Area Under a Speed-Time Graph

Find the distance travelled from the area under a speed-time graph, work out average speed, and find a missing time or speed from a given distance.

In the exam for Cambridge IGCSE Maths 0580, IGCSE Additional Maths 0606, Cambridge O Level Maths 4024 · 19 real past-paper questions below

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Distance Is the Area

On a speed–time graph, speed goes up the side and time goes along the bottom.

The space between the line and the time axis is the distance travelled. Steady speed makes a rectangle: $20 m/s \times 10 s = 200 m$ .

Shape under the line	Area
Rectangle (steady speed)	base $\times$ height
Triangle (from rest or to rest)	$\frac{1}{2} \times$ base $\times$ height
Trapezium (speed changes)	$\frac{1}{2} (a + b) \times h$

Rule

The area under the line is the distance travelled. The slope gives acceleration: that is in Speed-Time Graphs and Acceleration.

Key idea

In the 1300s, the French scholar Nicole Oresme drew speed against time and found the distance travelled as the area of a trapezium. It is one of the first known graphs of a real-life idea, made long before Galileo studied motion.

Total Distance: Split Into Shapes

The most common question: find the total distance for the whole journey. Split the area into triangles and rectangles, then add.

A car speeds up from rest to 10 m/s in 4 seconds, keeps 10 m/s until 10 seconds, then slows to rest at 12 seconds.

Worked example

1
$\frac{1}{2} \times 4 \times 10 = 20$ — From 0 to 4 s the speed rises from rest: a triangle. Its width is 4 s and its height is 10 m/s.
2
$10 - 4 = 6 s$ — The steady part runs from 4 s to 10 s. Its width is the end time minus the start time.
3
$6 \times 10 = 60$ — Steady speed makes a rectangle: width times height.
4
$12 - 10 = 2 s$ — The slowing part runs from 10 s to 12 s, so its width is 2 s.
5
$\frac{1}{2} \times 2 \times 10 = 10$ — Slowing to rest is another triangle.
6
$20 + 60 + 10 = 90 m$ — The total distance is the sum of the three areas.

Common mistake

Using the time reading as the width. The width is the time the part lasts: 10 − 4, not 10.

Key idea

The whole shape is also one trapezium: $\frac{1}{2} (12 + 6) \times 10 = 90$ . Same answer, a good check.

A Trapezium Under the Line

The 2-mark question: the speed changes but does not start or end at zero. The shape under that part is a trapezium.

Formula

distance = \frac{1}{2} (a + b) \times h

$a$ and $b$ are the two speeds (the parallel sides). $h$ is the time taken.

A bike goes from 6 m/s to 14 m/s in 5 seconds.

Worked example

1
$6 + 14 = 20$ — Add the two speeds, the parallel sides of the trapezium.
2
$\frac{1}{2} \times 20 = 10 m/s$ — Half of the sum is the middle speed of this part. It is the steady speed that would cover the same distance.
3
$10 \times 5 = 50 m$ — Distance is that speed times the time taken, 5 seconds.

Common mistake

Forgetting the half gives 100 metres, double the real answer.

Top Speed From the Acceleration

Sometimes the graph gives the acceleration, not the top speed. Work out the top speed first, then the areas. A car accelerates from rest at 2.5 m/s² for 6 s, then travels at a steady speed for 8 s.

Worked example

1
$2.5 \times 6 = 15 m/s$ — Acceleration is the speed gained each second. Over 6 seconds the speed gained is 2.5 × 6, and it started from rest, so this is the top speed.
2
$\frac{1}{2} \times 6 \times 15 = 45$ — Speeding up from rest is a triangle: width 6 s, height 15 m/s.
3
$8 \times 15 = 120$ — The steady part is a rectangle. Its width is the 8 s given, and its height is the top speed.
4
$45 + 120 = 165 m$ — Add the two areas for the total distance.

Common mistake

Using the acceleration as a height. 2.5 is metres per second per second, not a speed. Multiply it by the time first.

Units: km/h and Minutes

When the speed is in km/h but the time is in minutes, change the minutes to hours before you find any area. A graph shows speed rising from 40 km/h to 70 km/h in the first 12 minutes, then falling to 0 at 30 minutes.

Worked example

1
$12 \div 60 = 0.2 h$ — Speed is per hour, so the width must be in hours. 60 minutes make 1 hour.
2
$\frac{1}{2} (40 + 70) \times 0.2 = 11 km$ — The first part is a trapezium with parallel sides 40 and 70.
3
$30 - 12 = 18 min$ — The falling part runs from 12 min to 30 min.
4
$18 \div 60 = 0.3 h$ — Change this width into hours too.
5
$\frac{1}{2} \times 0.3 \times 70 = 10.5 km$ — Slowing to rest is a triangle with height 70 km/h.
6
$11 + 10.5 = 21.5 km$ — Add the areas. The distance is in km, because the speed is in km/h and the time in hours.

Common mistake

Using 12 and 18 minutes as the widths. The areas come out 60 times too big, because the speed is per hour.

Average Speed

The 2-mark next question: find the average speed for the whole journey.

Formula

average speed = \frac{total distance}{total time}

Worked example

1
$total distance = 90 m$ — This is the sum of the three areas in the earlier example.
2
$total time = 12 s$ — Read the time at the right-hand end of the graph. The journey started at 0.
3
$90 \div 12 = 7.5 m/s$ — Average speed is total distance divided by total time.

Common mistake

Adding the top and bottom speeds and halving. Always use the area, then divide by the time.

Finding a Missing Time or Speed

The 3–4 mark question: you are told the distance, and one time or speed is a letter. Write each area with the letter, add, and set it equal to the distance.

A car speeds up from rest to 8 m/s in 5 seconds. It then keeps 8 m/s until time $T$ . It travels 100 metres. Find $T$ .

Worked example

1
$\frac{1}{2} \times 5 \times 8 = 20$ — Speeding up from rest is a triangle: width 5 s, height 8 m/s.
2
$8 \times (T - 5)$ — The steady part is a rectangle. It starts at 5 s and ends at T, so its width is T − 5.
3
$20 + 8 (T - 5) = 100$ — The areas add up to the distance given.
4
$8 (T - 5) = 100 - 20 = 80$ — Take 20 away from both sides to isolate the rectangle.
5
$T - 5 = 80 \div 8 = 10$ — Divide both sides by 8.
6
$T = 10 + 5 = 15 s$ — Add 5 to both sides. Check: 20 + 8 × 10 = 100.

Common mistake

Writing the rectangle's width as $T$ . The steady part starts at 5 seconds, so its width is $T - 5$ .

Key idea

A missing speed works the same way. Rest to $V$ in 10 seconds, then $V$ for 20 seconds, 500 metres: $5 V + 20 V = 500$ , so $V = 20$ .

Key idea

Show every area and every line of the equation. Most of the marks are for the working, so a guessed answer scores about 1 mark.

Which Method? Read the Question

Look at what the question gives you and what it asks for.

You see	You do
Total distance	split into shapes, add the areas
Acceleration given, no top speed	top speed = acceleration × time, then areas
km/h with minutes	minutes ÷ 60 first, then areas
Speed changes, not from zero	trapezium $\frac{1}{2} (a + b) \times h$
Average speed	total area ÷ total time
Distance given, a letter on the graph	areas with the letter = distance, solve
Acceleration or deceleration	Speed-Time Graphs and Acceleration

Key idea

Check the units match. With km/h and minutes, change the minutes into hours first.

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