© nuffield foundation 2011 nuffield mathematics activity mean values
TRANSCRIPT
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.