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.

Cambridge IGCSE Physics (0625)

Motion

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Short Notes - Motion

Detailed Study Notes

Detailed notes on Motion, Forces and Energy for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Motion — Cambridge IGCSE 0625 Physics Extended (2026)

Distance, displacement, speed, velocity, acceleration. Use $s$ - $t$ and $v$ - $t$ graphs (gradient and area-under). The 0625 kinematics syllabus is $v = s/t$ , $a = \Delta v/\Delta t$ , motion graphs and free fall only — the suvat equations are not on the 0625 data sheet.

At a glance

Speed $= \dfrac{\text{distance}}{\text{time}}$ (scalar).
Velocity $= \dfrac{\text{displacement}}{\text{time}}$ (vector).
Acceleration $a = \dfrac{\Delta v}{\Delta t}$ , units $\text{m/s}^{2}$ .
$s$ - $t$ graph: gradient $=$ velocity. Flat $=$ stationary.
$v$ - $t$ graph: gradient $=$ acceleration; AREA UNDER $=$ distance travelled.
Free fall: $g \approx 9.8\,\text{m/s}^{2}$ (use $9.8$ or $10$ if told).
Terminal velocity: weight balanced by air resistance, net force = $0$ , $a = 0$ .

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, $v = 0$ .

Distance is the whole winding path (scalar); displacement is the straight A-to-B line with direction (vector).

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.

On a distance-time graph the gradient is the velocity; a steeper line means a faster object.

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: $40 + 120 + 50 = 210\,\text{m}$ .

Split the area under a velocity-time graph into triangles and rectangles to find the distance.

Graph	Gradient gives	Area under gives
distance–time ( $s$ - $t$ )	velocity	—
speed–time ( $v$ - $t$ )	acceleration	distance travelled

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).

On 0625 this is tested qualitatively (objects fall at the same rate regardless of mass) or through a speed-time graph — a falling object has a straight $v$ - $t$ line of gradient $g$ .

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.

When air resistance grows to equal the weight, the net force is zero and the speed stays constant.

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.

Going deeper for A*

Beyond the 0625 syllabus — the suvat equations let you calculate free-fall speeds and distances directly. For a coin dropped from rest, after $2\,\text{s}$ its speed is $v = u + at = 0 + 9.8 \times 2 = 19.6\,\text{m/s}$ , and the drop distance is $s = \tfrac{1}{2}gt^{2}$ . These equations ( $v = u + at$ , $v^{2} = u^{2} + 2as$ , $s = \tfrac{1}{2}gt^{2}$ ) are NOT on the 0625 data sheet and are not examinable on Cambridge 0625 — they belong to A-level (and some other-board IGCSE) kinematics. On 0625, get free-fall results from $a = \Delta v/\Delta t$ or the area under a speed-time graph instead.

Why air resistance increases with $v$ ? At low speeds, air resistance is roughly proportional to $v$ ; at higher speeds, to $v^{2}$ (turbulent). Either way: more $v \Rightarrow$ more drag, until drag balances weight.

Quick recap

Speed (scalar) vs velocity (vector).
Acceleration $= \Delta v / \Delta t$ .
$s$ - $t$ gradient = velocity.
$v$ - $t$ gradient = acceleration; area = distance.
Free fall: $g \approx 9.8\,\text{m/s}^{2}$ .
Terminal velocity: weight = air resistance.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Speed — distance per unit time (scalar).
Velocity — displacement per unit time (vector).
Acceleration — rate of change of velocity.
Free fall — fall under gravity alone (no air resistance).
Terminal velocity — constant velocity reached when weight equals air resistance.

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.

Take this whole topic with you

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.

Question type:

Question patterns to master — Motion

Almost every motion exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the speed / acceleration / final velocity — solvable with one motion equation ( $v=s/t$ , $a=\Delta v/\Delta t$ , $v=u+at$ , $v^{2}=u^{2}+2as$ ).

