What is the definition of motion in Physics?

In Physics, motion refers to the change in position of an object with respect to time and its reference point. It is a fundamental concept in the study of mechanics, which is a branch of Physics. Understanding motion is crucial for analyzing how forces affect the movement of objects, a key topic in the Cambridge IGCSE Physics curriculum.

How is speed different from velocity in the context of motion?

Speed is a scalar quantity that refers to how fast an object is moving, without considering its direction. It is calculated as the distance traveled divided by the time taken. Velocity, on the other hand, is a vector quantity that includes both the speed and direction of an object's motion. In the Cambridge IGCSE Physics syllabus, understanding the distinction between these two concepts is essential for solving problems related to motion.

What are the different types of motion covered in the Cambridge IGCSE Physics curriculum?

The Cambridge IGCSE Physics curriculum covers several types of motion, including linear motion, circular motion, and oscillatory motion. Linear motion refers to movement in a straight line, circular motion involves movement along a circular path, and oscillatory motion is characterized by repetitive back-and-forth movement, such as that of a pendulum. Each type of motion has unique characteristics and equations that describe its behavior.

How do you calculate acceleration in motion?

Acceleration is calculated as the rate of change of velocity with respect to time. It is a vector quantity, meaning it has both magnitude and direction. The formula for acceleration is a = (v - u) / t, where 'a' is acceleration, 'v' is the final velocity, 'u' is the initial velocity, and 't' is the time taken for the change in velocity. Understanding acceleration is crucial for analyzing motion in the Cambridge IGCSE Physics course.

What role do forces play in motion according to the Cambridge IGCSE Physics syllabus?

Forces are fundamental to understanding motion as they can cause an object to start moving, stop moving, change direction, or alter its speed. According to Newton's laws of motion, which are a key part of the Cambridge IGCSE Physics syllabus, the behavior of objects in motion is directly influenced by the forces acting upon them. These laws help explain how forces affect motion and are essential for solving related problems in Physics.

Why is "Reading area from $s$-$t$ or gradient from a $v$-$t$ misordered" a common mistake in Motion?

Why it happens: Mixing up the two graph types. How to avoid it: s-t → GRADIENT gives v. v-t → GRADIENT gives a, AREA gives s. Source: 0625/42 — every series

Why is "Treating speed and velocity as identical" a common mistake in Motion?

Why it happens: Used interchangeably in everyday speech. How to avoid it: Velocity needs a direction. If a vector answer is required, state the direction (e.g. 'east' or 'downward').

Why is "Writing positive acceleration when an object is slowing" a common mistake in Motion?

Why it happens: Calculating |v - u|/t. How to avoid it: Slowing → negative acceleration (taking the original direction as positive).

Why is "Mixing km/h with m/s without converting" a common mistake in Motion?

Why it happens: Question gives km/h; calculator wants m/s. How to avoid it: Convert at the start: 1\,km/h = 1000/3600\,m/s = 1/3.6\,m/s.

Motion, Forces and EnergyCambridge IGCSE Physics (0625)

Motion

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Motion — Cambridge IGCSE 0625 Physics Extended (2026)

Distance, displacement, speed, velocity, acceleration. Use $s$ - $t$ and $v$ - $t$ graphs (gradient and area-under) and the suvat-style equations on the data sheet.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

1.2 — Define speed and velocity. Calculate average speed using $\text{distance}/\text{time}$ .
1.2 — Define acceleration as change in velocity per unit time.
1.2 — Plot and interpret distance-time and speed-time graphs.
1.2 — Calculate distance from area under a speed-time graph.
1.3 — Describe motion of a falling object, including terminal velocity.

Speed, velocity and acceleration

Speed is scalar, velocity is vector. Acceleration is rate of change of velocity.

Speed. Distance covered per unit time. Scalar. $v = \dfrac{d}{t}.$

Velocity. Displacement per unit time. Vector — direction matters.

Acceleration. $a = \dfrac{\Delta v}{\Delta t} = \dfrac{v - u}{t}.$ Where $u$ is initial velocity, $v$ is final velocity, $t$ is the time taken. Units: $\text{m/s}^{2}$ .

