What is the difference between speed and velocity in the context of motion?

Speed is a scalar quantity that refers to how fast an object is moving, without considering its direction. Velocity, on the other hand, is a vector quantity that includes both the speed of the object and the direction of its motion. Understanding the distinction between these two is crucial in Edexcel International A Levels Physics, particularly when analyzing motion in Mechanics and Materials.

How do you calculate acceleration in motion problems?

Acceleration is calculated by the change in velocity divided by the time over which the change occurs. Mathematically, it is expressed as 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 this change. This concept is fundamental in the study of motion under the Edexcel International A Levels curriculum.

What are the equations of motion and how are they applied?

The equations of motion are a set of three equations that describe the relationship between displacement, velocity, acceleration, and time. They are: 1) v = u + at, 2) s = ut + 0.5at², and 3) v² = u² + 2as. These equations are used to solve various problems related to linear motion in Physics, especially in the Mechanics and Materials section of the Edexcel International A Levels.

What is the significance of Newton's First Law of Motion?

Newton's First Law of Motion, also known as the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. This principle is significant because it introduces the concept of inertia and explains why objects behave the way they do when forces are applied. It is a foundational concept in the study of motion for Edexcel International A Levels Physics students.

How does friction affect motion, and why is it important in Physics?

Friction is a force that opposes the relative motion or tendency of such motion of two surfaces in contact. It plays a crucial role in motion as it can either hinder or facilitate movement, depending on the context. Understanding friction is important in Physics because it affects how objects move and is a key factor in solving problems related to motion in the Mechanics and Materials section of the Edexcel International A Levels curriculum.

Why is "Putting the resultant (slanted) launch velocity straight into a SUVAT equation for a projectile" a common mistake in Motion?

Why it happens: Students treat the launch as a single one-dimensional motion instead of two independent ones. How to avoid it: **Always resolve first**: use u_x=u for the horizontal axis and u_y=u for the vertical axis, then apply SUVAT to each axis separately. The axes are linked only through the shared time t. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Confusing gradient and area on a velocity–time graph (e.g. reading the velocity off the axis as the distance)" a common mistake in Motion?

Why it happens: The two operations look similar and students forget which gives which quantity. How to avoid it: Remember: on a v–t graph the **gradient = acceleration** and the **area under the line = displacement**. To find a distance, calculate the AREA (triangle/trapezium or count squares). Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Swapping sin and cos when resolving (using $F\sin\theta$ for the horizontal component)" a common mistake in Motion?

Why it happens: Not checking which side of the right-angled triangle the angle is measured from. How to avoid it: When is measured **from the horizontal**, the **horizontal component is F** and the vertical is F. Sketch the triangle and check that √(F_x^2+F_y^2) recovers F. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Sign errors in vertical SUVAT — e.g. taking up as positive but still using $a=+9.81$" a common mistake in Motion?

Why it happens: No positive direction is defined before substituting, so u and a end up inconsistent. How to avoid it: **Define a positive direction first.** If up is positive, then for an object thrown up u>0 and a=-9.81\ m s^-2 (gravity acts down). Keep the sign convention consistent throughout the question. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Using SUVAT equations when the acceleration is not constant (e.g. with air resistance)" a common mistake in Motion?

Why it happens: SUVAT is the default tool and students apply it without checking the condition. How to avoid it: SUVAT is valid **only for constant acceleration**. If the v–t graph curves (changing gradient), use the **gradient of a tangent** for acceleration and the **area** for displacement instead. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "In Core Practical 1, quoting the gradient of the $s$–$t^2$ graph as $g$ (forgetting the factor of 2)" a common mistake in Motion?

Why it happens: Students stop at the gradient without comparing s=\tfrac12 gt^2 with y=mx. How to avoid it: The gradient equals \tfrac12 g, so **g = 2×gradient**. Also plot s against t^2 (not t) so the graph is a straight line in the first place. Source: Edexcel IAL Physics Unit 1 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Motion

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Short Notes - Motion

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Detailed Study Notes

Detailed notes on Mechanics and Materials for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Motion — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 1)

Displacement, velocity and acceleration; the four SUVAT equations and how to choose between them; reading and interpreting motion graphs (gradient and area); scalars versus vectors and resolving into components; projectile motion as two independent motions; and Core Practical 1 — determining g, the acceleration of free fall.

At a glance

Distance/speed are scalars (size only); displacement/velocity/acceleration are vectors (size and direction). Always quote a direction or sign for a vector.
Average velocity $= \dfrac{\text{displacement}}{\text{time}}$ ; instantaneous velocity is the gradient of the tangent to a displacement–time graph at that instant.
Four SUVAT equations (constant acceleration only): $v=u+at$ , $s=ut+\tfrac{1}{2}at^2$ , $v^2=u^2+2as$ , $s=\tfrac{1}{2}(u+v)t$ . List $s,u,v,a,t$ , pick the equation missing your unwanted variable.
Velocity–time graph: gradient $=$ acceleration; area under the line $=$ displacement. Displacement–time graph: gradient $=$ velocity.
Projectile motion = two independent motions: horizontal at constant velocity ( $a=0$ ), vertical under gravity ( $a=g=9.81\ \text{m s}^{-2}$ downwards). They share only the time $t$ .
Never put the resultant launch velocity into SUVAT — resolve into horizontal ( $u\cos\theta$ ) and vertical ( $u\sin\theta$ ) components and treat each axis separately.
Resolving a vector: horizontal component $=F\cos\theta$ , vertical component $=F\sin\theta$ (with $\theta$ measured from the horizontal). Resultant magnitude $=\sqrt{F_x^2+F_y^2}$ .
Core Practical 1: determine $g$ by free fall. Plotting $s$ against $t^2$ gives a straight line of gradient $\tfrac{1}{2}g$ , so $g=2\times\text{gradient}$ .

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

1.1 — Distinguish scalar quantities (distance, speed) from vector quantities (displacement, velocity, acceleration), and resolve a vector into two perpendicular components.
1.2 — Define velocity and acceleration, and distinguish instantaneous from average velocity.
1.3 — Use the equations for uniformly accelerated motion (SUVAT) in one dimension: $v=u+at$ , $s=ut+\tfrac{1}{2}at^2$ , $v^2=u^2+2as$ , $s=\tfrac{1}{2}(u+v)t$ .
1.4 — Interpret displacement–time, velocity–time and acceleration–time graphs; obtain velocity and acceleration from gradients and displacement from the area under a velocity–time graph.
1.5 — Analyse projectile motion by treating horizontal and vertical components of velocity independently.
CP1 — Determine the acceleration of free fall $g$ using a suitable method (e.g. light gates or video analysis) and evaluate sources of uncertainty.

Scalars, vectors and the language of motion

Get the vocabulary exact — a vector always needs a direction.

Mechanics begins with five quantities. Knowing which are scalars and which are vectors is worth easy marks and stops you making sign errors later.

A scalar has magnitude (size) only — e.g. distance, speed, mass, time, energy.
A vector has magnitude and direction — e.g. displacement, velocity, acceleration, force.

So distance is the total path length travelled, while displacement is the straight-line distance in a specified direction from start to finish. Walk 3 m east then 4 m west and your distance is 7 m but your displacement is 1 m west. Likewise speed is how fast you move; velocity is speed in a stated direction.

