The vertical axis is distance from the start, the horizontal axis is time, and — crucially — the gradient of the line is the speed:

A steeper line means a faster speed. This is different from a speed-time graph (where the gradient would be acceleration instead) — don't confuse the two.

• Sloping upward (positive gradient): moving away from the start at constant speed — steeper = faster.
• Flat/horizontal (zero gradient): stationary. The object isn't moving, but time is still passing — this segment still counts towards the total journey time.
• Sloping downward (negative gradient): returning towards the start. The gradient calculates as negative, but speed itself is always reported as positive — it's the magnitude of the rate, not the direction, that's being asked for.

A curved line means the speed itself is changing (acceleration or deceleration), but Foundation Tier questions are almost always built from straight-line segments.

1. Identify the start and end coordinates of that segment, $(t_1,d_1)$ and $(t_2,d_2)$.
2. Find the change in distance, $\Delta D = d_2-d_1$ (always use the positive distance covered, even on a downward-sloping "return" segment).
3. Find the change in time, $\Delta T = t_2-t_1$.
4. $\text{Speed}=\dfrac{\Delta D}{\Delta T}$.

Using the full length of a segment (a large "gradient triangle") rather than two nearby points reduces reading errors and makes your method clearer to an examiner.

The "total time" must include every stationary (flat) period — this is the single most common mistake in this topic. Averaging the individual segment speeds is not valid; you must use total distance over total time for the entire trip.

Because an hour has 60 minutes, not 100, you cannot treat "2 hours 40 minutes" as the decimal $2.40$. Convert the minutes separately: divide by 60.

Skipping this conversion and dividing by $2.40$ instead of $2.667$ is the classic "decimal time trap," and it silently corrupts every speed calculation built on it.

Break the narrative into phases, and find the start/end coordinates of each one before drawing anything:

"Neil left home at 12:00, travelled 44 miles in 90 minutes, stopped for 3 hours, then drove home in 2 hours."

• Phase 1: $(12{:}00, 0) \to (13{:}30, 44)$ — upward line.
• Phase 2 (rest): $(13{:}30,44)\to(16{:}30,44)$ — horizontal line.
• Phase 3 (return): $(16{:}30,44)\to(18{:}30,0)$ — downward line, ending back at distance $0$.

A cyclist's journey is shown on a distance-time graph: they travel 20 km in the first 1 hour 15 minutes, rest for 45 minutes, then return the 20 km home in 1 hour. Find their average speed for the whole journey.

1. Convert all times to hours: $1\text{ h }15\text{ min}=1.25\text{ h}$, rest $=0.75\text{ h}$, return $=1\text{ h}$.
2. Total distance $=20+20=40$ km (there-and-back, both legs count).
3. Total time $=1.25+0.75+1=3$ hours (rest period included).
4. Average speed $=40 \div 3 = 13.3$ km/h (3 s.f.).