6.0 Gradient and Area Under a Graph

8/3/2019 6.0 Gradient and Area Under a Graph

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Cha

pter

6

GRADIENT AND AREA

UNDER A GRAPH

GRADIENT AND AREA

UNDER A GRAPH6LEARNING OBJECTIVES

Understand and use the concept of quantity represented

by the gradient of a graph

Understand and use the concept of quantity represented

by the area under a graph

LEARNING OBJECTIVES

Chapter


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KEY TERMS

Rate - Kadar Gradient - Kecerunan Area under a graph - Luas di bawah graf Speed - laju Acceleration - Pecutan Deceleration - Nyahpecutan Distance-time graph - Graf jarak-masa Speed-time graph - Graf laju-masa Average speed - Laju purata Stationary - Pegun Rate of change of speed - Kadar perubahan laju

Constant/uniform speed - Laju malar/seragam


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REQUIRED BASIC MATHEMATICS

SKILLS

Able to calculate rates

Able to calculate the distance travelled and the speed of movingobject

Able to calculate average speed

Able to convert a speed in

km min-1 to km h-1

ms-1 to km h-1

Able to calculate the gradient of a straight line

Able to calculate the area of a triangle, rectangular, and trapezium Able to calculate the length of an unknown side of two similar

triangle


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REQUIRED BASIC MATHEMATICS SKILLS

Speed = Distance travelled

Time taken

Distanced travelled = Speed x Time takenAverage speed = Total distance travelled

Total time taken


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6.1 A The Quantity Represented by The Gradient of A Graph

0 12

34 Time (hours)

100

200

300

Distance(km

) Quantity on the

vertical axis is

distance, in km

Quantity on the

horizontal axis is

time, in hoursGradient =

Change in distance (in km)

Change in time (in hours)

The quantity represented

by the gradient of thegraph is the rate of change

of distance with respect to

time, i.e speed, in kmh-1

The Quantity Represented by The Gradient of A Graph


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6.1 A The Quantity Represented by The Gradient of A Graph

0 24

68 Time (s)

5

10

15

Speed (ms-1) Quantity on the

vertical axis is

speed, in m s-1

Quantity on the

horizontal axis is

time, in secondGradient =

Change in speed (in m s-1)

Change in time (in second)

The quantity represented

by the gradient of thegraph is the rate of change

of speed with respect to

time, i.e acceleration, in

m s-1

20

The Quantity Represented by The Gradient of A Graph


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6.1 B Distance-Time Graph

Time

Distance

The diagram shows the distance-time graph of a car(travels from point P to point Q)

A

B C

D

Section A to B indicates

the car moving at the

constant speed

Section B to C indicatesthe car at rest

Section C to D indicates

the car moving at slower

constant speed

Distance-Time Graph

E

Section D to E indicates

the car travels from

point Q back to point P


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6.1 C The Gradient of A Distance-Time Graph

The diagram shows a distance-time graph for a particle moving along

a straight line

The Gradient of A Distance-Time Graph

0 1 2 3 4 Time (hours)

100

200

300

Distance(km) The gradient can be

determined by choosing

two points on the line

Lets choose the point

A and B

Gradient = BC

AC

= 300 0

4 0

= 75 kmh-1

Therefore, the speed of

particle is 75 kmh-1

A

B

C


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6.1 C The Gradient of A Distance-Time Graph

The diagram shows a distance-time graph for the journey made by James

between two places

The Gradient of A Distance-Time Graph

0 1 2 3 4 Time (hours)

10

20

30

Distanc

e(km)

The gradient can be

determined by choosing

two points on the line

Lets choose the point

B and C

Gradient = ABAC

= 0 40

4 0

= -10 kmh-1

Therefore, James

travelled at a constant

speed of 10 kmh-1

between two places

40

A

B

C

The negative gradient

indicates that the distance

decreases with respect

to time


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6.1 D Finding the speed for a period of time

in a distance-time graph

The distance-time graph shows the movement of an object fora period of 10 seconds

Distance (m)

Time (s)

40

04 8 10

A B

C

For the first 4 seconds, the

movement of the object is

represented by the straight

line OA.

