# And today's question.

## How do we **describe **and **quantify **movement?

# What we already know

## SPEED = DISTANCE/TIME

## A normal journey.

From Sevilla to Huelva.

Distance is 92.9 km

Google maps says it should take 1 hour (without traffic)

Calculate the speed...

# SPEED = DISTANCE/TIME

92.9/1

speed is 92.9 km per hour.

## What information does this tell us?

- Not very much really.

Text

Did the car stop for petrol?

What was the fastest speed of the car?

What was the slowest speed of the car?

We know the car did not start at 92.9 km/h and then drive at a constant speed of 92.9 km/h until it reached Huelva then just stop.... but that seems to be what that information is telling us.

# Or if the police stopped the car, and it was travelling at 130 km/h, can you defend yourself by saying "but my average speed was 92 km/h"

## Car number 1

## Car number 2

if the speed limit is the same for both, which can drive the fastest **legally**?

But which one gets **faster**, * faster*?

a =\frac{v - u}{t}

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

Acceleration is equal to the change in velocity divided by the time taken.

or

Acceleration = (final velocity - initial velocity)/time taken

a = \frac{v-u}{t}

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

at=v-u

$at=v-u$

v = u+at

$v = u+at$

(ii)

v_{average}=\frac{v+u}{2}

$v_{average}=\frac{v+u}{2}$

s=v_{average}. t

$s=v_{average}. t$

s=\frac{v+u}{2}.t

$s=\frac{v+u}{2}.t$

s=\frac{v+u}{2}.t

$s=\frac{v+u}{2}.t$

s=\frac{u+at+u}{2}.t

$s=\frac{u+at+u}{2}.t$

s=\frac{2u+at}{2}.t

$s=\frac{2u+at}{2}.t$

s=ut +\frac{at^2}{2}

$s=ut +\frac{at^2}{2}$

# Your TWO equations to learn are then...

a = \frac{v-u}{t}

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

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

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

**Where:**

### = *initial velocity*

*= final velocity*

*= displacement*

*= time*

*= acceleration*

u

$u$

v

$v$

s

$s$

t

$t$

a

$a$