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Edexcel GCSE Physics (1PH0) · Topic 2 — Motion and forces
Topic 2 · Motion and forces

Why does a fast car take so long to stop?

Press the brake and the car keeps going — sometimes for the length of a football pitch. By the end of this mini-lesson you'll be able to predict motion with equations and graphs, explain it with Newton's laws, and use momentum and stopping distance to understand road safety.

velocity → friction Forces decide how the velocity changes.
Motion is described by numbers and direction; forces change that motion.

Tap Start — you'll answer questions and earn ⭐ as you go.

Idea 1 · Describing motion

Scalars vs vectors

Every physical quantity is one of two kinds:

  • Scalar — has size (magnitude) only. No direction. Examples: distance, speed, mass, energy.
  • Vector — has size and a specific direction. Examples: displacement, velocity, acceleration, force, weight, momentum.
The key pairs: distance (scalar, total path travelled) vs displacement (vector, straight line from start to finish, with direction). Speed (scalar) vs velocity (vector — speed in a stated direction).
Start / Finish 100 m 100 m distance = 400 m displacement = 0 m
One full lap of a 100 m square: distance 400 m, but displacement is 0 — you ended where you started.
Quick check

Scalar or vector?

?Which one of these is a vector quantity?
Idea 2 · Speed

Speed = distance ÷ time

v = s ÷ t(average) speed (m/s) = distance (m) ÷ time (s)

Rearranged: distance = average speed × time. Speed is usually measured in metres per second (m/s).

Typical everyday speeds you should recall (Edexcel 2.12): walking ≈ 1.5 m/s, running ≈ 3 m/s, cycling ≈ 6 m/s. The speed of sound in air ≈ 340 m/s; a gentle wind is a few m/s. A car in town ≈ 13 m/s (30 mph).
Measuring speed in the lab (2.11): use light gates connected to a data logger. The gate records the time for a card of known length to pass through, and the computer works out speed = card length ÷ time — far more precise than a stopwatch.
Sort it

Scalar or vector — three in a row

Tap the correct label for each quantity.

Idea 3 · Acceleration

Acceleration = change in velocity ÷ time

Acceleration tells you how quickly velocity changes. It is a vector.

a = (v − u) ÷ tacceleration (m/s²) = change in velocity (m/s) ÷ time taken (s)

Here v = final velocity, u = initial velocity. A negative acceleration (deceleration) means slowing down.

Free fall: near Earth's surface the acceleration due to gravity is g ≈ 10 m/s² (Edexcel 2.13). So a dropped object speeds up by about 10 m/s every second.
Worked example

A car speeds up from rest (u = 0) to v = 24 m/s in t = 12 s.

a = (24 − 0) ÷ 12 = 2 m/s²

Calculate

Your turn — acceleration

1A cyclist slows from 10 m/s to 4 m/s in 3 s. Calculate the acceleration (give the size; the minus sign just means slowing down).
m/s²
Hint: a = (v − u) ÷ t = (4 − 10) ÷ 3. A value of −2 (or 2) is accepted.
Idea 4 · Distance–time graphs

Gradient = speed

On a distance–time graph, the steepness (gradient) of the line is the speed. A horizontal line means stationary; a steeper line means faster.

distance → time → steep = fast flat = stopped gentle = slow gradient = speed
Steep line = high speed; flat line = at rest; gentler slope = lower speed.
Read it as a story: fast away, then a flat pause (stationary), then a slower journey back is impossible here — distance never decreases. (A displacement–time graph could come back down; distance–time cannot.)
Read the graph

Your turn — speed from a d–t graph

2On a distance–time graph a straight line rises from 0 m to 120 m over 8 s. What is the speed (the gradient)?
m/s
Hint: gradient = rise ÷ run = 120 ÷ 8.
Idea 5 · Velocity–time graphs

Gradient = acceleration; area = distance

The velocity–time graph is the most powerful tool in this topic. Two things to read off:

  • Gradient of the line = acceleration (a flat line = constant velocity, zero acceleration).
  • Area between the line and the time axis = distance travelled.
velocity → time → gradient = acceleration area = distance flat = constant velocity
Sloping up = accelerating. The shaded triangle's area gives the distance covered.
Worked example — distance from area

A car accelerates uniformly from rest to 20 m/s in 10 s.