Acceleration is the rate of change of velocity, $a=\dfrac{\Delta v}{\Delta t}$ , measured in $\text{m s}^{-2}$ . Because velocity is a vector, a body accelerates whenever its speed or its direction changes — a car going round a roundabout at constant speed is still accelerating.

Average vs instantaneous velocity

Average velocity over a time interval $=\dfrac{\text{displacement}}{\text{time taken}}$ .
Instantaneous velocity is the velocity at one moment — found from the gradient of the tangent to a displacement–time graph at that instant.

Adding and resolving vectors. Two vectors add tip-to-tail; for perpendicular vectors the resultant magnitude is $R=\sqrt{A^2+B^2}$ and its direction $\theta=\tan^{-1}\!\left(\dfrac{B}{A}\right)$ . The reverse — resolving a single vector $F$ at angle $\theta$ to the horizontal — splits it into a horizontal part $F\cos\theta$ and a vertical part $F\sin\theta$ .

A vector F at angle θ to the horizontal resolves into a horizontal component F cos θ and a vertical component F sin θ. The two components are perpendicular, so the original vector is recovered by Pythagoras: F = √((F cos θ)² + (F sin θ)²).

Scalar = size only (distance, speed); vector = size + direction (displacement, velocity, acceleration).
Average velocity = displacement ÷ time; instantaneous velocity = gradient of tangent on an s–t graph.
Resolve: horizontal = $F\cos\theta$ , vertical = $F\sin\theta$ (θ from the horizontal).
Resultant of perpendicular vectors: $R=\sqrt{A^2+B^2}$ .

See the full worked example for motion →

The SUVAT equations for uniform acceleration

Four equations, one decision: list s, u, v, a, t and pick the one that misses your spare variable.

When acceleration is constant, the four equations of uniformly accelerated motion (often called SUVAT) connect the five quantities:

$s$ = displacement, $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $t$ = time.

Equation	Variable it leaves out
$v = u + at$	$s$
$s = ut + \tfrac{1}{2}at^2$	$v$
$v^2 = u^2 + 2as$	$t$
$s = \tfrac{1}{2}(u+v)\,t$	$a$

The method that never fails:

Write down the five symbols $s,u,v,a,t$ and fill in the three you know (watch directions — choose a positive direction and give any opposing quantity a minus sign).
Identify the one you want.
Choose the equation that contains your four relevant quantities and omits the fifth (unwanted) one.
Substitute with units, rearrange, then evaluate to 3 significant figures.

These equations apply ONLY when $a$ is constant — including objects in free fall, where $a=g=9.81\ \text{m s}^{-2}$ acting downwards. They do not apply to non-uniform acceleration (e.g. a falling object with air resistance), where you must use graphs instead.

Sign convention. Decide up front which direction is positive. For a ball thrown upward, taking up as positive makes $u$ positive and $a=-9.81\ \text{m s}^{-2}$ (gravity acts downward). Mixing signs is the single biggest cause of lost marks here.

SUVAT applies only when acceleration is constant.
List $s,u,v,a,t$ ; choose the equation missing the unwanted variable.
Free fall: $a=g=9.81\ \text{m s}^{-2}$ downwards.
Define a positive direction and keep signs consistent throughout.

See the full worked example for motion →

Motion graphs — gradient and area

Gradient gives the next derivative up; area under a v–t line gives displacement.

Three graphs describe motion. The two operations you perform on them are gradient and area.

Displacement–time (s–t) graph

Gradient = velocity. A straight line ⇒ constant velocity. A horizontal line ⇒ stationary. A curve ⇒ changing velocity; read the instantaneous velocity from the gradient of a tangent.

Velocity–time (v–t) graph

Gradient = acceleration. A straight sloping line ⇒ constant acceleration; a horizontal line ⇒ constant velocity (zero acceleration).
Area between the line and the time axis = displacement. For a straight line this is a triangle/trapezium; for a curve, count squares or split into strips.

Acceleration–time (a–t) graph

Area under the line = change in velocity ( $\Delta v$ ).

On a velocity–time graph the gradient of the line gives the acceleration (Δv/Δt) and the area between the line and the time axis gives the displacement. Here a straight, positively sloping line means constant acceleration.

For a curved v–t graph the acceleration is changing: draw a large tangent at the point of interest and find its gradient, and find displacement from the area (counting squares or using the trapezium rule). When asked to determine a gradient from a graph, use a large triangle (at least half the line), read coordinates carefully, and state the units of the gradient.

On a displacement–time graph the gradient is the velocity. A steeper line means a higher speed; a horizontal line means the object is stationary; a negative gradient means motion back towards the start.

s–t graph: gradient = velocity (tangent for instantaneous velocity).
v–t graph: gradient = acceleration; area = displacement.
a–t graph: area = change in velocity.
Determine a gradient with a large triangle (≥ half the line) and state its units.

See the full worked example for motion →

Projectile motion — two independent motions

Horizontal at constant velocity, vertical under gravity; they share only the clock.

A projectile is any object moving under gravity alone (air resistance neglected) after being launched. The key insight is that its horizontal and vertical motions are completely independent and are linked only by the shared time $t$ .

Horizontal: no horizontal force (air resistance neglected), so $a_x=0$ and the horizontal velocity $u_x=u\cos\theta$ stays constant. Therefore $x = u_x t$ .
Vertical: the only force is gravity, so $a_y=g=9.81\ \text{m s}^{-2}$ downwards, and the vertical motion obeys SUVAT with $u_y=u\sin\theta$ .

A useful consequence: a ball thrown horizontally and a ball simply dropped from the same height hit the ground at the same time, because their vertical motions are identical.

The launch velocity u resolves into a constant horizontal component u cos θ and an initial vertical component u sin θ. The path is a parabola: gravity acts downward throughout, so the vertical velocity falls to zero at the top of the arc, then increases downward.

Standard projectile method

Resolve the launch velocity: $u_x=u\cos\theta$ , $u_y=u\sin\theta$ .
Vertical SUVAT (taking up as positive, $a=-g$ ): at maximum height $v_y=0$ ; the time to the top is $t_{\text{up}}=\dfrac{u_y}{g}$ , and for a level landing the time of flight is $2t_{\text{up}}$ .
Maximum height from $v_y^2=u_y^2-2g\,h$ with $v_y=0$ , giving $h=\dfrac{u_y^2}{2g}$ .
Range $=u_x\times(\text{time of flight})$ — a horizontal calculation that uses the time found from the vertical motion.

The motions meet only through $t$ : you almost always find the time from the vertical equations and then use it in the horizontal equation.

Horizontal: $a_x=0$ , velocity $u\cos\theta$ constant ⇒ $x=u_x t$ .
Vertical: $a_y=g$ downward, full SUVAT with $u_y=u\sin\theta$ .
At the top of the arc the vertical velocity is zero (horizontal velocity is not).
Time of flight from the vertical motion; range = horizontal velocity × time of flight.

See the full worked example for motion →

Core Practical 1 — measuring g, the acceleration of free fall

Drop, time, and plot s against t² so g = 2 × gradient.

Core Practical 1 determines the acceleration of free fall, $g$ . A common method releases a steel ball from rest and measures the fall distance $s$ and the time $t$ taken to fall it (using light gates and a data-logger, an electromagnet-and-trapdoor timer, or video/ticker-tape analysis).