How to approach it

List $s,u,v,a,t$ with units, pick the equation containing only your knowns plus the unknown, rearrange, substitute and attach the unit.

Common trap

Examiner reports flag dropping the negative sign on a deceleration — in $v^{2}=u^{2}+2as$ the sign of $a$ must be consistent for $s$ to come out positive.

Graph or diagram

Recognise it by

The question describes or shows a velocity-time motion in phases (constant, accelerating, decelerating) and asks for a distance or a quantity from the graph.

How to approach it

Remember: gradient of a $v\text{-}t$ graph gives acceleration, area under it gives distance. Split the shape into rectangles and triangles, find each area, then sum.

Common trap

Examiner reports flag reading gradient when distance is wanted, or vice versa — $s\text{-}t$ graphs give $v$ from the gradient, $v\text{-}t$ graphs give $s$ from the area.

Multi-step problem

Recognise it by

Several stages chained together — a unit conversion then $s=vt$ , or distance and displacement compared over a there-and-back trip.

How to approach it

Do the preparatory step first (convert km/h to m/s by dividing by $3.6$ , or find total distance and total time), then feed the result into the main formula.

Common trap

Examiner reports flag substituting km/h straight into $s=vt$ with seconds, and quoting the same value for average speed and average velocity when displacement is actually zero.

Show that / prove

Recognise it by

Describe, explain or sketch in words how a quantity changes — e.g. how a skydiver's velocity and acceleration evolve towards terminal velocity.

How to approach it

Argue from the resultant force at each stage ( $F_{\text{net}}=W-R$ ), state how it changes the acceleration, and finish with the explicit condition for the result (e.g. $R=W\Rightarrow a=0$ ).

Common trap

Examiner reports flag the vague phrase 'forces balance' — Cambridge wants the specific statement that 'air resistance equals weight'.

1Calculate average speed
CoreDirect calculation• 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
ExtendedDirect calculation• 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
ExtendedGraph or diagram• 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
$140\,\text{m}$
4Beyond 0625 — free-fall speed using $v = u + at$
ExtendedDirect calculation• g, free fall, beyond-0625
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
$25\,\text{m/s}$
Examiner tip
Enrichment beyond the Cambridge IGCSE 0625 syllabus. The suvat equation $v = u + at$ is not on the 0625 data sheet and is not examined on 0625 — it belongs to A-level (and other-board IGCSE) kinematics. On 0625, free fall is tested qualitatively or via $a = \Delta v/\Delta t$ and motion graphs.
5Convert km/h to m/s for a calculation
CoreMulti-step problem• 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)
ExtendedGraph or diagram• 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
ExtendedMulti-step problem• 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
ExtendedShow that / prove• 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'.
9Beyond 0625 — use $v^{2} = u^{2} + 2as$ to find stopping distance
Challenge• suvat, braking, beyond-0625
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
Enrichment beyond the Cambridge IGCSE 0625 syllabus. The suvat equation $v^{2} = u^{2} + 2as$ is not on the 0625 data sheet and is not examined on 0625 — it belongs to A-level (and other-board IGCSE) kinematics. On 0625, a braking/stopping-distance problem would be set up using the area under a speed-time graph instead.
10Beyond 0625 — drop time and horizontal range of a projectile
ChallengeMulti-step problem• free fall, synoptic, beyond-0625
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
Enrichment beyond the Cambridge IGCSE 0625 syllabus. Projectile motion and the suvat equation $s = \tfrac{1}{2}gt^{2}$ are not on the 0625 data sheet and are not examined on 0625 — they belong to A-level kinematics. Included here only to extend strong students.

