What are the equations of motion in kinematics?

In kinematics, the equations of motion describe the relationship between an object's displacement, velocity, acceleration, and time. The three primary equations are: 1) v = u + at, 2) s = ut + 0.5at², and 3) v² = u² + 2as, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration, 's' is the displacement, and 't' is the time.

How do you derive the first equation of motion?

The first equation of motion, v = u + at, can be derived from the definition of acceleration. Acceleration 'a' is the rate of change of velocity, given by a = (v - u) / t. Rearranging this formula gives v = u + at, which relates final velocity to initial velocity, acceleration, and time.

When should I use the second equation of motion?

The second equation of motion, s = ut + 0.5at², is used when you need to find the displacement of an object when the initial velocity, acceleration, and time are known. It is particularly useful in problems where the time duration of motion is given or needs to be calculated.

What is the significance of the third equation of motion?

The third equation of motion, v² = u² + 2as, is significant because it relates the velocities and displacement without involving time. This equation is particularly useful in scenarios where time is not given or is difficult to determine, allowing you to solve problems involving only velocities and displacement.

How do the equations of motion apply to free-fall problems?

In free-fall problems, the equations of motion apply by setting the acceleration 'a' to the acceleration due to gravity, 'g', which is approximately 9.8 m/s² on Earth. For example, if an object is dropped from rest, you can use these equations to calculate its velocity and displacement at any given time during its fall.

Why is "Using SUVAT with non-constant acceleration" a common mistake in Equations of motion?

Why it happens: Habit. How to avoid it: SUVAT requires CONSTANT acceleration. If a varies, use calculus or graphical methods. Source: 9702 Examiner Reports 2022-2024

Why is "Including gravity in horizontal motion of projectile" a common mistake in Equations of motion?

Why it happens: Treating motion as one direction. How to avoid it: Horizontal and vertical motions are INDEPENDENT. Gravity affects ONLY vertical. Horizontal velocity is constant (ignoring air resistance). Source: 9702 Examiner Reports 2022-2024

KinematicsCambridge International A Levels Physics (9702)

Equations of motion

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Equations of Motion Study Notes — Cambridge International A Level Physics 9702 Kinematics (2025-2027 syllabus)

SUVAT for constant acceleration. Position-time and velocity-time graphs. Projectile motion. Foundation for dynamics.

What you’ll learn

Mapped to the Cambridge IGCSE 9702 syllabus (2025-2027).

2.1.1 — Use SUVAT equations for uniform acceleration.
2.1.2 — Read and interpret position-time and velocity-time graphs.
2.1.3 — Solve projectile motion problems.

SUVAT equations

Constant acceleration only.

Five equations relate $s$ , $u$ , $v$ , $a$ , $t$ :

Equation	No
$v = u + at$	$s$
$s = ut + \frac{1}{2}at^2$	$v$
$v^2 = u^2 + 2as$	$t$
$s = \frac{1}{2}(u + v)t$	$a$
$s = vt - \frac{1}{2}at^2$	$u$

Pick the equation that contains the three known quantities plus the one you want to find.

Example. Car from rest, $a = 3.0$ m/s², $t = 8.0$ s. Find $v$ and $s$ .

$v = 0 + 3.0 \times 8.0 = 24$ m/s.
$s = 0 + \frac{1}{2}(3.0)(64) = 96$ m.

Cambridge tip. Identify u, v, a, s, t. Then pick equation.

Five SUVAT equations.
Constant acceleration only.
Pick equation matching knowns + unknown.

See the full worked example for equations of motion →

Motion graphs

Slope and area.

Position-time ( $s$ - $t$ ).

Slope = velocity.
Curving = changing velocity (acceleration).
Horizontal line = stationary.

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

Slope = acceleration.
Area under = displacement.
Negative area = motion backwards.

Acceleration-time ( $a$ - $t$ ).

Area under = change in velocity.

Example. $v$ - $t$ : accelerates 0→30 m/s in 10 s, then constant 30 m/s for 20 s.

Acceleration phase 1: $a = 3$ m/s².
Displacement = area under graph = $\frac{1}{2}(10)(30) + 20 \times 30 = 150 + 600 = 750$ m.

Cambridge tip. Mark scheme awards method marks for stating 'area = displacement' or 'slope = acceleration'.

$s$ - $t$ slope = $v$ .
$v$ - $t$ slope = $a$ , area = $s$ .
$a$ - $t$ area = $\Delta v$ .

See the full worked example for equations of motion →

Projectiles

Horizontal + vertical independent.

Setup. Object under gravity only (after launch).

Horizontal velocity CONSTANT (ignoring air resistance).
Vertical acceleration $= -g$ (taking up as positive).

Method.

Split initial velocity: $u_x = u\cos\theta$ , $u_y = u\sin\theta$ .
Apply SUVAT separately to each direction.
Time $t$ links the two.

Horizontal launch. $u_y = 0$ . Time to fall from height $h$ : $t = \sqrt{2h/g}$ . Range: $R = u_x \cdot t$ .

Example. Horizontal launch at 20 m/s from 80 m cliff.

$t = \sqrt{160/9.81} \approx 4.04$ s.
$R = 20 \times 4.04 \approx 80.8$ m.

Angled launch. Maximum range at $45°$ on level ground.

Cambridge tip. Always sketch + label axes.

Horizontal: constant velocity.
Vertical: $a = -g$ .
Independent motions, common $t$ .

