One of the most common examples of a linear relation is the relationship between distance and time when a travelling at a constant speed. For exmaple, if you are travelling 40 km/h then every hour you will travel 40 km.

A distance-time graph shows an object's distance from a fixed point over a period of time.

A rising line shows that the object's distance increases away from the reference point.

A falling line shows that the object's distance decreases towards the reference point.

A flat line shows the object isn't moving!

## Velocity

In distance vs time graphs, the distance should be represented on the vertical axis. Time would be the horizontal axis, and the velocity would be represented by the **slope** of the graph.

In order to calculate the slope, we can use the equation:

\(m = \cfrac{rise}{run} = \cfrac{y_2-y_1}{x_2-x_1} = \cfrac{d_2-d_1}{t_2-t_1} = v\)

From this equation we can see that the speed is distance over time:

\( v = \cfrac{d}{t} \)

## D vs T Graphs

To better demonstrate how to calculate rate of change on a distance-time graph refer to Timmy's shopping trip below.

Try and calculate the speed at each segment of Timm'y trip, when was he moving the fastest?

Show Answer
\( m_{AB} = \cfrac{y_2-y_1}{x_2-x_1}=\cfrac{200-0}{5-0}= 40 \; m/min\)

\( m_{BC} = \cfrac{1000-200}{10-5}= 160 \; m/min\)

\( m_{CD} = \cfrac{1400-1000}{15-10}= 80 \; m/min\)

\( m_{DE} = \cfrac{1400-1400}{20-15}= 0 \; m/min\)

\( m_{EF} = \cfrac{0-1400}{25-20}=-280 = 280 \; m/min\)

Timmy was moving the fastest during the last 5 min of his trip, he was likely sprinting back home. Timmy stopped during the 15th to 20th min of his trip, he likely reached the store and is stationary inside.