The formulae derived from the velocity-time graph: $v=u+at$ and $s=\bigg(\dfrac{u+v}{2}\bigg)t$ , can be used to create three more formulae. These formulae are essential for solving problems about objects moving in a straight line with constant acceleration. There are five formulae in total.

Equations

description	Equation
Velocity of an object with constant acceleration.	$v=u+at$
Displacement of an object over a certain time.	$s=\bigg(\dfrac{u+v}{2}\bigg)t$
Velocity of an object with constant acceleration.	$v^2 = u^2+2as$
Displacement of an object with constant acceleration using initial velocity.	$s=ut+\dfrac{1}{2}at^2$
Displacement of an object with constant acceleration using final velocity.	$s=vt-\dfrac{1}{2}at^2$

Variable definitions

QUANTITY NAME	SYMBOL	UNIT NAME	UNIT
$Displacement$	$s$	$Metre$	$m$
$Initial\ velocity$	$u$	$Metres\ per\ second$	$ms^{-1}$
$Final\ velocity$	$v$	$Metres\ per\ second$	$ms^{-1}$
$Acceleration$	$a$	$Metres\ per\ second\ squared$	$ms^{-2}$
$Time$	$t$	$Seconds$	$s$

Deriving formulae

To derive the new formulae, you can eliminate variables from the previously derived formulae.

First eliminate $t$ .

To do this, first identify a formula to use:

$v=u+at$

Rearrange to make $t$ the subject:

$t=\dfrac{v-u}{a}$

Now bring in the other derived formula:

$s=\bigg(\dfrac{u+v}{2}\bigg)t$

Substitute $t$ with the rearranged expression:

$s=\bigg(\dfrac{u+v}{2}\bigg)\bigg(\dfrac{v-u}{a}\bigg)=\dfrac{(u+v)(v-u)}{2a}$

Multiply both sides by $2a$ :

$2as=(u+v)(v-u)$

Expand the brackets:

$2as=v^2-u^2$

Rearrange to make $v^2$ the subject:

$\boxed{v^2=u^2+2as}$

Next, eliminate $v$ .

Using a previously derived formula:

$v=u+at$

Substitute this new expression for $v$ into the formula for displacement and simplify:

$s=(\dfrac{u+v}{2})t$

$s=(\dfrac{u+u+at}{2})t$

$s=(\dfrac{2u}{2}+\dfrac{at}{2})t$

$s=(u+\dfrac{1}{2}at)t$

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

Finally, eliminate $u$ .

Use a previously derived formula and make $u$ the subject:

$v=u+at,\ therefore\ u=v-at$

Substitute this into the new displacement equation and simplify:

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

$s=(v-at)t+\dfrac{1}{2}at^2$

$s=vt-at^2+\dfrac{1}{2}at^2$

$\boxed{s=vt-\dfrac{1}{2}at^2}$

SUVAT

These five formulae make up the 'SUVAT' formulae. These formulae are known as constant acceleration formulae. The formulae are:

v=u+at

s=\bigg(\dfrac{u+v}{2}\bigg)t

v^2 = u^2 +2as

s=ut+\dfrac{1}{2}at^2

s=vt-\dfrac{1}{2}at^2

You should be able to derive and use each of these formulae for problems about objects moving in a straight line with constant acceleration.

Example

A particle moves in a straight line from A to B with a constant acceleration of $4ms^{-2}$ . The velocity of the particle at point A is $2ms^{-1}$ and the velocity at B is $20ms^{-1}$ .

a: Find the distance from A to B.

b: Find the time it takes the particle to go from A to B.

First, identify the variables which are given: $u=2ms^{-1},\ v=20ms^{-1},\ a=4ms^{-2}$ .

a: The distance from A to B is also the displacement. The question does not give a value for $t$ , therefore use a formula which does not include it:

$v^2=u^2+2as$ 

Substitute known values:

$20^2 = 2^2+2(4)s$ 

$400=4+8s$ 

Rearrange to make $s$ the subject and solve:

$8s=396$

$s=\dfrac{396}{8}=49.5$ 

The distance from $A$ to $B$ is $\underline{49.5\ m}$ .

b: To avoid any potential errors in the working out for part a, use the equation which does not involve $s$ :

$v=u+at$ 

Substitute known values:

$20=2+4t$ 

Rearrange to make $t$ the subject and solve:

$4t=18$

$t=\dfrac{18}{4} = 4.5$ 

The particle takes $\underline{ 4.5\ s}$ to travel from $A$ to $B$ .

Learn with Basics

Length:

Unit 1

Speed, velocity and acceleration

Unit 2

Acceleration and velocity-time graphs

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Optional

Unit 3