Gradient of OA =

40 - 0

4 - 0

40 - 0

4 - 0

m

s

= 40 m

4 s

= 10 ms -1

Thus, the speed of the object

in the first 4 seconds is

10 ms -1.

Finding the speed for a period of time



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Distance (m)

Time (s)

40

04 8 10

A B

C

From A to B, the distance

remains constant.

8 - 4



The period of time when the

object is stationary is

4 seconds

Thus, the speed of the objectis zero from 4 seconds till

8 seconds.

4 8


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Distance (m)

Time (s)

40

04 8 10

A B

C

For the last 2 seconds, the

movement of the object is

represented by the straight

line BC.

Gradient of BC =

0 - 40

10 - 8

0 - 40

10 - 8

m

s

= - 40 m

2 s

= -20 ms -1

Thus, the speed of the object

in last 2 seconds is 20 ms -1.




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EXAMPLE 1

Distance (m)

Time (s)

24

6

6 12 15

The distance-time graph shows the

movement of an object for a period of

15 seconds.

Find

(a) its speed in the first 6 seconds,

(b) its speed in the last 3 seconds,

(c) the period of time when the

object is stationary,

(d) The average speed for the whole

journey.

0


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EXAMPLE 1

Distance (m)

Time (s)

24

6

6 12 15



15 seconds.

(a) Gradient of graph for the first6 seconds24- 6

6- 0

0

24 - 6

6 - 0

m

s

= 3 m s-1

=

Thus its speed in the first

6 seconds is 3 m s-1


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EXAMPLE 1

Distance (m)

Time (s)

24

6

6 12 15



15 seconds.

(b) Gradient of graph for the last3 seconds

0 - 24

15- 120

0 - 24

15 - 12

m

s

= -8 m s-1

=

Thus its speed in the last

3 seconds is 8 m s-1


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EXAMPLE 1

Distance (m)

Time (s)

24

6

6 12 15



15 seconds.

(c) The object is stationary from6 s till 12 s.

12- 60

Thus, the period of time when

the object is stationary is

6 seconds.


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EXAMPLE 1

Distance (m)

Time (s)

24

6

6 12 15



15 seconds.

(d) Average speed= total distance travelled

total time taken

0

= (18 + 24) m

15 s

= 2.8 m s-1


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EXAMPLE 2

Distance (m)

Time (min)

d

5670

9 15 32


whole journey of Cik Temah from her

house to the supermarket and back to

her house

0 47

(a) Calculate the period of time,

in minutes, during which Cik

Temah is at rest.

(b) Calculate the speed of Cik

Temah in the first 9 minutes.

(c) If the average speed in the

first 32 minutes is 312.5 m min-1,

find the value of d.

(d) Calculate the average speed of

the whole journey of 47 minutes.


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EXAMPLE 2

Distance (m)

Time (min)

d

5670

9 15 32




her house

0 4715 - 9

(a) 15 9 = 6 minutes

(a) Calculate the period of time,

in minutes, during which Cik

Temah is at rest.


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EXAMPLE 2

Distance (m)

Time (min)

d

5670

9 15 32




her house

0 47

5670 (b) 5670 m

9 min

= 630 m min-19

(b) Calculate the speed of Cik

Temah in the first 9 minutes.


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EXAMPLE 2

Distance (m)

Time (min)

d

5670

9 15 32




her house

0 47 (c) d = 312.532

d = 312.5 x 32

= 10 000

(c) If the average speed in the

first 32 minutes is 312.5 m min-1,find the value of d.


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EXAMPLE 2

Distance (m)

Time (min)

d

5670

9 15 32




her house

0 47

(c) 10 000 + 10 000

47

= 20 00047

= 425 25 m min-147

(d) Calculate the average speed of

the whole journey of 47 minutes.


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6.2 A The Quantity Represented By The Area Under A GraphThe Quantity Represented By The Area Under A Graph

A train travels at a uniform speed of 70 km h-1 for 3 hours.

Distance travelled = speed x time

= 70 km h-1 x 3 h

= 210 km

Speed (km h-1)

70

0 3 Time (h)

From the graph, the area under the

graph is = 70 x 3 = 210.

Thus, in a speed time graph, the area

under the graph represents the distancetravelled.


