Gradient & Area Under the Graf

8/6/2019 Gradient & Area Under the Graf

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5 marks


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BIL TOPIC FORM 03 04 05 06 07

Section A (52%) : Answer ALL questions (question 1 until 11)

1 REGION OF INEQUALITIES V 3 3 1

2 SET IV 1 1

3 SIMULTANEOUS EQUATIONS III 2 5 2 4 2

4 QUADRATIC EXPRESSION & EQUATIONS IV 1 7 1 3 3

5 LINES & PLANES IN 3-DIMENSIONS IV 4 3 4 2 4

6 THE STRAIGHT LINE IV 5 6 5 10 5

7 SOLID GEOMETRY - PERIMETER & AREA III 7 9 7 8 6

8 MATHEMATICAL REASONING IV 8 4 8 6 7

9 PROBABILITY II V 9 10 9 7 8

10 INVERSE MATRIX V 11 8 11 11 9

11 GRADIENT & AREA UNDER A GRAPH V 10 11 10 9 10

12 SOLID GEOMETRY (VOLUME) III 6 2 6 5 11

Section B (48%) : Answer FOUR questions from this section.

12 GRAPHS OF FUNCTIONS V 12 12 12 13 12

13 PLAN & ELEVATION V 15 15 15 15 13

14 EARTH AS A SPHERE V 16 16 16 16 14

15 TRANSFORMATION III V 13 13 13 12 15

16 STATISTICS IV 14 14 14 14 16


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Speed

Time

m = rate of change of speed= speed/time

= acceleration / deceleration

Constant/Uniform speed

acceleration

deceleration


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State the length of time, in s, that the particle moveswith uniform speed.

Speed (ms-1)

21

9

O 5 12 t

(t,21)

(t,0)1

Time (s)

7512 !


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takentime

speedinitialspeedfinal

speedofchangeofRate

-!

+ve acceleration

ve - deceleration

RateOfChangeOfSpeed

4

5 120

speed

Find the rateofchangeofspeedin

thefirst 5 seconds

Final speed = 4

Initial speed = 0

Rateofchangeofspeed

= = @ 0.8

5

04

5

04

5

04

5

04

5

4


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(a) State the length of time, in s, that the particlemoves with uniform speed.

(b) Calculate the rate of change of speed, in

ms

-1

, in the first 5 seconds.

Speed (ms-1)

21

9

O 5 12 t

(t,21)

(t,0)1

Time (s)


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Speed

Time

Area =

distance

travelled

Area under the graph is the distance

bb

2

1Area = ab

a(b+c)

b

Area = ab2

1

Area =

a a a

c


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A B

5

04 10 Time

speed

(i)Distanceduring thefirst 4 seconds

= area A

= x 4x 5 = 102

1

(ii)Distanceduring thelast 4 seconds= area B

= 6 x 5

= 30

(iii)Totaldistanceduring10 seconds

= area A + area B

= 10 + 30

= 40


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FindTheValueOfV(Speed)

4 10

v

2

0

speed

time

Distance travelled in the first 4 seconds

is 20 m

(2 + v) x4

= 202(2 + v) = 20

2 + v = 20/2

= 10

v = 10 2

= 8

2

1


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FindTheValueOfT(Time)

0 3 5 T

8

6

2

time

speed

TotaldistanceinT s is 45m.

Area A = (2 + 6)(3)/2 = 12

Area B = 2 X 6 = 12

Area C = (6 + 8)(T 5)/2

Totaldistance = 45

12 + 12 +7(T 5) = 45

24 +7(T 5) = 45

7(T 5) = 45 24

(T 5) = 21/7

(T - 5) = 3T = 3 + 5

= 8


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Calculate the value oft, if the total distance

travelled fortseconds is 148 metres.

Total distance= area under the graph

Speed (ms-1)

21

9

O 5 12 t

(t,21)

(t,0)1

18

18010

122192

197591

2

1148

!

!

v!

t

t

t

Stress on the correct values when substituting


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timeTotal

cetandisTotal=speedAverage

A B C

0 3 6 10

8

4

time

TotalDistance

= Area A + Area B + Area C

= (4)(3)/2 + (3)(4) + (4+8)(4)/2= 6 + 12 + 24

= 42

TotalTime = 10

AverageSpeed = 42/10

= 4.2


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Distance

Time

m = Rate of change of distance

= distance/ time

= speed

Object stops

Speed

Distance(km)

3

12

0

Distancefrom O toA = 12 0 = 12

Speedfrom O toA = Distance/Time

= 12/3

= 4 km/h


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AverageSpeed

Distance(km)

Time(minutes)

30 90

20

0

A

B

Totaldistancefrom O toB = 20

Total timefrom O toB

= 90 minutes

@Average speed(km/h)

=

= 13.33

60/90

20


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REMEMBER :1. Lengthoftimeis total time taken

2.AREA

of

trapeziu

m

3. Distance-timegraph , gradient = speed

equivalent to the rateofchangeof

distance

4. Speed-timegraph , gradient =

accelerationequivalent to the rateof

changeofspeed

GRADIENT AND AREA UNDER A GRAPH


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Gradient & Area Under the Graf