Because the ball starts from rest ( $u=0$ ) and accelerates uniformly at $g$ : $s = \tfrac{1}{2}g\,t^2$

You could compute $g=\dfrac{2s}{t^2}$ from a single drop, but a graphical method is far more reliable because it averages out random error and lets you spot anomalies:

Plot $s$ (y-axis) against $t^2$ (x-axis). This linearises the equation: comparing $s=\tfrac{1}{2}g\,t^2$ with $y=mx$ , the gradient is $\tfrac{1}{2}g$ .
Therefore $\boxed{g = 2 \times \text{gradient}}$ .
A straight line through the origin confirms $s\propto t^2$ and that acceleration is constant.

Plotting fall distance s against t² gives a straight line through the origin whose gradient is ½g. Reading the gradient from a large triangle and doubling it gives the acceleration of free fall, g.

Reducing and judging uncertainty

Reaction-time error dominates if you time by hand with a stopwatch — use light gates/electronic timing to remove it.
Repeat each drop and take a mean time to reduce random error.
Use a range of fall heights and a best-fit line so a single bad reading does not distort the result.
Measure the drop height with a metre rule to the nearest millimetre; air resistance is negligible for a small dense ball over short drops.
An accepted value is $g\approx9.81\ \text{m s}^{-2}$ ; compare your value and quote a percentage difference when evaluating accuracy.

From rest: $s=\tfrac{1}{2}gt^2$ , so a single drop gives $g=\dfrac{2s}{t^2}$ .
Better: plot $s$ vs $t^2$ → gradient = $\tfrac{1}{2}g$ → $g=2\times$ gradient.
Use light gates/electronic timing to remove reaction-time error.
Repeat and average; use several heights and a best-fit line.

See the full worked example for motion →

Quick recap

Distance and speed are scalars; displacement, velocity and acceleration are vectors needing a direction.
SUVAT ( $v=u+at$ , $s=ut+\tfrac12 at^2$ , $v^2=u^2+2as$ , $s=\tfrac12(u+v)t$ ) applies only for constant acceleration; list $s,u,v,a,t$ and pick the equation missing the unwanted one.
On a v–t graph: gradient = acceleration, area = displacement. On an s–t graph: gradient = velocity.
Projectiles split into independent horizontal (constant velocity) and vertical (acceleration $g$ ) motions sharing only $t$ .
Resolve before using SUVAT: $u_x=u\cos\theta$ , $u_y=u\sin\theta$ ; never plug the resultant velocity into SUVAT.
Core Practical 1: $s=\tfrac12 gt^2$ ; plot $s$ vs $t^2$ and $g=2\times$ gradient; remove reaction time with light gates.
Use $g=9.81\ \text{m s}^{-2}$ , keep full precision until the final step, and quote answers to 3 s.f. with units.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$v = u + at$ , $\ s = ut + \tfrac{1}{2}at^2$ , $\ v^2 = u^2 + 2as$ , $\ s = \tfrac{1}{2}(u+v)t$ — constant acceleration only.
v–t graph: gradient = acceleration; area under the line = displacement.
s–t graph: gradient = velocity (tangent for instantaneous velocity).
Resolve a vector: horizontal = $F\cos\theta$ , vertical = $F\sin\theta$ (θ from the horizontal).
Projectile: horizontal velocity constant ( $a=0$ ); vertical acceleration = $g=9.81\ \text{m s}^{-2}$ downwards; share only $t$ .
Maximum height $h=\dfrac{(u\sin\theta)^2}{2g}$ ; time of flight (level) $=\dfrac{2u\sin\theta}{g}$ ; range $=u\cos\theta\times$ time of flight.
Free-fall practical: $s=\tfrac{1}{2}gt^2$ ; plot $s$ vs $t^2$ ; $g = 2\times\text{gradient}$ .
Acceleration is the rate of change of velocity — a vector changes if speed OR direction changes.

How it’s examined

Motion is core Unit 1 content and appears on the Unit 1 paper (WPH11, Mechanics and Materials), which is structured and includes short calculations, ‘show that’ items, graph interpretation and a multi-part extended question. Typical formats: define scalar/vector or velocity/acceleration (1–2 marks, AO1); single- and multi-step SUVAT calculations where you must select the right equation and show working (2–4 marks, M and A marks); ‘determine the acceleration/velocity from the graph’ where you must draw a large triangle and give units (2–3 marks); a projectile problem requiring you to resolve, treat the axes separately, and combine via time (4–6 marks); and a practical/data question on Core Practical 1 (linearising $s=\tfrac12 gt^2$ , gradient = $\tfrac12 g$ , uncertainty). ALWAYS show the equation, substitution with units, and a 3 s.f. answer — ‘no working = risk zero’. For ‘show that’ work fully to the given value (one extra s.f.); for vectors quote a direction.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 1 examiner reports; Edexcel IAL Physics Unit 1 sample assessment materials and mark schemes. Last reviewed 2026-06-26.

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:

