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CHAPTER 14 GRADIENT AND AREA UNDER A GRAPH
Speed-time graph Distance-time graph
gradient represent rate of change of
speed the area under graph represent the
distance traveled
A
Gradient - represents speed
Distance traveled vertical axis
Area = ab Area =2
1ab Area =
2
1a(b+c)
14.1 Speed-Time Graph
A. Area
EXAMPLE EXERCISE (A) EXERCISE (B)
1.
Area = 342
1xx
= 6
1.
Gradient And Area Under A Graph 1
3
4
12
68
10
decelerationacceleration
constant/uniform speed
speed
0 time
Area =
distance travelled
distance
stationary
speed
time0
b
a
bb
a
c
a
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2.
Area = 12 x 4
= 48
3.
Area = 2
1(6 + 10) x 4
= 32
4.
Gradient And Area Under A Graph 2
12
4
9
7
8
13
5
8
12
16
10
8
8
6
1210
4
6
2 6
4
10
8
6
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B. Area Under Graph
Exercise 1
Example 2
Gradient And Area Under A Graph 3
Example 1
(i) Distance during the first 2 seconds= area A
=2
1x 2x 5 = 5
(ii) Distance during the last 4 seconds
= area B
= 4 x 5
= 20
(iii) Total distance during 6 seconds
= area A + area B
= 5 + 20
= 25
i) Distance during the first 4 seconds
ii) Distance during the last 6 seconds
iii) Total distance during 10 seconds
i) Distance in the first 2s= area A
=2
1x (3 + 6) x 2
= 9
ii) Distance in the last 6 s
= area B
= 6 x 3
= 18
iii) Total distance in 8 s
= area A + area B
= 9 + 18
=
speed
A B
5
02 6 Time
A B
5
04 10 Time
speed
0 2 8
3
6
Time
Speed
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Exercise 2
Example 3
Exercise 3
Gradient And Area Under A Graph 4
i) Distance during the first 3 seconds
ii) Distance during the last 6 seconds
iii) Total distance during 9 seconds
i) Distance during the first 2 seconds
= area A
=2
1x 2 x 5 = 5
ii) Distance during the last 2 seconds
= area C
=2
1x 2 x 5 = 5
iii) Distance during constant speed= area B
= 4 x 5 = 20
iv) Total distance
= area A + area B + area C= 5 + 20 + 5
= 30
i) Distance during the first 3 seconds
ii) Distance during the last 2 seconds
iii) Distance during constant speed
iv) Total distance
speed
0 3 9
speed
8
4
time
s
5
7
time
s
A C
B
5
2 6 80
speed
8
4
time
s
6
0 3 7 9
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Example 4
Exercise 4
Example 5
Gradient And Area Under A Graph 5
i) Distance during the first 5 seconds
= area A
=2
1x (2 + 4) x 5 = 15
ii) Distance during the last 7 seconds
= area B
=2
1x 4 x 7 = 14
iii) Total distance
= area A + area B
= 15 + 14
= 29
i) Distance during the first 10 seconds
ii) Distance during the last 4 seconds
iii) Total distance
i) Distance during the first 5 seconds
= area A
=21 x 5 x 6 = 15
ii) Distance during the last 2 seconds
= area C
=2
1x (10 + 6) x 2 = 16
iii) Total distance
= area A + area B + area C
= 15 + 30 + 16
= 61
2
4
5 12
2 BA
0
speed
Time
7
9
10 14
2
speed
0
Time
6
10
5 12
2BA C
100
speed
Time
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Exercise 5
Example 6
Exercise 6
Gradient And Area Under A Graph 6
i) Distance during the first 2 seconds
ii) Distance during the last 3 seconds
iii) Total distance
i) Distance in the first 2 seconds
= area A =2
1(2 + 5) x 2 = 7
ii) Distance in the last 6 seconds
= area C =2
1(5 + 8) x 6 = 39
iii) Total distance during 12 seconds
= area A + area B + area C= 7 + 20 + 39 = 66
i) Distance in the first 2 seconds
ii) Distance in the last 5 seconds
iii) Total distance during 10 seconds
speed
12
4
02 7 10
Time
8
5
2 A B C
0 2 6 12
speed
Time
7
4
2 A B C
0 2 5 10
speed
Time
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C. Rate Of Change Of Speed
takentime
speedinitial-speedfinal=speedofchangeofRate
Gradient And Area Under A Graph 7
+ve accelerationve - deceleration
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Gradient And Area Under A Graph 8
Example 1
Exercise 1i) Find the rate of change of speed in the first 3
seconds.