Worked. A car speeds up from $5\,\text{m/s}$ to $25\,\text{m/s}$ in $4\,\text{s}$ . Find acceleration.

$a = (25 - 5)/4 = 5\,\text{m/s}^{2}$ .

Negative acceleration = deceleration. Object slowing down.

Worked. A ball thrown up returns to the hand. Initial velocity $20\,\text{m/s}$ up; gravity gives $a = -9.8\,\text{m/s}^{2}$ (taking up as positive). At the top, $v = 0$ .

Speed = distance / time (scalar).
Velocity = displacement / time (vector).
Acceleration: $(v - u)/t$ .
Negative acceleration = deceleration.

Distance-time ( $s$ - $t$ ) graph

Gradient = velocity. Curve = changing velocity.

On an $s$ - $t$ graph:

Horizontal line → object stationary.
Straight slope → constant velocity.
Curve getting steeper → speeding up (accelerating).
Curve flattening → slowing down (decelerating).

Gradient = velocity. A steep slope means a fast object. $\text{velocity} = \dfrac{\Delta s}{\Delta t}.$

Worked. An $s$ - $t$ graph rises linearly from $(0, 0)$ to $(10\,\text{s}, 60\,\text{m})$ .

Gradient $= 60 / 10 = 6\,\text{m/s}$ — that's the velocity.

Tip. When a curve appears, calculate the INSTANTANEOUS velocity by drawing a tangent at that point and finding the gradient of the tangent.

Gradient of $s$ - $t$ = velocity.
Flat line = stationary.
Curve up = accelerating.
Curve down = decelerating.

Velocity-time ( $v$ - $t$ ) graph

Gradient = acceleration. Area under = distance travelled.

On a $v$ - $t$ graph:

Horizontal line → constant velocity (no acceleration).
Straight slope up → constant acceleration.
Straight slope down → constant deceleration.
Curve → changing acceleration.

Gradient = acceleration. $a = \dfrac{\Delta v}{\Delta t}.$

Area under the curve = distance travelled. Calculate area using rectangles, triangles, or trapeziums.

Worked. A $v$ - $t$ graph: object accelerates from $0$ to $20\,\text{m/s}$ over $4\,\text{s}$ , then constant at $20\,\text{m/s}$ for $6\,\text{s}$ , then decelerates to rest in $5\,\text{s}$ .

Acceleration phase area: $\dfrac{1}{2} \times 4 \times 20 = 40\,\text{m}$ .
Constant phase area: $6 \times 20 = 120\,\text{m}$ .
Deceleration phase area: $\dfrac{1}{2} \times 5 \times 20 = 50\,\text{m}$ .
Total distance: $40 + 120 + 50 = 210\,\text{m}$ .

Gradient of $v$ - $t$ = acceleration.
AREA UNDER = distance.
Trapezium area = $\dfrac{1}{2}(a + b) h$ .
Triangle area = $\dfrac{1}{2} b h$ .

Free fall and terminal velocity

Air-free: $g = 9.8\,\text{m/s}^{2}$ down. With air: terminal velocity when weight = air resistance.

Acceleration of free fall. Without air resistance, all objects fall with the same acceleration: $g \approx 9.8\,\text{m/s}^{2}$ (or $10$ if told).

Worked. A coin dropped from rest: after $2\,\text{s}$ , $v = u + at = 0 + 9.8 \times 2 = 19.6\,\text{m/s}$ .

Terminal velocity. A real falling object meets air resistance.

Initially: weight $>$ air resistance. Net force down → acceleration down.
As $v$ increases: air resistance increases.
Eventually: air resistance $=$ weight. Net force $= 0$ . No more acceleration. Constant velocity = terminal velocity.

On a $v$ - $t$ graph for a falling object: the curve rises steeply from rest, gradient decreases (deceleration of acceleration), and finally levels off horizontally at terminal velocity.

Worked. A skydiver opens a parachute at terminal velocity. Air resistance suddenly increases (parachute is large). New net force is UP → skydiver decelerates. They reach a new, smaller terminal velocity at which weight $=$ new air resistance.

