Learn9702Paper 2 · AS Level Structured QuestionsKinematicsEquations of motion and graphs

Kinematics · Subtopic 1 of 2

Equations of Motion and Graphs

For constant acceleration in a straight line, five quantities ( $s$ , $u$ , $v$ , $a$ , $t$ ) are linked by four equations. Choose the one that contains the three you know and the one you want. Pick a positive direction and stick to it: for vertical motion, $a=-9.81$ m s⁻² if up is positive. Graphs give the same information: gradients and areas.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH11AQA 7408 Paper 1≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Distinguish displacement from distance and velocity from speed.
Use the four equations of uniformly accelerated motion.
Handle vertical motion with a consistent sign convention.
Interpret gradients and areas of displacement–time and velocity–time graphs.

Pause and think

A ball is thrown straight up and caught at the same height $3$ s later. What was its average velocity? Its average speed?

Section 2 of 14

Core idea

Start hereFive quantities, four equations

Drop a stone off a cliff: how long until it lands, how fast is it going? Under constant acceleration, displacement, initial velocity, final velocity, acceleration and time are linked by four equations. Know three and the rest follow.

The big idea

Uniform acceleration: $v=u+at$ , $s=ut+\frac12at^2$ , $v^2=u^2+2as$ , $s=\frac12(u+v)t$ . $v$ – $t$ gradient = $a$ ; area = $s$ . $g=9.81$ m s $^{-2}$ .

1. Derivation

From the v–t graph: gradient a, area s

t (s) = 0.5

The trapezium area gives

s=\tfrac12(u+v)t

; substituting

v=u+at

gives

s=ut+\tfrac12at^2

; eliminating

t

gives

v^2=u^2+2as

From the straight-line v–t graph

$a=\frac{v-u}t\Rightarrow v=u+at$
Gradient.
$s=\tfrac12(u+v)t$
Trapezium area.
$s=\tfrac12(2u+at)t=ut+\tfrac12at^2$
Substitute $v$ .
$s=\frac{(u+v)}2\cdot\frac{(v-u)}a=\frac{v^2-u^2}{2a}$
Substitute $t$ ; difference of squares.
$v^2=u^2+2as$
Rearrange.

2. Examples

Worked exampleThrown down from 45 m

$u=5.0$ m s $^{-1}$ downward from 45 m.

Goal: Time and impact speed.
Strategy: Down positive; quadratic in $t$ .

$45=5t+4.905t^2$
$s=ut+\frac12at^2$ .
$t=\frac{-5+\sqrt{25+882.9}}{9.81}=2.56\text{ s}$
Positive root.
$v=5+9.81(2.56)=30.1\text{ m s}^{-1}$
$v=u+at$ .

Answer

2.56

30.1

m s

^{-1}

✓ Check

v^2=25+2(9.81)(45)=908

v=30.1

✓.

3. Measuring g

Measuring g: free fall timed with light gates

h=\tfrac12gt^2

from rest, so

h

against

t^2

is a straight line with gradient

\tfrac g2

. Gradient

4.9

gives

g=9.8

m s

^{-2}

Linearising the free-fall equation

$h=\tfrac12gt^2$
From rest.
$\underbrace{h}_{y}=\underbrace{\tfrac g2}_{m}\underbrace{t^2}_{x}$
Compare with $y=mx$ .

Plot

h

against

t^2

: gradient

\frac g2

. Light gates remove reaction-time error.

Common misconceptionMixed sign conventions Pick a positive direction once;

a=-9.81

if up is positive.

Pause and think

On a curved $v$ – $t$ graph, how do you find instantaneous acceleration?

Summary

Four suvat equations; uniform $a$ only.
Gradient $a$ , area $s$ .

Displacement: Vector distance from the start.

HistoryGalileo's odd numbers. Galileo found that a ball rolling from rest down a ramp covers distances in successive equal times in the ratio

1:3:5:7\ldots

. Adding them gives

1,4,9,16

: distance

\propto t^2

, exactly what

s=\tfrac12at^2

predicts.

Fun factThe feather and the hammer. In

1971

astronaut David Scott dropped a hammer and a falcon feather together on the Moon. With no air resistance, they hit the ground at exactly the same time. "Mr Galileo was correct," he said.

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