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Unit F5 Section 3
Applications of Graphs

In this section some applications of graphs are considered, particularly conversion graphs and graphs to describe motion.

The graph on the right can be used for converting US Dollars into and from Jamaican Dollars. (As currency exchange rates are continually changing, these rates might not be correct now.)

A distance-time graph of a car is shown opposite. The gradient of this graph gives the velocity (speed) of the car. The gradient is steepest from A to B, so this is when the car has the greatest speed. The gradient BC is zero, so the car is not moving.

The area under a velocity-time graph gives the distance travelled. Finding the shaded area on the graph shown opposite would give the distance travelled.

The gradient of this graph gives the acceleration of the car. There is constant acceleration from 0 to 20 seconds, then zero acceleration from 20 to 40 seconds (when the car has constant speed), constant deceleration from 40 to 50 seconds, etc.

Worked Examples

1

A temperature of 20 °C is equivalent to 68 °F and a temperature of 100 °C is equivalent to a temperature of 212 °F. Use this information to draw a conversion graph. Use the graph to convert:

Taking the horizontal axis as temperature in °C and the vertical axis as temperature in °F gives two pairs of coordinates, (20, 68) and (100, 212). These are plotted on a graph and a straight line drawn through the points.

(a)

30 °C to °Fahrenheit ,

Start at 30 °C, then move up to the line and across to the vertical axis, to give a temperature of about 86 °F .
(b)

180 °F to °Celsius.

Start at 180 °F , then move across to the line and down to the horizontal axis, to give a temperature of about 82 °C.
2

The graph shows the distance travelled by a girl on a bike.

Find the speed she is travelling on each stage of the journey.

For AB the gradient =
= 3.75

So the speed is 3.75 m/s.

Note

The units are m/s (metres per second), as m are the units for distance and s the units for time.

For BC the gradient =
= 10

So the speed is 10 m/s.

For CD the gradient is zero and so the speed is zero

For DE the gradient =
= 5

So the speed is 5 m/s.
3

The graph shows how the velocity of a bird varies as it flies between two trees. How far apart are the two trees?

The distance is given by the area under the graph. In order to find this area it has been split into three sections, A, B and C.

Area of A = × 6 × 6
= 18
Area of B = 6 × 6
= 36
Area of C = × 2 × 6
= 6
Total Area = 18 + 36 + 6
= 60

So the trees are 60 m apart. Note that the units are m (metres) because the units of velocity are m/s and the units of time are s (seconds).
4

The graph shows the velocity of a toy train for a time period of 14 seconds.

(a)

State

(i)

the maximum velocity attained by the train

6 m/s (from B to C)
(ii)

the total number of seconds, during which the train travelled at the maximum speed.

2 seconds (as B is reached at 3 seconds and C at 5 seconds)
(iii)

the time period during which the velocity is negative.

Velocity is negative from C (time 5 seconds) to E (time 9 seconds)

i.e. from 9 − 5 = 4 seconds
(b)

Calculate the acceleration of the train.

(i)

during the first 3 seconds

Acceleration is the gradient of AB = = 2 m/s2
(ii)

for the time period of x seconds where 6 ≤ x ≤ 9.

For 6 ≤ x ≤ 9, there is acceleration = = −2 m/s2

( or deceleration of 2 m/s2 )
(c)

Calculate the distance travelled

(i)

between 2 and 6 seconds

Distance travelled from 2 to 6 seconds

= 1 × + 2 × 6 + × 1 × 6

= 5 + 12 + 3

= 20 m
(ii)

over the entire journey of 14 seconds.

Distance travelled from A to D = × 2 × 4 + 20 = 24 m

By symmetry, distance travelled from D to G = 24 m

Distance travelled for G to H = × 2 × 4 = 4 m

So total distance travelled = 24 + 24 + 4 = 52 m
5
(a)

The velocity-time graph above, not drawn to scale, shows that a train stops at two stations, A and D. The train accelerates uniformly from A to B, maintains a constant speed from B to C and decelerates uniformly from C to D.

Using the information on the graph,

(i)

calculate, in ms−2, the train's acceleration

Acceleration = = 1 ms−2
(ii)

show that the train took 30 secs from C to D if it decelerated at ms−2 .

If it takes t seconds from C to D, then

= ⇒ t = 30 seconds
(iii)

If the time taken from A to D is 156 seconds, calculate the distance in metres between the two stations.

Time from B to C = 156 − (15 + 30) = 111 seconds

Total distance = area under graph
= ⎛
⎝ × 15 × 15 + 15 × 111 + × 15 × 30⎞
⎠ m
= ⎛
⎝ + 1665 + 225⎞
⎠ m
= m = 2002 m
(b)

An express train stops at station P and station Q. On leaving P, the train accelerates uniformly at the rate of 2 ms−2 for 60 seconds. Then it decelerates uniformly for 90 seconds and comes to a stop at Q.

(i)

Using 1 unit to represent 20 units on BOTH axes, sketch the velocity-time graph to represent the information above.
(ii)

Using your graph, determine

(a) the maximum velocity

(b) the rate at which the train decelerates.

(a) Maximum velocity = 120 ms−1

(b) Deceleration = = ms−2

Exercises

Use the approximation that 10 kg is about the same as 22 lbs to draw a graph for converting between pounds and kilograms. Use the graph to convert the following (to 1 d.p.):

(a)
6 lbs to kilograms, kg
(b)
8 lbs to kilograms, kg
(c)
5 kg to pounds, lbs
(d)
3 kg to pounds. lbs

Use the approximation that 10 gallons is about the same as 45 litres to draw a conversion graph. Use the graph to convert (to 1 d.p.):

(a)
5 gallons to litres, litres
(b)
30 litres to gallons. gallons

The graph shows how the distance travelled by a taxi increased.

(a)

How many times did the taxi stop?

(b)

Find the velocity of the taxi on each section of the journey.

AB: m/s

CD: m/s

EF: m/s

GH: m/s

(c)

On which part of the journey did the taxi travel fastest?

The graph below shows how the speed of an athlete varies during a race.

What was the distance of the race?

metres

Kevin runs at a constant speed for 10 seconds. He has then travelled 70 m. He then walks at a constant speed for 8 seconds until he is 86 m from his starting point.

(a)

Find the speed at which he runs and the speed at which he walks.

He runs at m/s and walks at m/s.
(b)

If he had covered the complete distance in the same time, with a constant speed, what would that speed have been?

m/s (to 2 d.p.)

Find the area under each graph below and state the distance that it represents.

(a)
miles
(b)
miles
(c)
mm
(d)
m

The speed-time graph below shows the journey of a car from 8:00 am to 11:00 am.

Using the graph, determine

(a)

the time at which the speed of the car was 40 km/h

a.m.
(b)

the distance the car travelled for the entire journey

km
(c)

the average speed of the car for the entire journey.

km/h (to 2 d.p.)