1Average speed vs average velocity (3 marks)
Getting startedMulti-step problem• scalars and vectors, AO2, Unit 1
Question
A jogger runs 30 m due east in 10 s, then 40 m due north in a further 10 s. Calculate (a) her average speed and (b) the magnitude of her average velocity for the whole run. (3 marks)
Step-by-step solution
1. Step 1
  Average speed uses total distance. Total distance $=30+40=70\ \text{m}$ in a total time of $20\ \text{s}$ .
  $\text{average speed} = \frac{\text{distance}}{\text{time}} = \frac{70}{20}$
2. Step 2
  Average velocity uses displacement (straight line from start to finish). The two legs are perpendicular, so use Pythagoras.
  $s = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = 50\ \text{m}$
3. Step 3
  Divide the displacement by the total time for the magnitude of the average velocity.
  $\text{average velocity} = \frac{50}{20} = 2.5\ \text{m s}^{-1}$
Answer
(a) Average speed $=3.5\ \text{m s}^{-1}$ . (b) Average velocity $=2.5\ \text{m s}^{-1}$ (directed roughly north-east, $53°$ N of E).
Examiner tip
Mark scheme: (1) average speed $=70/20=3.5\ \text{m s}^{-1}$ ; (2) displacement $=50\ \text{m}$ by Pythagoras; (3) average velocity $=2.5\ \text{m s}^{-1}$ . The two answers differ because speed uses path length (scalar) and velocity uses displacement (vector). A direction is needed for a complete velocity answer.
2Free fall from rest: finding the final velocity (2 marks)
Getting startedDirect calculation• SUVAT, free fall, Unit 1
Question
A stone is dropped from rest and falls freely for 2.5 s. Take $g = 9.81\ \text{m s}^{-2}$ and ignore air resistance. Calculate its velocity just before it lands. (2 marks)
Step-by-step solution
1. Step 1
  List SUVAT: $u=0$ , $a=g=9.81\ \text{m s}^{-2}$ , $t=2.5\ \text{s}$ , want $v$ (displacement $s$ not needed). The equation missing $s$ is $v=u+at$ .
  $v = u + at$
2. Step 2
  Substitute with units and evaluate.
  $v = 0 + 9.81 \times 2.5 = 24.525\ \text{m s}^{-1}$
Answer
$v = 24.5\ \text{m s}^{-1}$ (downwards), to 3 s.f.
Examiner tip
Mark scheme: (1) correct equation/substitution $v=9.81\times2.5$ (M); (2) $24.5\ \text{m s}^{-1}$ (A). Reaching $24.525$ then rounding to 3 s.f. is fine; quoting $24\ \text{m s}^{-1}$ (2 s.f.) risks losing the accuracy mark.
3Resolving a force into components (3 marks)
Getting startedDirect calculation• vectors, resolving, Unit 1
Question
A child pulls a sledge with a rope. The tension in the rope is 50 N and the rope makes an angle of 30° to the horizontal ground. Calculate the horizontal and vertical components of the tension. (3 marks)
Step-by-step solution
1. Step 1
  The angle is measured from the horizontal, so the horizontal component uses $\cos\theta$ .
  $F_x = F\cos\theta = 50\cos 30°$
2. Step 2
  Evaluate the horizontal component.
  $F_x = 50 \times 0.8660 = 43.3\ \text{N}$
3. Step 3
  The vertical component uses $\sin\theta$ .
  $F_y = F\sin\theta = 50\sin 30° = 50 \times 0.5 = 25.0\ \text{N}$
Answer
Horizontal component $=43.3\ \text{N}$ ; vertical component $=25.0\ \text{N}$ (upwards).
Examiner tip
Mark scheme: (1) horizontal uses cos (angle from horizontal); (2) $F_x=43.3\ \text{N}$ ; (3) $F_y=25.0\ \text{N}$ . The common error is swapping sin and cos. Check: $\sqrt{43.3^2+25.0^2}=50\ \text{N}$ recovers the original — a useful self-check.
4Displacement of an accelerating car (3 marks)
Building confidenceDirect calculation• Adapted from Edexcel IAL Physics Unit 1• SUVAT, AO2, Unit 1
Question
A car travelling at 5.0 m s⁻¹ accelerates uniformly at 2.0 m s⁻² for 6.0 s. Calculate the distance it travels in this time. (3 marks)
Step-by-step solution
1. Step 1
  List SUVAT: $u=5.0$ , $a=2.0$ , $t=6.0$ , want $s$ ( $v$ not needed). The equation missing $v$ is $s=ut+\tfrac12 at^2$ .
  $s = ut + \tfrac{1}{2}at^2$
2. Step 2
  Substitute the values with units.
  $s = (5.0 \times 6.0) + \tfrac{1}{2}(2.0)(6.0)^2$
3. Step 3
  Evaluate each term: $5.0\times6.0=30$ and $\tfrac12\times2.0\times36=36$ .
  $s = 30 + 36 = 66\ \text{m}$
Answer
$s = 66\ \text{m}$ .
Examiner tip
Mark scheme: (1) correct equation selected; (2) correct substitution (M); (3) $66\ \text{m}$ (A). A frequent slip is forgetting to square the time in the $\tfrac12 at^2$ term, giving $30+6=36\ \text{m}$ .
5Using v² = u² + 2as (no time given) (3 marks)
Building confidenceDirect calculation• SUVAT, AO2, Unit 1
Question
A cyclist accelerates uniformly from 8.0 m s⁻¹ at 2.0 m s⁻² over a distance of 50 m. Calculate her final velocity. (3 marks)
Step-by-step solution
1. Step 1
  List SUVAT: $u=8.0$ , $a=2.0$ , $s=50$ , want $v$ ; no time is given or wanted, so use the equation that omits $t$ : $v^2=u^2+2as$ .
  $v^2 = u^2 + 2as$
2. Step 2
  Substitute with units.
  $v^2 = 8.0^2 + 2(2.0)(50) = 64 + 200 = 264\ \text{m}^2\,\text{s}^{-2}$
3. Step 3
  Take the square root for $v$ (remember to root the whole right-hand side, not just one term).
  $v = \sqrt{264} = 16.2\ \text{m s}^{-1}$
Answer
$v = 16.2\ \text{m s}^{-1}$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) correct equation (omits $t$ ); (2) correct substitution → $v^2=264$ (M); (3) $v=16.2\ \text{m s}^{-1}$ (A). Choosing $v^2=u^2+2as$ avoids a two-stage calculation; many candidates needlessly find $t$ first and accumulate rounding error.
6Displacement from a velocity–time graph (4 marks)
Building confidenceGraph or diagram• velocity-time graph, area, AO2, Unit 1
Question
A train accelerates uniformly from rest to 20 m s⁻¹ in 5.0 s, travels at 20 m s⁻¹ for the next 10 s, then decelerates uniformly to rest in a further 5.0 s. By finding the area under the velocity–time graph, calculate the total displacement. (4 marks)
Step-by-step solution
1. Step 1
  On a v–t graph the area under the line = displacement. The graph is a trapezium: parallel sides are the bottom (total time $=20\ \text{s}$ ) and the top (constant-speed time $=10\ \text{s}$ ); the height is the velocity $20\ \text{m s}^{-1}$ .
2. Step 2
  Use the trapezium area formula (sum of the two parallel sides, halved, times the height).
  $s = \tfrac{1}{2}(a+b)\,h = \tfrac{1}{2}(20 + 10)(20)$
3. Step 3
  Evaluate.
  $s = \tfrac{1}{2}(30)(20) = 300\ \text{m}$
4. Step 4
  Check by splitting into shapes: accelerating triangle $\tfrac12(5)(20)=50$ , middle rectangle $10\times20=200$ , decelerating triangle $\tfrac12(5)(20)=50$ ; total $=50+200+50=300\ \text{m}$ . ✔
Answer
Total displacement $= 300\ \text{m}$ .