ii) Find the rate of change of speed in last 4
seconds.
Example 2
Exercise 2
Example 3
Find the rate of change of speed in the first 5
seconds
Final speed = 4
Initial speed = 0
Rate of change of speed = 5
04
= 5
4
@ 0.8
i) Find the rate of change of speed in the first 10
secondsFinal speed = 9
Initial speed = 7
Rate of change of speed = 2
79
= 1
ii) Find the rate of change of speed in the last 4
seconds
Final speed = 0
Initial speed = 9
Rate of change of speed =4
90=
4
9
i) Find the acceleration in the first 6 seconds
ii) Find the deceleration in the last 2 seconds
i) Find the rate of change of speed in the first 4
seconds
Final speed =
Initial speed =
Rate of change of speed =
ii) Find the rate of change of speed in the last 2
seconds
i) Find the rate of change of speed in the first 3
seconds
Final speed = 3
Initial speed = 6
Rate of change of speed =3
63=
3
3=-1
ii) Find the rate of change of speed in the last 2
secondsFinal speed = 0
Initial speed = 8
Rate of change of speed =2
80=
2
4= -2
i) Find the deceleration in the first 5 seconds
ii) Find the deceleration in the last 2 seconds
ii) Find the rate of change of speed in the first 6
seconds
ii) Find the rate of change of speed in the last 8
seconds
4
5 12
2
0
speed
time
6
3 7
2
0
speed
time
7
9
2
100
speed
time
4
8
2
6 80
speed
time
03 7 9
5
speed
time
04 10 12
6
speed
time
6
3
0 3 6 8 time
speed
10
5
05 10 12
speed
time0 6 10 18
14
20
24
speed
time
i) Find the rate of change of speed in the first 3
seconds
Final speed = 5
Initial speed = 0
Rate of change of speed =3
05=3
5
ii) Find the rate of change of speed in the last 2
seconds
Final speed = 0
Initial speed = 5
Rate of change of speed =2
50=2
5
14
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D. Find The Value Of V (Speed)
Example 1
Exercise 1
Distance travelled in 8 seconds is 40 m
Example 2
Gradient And Area Under A Graph 9
Distance travelledin 6 seconds is 30 m
Area A =2
1x 6 x v = 30
3v = 30
v =3
30= 10
Distance travelled in the first 4 seconds is 20 m
2
1(2 + v) x 4 = 20
2(2 + v) = 20
2 + v = 20/2
= 10v = 10 2
= 8
0
6
v
A
0
speed
time
v
0 8
speed
time
4 10
v
2
0
speed
time
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Exercise 2
Example 3
Exercise 3
Example 4
Gradient And Area Under A Graph 10
Distance travelled in the first 6 seconds is 24
m
Total distance travelled in 11 seconds is 60 m
Area A + Area B + Area C = 60
[2
1x 4 x 6 ]+ [3 x 6 ] + [
2
1x (6 + v) x 4 ] = 60
12 + 18 + 2(6 + v) = 60
2(6 + v) = 60 30
6 + v = 30/2
v = 15 6
= 9
Total distance travelled in first 14 seconds is 96 m
Total distance travelled in 12 s is 60 m
Area A + area B + area C = 60
[ 2
1
x (2 + 6) x 3 ] + [5 x 6] + [ 2
1
x (6 + v) x 4 ]= 8012 + 30 + 2(6 + v) = 80
2(6 + v) = 80-12-30
6 + v = 38/2
v = 19 6
= 13
6 12
v
3
0
speed
time
v
6
2
0 3 8 12
A B C
time
speed
4 8 14
v
6
0
A B C
time
speed
v
6
0 4 7 11
A B C
speed
time
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Exercise 4
E. Find The Value Of T (Time)
Example 1
Total distance during T s is 50m.