Model Answers — Motion

High-scoring sample answers for motion on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
Define acceleration, and state its SI unit.
Model answer
Acceleration is the rate of change of velocity (change in velocity per unit time). Its SI unit is the metre per second squared, $\text{m/s}^{2}$ .
Why this scores
One mark for 'rate of change of velocity' with the unit. 'Change in speed' is incomplete — acceleration is defined through velocity (a vector), so a direction change counts as an acceleration too.
Question 2
Paper 2/4 style2 marks
Explain the difference between distance and displacement, using a runner on a $400\,\text{m}$ circular track as an example.
Model answer
Distance is the total path length travelled (a scalar); displacement is the straight-line distance from start to finish in a stated direction (a vector). A runner who completes one full lap of a $400\,\text{m}$ track has travelled a distance of $400\,\text{m}$ , but because they finish exactly where they started their displacement is $0\,\text{m}$ .
Why this scores
One mark for the scalar-path-length vs vector-straight-line distinction; one mark for applying it to give $400\,\text{m}$ and $0\,\text{m}$ . Naming one as scalar and the other as vector also gains credit.
Question 3
Paper 4 structured style3 marks
A car speeds up uniformly from $5\,\text{m/s}$ to $25\,\text{m/s}$ in $8\,\text{s}$ . (a) Calculate its acceleration. (b) The car then brakes and its acceleration is $-4\,\text{m/s}^{2}$ . State what the negative sign tells you.
Model answer
(a) $a = \dfrac{\Delta v}{\Delta t} = \dfrac{25 - 5}{8} = \dfrac{20}{8} = \mathbf{2.5\,\text{m/s}^{2}}$ .

(b) The negative sign means the car is decelerating — its velocity is decreasing, i.e. the acceleration acts in the opposite direction to the motion.
Why this scores
Three marks: correct substitution into $a=\Delta v/\Delta t$ (1), value $2.5\,\text{m/s}^{2}$ (1), and a correct interpretation of the negative sign as slowing down/deceleration (1).
Question 4
Paper 4 graph style4 marks
A vehicle accelerates uniformly from rest to $20\,\text{m/s}$ in $8\,\text{s}$ , travels at a steady $20\,\text{m/s}$ for $10\,\text{s}$ , then decelerates uniformly to rest in $4\,\text{s}$ . Using the velocity–time graph, find (a) the acceleration in the first phase and (b) the total distance travelled.
Model answer
(a) Acceleration = gradient of the $v\text{-}t$ graph in the first phase: $a = \dfrac{20 - 0}{8} = \mathbf{2.5\,\text{m/s}^{2}}$ .

(b) Distance = area under the graph. Split it into a triangle, a rectangle and a triangle:

Acceleration phase: $\tfrac{1}{2} \times 8 \times 20 = 80\,\text{m}$

Steady phase: $20 \times 10 = 200\,\text{m}$

Deceleration phase: $\tfrac{1}{2} \times 4 \times 20 = 40\,\text{m}$

Total $= 80 + 200 + 40 = \mathbf{320\,\text{m}}$ .
Why this scores
Four marks: acceleration as gradient (1); recognising distance is the area (1); correct areas for all three sections (1); correct total (1). A frequent error is reading the gradient when the distance (area) is wanted — for a $v\text{-}t$ graph, gradient gives $a$ and area gives $s$ .
Question 5
Paper 4 extended-explanation style5 marks
A skydiver jumps from a balloon and falls without opening a parachute. Explain, in terms of the forces acting, how the skydiver's acceleration and velocity change until a terminal velocity is reached.
Model answer
1. Just after jumping, the only significant force is the weight (gravity) acting downward; air resistance is small because the speed is low, so the resultant force is large and downward and the skydiver accelerates at close to $g$ . 2. As the speed increases, the air resistance increases (it depends on speed). 3. The resultant force $= W - R$ therefore decreases, so the acceleration decreases — the skydiver is still speeding up, but less rapidly. 4. Eventually the air resistance grows equal to the weight ( $R = W$ ), so the resultant force is zero. 5. With zero resultant force the acceleration is zero, so the skydiver falls at a constant maximum speed — the terminal velocity.
Why this scores
Five marks: large resultant/near- $g$ at the start (1); air resistance increases with speed (1); resultant force and so acceleration decrease (1); $R = W$ so resultant is zero (1); constant terminal velocity (1). Cambridge wants the explicit statement 'air resistance equals weight', not just 'the forces balance'.
Question 6
Paper 4 multi-part structured style6 marks
A train accelerates uniformly from rest to $30\,\text{m/s}$ in $20\,\text{s}$ , runs at a steady $30\,\text{m/s}$ for $60\,\text{s}$ , then decelerates uniformly to rest in $40\,\text{s}$ . Find (a) the acceleration, (b) the deceleration, (c) the total distance travelled, and (d) the average speed for the whole journey.
Model answer
(a) Acceleration $= \dfrac{30 - 0}{20} = \mathbf{1.5\,\text{m/s}^{2}}$ .

