A distance–time graph plots distance from the start (vertical axis) against time (horizontal axis) — turning a journey into a shape you can read at a glance.

 Distance
    ^
    │      ┌───────┐
    │     /         \
    │    /           \
    │   /             \
    └──┴───────────────┴──► Time
      moving   stationary  returning

• A flat section means the object is stationary — time is passing, but no distance is being covered.
• A line sloping back down towards zero means the object is returning to its starting point.
• The steeper a section of the graph, the faster the object is travelling during that part of the journey — a nearly vertical line represents a very high speed, and a nearly horizontal line represents a very low one.

The gradient of a distance–time graph is the speed:

A cyclist travels $12\text{ km}$ in the first $40$ minutes. Find her speed in km/h.

1. Convert $40$ minutes to hours: $\dfrac{40}{60} = \dfrac{2}{3}\text{ h}$.
2. Divide distance by time: $\dfrac{12}{2/3} = 12 \times \dfrac{3}{2} = 18\text{ km/h}$.

A conversion graph is usually a straight line through the origin. Find the exchange rate as (target unit) $\div$ (starting unit), then multiply by whatever value you're converting.

A conversion graph shows \50 = £40$. Find how many pounds$\$135$ is worth.

1. Rate: £40 \div \50 = £0.8$ per dollar.
2. Multiply: $135 \times 0.8 = £108$.

Plot each given pair of values as a coordinate point, then join them — with a ruled straight line if the data is linear, or a smooth curve otherwise. Choose scales on each axis that use as much of the given grid as possible, since a graph squeezed into one corner is much harder to read values from accurately.

A car's fuel tank contains $50$ litres at the start of a trip, and empties at a steady rate of $8$ litres per hour. To draw this graph, calculate the fuel remaining at a few convenient times (e.g. $t=0,\ 2,\ 4,\ 6$ hours, giving $50, 34, 18, 2$ litres), plot these four points, then rule a single straight line through all of them, continuing it until it reaches zero.

A cyclist's journey: $12\text{ km}$ in $40$ minutes, rests for $20$ minutes, then returns home, covering the $12\text{ km}$ in $30$ minutes. Find her speed on the return journey, in km/h, and describe how the return section of the graph would compare to the outward section.

Solution

1. Convert $30$ minutes to hours: $\dfrac{30}{60}=0.5\text{ h}$.
2. Apply the speed formula:

The return speed ($24$ km/h) is faster than the outward speed ($18$ km/h) — on the graph, this shows up as the return section being steeper than the outward section, since a larger speed corresponds to a larger gradient over the same $12\text{ km}$ distance.