New QMaths 11B CD-ROM
Extra material
Area under a curve
If the speed of an object is constant, we know that
Speed = ⇒ Distance = speed × time
We can calculate distance travelled by multiplying the speed by the time. On a graph, this will be shown as the area under the line of the graph.
distance time
---A boat travels at a steady speed of 8 km/h. Draw a graph of speed against time. Show the distance travelled between 2 hours and 5 hours on the graph and find the distance travelled in this time.
Solution
Draw vertical (speed) and horizontal (time) axes. Since the boat travels at constant speed, the graph is a straight line parallel to the time axis.
Draw vertical lines from t= 2 and t= 5 to intersect the horizontal line.
Shade in the region bounded by the vertical lines, the horizontal line and the time axis.
The shaded region represents the distance travelled. Area= 3 × 8 = 24 km
Write the formula to calculate the distance. Distance= speed × time
Time from 2 to 5 hours = 3 hours. = 8 × 3
= 24 km
10 8 6 4 2 0 Speed (km/h)
2 4 6 8 9
1 3 5 7
Time (hours)
Example 1
A car travels from 10 am to 1 pm, with the driver steadily increasing the speed from 40 km/h at 10 am to 70 km/h at 1 pm. Use a graph to find the distance travelled.
Solution
Draw vertical (speed) and horizontal (time) axes.
We can draw a rectangle for each hour, or for each half-hour, as shown below.
For the one-hour intervals
If the driver increases speed steadily, then the average speed in the first hour will be about 45 km/h. In the second hour the average speed will be about 55 km/h, and in the third hour it will be about 65 km/h. The areas under each part can be added to give the total area, which is the total distance travelled. The width of each column is 1 h, so the total area is
A= 1 × 45 + 1 × 55 + 1 × 65 = 165 km
60 50 30 20 10 0 Speed (km/h) Time
12 noon 2 pm 10 am 70 40 60 50 30 20 10 0 Speed (km/h) Time
12 noon 2 pm 10 am
70
40
Example 2
The change in any quantity can be calculated (or estimated) using the area under a graph of the rate of change of that quantity. This applies to graphs of both simple and complicated shapes. With complicated shapes, such as shapes with curves, we can always get an approximate answer using rectangles. Smaller intervals will give more accurate results, and the exact area will give the exact answer.
For the half-hour intervals
The average speed in the first half-hour will be about 42.5 km/h. In the second half-hour it will be about 47.5 km/h, and so on. The width of each column is 0.5 h, so the total area is
A= 0.5 × 42.5 + 0.5 × 47.5 + … = 165 km
For smaller and smaller intervals
It would seem more accurate to reduce the time intervals to smaller amounts so that the average speed for each interval more closely reflects the steady increase. If this process is continued indefinitely, we just get the trapezium under the line.
The area of the trapezium is
A = × 3 × (40 + 70)
= 165 km
Write the result. The distance travelled is 165 km.
60 50
30 20 10 0
Speed (km/h)
Time
12 noon 2 pm 10 am
70
40
A
1 2
---The operator of a large dam controls the rate at which water is released. Between 9 am and 11 am, 30 ML/h is allowed out. From 11 am to 1 pm the flow rate is gradually increased to 70 ML/h. It is left at this rate until 4 pm, then reduced immediately to 40 ML/h. It stays at this rate until 6 pm. Use the area under the graph to find the amount of water released from the dam from 9 am to 6 pm.
Solution
Draw vertical (flow rate) and horizontal (time) axes. Transfer the information provided onto the graph as points.
Join the points with straight lines.
The area under the graph consists of three rectangles and a trapezium.
Mark the areas A1, A2, A3 and A4.
Calculate the total area by summing. Area=A1+A2+A3+A4
= 2 × 30 + + 3 × 70 + 2 × 40
= 60 + 100 + 210 + 80
50 40
20 10 0
Flo
w rate (ML/h)
Time 2 pm 60
30 70
noon
10 am 4 pm 6 pm
A1 A2 A3 A4
2×(30+ 70)
2
New QMaths 11B CD-ROM
Area under a curve continued
Exercise 1
Area under a curve
1 Between 9 am and 12 noon a car travels at a speed of 50 km/h, then it increases its speed to 65 km/h until
2 pm. Draw a graph and use the area to find the distance travelled.
2 A train travelling between suburban stations increases its speed from 0 to 17 m/s (60 km/h) in 12 s, travels
at this speed for 1 minute 30 s and then slows down to stop at the next station. The trip between the stations takes 2 minutes. Use a graph to find the distance between the stations.
3 A town reservoir is filled from several dams. The water is pumped at different rates from the dams.
Between 11 am and 1 pm the flow rate is 50 kL/h, from 1 pm to 4 pm it is 70 kL/h, but then it gradually decreases to 40 kL/h at 8 pm. Use a graph to find the total amount of water pumped in this time.
4 The power drawn by a water heater measures the rate at which energy is converted to heat. A water heater
has a maximum power rating of 3600 W (watts). It works at maximum power for 2 minutes, then it gradually decreases to 1200 W over the next minute. It then cuts out altogether for 2 minutes. When it cuts back in, it delivers the 2400 W for a further 3 minutes.
Given that 1 watt = 1 joule/second, use a graph to work out how much energy in kilojoules (kJ) has been
used to heat the water.
5 A lightning machine can build a voltage of 400 000 V before it discharges. As the charge is released, the
voltage drops uniformly to 0. The stored charge is about 3 C (coulombs). The voltage actually measures
the rate at which energy is released as the charge flows, and 1 volt = 1 joule/coulomb. Use a graph with
voltage on the vertical axis and charge on the horizontal axis to find the energy released at discharge.
6 A ball rolling down a slope accelerates from rest at 2 m/s2. This means that the slope of the velocity–time
graph is 2. Use a graph to find the distance the ball travels in 4 s. The graph shows the speed of an object dropped from a
plane. It was released at time t= 2 s. Estimate the distance it
fell in the next 12 s.
Solution
The distance can be estimated by counting squares under the graph.
Calculate the area of each square. Area of square= 2 s × 20 m/s
= 40 m
Count the squares between 2 s and 14 s. Total squares≈ 12
Calculate the total area. Total area≈ 40 × 12 = 500 m
Write the result. The object fell about 500 m in the next 12 s.
100 80
40 20 0
Speed (m/s)
Time (s) 8 12 60
4 16
2 6 10 14 18
1 2
---1 2
---Example 4
7 The graph shows the rate at which water flows from a leak in a storage dam. As the level drops, the rate of flow decreases. Use the graph to find the total amount of water that leaks out in the first 4 hours.
8 The graph shows the acceleration (rate of change of
velocity) of a bullet as it is fired from the barrel of a rifle. Use the graph to calculate the speed at which it leaves
the rifle (the muzzle velocity).
50 40
20 10 0
W
ater loss (kL/h)
Time (hours)
4 6
30
2
1 3 5
50 000 40 000
20 000 10 000 0
Acceleration (m/s
2)
Time (ms)
2 3
30 000
1 4 5
New QMaths 11B CD-ROM
Area under a curve continued
Exercise
Answers
1 265 km
2 1785 m
3 530 kL
4 1008 kJ
5 600 kJ
6 16 m
7 About 115 kL
8 About 220 m/s