Examiner tip
Mark scheme: (1) area under graph = displacement; (2) correct method (trapezium OR split into triangle + rectangle + triangle); (3) substitution; (4) $300\ \text{m}$ . Reading $20\ \text{m s}^{-1}$ off the velocity axis as the ‘distance’ is a common, costly error — you must find the AREA.
7Determining g from an s–t² graph gradient (4 marks)
Building confidenceGraph or diagram• Adapted from Edexcel IAL Physics Unit 1 (Core Practical 1)• Core Practical 1, graph, AO3, Unit 1
Question
In a free-fall experiment a steel ball is released from rest and the fall distance s is plotted against t². The best-fit line passes through the origin, and a large gradient triangle gives Δs = 2.45 m for Δ(t²) = 0.500 s². Determine a value for the acceleration of free fall, g. (4 marks)
Step-by-step solution
1. Step 1
  From rest, $s=\tfrac12 g t^2$ . Comparing with $y=mx$ (with $y=s$ , $x=t^2$ ), the gradient equals $\tfrac12 g$ .
  $s = \tfrac{1}{2}g\,t^2 \;\Rightarrow\; \text{gradient} = \tfrac{1}{2}g$
2. Step 2
  Calculate the gradient from the triangle, with units.
  $\text{gradient} = \frac{\Delta s}{\Delta(t^2)} = \frac{2.45}{0.500} = 4.90\ \text{m s}^{-2}$
3. Step 3
  Since the gradient is $\tfrac12 g$ , double it to get $g$ .
  $g = 2 \times \text{gradient} = 2 \times 4.90$
4. Step 4
  Evaluate the final value.
  $g = 9.80\ \text{m s}^{-2}$
Answer
$g = 9.80\ \text{m s}^{-2}$ — very close to the accepted $9.81\ \text{m s}^{-2}$ .
Examiner tip
Mark scheme: (1) gradient $=\tfrac12 g$ (linking the equation to $y=mx$ ); (2) gradient $=4.90\ \text{m s}^{-2}$ from a large triangle; (3) $g=2\times$ gradient; (4) $g=9.80\ \text{m s}^{-2}$ . Forgetting the factor of 2 (quoting $g=4.90$ ) is the dominant error; a small gradient triangle that loses precision is the next.
8Range of a projectile launched at an angle (6 marks)
StretchMulti-step problem• Adapted from Edexcel IAL Physics Unit 1• projectile, AO2, Unit 1
Question
A ball is kicked from level ground at 20 m s⁻¹ at 30° above the horizontal. Take $g = 9.81\ \text{m s}^{-2}$ and ignore air resistance. Calculate (a) the time of flight and (b) the horizontal range. (6 marks)
Step-by-step solution
1. Step 1
  Resolve the launch velocity into components.
  $u_x = 20\cos 30° = 17.32\ \text{m s}^{-1}, \quad u_y = 20\sin 30° = 10.0\ \text{m s}^{-1}$
2. Step 2
  Vertical motion (take up as positive, $a=-9.81\ \text{m s}^{-2}$ ). For a level landing the vertical displacement is zero; the time of flight is twice the time to the top, where $v_y=0$ . Use $t_{\text{up}}=u_y/g$ .
  $t_{\text{flight}} = \frac{2u_y}{g} = \frac{2 \times 10.0}{9.81}$
3. Step 3
  Evaluate the time of flight.
  $t_{\text{flight}} = 2.039\ \text{s}$
4. Step 4
  Horizontal motion has constant velocity, so range $=u_x\times t_{\text{flight}}$ . The two axes are linked only through the shared time.
  $R = u_x \times t_{\text{flight}} = 17.32 \times 2.039$
5. Step 5
  Evaluate the range.
  $R = 35.3\ \text{m}$
Answer
(a) Time of flight $= 2.04\ \text{s}$ . (b) Range $= 35.3\ \text{m}$ .
Examiner tip
Mark scheme: (1) resolve $u_x=17.3$ , $u_y=10.0$ ; (2) vertical SUVAT for time (M); (3) $t=2.04\ \text{s}$ (A); (4) range = horizontal velocity × time (M); (5) $35.3\ \text{m}$ (A). The classic error is putting $20\ \text{m s}^{-1}$ (the resultant) into a vertical equation — you MUST use the resolved components and treat each axis separately.
9Horizontal launch from a cliff (6 marks)
StretchMulti-step problem• projectile, horizontal projection, AO2, Unit 1
Question
A stone is thrown horizontally at 15 m s⁻¹ from the top of a vertical cliff 45 m high. Take $g = 9.81\ \text{m s}^{-2}$ and ignore air resistance. Calculate (a) the time taken to reach the sea below and (b) the horizontal distance from the base of the cliff where it lands. (6 marks)
Step-by-step solution
1. Step 1
  The launch is horizontal, so the initial vertical velocity is zero ( $u_y=0$ ); the horizontal velocity is $15\ \text{m s}^{-1}$ throughout.
2. Step 2
  Vertical motion: use $s=u_y t+\tfrac12 g t^2$ with $u_y=0$ and $s=45\ \text{m}$ downward (take down as positive here).
  $45 = 0 + \tfrac{1}{2}(9.81)t^2$
3. Step 3
  Rearrange for $t$ .
  $t^2 = \frac{2 \times 45}{9.81} = 9.174\ \text{s}^2$
4. Step 4
  Take the square root.
  $t = \sqrt{9.174} = 3.029\ \text{s}$
5. Step 5
  Horizontal motion is at constant velocity, so distance $=$ horizontal velocity $\times t$ .
  $x = 15 \times 3.029 = 45.4\ \text{m}$
Answer
(a) Time $= 3.03\ \text{s}$ . (b) Horizontal distance $= 45.4\ \text{m}$ .
Examiner tip
Mark scheme: (1) recognise $u_y=0$ ; (2) vertical $s=\tfrac12 gt^2$ (M); (3)/(4) $t=3.03\ \text{s}$ (A); (5) horizontal $x=v\times t=45.4\ \text{m}$ (M + A). The vertical and horizontal motions are independent — only the time links them. Using the 15 m s⁻¹ in the vertical equation is wrong.
10Maximum height of a vertically thrown ball (6 marks)
StretchMulti-step problem• SUVAT, free fall, AO2, Unit 1
Question
A ball is thrown vertically upward with an initial velocity of 12 m s⁻¹. Take $g = 9.81\ \text{m s}^{-2}$ and ignore air resistance. Calculate (a) the maximum height reached and (b) the total time the ball is in the air before returning to its starting point. (6 marks)
Step-by-step solution
1. Step 1
  Take up as positive, so $u=+12\ \text{m s}^{-1}$ and $a=-9.81\ \text{m s}^{-2}$ . At the maximum height the velocity is momentarily zero, $v=0$ .
2. Step 2
  Use $v^2=u^2+2as$ (omits $t$ ) to find the maximum height $s$ .
  $0 = 12^2 + 2(-9.81)s$
3. Step 3
  Rearrange for $s$ .
  $s = \frac{-144}{2 \times (-9.81)} = \frac{144}{19.62} = 7.34\ \text{m}$
4. Step 4
  For the time, the time up is found from $v=u+at$ with $v=0$ .
  $0 = 12 + (-9.81)t_{\text{up}} \;\Rightarrow\; t_{\text{up}} = \frac{12}{9.81} = 1.223\ \text{s}$
5. Step 5
  By symmetry the time down equals the time up, so the total time aloft is twice $t_{\text{up}}$ .
  $t_{\text{total}} = 2 \times 1.223 = 2.45\ \text{s}$
Answer
(a) Maximum height $= 7.34\ \text{m}$ . (b) Total time in the air $= 2.45\ \text{s}$ .
Examiner tip
Mark scheme: (1) $v=0$ at the top and correct signs; (2)/(3) max height $=7.34\ \text{m}$ (M + A); (4) time up $=1.22\ \text{s}$ (M + A); (5) total $=2.45\ \text{s}$ by symmetry. Sign errors (treating $g$ as $+9.81$ while $u$ is up) are the usual cause of lost marks — define a positive direction first.

