IGCSE Physics · Unit 1: Motion, Forces and Energy · Core + Extended
Motion CoreExtended adds acceleration formula & terminal velocity
This note covers how physicists describe movement precisely: the difference between speed and velocity, how to read distance-time and speed-time graphs like a physicist rather than just plotting points, and what really happens to a falling object once air resistance gets involved. Graphs are the heart of this topic — almost every exam question here is really a graph-reading question in disguise.
Why this matters
Once you can read a distance-time or speed-time graph fluently — spotting rest, constant speed, acceleration and deceleration at a glance — a huge chunk of the Motion, Forces and Energy topic becomes much easier, because velocity and acceleration graphs reappear constantly in momentum, forces, and even electromagnetic induction later in the syllabus.
1. Speed, Velocity and Average Speed
Speed is defined as distance travelled per unit time. It is a scalar — no direction is involved.
v = s / t
Velocity is speed in a given direction — it is the vector version of speed. Two objects can have the same speed but different velocities if they are moving in different directions.
Because real journeys rarely happen at a perfectly constant speed, we often use average speed instead:
average speed = total distance travelled / total time taken
Worked example — basic speed
A cyclist travels 300 m in 25 s. Find her speed.
v = s / t = 300 / 25 = 12 m/s
v = s / t = 300 / 25 = 12 m/s
Worked example — average speed over a journey with a stop
A car travels 40 km in 30 minutes, stops for 15 minutes, then travels a further 20 km in 20 minutes.
Total distance = 40 + 20 = 60 km
Total time = 30 + 15 + 20 = 65 minutes = 65/60 hours ≈ 1.083 h
Average speed = 60 ÷ 1.083 ≈ 55.4 km/h
Notice the stationary 15 minutes still counts in the total time — it just doesn’t add any distance.
Total distance = 40 + 20 = 60 km
Total time = 30 + 15 + 20 = 65 minutes = 65/60 hours ≈ 1.083 h
Average speed = 60 ÷ 1.083 ≈ 55.4 km/h
Notice the stationary 15 minutes still counts in the total time — it just doesn’t add any distance.
Try it — Speed and Average Speed
- A runner covers 100 m in 12.5 s. Calculate her speed.
- A train travels 180 km in 2 hours, including a 20-minute stop at a station. Calculate its average speed for the whole journey in km/h.
- Explain, using the terms speed and velocity, why two cars travelling at 60 km/h on opposite sides of a straight road do not have the same velocity.
Show answers
- v = 100/12.5 = 8 m/s
- Total time = 2 hours = 120 minutes (the 20-minute stop is already included in the 2 hours). Average speed = 180/2 = 90 km/h
- Speed only describes how fast they are travelling (60 km/h each, so their speeds are equal), but velocity also includes direction. Since the cars are travelling in opposite directions along the road, their velocities are different — in fact opposite.
2. Distance–Time Graphs
A distance-time graph plots distance travelled (y-axis) against time (x-axis). The gradient (slope) of the graph at any point gives the speed at that point.
| Shape of graph | What it means |
|---|---|
| Horizontal line (zero gradient) | object is at rest |
| Straight line, constant gradient | object moving at constant speed |
| Curve getting steeper | object is accelerating |
| Curve getting less steep | object is decelerating |
Worked example — speed from a distance-time graph
A straight-line section of a distance-time graph rises from 0 m at t = 0 s to 60 m at t = 15 s.
Speed = gradient = change in distance / change in time = 60 / 15 = 4 m/s
Speed = gradient = change in distance / change in time = 60 / 15 = 4 m/s
3. Speed–Time Graphs
A speed-time graph plots speed (y-axis) against time (x-axis). Two key skills here:
- The area under the graph gives the distance travelled — this works for both constant speed and constant acceleration sections.
- The gradient of the graph gives the acceleration (Extended, see Section 4).
| Shape of graph | What it means |
|---|---|
| Horizontal line (zero gradient) | constant speed (zero acceleration) |
| Straight line sloping upward | constant acceleration |
| Straight line sloping downward | constant deceleration |
| Curve with changing steepness | changing acceleration |
Worked example — distance from a speed-time graph (constant speed)
An object moves at a constant 8 m/s for 10 s, shown as a horizontal line on a speed-time graph.