Area A = (4)(10)/2 = 20
Area B = (T 4)(10)
Total distance = 20 + (T 4)(10) = 50
(T 4)(10) = 2050 10 T 4 = 30
10 T = 30 + 4010 T = 70
T = 70 / 10
T = 7
Exercise 1.
Total distance during Ts is 56 m.
Example 2.
Total distance in Ts is 58m.
Area A = 282142/)4)(86( ==+
Area B = (T 4)6
Total distance = 28 + (T 4)6 = 58
(T 4)6 = 58 28= 30
(T 4) = 30/6
Gradient And Area Under A Graph 11
Total distance travelled in 14 s is 140 mv
10
4
0 5 12 14 time
speed
time
0 4 T
A B
10
time
speed
0 4 T
A B
10
time
speed
04 T
8
6
A B
time
speed
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= 5
T = 5 + 4
T = 9
Exercise 2Total distance in T s is 65m.
Example 3.
Total distance in T s is 55m.
Area A = (3)(2) / 2 = 3
Area B = (4)(3) = 12
Area C = ( 5 + 3)(T 6)/2
Total distance = 3 +12 + (8)( T 6)/2 = 55
15 + 4(T 6) = 55
4(T 6) = 55 15
= 40T 6 = 40/4
= 10
T = 10 + 6
= 16
Exercise 3.
Total distance in T s is 54m.
Example 4.
Total distance in T s is 45m.
Area A = (2 + 6)(3)/2 = 12
Area B = 1262 =
Area C = (6 + 8)(T 5)/2Total distance = 12 + 12 +7(T 5) = 45
24 +7(T 5) = 45
Gradient And Area Under A Graph 12
05 T
6
4
A B
time
speed
0 3 5 T
8
6
2
time
speed
0 3 6 T
4
7
A B C
time
speed
0 2 6 T
3
5
A B C
time
speed
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7(T 5) = 45 24
(T 5) = 21 / 7 = 3
T = 3 + 5 = 8
Exercise 4.
Total distance in T s is 70m.
F. Average Speed
timeTotal
cetandisTotal=speedAverage
Example 1
Total Distance = Area A + Area B
= (2)(8) / 2 + (8)(2)= 8 + 16
= 24
Total Time = 4
Average Speed = 24 / 4
= 6
Exercise 1
Find the average speed in the first 4 seconds
Example 2
Total Distance = Area A + Area B
= (6+8)4/2 + (6)(8)/2
Gradient And Area Under A Graph 13
8
6
2
04 6 T
time
speed
0 2 4
8
A B
speed
time
6
A B
0 2 4time
speed
8
6
A B
0 4 10time
speed
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= 28 + 24
= 52
Total Time = 10
Average Speed = 52 / 10
= 5.2
Exercise 2
Find the average speed in the first 6 seconds.
Example 1
Total Distance = Area A + Area B + Area C
= (2)(6) / 2 + (2)(6) + (3)(6)/2
= 6 + 12 + 9
= 27
Total Time = 7
Average Speed = 27 / 7
= 3.86
10
A B C
Exercise 3
Find the average speed in the first 8 second.
Example 1
Total Distance = Area A + Area B + Area C
= (4)(3)/2 + (3)(4) + (4+8)(4)/2
= 6 + 12 + 24
= 42
Total Time = 10
Average Speed = 42 / 10= 4.2
Gradient And Area Under A Graph 14
0 3 6
4
2
A B
time
speed
A B C
0 3 6 10
8
4
time
speed
0 4 6 8
speed
time
0 2 4 7
6
A B C
time
speed
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Exercise 4
Find the average speed in the first 10 seconds.