Model Answers — Motion

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

Question 1
2 marks
Q (2 marks). State the difference between a scalar quantity and a vector quantity, and give one example of each.
Model answer
A scalar quantity has magnitude (size) only, whereas a vector quantity has both magnitude and direction.

An example of a scalar is distance (or speed, mass, energy); an example of a vector is displacement (or velocity, acceleration, force).
Why this scores
Why this scores 2/2. (1) Scalar = magnitude only, vector = magnitude AND direction; (2) one correct example of each. Saying a vector ‘can be positive or negative’ alone is not enough — the defining feature is that it has a direction.
Question 2
2 marks
Q (2 marks). Explain how you would find the instantaneous velocity of an object at a particular time from its displacement–time graph.
Model answer
Draw a tangent to the curve at the point corresponding to that time, then find the gradient of the tangent. The gradient of a displacement–time graph is the velocity, so the gradient of the tangent gives the instantaneous velocity at that instant.
Why this scores
Why this scores 2/2. (1) draw a tangent at the point; (2) gradient of the tangent = velocity. Average velocity (a chord between two points) would not be credited here — the question asks for the instantaneous value, which needs a tangent.
Question 3
3 marks
Q (3 marks). A coin is dropped from rest down a deep well and falls freely for 1.8 s before hitting the water. Take $g = 9.81\ \text{m s}^{-2}$ . Calculate the depth of the well, showing your working.
Model answer
Taking downward as positive, the coin starts from rest so $u = 0$ , with $a = g = 9.81\ \text{m s}^{-2}$ and $t = 1.8\ \text{s}$ . The final velocity is not wanted, so the appropriate equation is:

$s = ut + \tfrac{1}{2}at^2$

Substituting:

$s = (0)(1.8) + \tfrac{1}{2}(9.81)(1.8)^2 = \tfrac{1}{2}(9.81)(3.24) = 15.9\ \text{m}$

The well is therefore about 15.9 m deep (to 3 significant figures).
Why this scores
Why this scores 3/3. (1) correct equation selected with $u=0$ ; (2) correct substitution (M); (3) $15.9\ \text{m}$ to 3 s.f. (A). Squaring the time is essential — $(1.8)^2=3.24$ , not $3.6$ . Full working protects both marks if a small arithmetic slip occurs.
Question 4
4 marks
Q (4 marks). A car starts from rest and accelerates uniformly, reaching 24 m s⁻¹ after 8.0 s. (a) Calculate its acceleration. (b) Calculate the distance it travels in this time.
Model answer
(a) The acceleration is the rate of change of velocity. With $u = 0$ , $v = 24\ \text{m s}^{-1}$ and $t = 8.0\ \text{s}$ :

$a = \frac{v - u}{t} = \frac{24 - 0}{8.0} = 3.0\ \text{m s}^{-2}$

(b) The displacement equals the average velocity multiplied by the time (the area under the velocity–time graph):

$s = \tfrac{1}{2}(u + v)\,t = \tfrac{1}{2}(0 + 24)(8.0) = 96\ \text{m}$

The car travels 96 m.
Why this scores
Why this scores 4/4. (a) 2 marks: $a=(v-u)/t$ with substitution → $3.0\ \text{m s}^{-2}$ . (b) 2 marks: any valid SUVAT route (here $s=\tfrac12(u+v)t$ , equivalently the area under the v–t graph) → $96\ \text{m}$ . Using $v^2=u^2+2as$ or $s=ut+\tfrac12 at^2$ with $a=3.0$ would give the same answer and full credit.
Question 5
5 marks
Q (5 marks). A hiker walks 600 m due east and then 800 m due north. (a) Calculate the magnitude of her resultant displacement. (b) Calculate the direction of this displacement as an angle measured from the east (the first leg). (c) State, with a reason, whether her average speed is greater than the magnitude of her average velocity.
Model answer
(a) The two legs are perpendicular, so the resultant displacement is found by Pythagoras:

$R = \sqrt{600^2 + 800^2} = \sqrt{360000 + 640000} = \sqrt{1000000} = 1000\ \text{m}$

(b) The angle $\theta$ north of east is:

$\theta = \tan^{-1}\!\left(\frac{800}{600}\right) = \tan^{-1}(1.333) = 53.1°\ \text{N of E}$

(c) Her average speed is greater than the magnitude of her average velocity. Average speed uses the total distance travelled ( $600 + 800 = 1400\ \text{m}$ ), whereas average velocity uses the straight-line displacement ( $1000\ \text{m}$ ). Because the displacement is smaller than the path length, the average velocity is smaller than the average speed (for the same total time).
Why this scores
Why this scores 5/5. (a) 2 marks: Pythagoras method and $1000\ \text{m}$ . (b) 2 marks: $\tan^{-1}(800/600)$ and $53.1°$ (a direction stated). (c) 1 mark: speed uses distance (1400 m) > velocity uses displacement (1000 m). A bald number with no direction in (b) would be incomplete for a vector.
Question 6
4 marks
Q (4 marks). A motorcyclist decelerates uniformly from 30 m s⁻¹ to 12 m s⁻¹ while travelling 126 m. Show that the magnitude of the deceleration is about 3 m s⁻².
Model answer
The time is neither given nor wanted, so use $v^2 = u^2 + 2as$ with $u = 30\ \text{m s}^{-1}$ , $v = 12\ \text{m s}^{-1}$ and $s = 126\ \text{m}$ :

$12^2 = 30^2 + 2a(126)$

$144 = 900 + 252a$

Rearranging for $a$ :

$a = \frac{144 - 900}{252} = \frac{-756}{252} = -3.00\ \text{m s}^{-2}$

The negative sign shows the acceleration opposes the motion, so the magnitude of the deceleration is $3.00\ \text{m s}^{-2}$ , as required.
Why this scores
Why this scores 4/4. (1) correct equation chosen ( $v^2=u^2+2as$ , omits $t$ ); (2) correct substitution; (3) rearrangement to $a=-756/252$ ; (4) $a=-3.00\ \text{m s}^{-2}$ , with the negative sign interpreted as deceleration. For a ‘show that’, you must work through to the value (here to one extra s.f., $3.00$ ) and not simply assert it.
Question 7
Adapted from Edexcel IAL Physics Unit 16 marks
Q (6 marks). A stone is launched from ground level at 25 m s⁻¹ at an angle of 40° above the horizontal. Take $g = 9.81\ \text{m s}^{-2}$ and neglect air resistance. (a) Calculate the maximum height reached. (b) Calculate the horizontal range. (c) Explain why the horizontal velocity remains constant throughout the flight.
Model answer
First resolve the launch velocity: $u_x = 25\cos 40° = 19.15\ \text{m s}^{-1}$ and $u_y = 25\sin 40° = 16.07\ \text{m s}^{-1}$ .

(a) Vertically, at the maximum height $v_y = 0$ . Using $v_y^2 = u_y^2 - 2gh$ :

$0 = (16.07)^2 - 2(9.81)h \;\Rightarrow\; h = \frac{258.3}{19.62} = 13.2\ \text{m}$

(b) The time of flight (level landing) is twice the time to the top:

$t = \frac{2u_y}{g} = \frac{2 \times 16.07}{9.81} = 3.276\ \text{s}$

The horizontal range is the (constant) horizontal velocity times this time:

$R = u_x \times t = 19.15 \times 3.276 = 62.7\ \text{m}$

(c) Once launched, the only force acting on the stone is its weight, which acts vertically downward (air resistance is neglected). There is no horizontal force, so by Newton's first law there is no horizontal acceleration and the horizontal component of velocity stays constant throughout the flight.
Why this scores
Why this scores 6/6. Resolve $u_x=19.2$ , $u_y=16.1$ . (a) 2 marks: $v_y=0$ at top and $h=13.2\ \text{m}$ . (b) 2 marks: time of flight $3.28\ \text{s}$ and $R=62.7\ \text{m}$ . (c) 2 marks: weight acts vertically → no horizontal force → no horizontal acceleration → constant horizontal velocity. Treating the axes independently and using the resolved components (not 25 m s⁻¹) is essential.
Question 8
Adapted from Edexcel IAL Physics Unit 1 (Core Practical 1)6 marks
Q (6 marks). A student measures the acceleration of free fall by releasing a steel ball from an electromagnet and timing its fall through a known height to a trapdoor switch. (a) Starting from $s = \tfrac{1}{2}gt^2$ , explain how plotting $s$ against $t^2$ allows $g$ to be found. (b) Give two ways the student could reduce the uncertainty in the value of $g$ obtained, and for each explain how it helps.
Model answer
(a) Because the ball is released from rest, $u = 0$ and the fall obeys $s = \tfrac{1}{2}g t^2$ . Comparing this with the equation of a straight line through the origin, $y = mx$ (with $y = s$ and $x = t^2$ ), the gradient $m$ equals $\tfrac{1}{2}g$ . So if $s$ is plotted on the $y$ -axis against $t^2$ on the $x$ -axis, the points lie on a straight line through the origin, and:

$g = 2 \times \text{gradient}$

Using the whole best-fit line (a large gradient triangle) averages out random errors, giving a more reliable value than a single drop.