Distance = area of triangle = ½ × base × height = ½ × 10 × 20 = 100 m

Read the graph

Your turn — distance from a v–t graph

3A van travels at a constant 18 m/s for 12 s. On its velocity–time graph, what distance does the area under the line represent?
m
Hint: constant velocity → the area is a rectangle = velocity × time = 18 × 12.
Idea 6 · The "no-time" equation

v² − u² = 2 × a × x

When you know the distance but not the time, this equation of uniform motion links velocity, acceleration and distance directly.

v² − u² = 2 a x(final velocity)² − (initial velocity)² = 2 × acceleration (m/s²) × distance (m)
Worked example

A car starts from rest (u = 0) and accelerates at 3 m/s² over a distance of 24 m. Find v.

v² = u² + 2ax = 0 + 2 × 3 × 24 = 144

v = √144 = 12 m/s

Calculate

Your turn — v² − u² = 2ax

4A motorbike accelerates from rest at 4 m/s² over 50 m. Calculate its final velocity v.
m/s
Hint: v² = u² + 2ax = 0 + 2 × 4 × 50 = 400, then v = √400.
Idea 7 · Newton's first law

No resultant force → no change in motion

Newton's first law: an object stays at rest, or keeps moving at constant velocity, unless acted on by a resultant force.

  • If the resultant force is zero → the object is at rest or moving at constant velocity (forces are balanced).
  • If the resultant force is not zero → the object's speed and/or direction changes (forces are unbalanced).
Analogy — the seatbelt: when a car stops suddenly, your body "wants" to keep moving forward at constant velocity (Newton's 1st law). The seatbelt provides the resultant force that changes your motion. This tendency to keep doing what you're doing is called inertia.
Quick check

Balanced forces

?A skydiver falls at a steady terminal velocity. What is the resultant force on her?
Idea 8 · Newton's second law

F = m × a

Newton's second law: the resultant force on an object equals its mass times its acceleration.

F = m aresultant force (N) = mass (kg) × acceleration (m/s²)

Bigger force → bigger acceleration. Bigger mass → smaller acceleration for the same force.

2 kg 60 N drive 20 N friction resultant = 60 − 20 = 40 N → a = 40 ÷ 2 = 20 m/s²
Find the resultant force first, then use F = ma.
Weight is a force, so it also fits F = ma: weight = mass × gravitational field strength, W = m × g (with g ≈ 10 N/kg). Weight is measured with a newtonmeter (force meter), and is bigger where g is bigger.
Calculate

Your turn — F = ma

5A 1200 kg car accelerates at 2.5 m/s². Calculate the resultant force needed.
N
Hint: F = m × a = 1200 × 2.5.
Idea 9 · Core Practical & inertia

Investigating F, m and a

Core Practical (2.19): investigate the link between force, mass and acceleration using a trolley on a runway, pulled by a hanging mass over a pulley, timed with light gates.

  • Vary the force (move masses onto the hanger) with total mass kept constant → acceleration is directly proportional to force.
  • Vary the mass (add masses to the trolley) with force kept constant → acceleration is inversely proportional to mass.
HT — inertial mass (2.22): inertial mass measures how hard it is to change the velocity of an object. It is defined as the ratio of force ÷ acceleration (rearranged from F = ma). A large inertial mass resists changes in motion strongly.
Idea 10 · Newton's third law

Every action has an equal, opposite reaction

Newton's third law: when object A pushes on object B, object B pushes back on A with a force that is equal in size and opposite in direction. The two forces act on different objects, so they never cancel out.

