© nuffield foundation 2011 nuffield mathematics activity mean values

9
© Nuffield Foundation 2011 Nuffield Mathematics Activity Mean values

Upload: amelia-simon

Post on 23-Dec-2015

224 views

Category:

Documents


1 download

TRANSCRIPT

© Nuffield Foundation 2011

Nuffield Mathematics Activity

Mean values

The speed of a car varies with time.

How can you work out the average speed of the car?Can you work it out from a speed-time graph?

Speed–time graph

Think aboutCan you describe what happens?

Can you find the average speed?

02468

101214

0 1 2 3 4 5 6 7 8 9 10

Spee

d (m

/s)

Time (s)

Car journey along a road in town

This activity is about finding mean values.

Mean speed

Car travelling at 15 m s–1

Area =

The distance travelled in 4 seconds is 60 metres

4 × 15 = 60

v m s–1

t seconds0 4

15

taken time Totaltravelled distance Total

where distance travelled = area under the speed–time graph

t seconds

v m s–1

0 2 4 6 8 10

2

4

6

8

10

12

14

Car travelling along road in town

Think about…The areas under this graph are triangles and trapezia. Can you recall the formulae for these areas?

b

h

Area of a triangle = 2

hb

A B C D E

t seconds

v m s–1

0 2 4 6 8 10

2

4

6

8

10

12

14Area of A = 2

82.8

Area of C =

22.713.78

= 29.295

Area of B = 0.68 = 4.8

= 11.2

Area of E = = 11.135

Area of D = = 26.8

213.11.7

Car travelling along road in town

= 83.23Total area

Mean speed taken time Totaltravelled distance Total

9.883.23 = 8.5 m s–1 (to 1dp)

2

213.113.7

Example

10

040.0130.1 dtttArea =

10

4

40.1

0

t5

50.01t

100 0.025t 4 – 0.002t 5

– 200 = 250

Distance = 50 metres

Mean speed

Integration allows you to find the area under a curve.

The speed of a cyclist along a road can be modelled by the function v = 0.1t 3 – 0.01t 4

1050 = 5 m s–1

v

t0

v = 0.1t 3- 0.01t 4

10

Using Integration

6.1-8.113.7-13.1

Integration gives the area under a graph ba dxyA =

(6.1, 13.7) (8.1, 13.1)v

t0

D

For section D:

Gradient, m = –0.3

Using (6.1, 13.7) in y = mx + c

13.7 = - 0.3 × 6.1 + c c = 15.53

Distance =

8.16.1 0.3-15.53 dtt

8.1

6.1

20.315.53

2

tt

= 115.9515 – 89.1515 = 26.8 For a linear graph this method takes longer.

Reflect on your work

Can you summarise the method for finding the mean value of a quantity from its graph?

In cases where the graph consists of straight lines, you can use either geometrical formulae to find the area or integration.

Which of these methods do you prefer?Why?

Mean values