a straight line: F = k e, where k is the spring constant in N/m. A 6.0 N load stretching a spring 0.24 m gives k = 6.0 ÷ 0.24 = 25 N/m. Loading and unloading traces that retrace each other mean elastic deformation (it returns to shape); past the limit of proportionality the graph curves and the spring may be permanently stretched.
Practical pitfalls
In the acceleration practical, never change the trolley's mass while varying force — moving weights between the hanger and the trolley keeps the total constant. In the spring practical, read the ruler at eye level with the spring at rest, and check the unloading readings for permanent stretch.
Notes · 4 of 4
Stopping distances and terminal velocity
Reaction, brakes, and why speed is squared.
Stopping distance = thinking distance + braking distance. The thinking distance is covered during the driver's reaction time, so it grows with speed and with anything that slows reactions: tiredness, alcohol, drugs, distractions. At 30 m/s with a 0.80 s reaction time, the car travels 30 × 0.80 = 24 m before the brake pedal even moves. The braking distance then grows with speed, the vehicle's mass, worn brakes and tyres, and slick roads.
The speed-squared rule has an energy reason: the brakes must absorb the car's kinetic energy, ½ m v². An 800 kg car at 15 m/s carries ½ × 800 × 15² = 90 000 J; against a braking force of 3 000 N the stopping distance is 90 000 ÷ 3 000 = 30 m. Double the speed and the kinetic energy — and so the braking distance — quadruples. Very hard braking also risks overheated brakes and injured passengers, which is why huge decelerations are dangerous.
Terminal velocity completes the picture. A skydiver accelerates at first, but drag grows with speed until it balances the weight — resultant zero, so by Newton's first law the speed stops changing. Opening the parachute raises the drag sharply, slowing the fall until a new, slower balance is reached.
Worked example — a full emergency stop
Q: A van travels at 24 m/s. Reaction time 0.90 s; deceleration 6.0 m/s². Find the stopping distance. A: Thinking = 24 × 0.90 = 21.6 m. Braking: v² = u² − 2as with v = 0, so s = 24² ÷ (2 × 6.0) = 576 ÷ 12 = 48 m. Stopping distance = 21.6 + 48 = 69.6 m — and wet roads would stretch the braking half further.
Revise
Key terms
The words examiners expect you to use precisely.
- Scalar
- A quantity with size only — speed, distance, mass, time, energy.
- Vector
- A quantity with size and direction — velocity, displacement, force, acceleration.
- Weight
- The gravitational force on an object, W = m g in newtons, acting through the centre of mass.
- Resultant force
- The single force with the same effect as all the forces acting combined.
- Acceleration
- The rate of change of velocity, a = (v − u) ÷ t, in m/s².
- Newton's first law
- An object stays at rest or at constant velocity unless a resultant force acts.
Show 6 more key termsShow fewer key terms
- Newton's second law
- Resultant force = mass × acceleration: F = m a.
- Newton's third law
- Interacting objects exert equal, opposite forces of the same type on each other.
- Spring constant (k)
- Force per unit extension, F = k e, in N/m; measured from the linear region of a force–extension graph.
- Thinking distance
- The distance travelled during the driver's reaction time: speed × reaction time.
- Braking distance
- The distance travelled while the brakes bring the vehicle to rest; grows with the square of the speed.
- Terminal velocity
- The steady speed reached when drag balances weight and the resultant force is zero.
Revise
Common mistakes
Where students most often lose marks in this chapter — and what to do instead.
Mistake: Using mass in kilograms inside W = m g and quoting newtons as kg
Weight is a force in newtons; mass is the kilogram amount of matter.
Instead: A 25 kg dog weighs 25 × 10 = 250 N; its mass is unchanged on the Moon even though its weight falls.
Mistake: Believing steady velocity needs a constant forwards force
That predicts continuous acceleration.
Instead: Constant velocity means the resultant force is zero — thrust exactly balanced by drag.
Show 10 more mistakesShow fewer mistakes
Mistake: Reading acceleration from a distance–time graph
The d–t gradient is speed; acceleration lives on the v–t graph.
Instead: d–t gradient → speed; v–t gradient → acceleration; v–t area → distance.
Mistake: Taking the area under a distance–time graph
Areas belong to velocity–time graphs.
Instead: Distance comes from the v–t area (rectangles and triangles).
Mistake: Putting Newton's third law pairs on the same object
Equal-and-opposite forces on one object would cancel to nothing, ever.
Instead: Third-law pairs act on different objects: tyre pushes road, road pushes tyre.