14.2 Distance-Time Graph
A. Distance And Speed
Example 1
Distance from O to A = 12 0 = 12
Speed from O to A = Distance / Time
= 12/3
= 4 km/h
Exercise 1
Find:
i) Distance from O to A
ii) Speed from O to A
Example 2
Distance from O to A = 12 0 = 12
Speed from O to A = Distance / Time
Gradient And Area Under A Graph 15
Time(hrs)
Distance(km)
3
12
0
A
Time(minutes)
Distance(km)
60
12
0
A
Distance(km)
4
12
0Time(hrs)
A
A B C
0 4 7 10
8
11
time
speed
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=60/60
12
= 12 km/h
Exercise 2
Find the speed from O to A
B. Average Speed
Example 1
Total distance from O to B = 20
Total time from O to B = 90 minutes
Average speed (km/h) =
60/90
20
= 13.33
Exercise 1
Find the average speed in km/h from O to B
Exercise 2
Find the average speed in km/h from O to B.
Gradient And Area Under A Graph 16
40
16
0
Time(minutes)
A
Distance(km)
Distance(km)
Time(minutes)
30 90
20
0
A
B
Time(minutes)
Distance(km)
10 40
30
0
A
B
10
Distance(km)
20 90
80
0
A
B
30
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Exercise 3
Find the average speed in km/h from O to B.
Questions Based On Examination Format
Gradient And Area Under A Graph 17
Distance(km)
3 10
8
0
A
B
4
Time(minutes)
Time(hrs)
0 4 T
10
14
Speed (m s-1)
Time (s)
Example 1
The diagram shows the speed-time graph of a particle over a period
ofTseconds. Given that the total distance travelled in Tseconds is
90 m. Calculate
a) its acceleration in the first 4 seconds.b) the value ofT
Solution
a)time
speedinitialspeedFinal =
4
1014=4
4= 1
b) area A + area B = 90
9014)4(4)1014(2
1=++ xTxx
48 + (T 4) x 14 = 90
(T - 4) x 14 = 90 - 48
T - 4 = 42/14
= 3 + 4 = 7
0 4 T
12
20
Speed (m s-1)
Time (s)
Exercise 1
The diagram shows the speed-time graph of a particle over a period
ofTseconds. Given that the total distance travelled in Tseconds is
124 m. Calculate
c) its acceleration in the first 4 seconds.
d) the value of T
Solution
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Gradient And Area Under A Graph 18
07 11
u
20
Speed (m s-1)
Time (s)
Example 2
The diagram shows the speed-time graph of a particle
over a period of 11 seconds. Calculate the value ofu in
each of the following cases
Its acceleration in the first 7 seconds is 2 m s-2.
The total distance travelled during the 11 seconds
is 138m
Solution
a) = 2
20 - u = 14
u = 6
b) x (20 + u) x 7 + x 4 x 20 = 138x (20 + u) x 7 = 138 - 40
(20 + u) x 7 = 98(2)
20 + u = 196/7
u = 28 20
= 8
06 10
u
8
Speed (m s-1)
Time (s)
Exercise 2
The diagram shows the speed-time graph of a
particle over a period of 10 seconds. Calculate the
value ofu in each of the following cases
Its acceleration in the first 6 seconds is 0.5 ms-2.The total distance traveled in the 10 seconds is 58 m
Solution
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Gradient And Area Under A Graph 19
07 9 17
12
20
u
Time (s)
Speed (m s-1)Example 3
The diagram shows the speed-time graph of an object
over a period of 17 seconds. Calculate
a) its deceleration in the first 7 seconds
b) the value ofu, if the distance travelled in the last
10 seconds is 87 m.