Examples: you push the ground backward, the ground pushes you forward (walking). A gun pushes the bullet forward; the bullet pushes the gun back (recoil). At HT this links to conservation of momentum in collisions.
HT — circular motion (2.20–2.21): an object moving in a circle at constant speed still has a changing velocity, because its direction keeps changing. That change needs a resultant force called the centripetal force, which always points towards the centre of the circle.
Idea 11 · Momentum (Higher Tier)

p = m × v

HT — momentum (2.24): momentum is "mass in motion" — a vector. Its unit is the kilogram metre per second (kg m/s).

p = m vmomentum (kg m/s) = mass (kg) × velocity (m/s)
BEFORE at rest AFTER stuck together, slower total momentum before = total momentum after
In a collision, total momentum is conserved (no external force).
HT — conservation of momentum (2.23, 2.25): in any collision or explosion with no external force, total momentum before = total momentum after. This follows directly from Newton's third law.
HT — force as rate of change of momentum (2.26): Newton's second law can be written F = (mv − mu) ÷ t — force equals change in momentum ÷ time. A longer collision time means a smaller force (the physics behind crumple zones and airbags).
Calculate · HT

Your turn — momentum

6A 0.5 kg football is kicked at 24 m/s. Calculate its momentum.
kg m/s
Hint: p = m × v = 0.5 × 24.
Idea 12 · Stopping distance

Stopping = thinking + braking

The stopping distance of a vehicle is the sum of two parts (Edexcel 2.28):

stopping = thinking + brakingthinking distance (during reaction) + braking distance (while brakes act)
thinking braking distance total stopping distance Braking distance grows with the square of speed — it dominates at high speed.
Thinking distance (reaction) + braking distance (deceleration) = stopping distance.

Thinking distance is how far the car travels during the driver's reaction time (2.27). It is increased by anything that slows reactions (2.30): tiredness, alcohol and other drugs, distractions (e.g. phones), and higher speed.

Quick check

What affects braking distance?

?Which factor increases the braking distance (not the thinking distance)?
Idea 13 · Speed, energy & danger

Why high speed is so dangerous

Factors affecting total stopping distance (2.29): the vehicle's speed, its mass, the driver's reaction time, the state of the brakes, the state of the road, and the friction between tyre and road.

Braking distance ∝ speed² (2.33-P): the work done by the brakes to stop the car equals the car's kinetic energy. Because kinetic energy depends on speed squared, doubling your speed gives four times the braking distance. This is why small speed increases are so risky (2.32-P estimates this over a range of speeds).
Dangers of large decelerations: stopping a fast, heavy vehicle very quickly means a very large braking force (F = ma). Large forces can throw occupants forward, overheat the brakes, and cause skidding. Crumple zones, seatbelts and airbags increase the collision time, reducing the force (F = Δp ÷ t).
Quick check

Double the speed…

?A car's speed doubles. Roughly what happens to its braking distance (everything else the same)?
Recap

What you've nailed

Scalars vs vectors: scalar = size only (distance, speed, mass); vector = size + direction (displacement, velocity, acceleration, force, momentum).

Equations: v = s ÷ t · a = (v − u) ÷ t · v² − u² = 2ax · F = ma · W = mg · p = mv (HT) · F = (mv − mu) ÷ t (HT).

Graphs: distance–time gradient = speed; velocity–time gradient = acceleration and area = distance.

Newton's laws: 1st — no resultant force ⇒ constant velocity (inertia); 2nd — F = ma; 3rd — equal & opposite forces on different objects.

Safety: stopping = thinking + braking; braking ∝ speed²; longer collision time ⇒ smaller force (crumple zones, airbags).

Recall: g ≈ 10 m/s²; typical speeds (walk 1.5, run 3, cycle 6, sound 340 m/s); Core Practical = trolley + light gates for F, m, a.

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