Gradient and Area

8/6/2019 Gradient and Area

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TOPIC : GRADIENT AND AREA UNDER A GRAPH

1 Diagram 1 shows the speed-time graph of a particle for a period of 15 s.

DIAGRAM 1

(a) State the distance, in m, 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] Answer:

(a)

(b)

(c)

Speed (ms-1)

Time (s)

23

k

5

0 6 10 15


2/18

2. Diagram 2 shows the speed-time graph of two particles, and for a period of 8s.

DIAGRAM 2

The graph OKNMrepresents the movement of particle and the graph JKL

represents the movement of particle .Both particles start moving at the same time.

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

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

(c) Calculate the difference in distance, in m, of particle and particle for a periodof 8 s.

[6 marks] Answer:

(a)

(b)

(c)

Time (s)

Speed (ms-1)

O

4

8

K

N M

8

L

6

J

3


3/18

3. Diagram 3 shows the distance-time graph of the journeys taken by Ali and Fuad.

DIAGRAM 3

The straight line OB represents Alis journey from town X to town Y, while the straightline FG represents Fuads journey from town Y to town X.

Ali and Fuad uses the same route.

(a) State the distance, in km, of town Y from town X.(b) Find the time Ali and Fuad meet each other during their journey.(c) Find the distance when they meet from town Y.(d) Calculate Fuads speed.

[6 marks] Answer:

(a)

(b)

(c)

(d)

Distance (km)

Time

105

60

O0700 0730 0800 0830 0900 0930

F

G

B


4/18

4. Diagram 4 shows the speed-time graph of a motorcyclist in a period of 30seconds.Given that the total distance travelled by the motorcyclist is 525 m.

Calculate,

(a) the rate of change of speed in the last 5 second,

(b) the duration of uniform speed,

(c) the value ofv.

[6 marks]

Answer:

(a)

(b)

(c)

time (s)

Speed (m s-1)

20

v

10 25 30

DIAGRAM 4


5/18

5. Diagram 5 shows a velocity-time graph for a particle.

velocity ( m s-1)

20

(a) State the time, in s, the particle moves with constant velocity.

(b) Calculate the accleration, in m s-2, of the particle in the last 5 seconds.

(c) Find the value of uif the total distance travelled after 15 seconds is 190 m.[ 6 marks ]

Answer:

(a)

(b)

(c)

15 25 30 time (s)

u

O

DIAGRAM 5


6/18

6

Diagram 6 shows the speed-time graph of a particle in a duration of 20 seconds.

(a) Find the speed of the particle at the third second.(b) Find the value oftif the distance travelled by the particle at constant speed is 45 m.(c) Calculate the average speed of the particle within the 20 seconds.

[6 marks]

Answer: (a)

(b)

(c)

Diagram 6


7/18

7

Diagram 7 shows the speed-time graph of a particle for a period of t seconds.(a) Calculate the rate of change of speed for the particle in the first 5 seconds, in m s -2.(b) It is given that the total distance traveled by the particle is 220 m, find the value of t.

[5 marks]

Answer, (a)

8,

(b)

Diagram 7


8/18

Diagram 8

Diagram 8 shows the distance-time graph for the movement of a car for a period of half anhour.(a) State the time, in minute, when the car is resting.(b) Calculate

(i) the speed, in km/minute, of the car for the first 10 minutes,

(ii) the average speed, in km/hour, of the car for the whole journey.[5 marks]

Answer, (a)

(b) (i)

(ii)


9/18

9.

Diagram 9

Diagram 9 shows the speed-time graph for the movement of a particle for a period of 30seconds.(a) Calculate the rate of change of speed in the last 15 s.(b) The total distance travelled by the particle in 30 s is 60 m. Calculate

(i) the value oft,

(ii) the average speed, in m s-1

, of the particle for the period of 30 seconds.[6marks]

Answer: (a)

(b) (i)

(ii)

10.