Mistake: Changing the trolley's mass while varying force in the acceleration practical
Then the a–F graph cannot isolate the force effect.
Instead: Keep total mass constant — move weights between hanger and trolley — and vary only the pulling force.
Mistake: Skipping the friction-tilt step
Friction then silently eats part of the applied force.
Instead: Tilt the runway so the component of gravity cancels friction before measuring.
Mistake: Using F = k e beyond the spring's linear region
Past the limit of proportionality the graph curves and k is no longer constant.
Instead: Quote k only from the straight-line region; note where the data depart from it.
Mistake: Blaming alcohol for longer braking distance
Alcohol slows reactions; the brakes do not weaken.
Instead: Reaction time rises, so thinking distance lengthens; braking distance answers to speed, mass and grip.
Mistake: Saying double the speed means double the braking distance
Kinetic energy grows with v², not v.
Instead: Braking distance quadruples when speed doubles (18 m at 12 m/s becomes 72 m at 24 m/s).
Mistake: Quoting a spring constant in N/cm
The unit asked for is N/m; mixing them is out by a factor of 100.
Instead: Convert: 2 N per 8 cm = 2 ÷ 0.08 = 25 N/m.
Mistake: Saying forces vanish at terminal velocity
Weight and drag both still act.
Instead: They balance exactly — zero resultant — which is why the speed stops changing.
Check yourself
Quick quiz
20 questions with instant explanations. About 5 minutes.
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Explain forces, weight and resultants
Which pair are both vector quantities?
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Practise
Topical questions
12 exam-style questions with model answers and mark schemes.
A set of structured and calculation questions matched to this chapter, in three tiers, each with a model answer, mark scheme and examiner tip. Higher-tier questions are tagged.
Practise topical questionsExam practice
Practice paper
A 30-mark paper in the style of a Combined Science paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Chapter 27 Practice Paper
30 marks · 40 min · 5 pages
6 multiple-choice questions and structured questions with answer lines. Scalars and vectors, weight and resultant forces, distance–time and velocity–time graphs, acceleration, Newton's three laws, the acceleration and force–extension required practicals, stopping distances with the speed-squared rule, and terminal velocity.
Mark scheme
PROChapter 27 Practice Paper — mark scheme
Answers · Marking points · 3 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
How does this chapter fit the real exams?
Physics Paper 1
1 h 15 · 70 marks (Trilogy)
Energy, electricity, particle model, atomic structure (chapters 21–26)
Physics Paper 2
1 h 15 · 70 marks (Trilogy)
Forces, waves, magnetism and electromagnetism (chapters 27–30)
Tiers
Foundation (grades 1-1 to 5-5) · Higher (4-3 to 9-9)
Two grades are awarded together; the 21 required practicals are examined on the written papers
Want a full paper? Try the Biology Paper 1-style (Foundation), Biology Paper 1-style (Higher), Biology Paper 2-style (Foundation), Biology Paper 2-style (Higher), Chemistry Paper 1-style (Foundation), Chemistry Paper 1-style (Higher), Chemistry Paper 2-style (Foundation), Chemistry Paper 2-style (Higher), Physics Paper 1-style (Foundation), Physics Paper 1-style (Higher), Physics Paper 2-style (Foundation), Physics Paper 2-style (Higher), Required practicals practice sets.
Questions
Frequently asked questions
How do I read motion graphs without mixing them up?
Ask what the axes are. Distance–time: gradient is speed, horizontal means rest. Velocity–time: gradient is acceleration, horizontal means constant velocity, and the area under the line is the distance. Split areas into rectangles and triangles when the shape is complex.
What makes forces a Newton's third law pair?
Both forces must be equal in size, opposite in direction, of the same type, and acting on different objects. The tyre pushes the road backwards; the road pushes the tyre forwards. Two forces on the same object are never a third-law pair.
Why does the acceleration practical tilt the runway?
To cancel friction. A gentle tilt makes gravity's component along the track balance the frictional drag, so the hanging weight provides the whole resultant force. The a–F graph is then a clean straight line through the origin.
Why exactly does braking distance grow with speed squared?
The brakes remove kinetic energy, and KE = ½ m v². Double the speed and the energy quadruples; since the braking force is roughly fixed, work = force × distance means the distance quadruples too. Thinking distance, by contrast, only grows in proportion to speed.
What is happening at terminal velocity?
Drag grows with speed until it exactly balances weight. The resultant force is then zero, so by Newton's first law the velocity stops changing — the skydiver falls at a constant speed until the parachute changes the balance.
Up next
Chapter 28: Work, energy and momentum