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6.2 B Finding The Area Under A GraphFinding The Area Under A Graph

The area under a graph depends on the types of geometrical shapesformed on the graph such as triangle, a rectangle, a trapezium or

combination of these shapes.

Use the formulae for the areas of geometrical shapes to find the area

under a graph.

Rectangle Triangle Trapezium

a

b

a

bba

c

Area = a x b Area = x a x b Area = x (a + b) x c


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EXAMPLE Find the area under the following graph.

y

x

25

15

0 10 30 40


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EXAMPLE

A B C

Find the area under the following graph.

y

x

25

15

0 10 30 40

Area under the graph

= Area of trapezium A +

Area of rectangle B +

Area of triangle C

= [ x (15 + 25) x 10 ] +

(20 x 15) + ( x 10 x 15)= 200 + 300 + 75

= 575 unit2


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EXAMPLE

A B

Find the area under the following graph.

y

x

25

15

0 10 30 40

Area under the graph

= Area of trapezium A +

Area of trapezium B

= [ x (15 + 25) x 10 ] +

[ x (20 + 30) x 15]

= 200 + 375

= 575 unit2


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6.2 CDetermine the distance by finding the area under

a speed-time graph

Determine the distance by finding the area under

a speed-time graph

EXAMPLE Speed (km h-1)

Time (min)

100

70

30

020 30 60

The speed-time graph shows the journey of a car for a period of I hour

Calculate

(a) The distance, in km, travelled by the car at a uniform speed.

(b) Total distance, in km, travelled by the car.


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a speed-time graph


a speed-time graph

EXAMPLE Speed (km h-1)

Time (min)

100

70

30

020 30 60

(a) Distance travelled at uniform speed

= 30 x 10

60= 5 km

Convert minutes to hour


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a speed-time graph


a speed-time graph

EXAMPLE

Speed(kmh

-1)

Time (min)

100

70

30

020 30 60

A B C [ x (70 + 30) x 20 ]60

+

=

=

Total distance travelled

=

Area of trapezium A +

Area of rectangle B +

Area of trapezium C

5 + [ x (30 + 100)] x 30

6054 1 km

6


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6.2 DSolve problems involving gradient and

area under a graph

Solve problems involving gradient and

area under a graph

EXAMPLE Speed (m s-1)

Time (s)

40

06 15 t

The speed-time graph shows the movement of a particle for a period of

t seconds. Calculate(a) the acceleration, in m s-2, in the first 6 seconds.

(b) the distance travelled at a uniform speed.

(c) the value of t, if the average speed for the period of t seconds is 29 m s-1.


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area under a graph


area under a graph


Time (s)

40

06 15 t

(a) Acceleration in the first 6 seconds

= gradient of graph

40

6

= 40

6=

6 2 m s-2

3


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area under a graph


area under a graph


Time (s)

40

06 15 t

(b) Distance travelled at a uniform speed

= area of rectangle A

= =

A

9 x 40 360 m


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area under a graph


area under a graph

EXAMPLE

Spe

ed(ms

-1)

Time (s)

40

06 15 t

t

40

9(c) Total distance travelled

= Area of trapezium

= x (t + 9) x 40= 20t + 180

Average speed = 29 m s-1

20t + 180

t= 29

20t + 180 = 29t

9t = 180

t = 20


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Speed (ms-1)

Time (s)

18

0

13 17DIAGRAM 66

Diagram 6 shows the speed-time graph

of a particle for a period of 17 s.

(a) State the length of time, in s, that

particle moves with uniform speed

(b) Calculate the rate of change of speed,

in m s-2, in the last 4 s.

(c) Calculate the value ofu, if the total

distance travelled in the first 13 s

is 195 m.

u


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Speed (ms-1)

Time (s)

18

06 13 1713 - 6

10. ( a ) 7 s P1

Answer:

u


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Speed (ms-1)

Time (s)

18

0 6 13 17

0 - 18

17 - 13

10. ( b ) 0 18 = -9

17- 13 2 K1N1

Answer:

u


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Speed (ms-1)

Time (s)

18

06 13 17

10 (c)

Bu

1 x ( u + 18 ) x 6 + 7 x 18 = 195

23u + 54 + 126 = 195

u = 5

K2

N1

Answer:

A

3u = 195 - 54 126 = 15


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area under a graph


area under a graph

Diagram 7 shows the speed-time graph of a particle for a period of 8 s.