10/18

Diagram 10

Diagram 10 shows the distance-time graph of a car for the period ofTs. Find(a) the speed of the car for the first 12 s,(b) the time during which the car is at rest,

(c) the value of T if the average speed for the whole journey of the car is 133

1ms -1.

Answer: (a)

[6 marks]

(b)

c


11/18

11.

Diagram 11

In Diagram 11, the graph OPQ represents the journey of a car in 10 s and the graph OABrepresents the journey of a motorcycle in Ts. Find(a) the rate of change of speed of the car in the last 4 s,

(b) the distance travelled by the car in 10s,(c) the value ofT, if the distance travelled by the car in 10 s and the distance travelled by the

motorcycle in Ts are the same.[5marks]

Answer, (a)

(b )

(c)


12/18

SKEMA JAWAPAN : GRADIENT AND AREA UNDER A GRAPH

1 ( a ) 20

( b )60

523

3 atau nyahpecutan 3 atau awapecutan 3

( c ) ( ) ( ) 139k552

1545236

2

1=++++

k = 9

1

11

21

6

2 (a ) 2 saat

(b)06

08

3

4atausetara

( c ) ( )8282

1+

( ) 848282

1+

8

1

1

1

1

1

1 6

3 ( a ) 105 km

( b ) 0800 a.m

( c ) 105 60 = 45km

( d ) jkm /425.2

105=

1

1

11

11 6


13/18

4(a)

2530

200

24 ms

(b) 15 s

(c) ( ) ( ) 5252015202

12010

2

1=+++ v

v = 15 ms-1

1

1

1

2

16

.

5 (a) 15

(b)3025

020

4

(c) 1902052

110)20(

2

1=++ u

u= 8

1

1

1

2

16


14/18

6(a)

5

10=3

v

v =5

30= 6 ms -1

(b) ( t 5 ) x 10 = 45

t -5 = 4.5t = 9.5 s

(c)2

1x 5 x10 + 45 +

2

1x 6.5 (10 +22) +

2

1x 4 x 22

25 + 45 +104 + 44 = 218

Average speed =

20

218= 10.9 ms-1

1

1

1

1

1

1

6


15/18

8 (a) 15 - 10 = 5 minutes

(b) (i)min10

5km= 0.5 km / minute

(ii)min30

)1055( km++= 20 / 0.5 hour

= 40 km / hour

1

1 , 1

1

1

5

7(a)

5

138= -1 ms-2

(b) 2

1

x 5 ( 8 + 13 ) + 10 x 8 + 2

1

x ( t 15 ) x ( 8 + 17 ) =

220

52.5 + 80 +2

25( t -15 ) = 220

2

25( t 15 ) = 87.5

( t 15 ) = 7

t = 22 s

1 , 1

2

1 5


16/18

9(a) Rate of change of speed =

15

2036 = 1.07 ms -2

(b) (i ) Total distance

=2

1x 20 x t + ( 15 t ) x 20 +

2

1(20 +36 ) x 15

= 10 t + 300 20 t + 420

= 720 10 t

Therefore , 720 10 t = 660

10 t = 60

t = 6 s

(ii ) Average speed =30

660= 22 ms -1

1 , 1

2

1

1 6


17/18

10(a) speed of the car for the first 12 s =

12

150= 12. 5 ms -1

(b) Time = 20 12 = 8 s

(c) total distance traveled = 400 m

T

400= 13

3

1

T = 30 s

1 , 1

1 , 1

1

1

6


18/18

11 (a) Rate of change of speed of the car in the last 4 s

=4

20= -5 ms-2

(b) Distance traveled in 10 s =

2

110 x 20 = 100m

(c) Distance traveled by motor cycle in T s

=2

1x 10 x10 + 10 x ( T -10 )

= 50 + 10 T 100

= 10 T - 50

Therefore, 10 T 50 = 100

10 T = 150

T = 15 s

1 , 1

1

1

15

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Gradient and Area