Solution
a) 7
2012
= 7
8
b)2
1x (u + 12) x 2 +
2
1x u x 8 = 87
(u + 12) + 4u = 87
5u = 87 12
u = 15
04 10 14
15
29
u
Time (s)
Speed (m s-1)
Exercise 3
The diagram shows the speed-time graph of an object
over a period of 14 seconds. Calculate
a) its deceleration during the first 4 seconds
b) the value ofu, if the distance travelled in the last
10 seconds is 150 m.
Solution
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Gradient And Area Under A Graph 200 t 5 6
144
250
Distance from A (km)
Time(hrs)
P
QR
S
Exercise 4
In the diagram, OPQ is the distance-time graph of a car
traveling from town A to town B. The straight line RPS
represents the distance-time graph of a van traveling from
town B to town A. Calculate the
a) average speed, in km h-1, of the car from town A to B
b) value oftif the van travelled at uniform speed.
Solution
0 t 3 4
144
240
Distance from A (km)
Time(hrs)
P
QR
Example 4
In the diagram, OPQ is the distance-time graph of a car
traveling from town A to town B. The straight line
RPS represents the distance-time graph of a van
traveling from town B to town A. Calculate the
average speed, in km h-1
, of the car from town Ato B
value oftif the van traveled at uniform speed.
Solution
a) = = 60 km h-1
b) = 80
80t= 144
t= 1.2
S
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PAST YEAR QUESTIONS
1. Nov 2003
Speed (ms-1)
2. Nov 2004
Gradient And Area Under A Graph 21
Diagram shows the speed-time graph of a particle for a
period of 17 s.
State the length of time, in s, that particle moveswith uniform speed.
b) Calculate the rate of change of speed,
in ms, in the last 4 s
c) Calculate the value of u , if the total distance
traveled in the first 13 s is 195 m.
[ 6 marks]0 6 13 17
18
u
Time (s)
Diagram shows the distance-time graph for the
journeys of a car and a bus. The graphJKLM
represents the journey of the car and the graphJLN
represents the journey of the bus. Both vehicles
depart from town Tat the same time and travel along
the same road.
State the length of time, in hours, during
which the car is stationary.
Calculate the average speed, in km h,
of the car over the 2 hour period.
At a certain time during the journey, both
vehicles are at the same location.
0
K L
J
M
N
48
120
170
0.5 0.8 2.0
Distance from town T (km)
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3. Nov 2005
.
4. July 2006
5. Nov 2006, Q9
Diagram 4
shows the speed-
time graph for the
movement of
a particle for a
period of 20 seconds.
Gradient And Area Under A Graph 22
State the time taken by the bus to reach
that location from town T.
[6 marks]
0 5 12 t
1
9
21
Speed (ms-1)
Time (s)
Diagram shows the speed-time graph of a particle for a
period oftseconds.
State the length of time, in s, that the particle moves
with uniform speed.
Calculate the rate of change of speed, in ms-2, in the
first 5 seconds.
Calculate the value of t, if the total distance traveled for
the period of t seconds is 148 meters.
[6 marks]
016 22 30
u
25
Speed (ms-1)
Time (s)
Diagram shows the speed-time graph for the movement
of a particle for a period of 30seconds.
State the length of time, ins, for which the particle
moves with uniform speed.Calculate the rate of change of speed, in ms-2, in the last
8 seconds.
Calculate the value of u, if the distance traveled in the
last 14 seconds is 229 m.
[6 marks]
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(a) State the uniform speed, in ms-1, of the particle.
(b) The distance traveled by the particle with uniform speed is 84m.
Calculate
(i) the value of t,
(ii) the average speed, in ms-1, of the particle for the period of 20 seconds.
[ 6 marks ]
8 Julai 2007.
Diagram 3 shows the speed-time graph for the movement of a particle fora
period of 20 seconds.
(a) State the uniform speed, in ms-1, of the particle.
(b) Calculate the rate of change of speed, in ms-2, in the last 5 seconds.
