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Cambridge IGCSE Physics 0625: Speed, Velocity and Acceleration Notes

Speed tells you how fast something moves, velocity tells you how fast and in which direction, and acceleration tells you how quickly the velocity changes. These notes cover motion in the Cambridge IGCSE Physics (0625) syllabus, with clear formulas, distance–time and speed–time graphs, worked examples, practice questions with answers and exam tips.

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Written by Ms. Humaira Asim, a science teacher with 15+ years of teaching experience. Written for the Cambridge IGCSE Physics 0625 syllabus.

Core or Extended? Most of this page is Core content. Finding speed from the gradient of a distance–time graph, acceleration from the gradient of a speed–time graph, and terminal velocity are Extended content. Check your own syllabus to see which parts apply to you.
Cambridge IGCSE Physics 0625 speed, velocity and acceleration notes banner with a speed-time graph

Speed and average speed

Speed is the distance an object travels per unit time. Most objects do not move at a steady speed for a whole journey, so in exam questions you usually calculate the average speed.

average speed = total distance ÷ total timev = s ÷ t  ·  speed in m/s, distance in m, time in s

You can rearrange the formula: distance = speed × time, and time = distance ÷ speed.

Always use the total distance and the total time, including any time the object was stopped. Do not simply average the speeds of the different parts of a journey.

Converting km/h and m/s

The SI unit of speed is metres per second (m/s). Speeds in everyday life are often given in kilometres per hour (km/h).

To convertDo thisExample
km/h to m/sdivide by 3.672 km/h ÷ 3.6 = 20 m/s
m/s to km/hmultiply by 3.615 m/s × 3.6 = 54 km/h

Why 3.6? 1 km/h = 1000 m ÷ 3600 s = 1 ÷ 3.6 m/s.

Speed and velocity

Speed is a scalar: it has size (magnitude) only. Velocity is a vector: it is speed in a stated direction. For example, “20 m/s” is a speed, while “20 m/s due north” is a velocity.

This means an object can have a constant speed but a changing velocity. A car driving around a roundabout at a steady 10 m/s has a changing velocity, because its direction keeps changing.

Acceleration

Acceleration is the change in velocity per unit time.

acceleration = change in velocity ÷ time takena = Δv ÷ Δt = (v − u) ÷ t  ·  u = starting velocity, v = final velocity, a in m/s2
  • The unit of acceleration is metres per second squared, m/s2. An acceleration of 3 m/s2 means the velocity increases by 3 m/s every second.
  • If an object slows down, its acceleration is negative. This is called deceleration. A change from 20 m/s to 8 m/s in 4 s is an acceleration of −3 m/s2, or a deceleration of 3 m/s2.
  • “Starts from rest” means the starting velocity u = 0.

Distance–time graphs

A distance–time graph shows how far an object has travelled at each moment. Its shape tells you how the object is moving.

Shape of lineWhat it means
Horizontal (flat) lineObject is at rest (stationary)
Straight sloping lineConstant speed
Steeper straight lineFaster constant speed
Curve getting steeperSpeeding up (accelerating)
Curve getting less steepSlowing down (decelerating)
distancetime
At rest
distancetime
Constant speed
distancetime
Speeding up
distancetime
Slowing down
speed = gradient of a distance–time graphgradient = change in distance ÷ change in time (Extended)

For a curved line, draw a tangent to the curve at the point you want and find the gradient of the tangent. This gives the speed at that instant.

Speed–time graphs

A speed–time graph shows the speed of an object at each moment.

Shape of lineWhat it means
Horizontal lineConstant speed (zero acceleration)
Horizontal line on the time axisObject is at rest
Straight line sloping upConstant acceleration
Straight line sloping downConstant deceleration
Curve getting steeper or less steepChanging acceleration
speedtime
Constant speed
speedtime
Constant acceleration
speedtime
Constant deceleration
speedtime
At rest
acceleration = gradient of a speed–time graphgradient = change in speed ÷ change in time (Extended)
distance travelled = area under a speed–time graphsplit the area into rectangles and triangles, then add them up

Remember the areas: rectangle = base × height, and triangle = ½ × base × height.

Free fall and terminal velocity

Near the Earth’s surface, an object falling freely (with no air resistance) has a constant acceleration of about 9.8 m/s2. This is the acceleration of free fall, g. All objects fall with this same acceleration if air resistance is negligible, whatever their mass.

With air resistance (Extended):

  1. When an object, such as a skydiver, first starts to fall, its speed is low, so air resistance is small. The weight is much larger, so it accelerates at close to 9.8 m/s2.
  2. As its speed increases, air resistance increases, so the resultant (overall) downward force decreases. The acceleration decreases.
  3. Eventually air resistance equals weight. The resultant force is zero, so the acceleration is zero and the object falls at a constant speed. This speed is called the terminal velocity.
terminal velocity123timespeed
1 Steep: large acceleration, little air resistance. 2 Less steep: air resistance increasing, acceleration decreasing. 3 Flat: air resistance = weight, so constant speed (terminal velocity).

On a speed–time graph, this appears as a curve that starts steep, gets less steep and then levels off to a horizontal line.

Worked examples

Example 1: average speed

A cyclist travels 150 m in 6.0 s. Calculate her average speed.

  1. average speed = distance ÷ time
  2. = 150 ÷ 6.0
  3. = 25 m/s

Example 2: speed in m/s from km and hours

A train travels 240 km in 3.0 hours. Calculate its average speed in m/s.

