Chapter 14 II Gradient & Area Under Graph ENHANCE

7/28/2019 Chapter 14 II Gradient & Area Under Graph ENHANCE

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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.


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


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






= area A

=2

1x 2 x 5 = 5


= 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



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



= area A

=2

1x (2 + 4) x 5 = 15


= area B

=2

1x 4 x 7 = 14

iii) Total distance

= area A + area B

= 15 + 14

= 29



iii) Total distance


= area A

=21 x 5 x 6 = 15


= 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




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


= area A + area B + area C= 7 + 20 + 39 = 66

i) Distance in the first 2 seconds

ii) Distance in the last 5 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


+ve accelerationve - deceleration


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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


seconds

Final speed =

Initial speed =

Rate of change of speed =


seconds


seconds

Final speed = 3

Initial speed = 6


63=

3

3=-1


secondsFinal speed = 0

Initial speed = 8


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


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


seconds

Final speed = 5

Initial speed = 0


05=3

5


seconds

Final speed = 0

Initial speed = 5


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


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


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


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.


Example 4.


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


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.


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


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


= 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


= 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


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


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.


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


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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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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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


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.


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 .


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


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.


[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.


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



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14.2 DISTANCE-TIME SPEED

DISTANCE AND SPEED



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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

Documents

Chapter 14 II Gradient & Area Under Graph ENHANCE