(b) Two improvements:

Use electronic timing (electromagnet release with a data-logger/trapdoor switch) rather than a hand-operated stopwatch. This removes the human reaction-time error (about 0.2 s), which would otherwise be a large systematic/random error on the short fall times.

Repeat each drop several times and use the mean time, and take readings over a range of heights. Averaging reduces the effect of random errors in individual timings, and several heights let a best-fit line be drawn so a single anomalous reading does not distort the gradient.
Why this scores
Why this scores 6/6. (a) 3 marks: $u=0$ so $s=\tfrac12 gt^2$ ; compare with $y=mx$ → gradient $=\tfrac12 g$ ; $g=2\times$ gradient. (b) 3 marks: two valid improvements EACH with a reason — electronic timing removes reaction time; repeats/mean and multiple heights reduce random error and allow a best-fit line. Vague answers (‘measure more carefully’) without the mechanism are not credited.
Question 9
Adapted from Edexcel IAL Physics Unit 16 marks
Q (6 marks). A car's motion is recorded on a velocity–time graph. From $t = 0$ to $t = 10\ \text{s}$ the velocity rises uniformly from 0 to 20 m s⁻¹; from $t = 10\ \text{s}$ to $t = 30\ \text{s}$ it stays at 20 m s⁻¹; from $t = 30\ \text{s}$ to $t = 40\ \text{s}$ it falls uniformly to rest. (a) Calculate the acceleration during the first 10 s. (b) Calculate the total distance travelled. (c) Describe and explain the motion of the car in the final 10 s in terms of the graph.
Model answer
(a) Acceleration is the gradient of the velocity–time graph in the first phase:

$a = \frac{\Delta v}{\Delta t} = \frac{20 - 0}{10 - 0} = 2.0\ \text{m s}^{-2}$

(b) The distance is the total area under the graph, which is a trapezium with parallel sides 40 s (the whole time at the base) and 20 s (the constant-speed top from 10 s to 30 s), and height 20 m s⁻¹:

$s = \tfrac{1}{2}(40 + 20)(20) = \tfrac{1}{2}(60)(20) = 600\ \text{m}$

(Equivalently: accelerating triangle $\tfrac12(10)(20)=100\ \text{m}$ , plus rectangle $20\times20=400\ \text{m}$ , plus decelerating triangle $\tfrac12(10)(20)=100\ \text{m}$ , giving $600\ \text{m}$ .)

(c) In the final 10 s the line slopes downward to the time axis, so the velocity is decreasing: the car is decelerating uniformly (constant negative gradient) from 20 m s⁻¹ to rest. The gradient there is $\dfrac{0 - 20}{40 - 30} = -2.0\ \text{m s}^{-2}$ , so the car slows at $2.0\ \text{m s}^{-2}$ until it stops at $t = 40\ \text{s}$ .
Why this scores
Why this scores 6/6. (a) 2 marks: gradient method and $2.0\ \text{m s}^{-2}$ . (b) 2 marks: area = displacement, trapezium (or split shapes) → $600\ \text{m}$ . (c) 2 marks: negative gradient → decreasing velocity / uniform deceleration, quoting $-2.0\ \text{m s}^{-2}$ and that it reaches rest. Linking the graph feature (gradient/area) to the physical quantity is the key examiner discriminator.

Key Formulae — Motion

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

First SUVAT equation
$v = u + at$
$v$
final velocity / m s⁻¹
$u$
initial velocity / m s⁻¹
$a$
acceleration / m s⁻² (constant)
$t$
time / s
When to use
Constant acceleration when displacement $s$ is neither known nor wanted. In the Edexcel formulae booklet (Appendix 7).
Example
Dropped from rest for 2.5 s: $v=0+9.81\times2.5=24.5\ \text{m s}^{-1}$ .
Second SUVAT equation
$s = ut + \tfrac{1}{2}at^2$
$s$
displacement / m
$u$
initial velocity / m s⁻¹
$a$
acceleration / m s⁻²
$t$
time / s
When to use
Constant acceleration when the final velocity $v$ is neither known nor wanted. In the formulae booklet.
Example
$u=5.0,\ a=2.0,\ t=6.0$ : $s=30+36=66\ \text{m}$ .
Third SUVAT equation
$v^2 = u^2 + 2as$
$v$
final velocity / m s⁻¹
$u$
initial velocity / m s⁻¹
$a$
acceleration / m s⁻²
$s$
displacement / m
When to use
Constant acceleration when time $t$ is neither known nor wanted — avoids a two-stage calculation. In the formulae booklet.
Example
$u=8.0,\ a=2.0,\ s=50$ : $v^2=264$ , so $v=16.2\ \text{m s}^{-1}$ .
Fourth SUVAT equation
$s = \tfrac{1}{2}(u+v)\,t$
$s$
displacement / m
$u$
initial velocity / m s⁻¹
$v$
final velocity / m s⁻¹
$t$
time / s
When to use
Constant acceleration when the acceleration $a$ is neither known nor wanted (equivalently, displacement = average velocity × time). In the formulae booklet.
Example
$u=0,\ v=20,\ t=5$ : $s=\tfrac12(20)(5)=50\ \text{m}$ .
Acceleration (definition)
$a = \frac{\Delta v}{\Delta t}$
$a$
acceleration / m s⁻²
$\Delta v$
change in velocity / m s⁻¹
$\Delta t$
time interval / s
When to use
Defining acceleration, or reading it as the gradient of a velocity–time graph. Must be recalled.
Example
Velocity rises by 20 m s⁻¹ in 5 s: $a=20/5=4.0\ \text{m s}^{-2}$ .
Resolving a vector
$F_x = F\cos\theta, \quad F_y = F\sin\theta$
$F$
magnitude of the vector
$\theta$
angle measured from the horizontal
$F_x$
horizontal component
$F_y$
vertical component
When to use
Splitting any vector (force, velocity) into two perpendicular components before applying SUVAT or balancing forces. Must be recalled.
Example
$50\ \text{N}$ at $30°$ : $F_x=43.3\ \text{N}$ , $F_y=25.0\ \text{N}$ .
Resultant of perpendicular vectors
$R = \sqrt{F_x^2 + F_y^2}, \quad \theta = \tan^{-1}\!\left(\tfrac{F_y}{F_x}\right)$
$R$
magnitude of the resultant
$F_x,\ F_y$
perpendicular components
$\theta$
angle of resultant above the horizontal
When to use
Combining two perpendicular vectors (e.g. components, or two legs of a journey) into a single resultant. Must be recalled.
Example
Components $43.3$ and $25.0$ : $R=\sqrt{43.3^2+25.0^2}=50.0$ .