Free fall: $a = g \approx 9.8\,\text{m/s}^{2}$ .
Terminal velocity: weight = air resistance.
$v$ - $t$ graph: levels off at terminal velocity.
Parachute: jumps to a new, lower terminal velocity.

How it’s examined

Motion appears every Paper 4 (5-8 marks), typically a multi-part graph question: read off values, calculate gradient (velocity or acceleration), find area-under (distance), and discuss terminal velocity. Examiner reports flag confusing $s$ - $t$ with $v$ - $t$ graphs and forgetting that area-under- $v$ - $t$ is distance.

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (1.2-1.3); 0625/42 Oct/Nov 2024 — Q3 ($v$-$t$ graph, terminal velocity); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

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Step-by-step worked examples — Motion

Step-by-step solutions to past-paper-style questions on motion, written exactly the way a tutor would explain them at the board.

1Calculate average speed
Core• speed
Question
A car travels $120\,\text{m}$ in $8\,\text{s}$ . Find its average speed.
Step-by-step solution
1. Step 1
  Average speed $= \dfrac{\text{distance}}{\text{time}}$ .
  $v = \dfrac{120}{8} = 15\,\text{m/s}$
Answer
$15\,\text{m/s}$
2Calculate acceleration
Extended• Adapted from 0625/42 May/Jun 2024 Q3• acceleration
Question
A cyclist accelerates from $4\,\text{m/s}$ to $10\,\text{m/s}$ in $3\,\text{s}$ . Find the acceleration.
Step-by-step solution
1. Step 1
  $a = \dfrac{\Delta v}{\Delta t} = \dfrac{v - u}{t}$ .
  $a = \dfrac{10 - 4}{3} = 2\,\text{m/s}^{2}$
Answer
$2\,\text{m/s}^{2}$
3Read distance from a velocity-time graph
Extended• v-t graph
Question
A vehicle moves at $20\,\text{m/s}$ for $5\,\text{s}$ , then decelerates uniformly to rest in $4\,\text{s}$ . Find the total distance.
Step-by-step solution
1. Step 1
  Distance = AREA under the $v\text{-}t$ graph.
2. Step 2
  Constant section: rectangle.
  $A_{1} = 20 \times 5 = 100\,\text{m}$
3. Step 3
  Decelerating section: triangle.
  $A_{2} = \dfrac{1}{2} \times 4 \times 20 = 40\,\text{m}$
4. Step 4
  Total.
  $100 + 40 = 140\,\text{m}$
Answer
$140\,\text{m}$
4Free-fall acceleration
Extended• g, free fall
Question
An object is dropped and falls for $2.5\,\text{s}$ . Ignoring air resistance, find its speed just before hitting the ground. Take $g = 10\,\text{m/s}^{2}$ .
Step-by-step solution
1. Step 1
  Use $v = u + at$ with $u = 0$ .
  $v = 0 + 10 \times 2.5 = 25\,\text{m/s}$
Answer
$25\,\text{m/s}$
5Convert km/h to m/s for a calculation
Core• units, conversion
Question
A train moves at a steady $108\,\text{km/h}$ . How far does it travel in $25\,\text{s}$ ?
Step-by-step solution
1. Step 1
  Convert km/h to m/s by dividing by $3.6$ .
  $v = \dfrac{108}{3.6} = 30\,\text{m/s}$
2. Step 2
  Use $s = vt$ .
  $s = 30 \times 25 = 750\,\text{m}$
Answer
$750\,\text{m}$
Examiner tip
The examiner report flags candidates often substitute $108$ straight into $s = vt$ — that mixes km/h with seconds and gives a wrong-by-3.6 answer.
6Distance from a two-phase velocity-time graph (accelerate then cruise)
Extended• Adapted from 0625/42 Oct/Nov 2023 Q3• v-t graph, area
Question
A cyclist accelerates uniformly from rest to $8\,\text{m/s}$ in $4\,\text{s}$ , then cruises at $8\,\text{m/s}$ for a further $10\,\text{s}$ . Find the total distance.
Step-by-step solution
1. Step 1
  Distance under acceleration phase = area of triangle.
  $A_{1} = \dfrac{1}{2} \times 4 \times 8 = 16\,\text{m}$
2. Step 2
  Distance under cruise phase = area of rectangle.
  $A_{2} = 8 \times 10 = 80\,\text{m}$
3. Step 3
  Sum.
  $s = 16 + 80 = 96\,\text{m}$