(a)The rate of change of speed of the particle in the first ts is 3 m s-2.

5

(b) The total distance travelled by the particle during the period of 8 s is 75 m.

Speed (m s-1)

Time (s)

12

9

0 t 8

Find the value of t which satisfies each of the following cases.


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area under a graph


area under a graph

Speed (m s-1)

Time (s)

12

9

0t 8

t - 0

12 - 9

12 9

t - 0

(a)

= 3

53

t= 3

5

3t = 15

t = 5


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area under a graph


area under a graph

Speed (m s-1)

Time (s)

12

9

0t 8

A B

Area under the graph = 75

(9 + 12)(t) +

(b)

(8 - t)(12) = 75

(9 + 12)(t) + (8 t)(12) = 75

2

21t + (8 t)(12) = 150

9t + 96 = 150

9t = 150 - 96

9t = 54

t = 6


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GRADIENT AND AREA

UNDER A GRAPH


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Diagram 1 shows the distance-time graph of a taxi.

20

50

Distance (km)

Time (minutes)0 30 60 100 t

The taxi starts from station P to station Q. After stopping for 30 minutes atstation Q, it travels to station R before returning to station P. Calculate

(a) the distance, in km, from Q to R,

(b) its speed, in km h-1, from P to Q,

(c) the value of t, in minutes, if the speed of the taxi from station R to station P

60 km h-1

. [6 marks]

Diagram 1


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Answer:

20

50

Distance (km)

Time (minutes)0 30 60 100 t

(a)

20

50

50 20 = 30 km

20

30

60

P1

(b) 20 (30)

60

= 20 x (60)

30

= 40 km h-1

K1

N1

(c)

0-50

t 100

60

50 (t 100) = 60

60

50 x (60) = 60

t - 100

t 100 = 50 x 60

60

t = 50 + 100

= 150

K2

N1


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Diagram 2 shows the speed-time graph of a particle for a period

of 15 s.

Speed (ms-1)

Time (s)

23

k

5

0 6 10 15

DIAGRAM 2

(a) State the length of time, in s, the particle moves with constant speed.

(b) Calculate the rate of change of speed, in ms-2, in the first 6 s.

(c) Calculate the value of k, if the total distance travelled in the first 15 s is 139m.

[6 marks]


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Speed (ms-1)

Time (s)

23

k

5

0 6 10 15

Answer: (a) 10 6 = 4 s

(b)

5 - 23

6

5 23 = -3 m/s2

6

P1

K1N1

(a) State the distance, in m, the particle moves with constant speed.



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Speed (ms-1)

Time (s)

23

k

5

0 6 10 15

Answer: (c)

A B C

1 x (23 + 5) x 6 + 5 x 4 + 1 x (5 + k) x 5 = 139

2 2

104 + 5 x (5 + k) = 139

2

25 + 5k = 70

k =10

K2

N1

(c) Calculate the value of k, if the total distance

travelled in the first 15 s is 139m.

Di 1 h th d ti h f ti l f


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Speed (ms-1)

Time (s)

23

k

5

06 10 15

Diagram 1 shows the speed-time graph of a particle for

a period of 15 s.

(a) State the distance, in m, the particle moves with constant speed.


(c) Calculate the value of k, if the total distance travelled in the first

15 s is139m.

DIAGRAM 1


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Speed (ms-1)

Time (s)

23

k

5

06 10 15

(a) The distance, in m, the particle moves with constant speed

4

5

4

5

x = 20 m


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Speed (ms-1)

Time (s)

23

k

5

06 10 15

DIAGRAM 1


6

5-235-23

6

= -3 ms-2


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Speed (ms-1)

Time (s)

23

k

5

0 6 10 15

DIAGRAM 1

(c) Calculate the value of k, if the total distance travelled in the first

15 s is139m.

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1546523

2

1!vvvvv k

K= 9

65232

1vv 54v 55

2

1vv k


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6.0 Gradient and Area Under a Graph

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