(c) Given that the distance traveled by the particle in the first 15
seconds is 408 meters, calculate the value of t.
9 November 2007.
Diagram 5 shows the distance-time graph for the journeys of a bus and
a taxi .
Gradient And Area Under A Graph 23
0 t 20 Time (s)
Speed (m s-1)
36
15
24
Diagram 3
[6 marks]
0 65 150
30
Time (minutes)
Distance (km)
90
30
J
RQ
KP
Diagram 5
S
40
TownB
TownA
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The graph PQRS represents the journey of the bus from townA to townB.
The graph JK represents the journey of the taxi from B to town A.
The bus leaves town A and the taxi leaves town B at the same and they
travel along the same road.
(a) State the length of time, in minutes, during which the car is stationary.
(b) (i) If the journey starts at 9.00 a.m., at what time do the vehicles
meet ?
(ii) Find the distance, in km, from town B when the vehiclesmeet .
(c) Calculate the average speed, in km h-1, of the bus for the whole journey.
10 Jun 2008.
Diagram 10 shows a speed-time graph for the movement of a particle fora
period of 25 s.
(a) State the duration of time s, for which the particle moves with uniform
Gradient And Area Under A Graph 24
Diagram 10
0
u
16 Time (s)
Speed (m s-1)
28
12
10
25
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speed.
(b) Calculate the rate of change of speed, in ms-2, in the first 12 s.
(c) Calculate the value of u, if the total distance traveled in the period
of 25 s is 438 m.
Gradient And Area Under A Graph 25
[6 marks]
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11 November 2008.
Diagram 9 shows the speed-time graph for the movement of a particle fora period of t
seconds.
(a) State the uniform speed, in m s1 , of the particle.
(b) Calculate the rate of change of speed, in ms-2, in the first 4 seconds.
(c) The total distance traveled in
tseconds is 184 meters.
Gradient And Area Under A Graph 26
Diagram 9
[6 marks]
0 Time (s)
Speed (m s-1)
20
4
12
t
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ANSWERS
14.1 Speed-time graph
Area
Exercise 1
A. 1. 40 B. 1. 36
2. 104 2. 63
3. 60 3. 104
4. 64 4. 78
Area under Graph Rate of change speed
1. i) 10 1. 2
ii) 30
iii) 40 2. i)3
1
ii) -42. i) 12
ii) 30 3. i)3
2@ 0.67
iii) 42 ii) -3
3. i) 9 4. i) -1
ii) 6 ii)2
5@ -2.5
iii) 24
iv) 39 Find the value of v
4. i) 80 1. 10
ii) 18 2. 5
iii) 98 3. 14
4. 25
5. i) 4
ii) 24 Find the value of t
iii) 48
1. 7.6
6. i) 6 2. 15
ii) 27.5 3. 12
iii) 45.5 4. 12
Average speed
1.2
9@ 4.5 3.
4
25@ 6.25
2.3
5@ 1.67 4. 8.35
Gradient And Area Under A Graph 27
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14.2 DISTANCE-TIME SPEED
DISTANCE AND SPEED
Gradient And Area Under A Graph 28
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1 3 2 24
AVERAGE SPEED
1 45 2 80 3
5
4@ 0.8
SPM Format questions1a 2 1b 7 2a 5 2b 6
3a
2
7@ -3.5
3b 21 4a 41/67 4b 2.88
PAST YEAR QUESTIONS
2003 (a) 7 b -4.5 c 5
2004 (a) 0.3 b 85 c (i) 48 C(ii) 0.8
2005 (a) 7 b 1.6 c 16
2006
(J)
(a) 16 b 3.125 c 18
2006
(N)
(a) 14 b (i) 18
(ii) 16
2007
(J)
(a) 24 b 4.8 c 8
2007
(N)
(a) 35 b (i) 9.40 a.m (ii) 60 c 36
2008
(J)
(a) 9 b 1.5 c 14
2008
(N)
(a) 20 b 2 c 10