Answer
$96\,\text{m}$
7Average speed vs average velocity for a return trip
Extended• average, displacement
Question
A runner jogs $400\,\text{m}$ east in $80\,\text{s}$ , then $400\,\text{m}$ west back to the start in $120\,\text{s}$ . Find the average speed and the average velocity.
Step-by-step solution
1. Step 1
  Total distance = $400 + 400 = 800\,\text{m}$ . Total time = $80 + 120 = 200\,\text{s}$ .
2. Step 2
  Average speed uses total distance (scalar).
  $\bar{v}_{\text{speed}} = \dfrac{800}{200} = 4.0\,\text{m/s}$
3. Step 3
  Average velocity uses displacement (vector). Net displacement is zero — runner returns to start.
  $\bar{v}_{\text{vel}} = \dfrac{0}{200} = 0\,\text{m/s}$
Answer
Average speed $= 4.0\,\text{m/s}$ . Average velocity $= 0$ .
Examiner tip
The examiner report flags candidates often quote the same value for both — the trick is that displacement is zero whenever the runner ends at the starting point.
8Terminal velocity — force argument for a skydiver
Extended• terminal velocity, air resistance
Question
A skydiver falls from rest. Sketch (in words) how their acceleration and velocity change, and state the condition for terminal velocity.
Step-by-step solution
1. Step 1
  At release, only weight acts. Air resistance is zero, so acceleration $a = g$ .
2. Step 2
  As $v$ increases, air resistance increases. Resultant force = $W - R$ decreases, so acceleration decreases (but speed still rises).
  $F_{\text{net}} = W - R$
3. Step 3
  When $R = W$ , resultant force is zero, so acceleration is zero — the skydiver moves at constant terminal velocity.
  $R = W \Rightarrow a = 0$
Answer
Terminal velocity is reached when air resistance equals weight; net force is zero so acceleration is zero and velocity is constant.
Examiner tip
The 2024 mark scheme awards method marks for explicitly stating 'air resistance equals weight' — not just 'forces balance'.
9Use $v^{2} = u^{2} + 2as$ to find stopping distance
Extended• Adapted from 0625/42 May/Jun 2023 Q4• suvat, braking
Question
A car travelling at $20\,\text{m/s}$ brakes with a uniform deceleration of $5\,\text{m/s}^{2}$ until it stops. Find the stopping distance.
Step-by-step solution
1. Step 1
  List values. $u = 20$ , $v = 0$ , $a = -5\,\text{m/s}^{2}$ .
2. Step 2
  Use $v^{2} = u^{2} + 2as$ , rearrange for $s$ .
  $s = \dfrac{v^{2} - u^{2}}{2a} = \dfrac{0 - 400}{2 \times (-5)}$
3. Step 3
  Evaluate.
  $s = \dfrac{-400}{-10} = 40\,\text{m}$
Answer
$40\,\text{m}$
Examiner tip
The examiner report flags candidates often drop the negative sign on $a$ when decelerating — but $v^{2} = u^{2} + 2as$ requires the sign to be consistent for $s$ to come out positive.
10A* — drop time and horizontal range of a projectile
Challenge• free fall, synoptic
Question
A ball is rolled horizontally off a $1.25\,\text{m}$ -high table at $3.0\,\text{m/s}$ . Take $g = 10\,\text{m/s}^{2}$ and ignore air resistance. Find the time to hit the floor and the horizontal distance travelled.
Step-by-step solution
1. Step 1
  Vertical motion: $u = 0$ , $s = 1.25\,\text{m}$ , $a = g$ . Use $s = \tfrac{1}{2} g t^{2}$ .
  $t = \sqrt{\dfrac{2s}{g}} = \sqrt{\dfrac{2 \times 1.25}{10}}$
2. Step 2
  Evaluate $t$ .
  $t = \sqrt{0.25} = 0.5\,\text{s}$
3. Step 3
  Horizontal motion is independent and at constant $3.0\,\text{m/s}$ .
  $x = v_{x} t = 3.0 \times 0.5 = 1.5\,\text{m}$
Answer
Time of flight $= 0.5\,\text{s}$ . Horizontal range $= 1.5\,\text{m}$ .
Examiner tip
The 2024 mark scheme awards method marks for treating vertical and horizontal motion independently — the horizontal speed does not affect the fall time.