  1. Convert to SI units: 240 km = 240 000 m, and 3.0 h = 3.0 × 3600 = 10 800 s
  2. average speed = 240 000 ÷ 10 800
  3. = 22 m/s (2 s.f.)

Example 3: acceleration

A car speeds up from 5.0 m/s to 25 m/s in 4.0 s. Calculate its acceleration.

  1. a = (v − u) ÷ t
  2. = (25 − 5.0) ÷ 4.0 = 20 ÷ 4.0
  3. = 5.0 m/s2

Example 4: deceleration

A bus travelling at 12 m/s stops in 3.0 s. Calculate its deceleration.

  1. a = (v − u) ÷ t = (0 − 12) ÷ 3.0
  2. = −4.0 m/s2
  3. deceleration = 4.0 m/s2

Example 5: finding the final velocity

A motorbike starts from rest and accelerates at 2.5 m/s2 for 8.0 s. Calculate its final velocity.

  1. Rearrange a = (v − u) ÷ t to give v = u + at
  2. v = 0 + (2.5 × 8.0)
  3. = 20 m/s

Example 6: gradient of a distance–time graph (Extended)

A straight line on a distance–time graph goes from (0 s, 0 m) to (5.0 s, 40 m). Calculate the speed.

  1. gradient = change in distance ÷ change in time
  2. = (40 − 0) ÷ (5.0 − 0)
  3. = 8.0 m/s

Example 7: area under a speed–time graph

A car accelerates uniformly from rest to 12 m/s in 4.0 s, then travels at a constant 12 m/s for 6.0 s. Calculate the total distance travelled and the average speed.

02468100481224 m72 mtime / sspeed / (m/s)
Distance = area of triangle + area of rectangle = 24 m + 72 m = 96 m
  1. First part (triangle): ½ × 4.0 × 12 = 24 m
  2. Second part (rectangle): 6.0 × 12 = 72 m
  3. Total distance = 24 + 72 = 96 m
  4. Average speed = 96 ÷ (4.0 + 6.0) = 9.6 m/s

Common mistakes

  • Mixing units. Convert km to m, and minutes or hours to seconds, before you calculate.
  • Averaging the speeds. Average speed is total distance ÷ total time, not the mean of the speeds.
  • Confusing the two graphs. A flat line on a distance–time graph means stopped. A flat line on a speed–time graph means constant speed. Read the axis labels first.
  • Wrong unit for acceleration. It is m/s2, not m/s.
  • Forgetting the ½ when finding the area of a triangle under a speed–time graph.
  • Saying “velocity” when there is no direction. If a question asks for a velocity, give a direction too.

Practice questions

Try each question first, then open it to check your answer.

Q1. A runner completes 400 m in 50 s. Calculate her average speed.

400 ÷ 50 = 8.0 m/s

Q2. Convert 90 km/h into m/s.

90 ÷ 3.6 = 25 m/s

Q3. A lorry travels at 15 m/s for 2.0 minutes. How far does it travel?

Convert the time: 2.0 min = 120 s. Distance = 15 × 120 = 1800 m (1.8 km)

Q4. A bus speeds up from rest to 15 m/s in 10 s. Calculate its acceleration.

a = (15 − 0) ÷ 10 = 1.5 m/s2

Q5. A car travelling at 30 m/s brakes with a deceleration of 6.0 m/s2. How long does it take to stop?

Time = change in velocity ÷ deceleration = 30 ÷ 6.0 = 5.0 s

Q6. On a speed–time graph, an object speeds up from 0 to 20 m/s in 5.0 s, then slows down steadily to rest in a further 10 s. Calculate the total distance travelled.

First triangle: ½ × 5.0 × 20 = 50 m. Second triangle: ½ × 10 × 20 = 100 m. Total = 150 m

Q7. A car drives around a circular track at a constant speed. Explain why its velocity is changing.

Velocity is a vector, so it includes direction. The car’s direction is always changing as it goes around the track, so its velocity changes even though its speed stays the same.

Q8 (Extended). Explain why a skydiver reaches a terminal velocity.

As the skydiver’s speed increases, air resistance increases. The resultant downward force (weight − air resistance) gets smaller, so the acceleration decreases. When air resistance equals weight, the resultant force is zero, so the acceleration is zero and the skydiver falls at a constant (terminal) velocity.

Exam tips

  • Write the formula first, then substitute the numbers, then give the answer with a unit. Method marks are often given even if the final answer is wrong.
  • When reading values from a graph, use large triangles for gradients. This makes your answer more accurate.
  • Give answers to the same number of significant figures as the data, usually 2 or 3.
  • For “describe the motion” questions, use the key words: at rest, constant speed, accelerating, decelerating, and quote values from the graph.

Frequently asked questions

Is speed the same as velocity?

No. Speed only tells you how fast something is moving. Velocity also tells you the direction, so it is a vector quantity.

What does a negative acceleration mean?

It means the object is slowing down. This is usually called a deceleration.

What is the value of g in IGCSE Physics?

The acceleration of free fall near the Earth’s surface is about 9.8 m/s2. Use the value given in the question paper.

Need help with Physics calculations?

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This topic is also part of GCSE and IGCSE Combined Science: see all Combined Science notes. More Physics notes: Forces, resultant force and vectors · Physics notes topic list. These notes follow the Cambridge IGCSE Physics 0625 syllabus. Always check the current syllabus on the Cambridge International website.

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