Distance = area under graph = base × height = 10 × 8 = 80 m
Distance = area under graph = base × height = 10 × 8 = 80 m
Worked example — distance from a speed-time graph (constant acceleration from rest)
An object accelerates uniformly from rest to 12 m/s over 6 s, shown as a straight sloped line on a speed-time graph.
This area is a triangle: distance = ½ × base × height = ½ × 6 × 12 = 36 m
This area is a triangle: distance = ½ × base × height = ½ × 6 × 12 = 36 m
Worked example — distance from a speed-time graph (trapezium: speeds up, then constant)
An object accelerates from 0 to 10 m/s in the first 4 s, then travels at a constant 10 m/s for a further 6 s.
Area of triangle (0–4 s) = ½ × 4 × 10 = 20 m
Area of rectangle (4–10 s) = 6 × 10 = 60 m
Total distance = 20 + 60 = 80 m
Split the shape into simple triangles and rectangles, find each area separately, then add them together.
Area of triangle (0–4 s) = ½ × 4 × 10 = 20 m
Area of rectangle (4–10 s) = 6 × 10 = 60 m
Total distance = 20 + 60 = 80 m
Split the shape into simple triangles and rectangles, find each area separately, then add them together.
Try it — Reading Graphs
- A distance-time graph shows a straight line from (0 s, 0 m) to (20 s, 100 m). Find the speed.
- A speed-time graph shows an object travelling at a constant 15 m/s for 8 s. Find the distance travelled.
- An object accelerates uniformly from rest to 20 m/s in 5 s, then decelerates uniformly to rest over the next 3 s. Sketch the shape of the speed-time graph in words, and calculate the total distance travelled.
Show answers
- speed = gradient = 100/20 = 5 m/s
- distance = area = 15 × 8 = 120 m
- The graph would show a straight line rising from (0,0) to (5,20), then a straight line falling from (5,20) to (8,0) — an overall triangle shape. Total distance = area of triangle = ½ × base × height = ½ × 8 × 20 = 80 m
4. Acceleration Extended only
Acceleration is defined as the change in velocity per unit time.
a = Δv / Δt
Acceleration can be calculated from the gradient of a speed-time graph, in exactly the same way that speed is found from the gradient of a distance-time graph. A deceleration is simply a negative acceleration — the object is slowing down, and this shows up as a negative value when you use the equation, or as a downward-sloping line on a speed-time graph.
Worked example — calculating acceleration
A cyclist speeds up from 2 m/s to 14 m/s in 4 s.
a = Δv / Δt = (14 − 2) / 4 = 12 / 4 = 3 m/s²
a = Δv / Δt = (14 − 2) / 4 = 12 / 4 = 3 m/s²
Worked example — calculating deceleration
A car slows from 25 m/s to 5 m/s in 8 s.
a = Δv / Δt = (5 − 25) / 8 = −20 / 8 = −2.5 m/s²
This is a deceleration of 2.5 m/s² — the negative sign shows the velocity is decreasing.
a = Δv / Δt = (5 − 25) / 8 = −20 / 8 = −2.5 m/s²
This is a deceleration of 2.5 m/s² — the negative sign shows the velocity is decreasing.
Worked example — acceleration from a speed-time graph gradient
A speed-time graph shows a straight line rising from 5 m/s at t = 0 s to 35 m/s at t = 10 s.
a = gradient = (35 − 5) / (10 − 0) = 30 / 10 = 3 m/s²
a = gradient = (35 − 5) / (10 − 0) = 30 / 10 = 3 m/s²
Try it — Acceleration
- A ball rolling down a ramp speeds up from 0 to 6 m/s in 3 s. Calculate its acceleration.
- A skater slows from 9 m/s to 3 m/s in 4 s. Calculate the acceleration, and state whether this is an acceleration or a deceleration.
- A speed-time graph shows a straight line falling from 20 m/s at t = 2 s to 0 m/s at t = 10 s. Find the acceleration.
Show answers
- a = (6−0)/3 = 2 m/s²
- a = (3−9)/4 = −6/4 = −1.5 m/s². Since the value is negative, this is a deceleration of 1.5 m/s².
- a = (0−20)/(10−2) = −20/8 = −2.5 m/s²
5. Acceleration of Free Fall
Near the Earth’s surface, the acceleration of free fall, g, is approximately constant and equal to about 9.8 m/s². This means that, ignoring air resistance, every object speeds up by roughly 9.8 m/s for every second it falls, regardless of its mass.
Worked example — speed after falling
A ball is dropped from rest and falls freely (no air resistance) for 3 s.