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 Cambridge IGCSE sitting.

Scalar quantity
Examiner keyword
A quantity that has magnitude (size) only and no direction. Examples: distance, speed, mass, time, energy, temperature.
Vector quantity
Examiner keyword
A quantity that has both magnitude and direction. Examples: displacement, velocity, acceleration, force, momentum.
Distance
A scalar measure of the total path length travelled by an object, regardless of direction.
Displacement
Examiner keyword
A vector quantity: the straight-line distance from the start point to the end point, in a specified direction.
Speed
A scalar quantity equal to the rate of change of distance (distance travelled per unit time).
Velocity
Examiner keyword
A vector quantity equal to the rate of change of displacement; speed in a stated direction. Average velocity = displacement ÷ time.
Acceleration
Examiner keyword
A vector quantity equal to the rate of change of velocity, $a=\dfrac{\Delta v}{\Delta t}$ (unit m s⁻²). An object accelerates if its speed OR its direction changes.
Instantaneous velocity
Examiner keyword
The velocity of an object at a single instant — found from the gradient of the tangent to a displacement–time graph at that point.
Average velocity
The total displacement divided by the total time taken: $\bar{v}=\dfrac{\text{displacement}}{\text{time}}$ . May be smaller than average speed if the path is not straight.
Uniform (constant) acceleration
Examiner keyword
Acceleration that does not change in magnitude or direction with time. The four SUVAT equations apply only under this condition.
Free fall / acceleration of free fall (g)
Examiner keyword
Motion under gravity alone (air resistance neglected); all objects accelerate downward at $g=9.81\ \text{m s}^{-2}$ near the Earth's surface, independent of their mass.
Projectile
Examiner keyword
An object moving freely under gravity after being launched. Its horizontal motion (constant velocity) and vertical motion (acceleration g) are independent and linked only by the shared time.
Resolving a vector
Examiner keyword
Splitting a single vector into two perpendicular components: horizontal $=F\cos\theta$ and vertical $=F\sin\theta$ , where $\theta$ is measured from the horizontal.
Resultant vector
Examiner keyword
The single vector that has the same effect as two or more vectors combined. For perpendicular vectors, magnitude $=\sqrt{F_x^2+F_y^2}$ and direction $=\tan^{-1}(F_y/F_x)$ .
Area under a velocity–time graph
Examiner keyword
Equal to the displacement of the object. (The gradient of the same graph gives the acceleration.)
Gradient of a displacement–time graph
Equal to the velocity of the object. For a curve, the gradient of the tangent gives the instantaneous velocity.

Common Mistakes and Misconceptions — Motion

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

✕Putting the resultant (slanted) launch velocity straight into a SUVAT equation for a projectile
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students treat the launch as a single one-dimensional motion instead of two independent ones.
How to avoid it
Always resolve first: use $u_x=u\cos\theta$ for the horizontal axis and $u_y=u\sin\theta$ for the vertical axis, then apply SUVAT to each axis separately. The axes are linked only through the shared time $t$ .
✕Confusing gradient and area on a velocity–time graph (e.g. reading the velocity off the axis as the distance)
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
The two operations look similar and students forget which gives which quantity.
How to avoid it
Remember: on a v–t graph the gradient = acceleration and the area under the line = displacement. To find a distance, calculate the AREA (triangle/trapezium or count squares).
✕Swapping sin and cos when resolving (using $F\sin\theta$ for the horizontal component)
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Not checking which side of the right-angled triangle the angle is measured from.
How to avoid it
When $\theta$ is measured from the horizontal, the horizontal component is $F\cos\theta$ and the vertical is $F\sin\theta$ . Sketch the triangle and check that $\sqrt{F_x^2+F_y^2}$ recovers $F$ .
✕Sign errors in vertical SUVAT — e.g. taking up as positive but still using $a=+9.81$
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
No positive direction is defined before substituting, so $u$ and $a$ end up inconsistent.
How to avoid it
Define a positive direction first. If up is positive, then for an object thrown up $u>0$ and $a=-9.81\ \text{m s}^{-2}$ (gravity acts down). Keep the sign convention consistent throughout the question.
✕Using SUVAT equations when the acceleration is not constant (e.g. with air resistance)
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
SUVAT is the default tool and students apply it without checking the condition.
How to avoid it
SUVAT is valid only for constant acceleration. If the v–t graph curves (changing gradient), use the gradient of a tangent for acceleration and the area for displacement instead.
✕In Core Practical 1, quoting the gradient of the $s$ – $t^2$ graph as $g$ (forgetting the factor of 2)
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students stop at the gradient without comparing $s=\tfrac12 gt^2$ with $y=mx$ .
How to avoid it
The gradient equals $\tfrac12 g$ , so $g = 2\times\text{gradient}$ . Also plot $s$ against $t^2$ (not $t$ ) so the graph is a straight line in the first place.

Motion — frequently asked questions

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

Motion

Short Notes - Motion

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Average speed vs average velocity (3 marks)

2Free fall from rest: finding the final velocity (2 marks)

3Resolving a force into components (3 marks)

4Displacement of an accelerating car (3 marks)

5Using v² = u² + 2as (no time given) (3 marks)

6Displacement from a velocity–time graph (4 marks)

7Determining g from an s–t² graph gradient (4 marks)

8Range of a projectile launched at an angle (6 marks)

9Horizontal launch from a cliff (6 marks)

10Maximum height of a vertically thrown ball (6 marks)

First SUVAT equation

Second SUVAT equation

Third SUVAT equation

Fourth SUVAT equation

Acceleration (definition)

Resolving a vector

Resultant of perpendicular vectors

Scalar quantity

Vector quantity

Distance

Displacement

Speed

Velocity

Acceleration

Instantaneous velocity

Average velocity

Uniform (constant) acceleration

Free fall / acceleration of free fall (g)

Projectile

Resolving a vector

Resultant vector

Area under a velocity–time graph

Gradient of a displacement–time graph

✕Putting the resultant (slanted) launch velocity straight into a SUVAT equation for a projectile

✕Confusing gradient and area on a velocity–time graph (e.g. reading the velocity off the axis as the distance)

✕Swapping sin and cos when resolving (using Fsin⁡θF\sin\thetaFsinθ for the horizontal component)

✕Sign errors in vertical SUVAT — e.g. taking up as positive but still using a=+9.81a=+9.81a=+9.81

✕Using SUVAT equations when the acceleration is not constant (e.g. with air resistance)

✕In Core Practical 1, quoting the gradient of the sss–t2t^2t2 graph as ggg (forgetting the factor of 2)

Motion — frequently asked questions

✕Swapping sin and cos when resolving (using $F\sin\theta$ for the horizontal component)

✕Sign errors in vertical SUVAT — e.g. taking up as positive but still using $a=+9.81$

✕In Core Practical 1, quoting the gradient of the $s$ – $t^2$ graph as $g$ (forgetting the factor of 2)