See the full worked example for equations of motion →

How it’s examined

Kinematics appears on every Paper 1 and Paper 2. Most-tested: SUVAT (5-7 marks), graph interpretation (6 marks), projectile (10 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

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Step-by-step worked examples — Equations of motion

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

1SUVAT problem (5 marks)
Extended• Adapted from 9702/22 May/Jun 2024• SUVAT
Question
Car accelerates from rest at $3.0 \text{ m/s}^2$ . Find its velocity and displacement after 8.0 s. (5 marks)
Step-by-step solution
1. Step 1
  Identify variables. $u = 0$ , $a = 3.0$ m/s², $t = 8.0$ s. Find $v, s$ .
2. Step 2
  $v = u + at$ .
  $v = 0 + 3.0 \times 8.0 = 24 \text{ m/s}$
3. Step 3
  $s = ut + \frac{1}{2}at^2$ .
  $s = 0 + \dfrac{1}{2}(3.0)(64) = 96 \text{ m}$
Answer
$v = 24$ m/s; $s = 96$ m.
2Projectile (7 marks)
Extended• projectile
Question
Ball projected HORIZONTALLY at 20 m/s from a cliff 80 m high. Find time to hit ground and horizontal range. Take $g = 9.81$ . (7 marks)
Step-by-step solution
1. Step 1
  Vertical: $u_y = 0$ , $a = g$ , $s_y = 80$ . Use $s = \frac{1}{2}gt^2$ .
  $80 = \dfrac{1}{2}(9.81)t^2 \Rightarrow t = \sqrt{\dfrac{160}{9.81}} \approx 4.04 \text{ s}$
2. Step 2
  Horizontal: no acceleration. Range $= u_x \times t$ .
  $R = 20 \times 4.04 \approx 80.8 \text{ m}$
Answer
$t \approx 4.04$ s; $R \approx 80.8$ m.
3Velocity-time graph (6 marks)
Extended• v-t graph
Question
Object accelerates uniformly from 0 to 30 m/s in 10 s, then maintains 30 m/s for 20 s. Find acceleration and total displacement. (6 marks)
Step-by-step solution
1. Step 1
  Acceleration during first 10 s.
  $a = \dfrac{\Delta v}{\Delta t} = \dfrac{30 - 0}{10} = 3 \text{ m/s}^2$
2. Step 2
  Displacement = area under v-t graph.
3. Step 3
  First phase (triangle): $\frac{1}{2}(10)(30) = 150$ m.
4. Step 4
  Second phase (rectangle): $20 \times 30 = 600$ m.
5. Step 5
  Total.
  $s = 150 + 600 = 750 \text{ m}$
Answer
$a = 3$ m/s²; total displacement $= 750$ m.

Key Formulae — Equations of motion

The formulae you need to memorise for equations of motion on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

SUVAT — $v = u + at$
$v = u + at$
$u, v$
initial, final velocity
$a$
constant acceleration
$t$
time
When to use
Constant acceleration. No displacement involved.
SUVAT — $s = ut + \tfrac{1}{2}at^2$
$s = ut + \tfrac{1}{2}at^2$
$s$
displacement
When to use
Displacement from initial velocity, acceleration, and time.
SUVAT — $v^2 = u^2 + 2as$
$v^2 = u^2 + 2as$
When to use
Velocity-displacement relationship. No time needed.
SUVAT — $s = \tfrac{1}{2}(u + v)t$
$s = \tfrac{1}{2}(u + v)t$
When to use
Displacement from mean velocity × time.
Graph interpretations
$v = \dfrac{ds}{dt}, \quad a = \dfrac{dv}{dt}, \quad s = \int v \, dt$
When to use
Slope of $s$ - $t$ = velocity. Slope of $v$ - $t$ = acceleration. Area under $v$ - $t$ = displacement.

Key Definitions and Keywords — Equations of motion

Definitions to memorise and the exact keywords mark schemes credit for equations of motion answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Displacement
Examiner keyword
Vector from a starting point to current position. Units: m.
Velocity
Examiner keyword
Vector. Rate of change of displacement. $v = ds/dt$ .
Acceleration
Examiner keyword
Vector. Rate of change of velocity. $a = dv/dt$ .
Projectile
Examiner keyword
Object moving under gravity ONLY (after launch). Horizontal velocity constant; vertical undergoes uniform acceleration $g$ .

Common Mistakes and Misconceptions — Equations of motion

The traps other students keep falling into on equations of motion questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Using SUVAT with non-constant acceleration
9702 Examiner Reports 2022-2024
Why it happens
Habit.
How to avoid it
SUVAT requires CONSTANT acceleration. If $a$ varies, use calculus or graphical methods.
✕Including gravity in horizontal motion of projectile
9702 Examiner Reports 2022-2024
Why it happens
Treating motion as one direction.
How to avoid it
Horizontal and vertical motions are INDEPENDENT. Gravity affects ONLY vertical. Horizontal velocity is constant (ignoring air resistance).

Practice questions

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Past paper style quiz

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Equations of motion — frequently asked questions

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

Equations of motion

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1SUVAT problem (5 marks)

2Projectile (7 marks)

3Velocity-time graph (6 marks)

SUVAT — v=u+atv = u + atv=u+at

SUVAT — s=ut+12at2s = ut + \tfrac{1}{2}at^2s=ut+21​at2

SUVAT — v2=u2+2asv^2 = u^2 + 2asv2=u2+2as

SUVAT — s=12(u+v)ts = \tfrac{1}{2}(u + v)ts=21​(u+v)t

Graph interpretations

Displacement

Velocity

Acceleration

Projectile

✕Using SUVAT with non-constant acceleration

✕Including gravity in horizontal motion of projectile

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Equations of motion — frequently asked questions

SUVAT — $v = u + at$

SUVAT — $s = ut + \tfrac{1}{2}at^2$

SUVAT — $v^2 = u^2 + 2as$

SUVAT — $s = \tfrac{1}{2}(u + v)t$