Key Formulae — Motion

The formulae you need to memorise for motion on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Speed
$v = \dfrac{s}{t}$
$s$
distance travelled
$t$
time taken
When to use
Constant or average speed.
Acceleration
$a = \dfrac{\Delta v}{\Delta t}$
$\Delta v$
change in velocity
$\Delta t$
time interval
When to use
Uniform acceleration problems.
Graph readings
$\text{Gradient of } s\text{-}t = v;\ \text{Gradient of } v\text{-}t = a;\ \text{Area under } v\text{-}t = s$
When to use
Pull motion quantities from a graph rather than computing.
Free fall (no air resistance)
$g \approx 9.8 \,\text{m/s}^{2} \text{ (use } 10 \text{ unless told otherwise)}$
When to use
Vertical motion problems near Earth's surface.

Key Definitions and Keywords — Motion

Definitions to memorise and the exact keywords mark schemes credit for motion answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Distance
Examiner keyword
Total path length travelled. Scalar.
Displacement
Examiner keyword
Straight-line distance from start to finish, with direction. Vector.
Speed
Examiner keyword
Rate at which distance is covered. Scalar; SI unit $\text{m/s}$ .
Velocity
Examiner keyword
Speed in a stated direction. Vector; SI unit $\text{m/s}$ .
Acceleration
Examiner keyword
Rate of change of velocity. Vector; SI unit $\text{m/s}^{2}$ .
Deceleration
Negative acceleration — the object slows down.

Common Mistakes and Misconceptions — Motion

The traps other students keep falling into on motion questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Reading area from $s$ - $t$ or gradient from a $v$ - $t$ misordered
0625/42 — every series
Why it happens
Mixing up the two graph types.
How to avoid it
$s$ - $t$ → GRADIENT gives $v$ . $v$ - $t$ → GRADIENT gives $a$ , AREA gives $s$ .
✕Treating speed and velocity as identical
Why it happens
Used interchangeably in everyday speech.
How to avoid it
Velocity needs a direction. If a vector answer is required, state the direction (e.g. 'east' or 'downward').
✕Writing positive acceleration when an object is slowing
Why it happens
Calculating $|v - u|/t$ .
How to avoid it
Slowing → negative acceleration (taking the original direction as positive).
✕Mixing km/h with m/s without converting
Why it happens
Question gives km/h; calculator wants m/s.
How to avoid it
Convert at the start: $1\,\text{km/h} = \dfrac{1000}{3600}\,\text{m/s} = \dfrac{1}{3.6}\,\text{m/s}$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Video lesson

Short walkthrough of the concepts students most often get stuck on.

Motion — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Motion

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Calculate average speed

2Calculate acceleration

3Read distance from a velocity-time graph

4Free-fall acceleration

5Convert km/h to m/s for a calculation

6Distance from a two-phase velocity-time graph (accelerate then cruise)

7Average speed vs average velocity for a return trip

8Terminal velocity — force argument for a skydiver

9Use v2=u2+2asv^{2} = u^{2} + 2asv2=u2+2as to find stopping distance

10A* — drop time and horizontal range of a projectile

Speed

Acceleration

Graph readings

Free fall (no air resistance)

Distance

Displacement

Speed

Velocity

Acceleration

Deceleration

✕Reading area from sss-ttt or gradient from a vvv-ttt misordered

✕Treating speed and velocity as identical

✕Writing positive acceleration when an object is slowing

✕Mixing km/h with m/s without converting

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Motion — frequently asked questions

9Use $v^{2} = u^{2} + 2as$ to find stopping distance

✕Reading area from $s$ - $t$ or gradient from a $v$ - $t$ misordered