Speed gained = a × t = 9.8 × 3 = 29.4 m/s
Speed gained = a × t = 9.8 × 3 = 29.4 m/s
6. Falling With and Without Air Resistance: Terminal Velocity Extended only
Without air resistance, a falling object accelerates at a constant g = 9.8 m/s² all the way down — its speed-time graph is a straight line with constant gradient. In real life, though, air resistance acts on a falling object, and this changes everything.
As a falling object speeds up, air resistance (drag) increases. Eventually air resistance becomes equal in size to the object’s weight, so the resultant force on the object becomes zero. From that point on, the object stops accelerating and falls at a constant maximum speed called the terminal velocity.
Describing the motion — the four stages
1. Just released: speed is low, so air resistance is small; weight is much bigger than air resistance, so acceleration is close to g.
2. Speeding up: as speed increases, air resistance increases too, so the resultant force (and hence the acceleration) gets smaller — the object is still accelerating, but less and less.
3. Terminal velocity reached: air resistance now equals weight; resultant force = 0, so acceleration = 0. The object falls at a constant, maximum speed.
4. On a speed-time graph: the line starts steep (high acceleration), curves and becomes less steep (decreasing acceleration), then flattens into a horizontal line (constant terminal velocity).
2. Speeding up: as speed increases, air resistance increases too, so the resultant force (and hence the acceleration) gets smaller — the object is still accelerating, but less and less.
3. Terminal velocity reached: air resistance now equals weight; resultant force = 0, so acceleration = 0. The object falls at a constant, maximum speed.
4. On a speed-time graph: the line starts steep (high acceleration), curves and becomes less steep (decreasing acceleration), then flattens into a horizontal line (constant terminal velocity).
Try it — Terminal Velocity
- Explain why a skydiver’s acceleration decreases as their falling speed increases, before they open their parachute.
- Sketch (in words) the shape of a speed-time graph for a skydiver from the moment they jump until they reach terminal velocity.
- Explain what happens to a skydiver’s speed and acceleration immediately after they open their parachute.
Show answers
- As falling speed increases, air resistance increases too. Since air resistance acts upward against weight, the resultant (net) downward force gets smaller as speed increases, and a smaller resultant force means a smaller acceleration.
- The graph starts with a steep upward slope (high initial acceleration, close to g), which gradually becomes less steep as air resistance builds up (decreasing acceleration), before flattening into a horizontal line once air resistance equals weight (constant terminal velocity, zero acceleration).
- Opening the parachute suddenly increases air resistance a great deal, so air resistance becomes much greater than weight. This creates a large resultant force acting upward, causing the skydiver to decelerate rapidly, until a new, much lower terminal velocity is reached.
Common mistakes to watch for
- Confusing speed and velocity — remember velocity always needs a direction attached.
- Reading the gradient of a distance-time graph as acceleration instead of speed — gradient of distance-time graph = speed; gradient of speed-time graph = acceleration.
- Forgetting that a horizontal line on a speed-time graph means constant speed (zero acceleration), not that the object is at rest.
- Calculating the area under a speed-time graph as a single rectangle when the shape is actually a triangle or trapezium — always split it into simple shapes first.
- Forgetting the negative sign when calculating a deceleration, or saying “the acceleration is negative” without recognising this simply means the object is slowing down.
- Assuming a falling object accelerates at a constant 9.8 m/s² all the way to the ground when air resistance is present — this is only true in the absence of air resistance.
- Thinking terminal velocity means the object stops — it keeps moving at a constant (non-zero) maximum speed, it just stops accelerating.
7. Quick-Fire Challenge Round
True or False?
- The gradient of a distance-time graph gives the acceleration of the object.
- A horizontal line on a speed-time graph means the object is not accelerating.
- Deceleration is a type of negative acceleration.
- A falling object reaches terminal velocity when its weight becomes greater than air resistance.
Show answers
- False — the gradient of a distance-time graph gives speed. The gradient of a speed-time graph gives acceleration.
- True — a horizontal line means speed is constant, so acceleration is zero.
- True — a deceleration is simply an acceleration with a negative value, meaning the object is slowing down.
- False — terminal velocity is reached when air resistance becomes equal to weight, making the resultant force (and acceleration) zero, not when weight is greater.
Spot the Error
- “A car travels 120 km in 3 hours including a 30-minute stop. Its average speed is 120 ÷ 2.5 = 48 km/h, since we should ignore the stopped time.”