(b) Deceleration $= \dfrac{30 - 0}{40} = \mathbf{0.75\,\text{m/s}^{2}}$ (a deceleration of $0.75\,\text{m/s}^{2}$ ).

(c) Distance = area under the $v\text{-}t$ graph:

Speeding up: $\tfrac{1}{2} \times 20 \times 30 = 300\,\text{m}$

Steady: $30 \times 60 = 1800\,\text{m}$

Slowing down: $\tfrac{1}{2} \times 40 \times 30 = 600\,\text{m}$

Total $= 300 + 1800 + 600 = \mathbf{2700\,\text{m}}$ .

(d) Average speed $= \dfrac{\text{total distance}}{\text{total time}} = \dfrac{2700}{20 + 60 + 40} = \dfrac{2700}{120} = \mathbf{22.5\,\text{m/s}}$ .
Why this scores
Six marks: acceleration (1), deceleration (1), the three areas (1), total distance (1), total time = 120 s (1), average speed (1). A common slip is dividing the total distance by only the cruising time — the average speed uses the whole journey 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: $1\,\text{km/h} = \dfrac{1000}{3600}\,\text{m/s} = \dfrac{1}{3.6}\,\text{m/s}$ .

Motion — frequently asked questions

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

Graph

Gradient gives

Area under gives

distance–time (

s

t

)

velocity

—

speed–time (

v

t

)

acceleration

distance travelled

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).

On 0625 this is tested qualitatively (objects fall at the same rate regardless of mass) or through a speed-time graph — a falling object has a straight $v$ - $t$ line of gradient $g$ .

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.

When air resistance grows to equal the weight, the net force is zero and the speed stays constant.

Motion

Short Notes - Motion

Detailed Study Notes

At a glance

What you’ll learn

Speed, velocity and acceleration

Distance-time (sss-ttt) graph

Velocity-time (vvv-ttt) graph

Free fall and terminal velocity

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Motion

1Calculate average speed

2Calculate acceleration

3Read distance from a velocity-time graph

4Beyond 0625 — free-fall speed using v=u+atv = u + atv=u+at

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

9Beyond 0625 — use v2=u2+2asv^{2} = u^{2} + 2asv2=u2+2as to find stopping distance

10Beyond 0625 — 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

Motion — frequently asked questions

Motion

Short Notes - Motion

Detailed Study Notes

At a glance

What you’ll learn

Speed, velocity and acceleration

Distance-time (sss-ttt) graph

Velocity-time (vvv-ttt) graph

Free fall and terminal velocity

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Motion

1Calculate average speed

2Calculate acceleration

3Read distance from a velocity-time graph

4Beyond 0625 — free-fall speed using v=u+atv = u + atv=u+at

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

9Beyond 0625 — use v2=u2+2asv^{2} = u^{2} + 2asv2=u2+2as to find stopping distance

10Beyond 0625 — drop time and horizontal range of a projectile

Speed

Acceleration

Graph readings

Free fall (no air resistance)

Distance

Displacement

Speed

Velocity

Acceleration

Deceleration

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

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

4Beyond 0625 — free-fall speed using $v = u + at$

9Beyond 0625 — use $v^{2} = u^{2} + 2as$ to find stopping distance

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

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

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

4Beyond 0625 — free-fall speed using $v = u + at$

9Beyond 0625 — use $v^{2} = u^{2} + 2as$ to find stopping distance