- “An object with a speed-time graph that is a straight horizontal line is not moving.”
- “A skydiver at terminal velocity has zero speed.”
Show answers
- The stopped time should be included in the total time used for average speed, since average speed is total distance ÷ total time for the whole journey. Correct: 120 ÷ 3 = 40 km/h.
- A horizontal line on a speed-time graph means constant, non-zero speed (unless that constant value is zero) — it does not by itself mean the object is stationary. A horizontal line at zero speed would mean the object is at rest; a horizontal line at, say, 10 m/s means the object is moving steadily.
- Terminal velocity means the skydiver has reached their maximum, constant falling speed — it is not zero. It is the acceleration that becomes zero at terminal velocity, not the speed.
8. Mixed Practice — Bringing It All Together
- A cyclist travels 45 km in 1.5 hours. Calculate her average speed in km/h.
- A distance-time graph shows a straight line from (0 s, 0 m) to (12 s, 84 m). Calculate the speed.
- A speed-time graph shows an object accelerating uniformly from 0 to 16 m/s in 4 s, then travelling at a constant 16 m/s for 5 s. Calculate the total distance travelled.
- A car’s velocity changes from 10 m/s to 30 m/s in 5 s. Calculate its acceleration.
- Describe, in terms of the forces acting, why a falling skydiver eventually stops accelerating.
Show answers
- 45 ÷ 1.5 = 30 km/h
- 84/12 = 7 m/s
- Triangle: ½ × 4 × 16 = 32 m. Rectangle: 5 × 16 = 80 m. Total = 32 + 80 = 112 m
- a = (30−10)/5 = 20/5 = 4 m/s²
- As the skydiver’s falling speed increases, air resistance increases with it. Once air resistance becomes equal in size to the skydiver’s weight, the resultant force acting on them becomes zero, so their acceleration becomes zero and they fall at a constant terminal velocity.
Practice at Home
A large set again — take it steadily across a few sittings. Parts C and D are Extended-only content.
Part A — Speed and average speed
- A jogger runs 3 km in 18 minutes. Find her average speed in km/h.
- A lorry travels 250 km in 4 hours, including two 15-minute rest stops. Find its average speed for the whole journey.
- Explain the difference between speed and velocity, giving an everyday example of each.
Part B — Reading graphs
- A distance-time graph shows a straight line from (0 s, 0 m) to (25 s, 150 m). Find the speed.
- A speed-time graph shows an object at a constant 6 m/s for 12 s. Find the distance travelled.
- Sketch (in words) the shape of a distance-time graph for an object that starts at rest, speeds up, then travels at constant speed.
Part C — Acceleration (Extended)
- A car accelerates from 8 m/s to 32 m/s in 6 s. Calculate its acceleration.
- A cyclist decelerates from 12 m/s to 4 m/s in 4 s. Calculate the deceleration.
- A speed-time graph shows a straight line from 10 m/s at t = 0 s to 40 m/s at t = 5 s. Find the acceleration, and calculate the distance travelled over this time using the area under the graph.
Part D — Free fall, air resistance and terminal velocity (Extended)
- State the approximate value of the acceleration of free fall near the Earth’s surface.
- A stone is dropped from rest and falls freely for 2.5 s with no air resistance. Calculate its speed at the end of this time.
- Explain, step by step, how a falling raindrop’s acceleration changes from the moment it forms in a cloud until it reaches the ground, assuming it reaches terminal velocity before landing.
Key Vocabulary Recap
Speed — distance travelled per unit time (scalar)
Velocity — speed in a given direction (vector)
Acceleration — change in velocity per unit time
Deceleration — a negative acceleration; the object is slowing down
Gradient — the slope of a graph; gives speed on a distance-time graph, or acceleration on a speed-time graph
Terminal velocity — the constant maximum speed reached by a falling object once air resistance equals its weight
Syllabus Reference
This note covers Cambridge IGCSE Physics (0625) section 1.2, Motion — Core points 1-8 (defining and calculating speed and average speed, sketching and interpreting distance-time and speed-time graphs, calculating speed from a gradient, calculating distance from the area under a speed-time graph, and the approximate value of g) and Supplement points 9-13 (defining and calculating acceleration, determining constant vs changing acceleration from a graph, calculating acceleration from a gradient, deceleration as negative acceleration, and the motion of falling objects with and without air/liquid resistance including terminal velocity).