part2 edmet

4
A car travels along a road and its velocity-time function is illustrated in Diagram 1. The straight line PQ is parallel to the straight line RS. 80 . 20 A B C D E F -80 . V = 60t + 20 Let y = 60x + 20 If x = 0 y = 60x + 20 y = 60(0) + 20 y = 20 If x = 1 y = 60x + 20 y = 60(1) + 20 y = 80 (a) From the graph, find (i) the acceleration of the car in the first hour, Acceleration = area under graph Equation given ; v = 60t + 20 Let y = 60x + 20 Gradient = 60 Acceleration = 60kmh -2 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 v (km/h) t (h) P R S Diagram 1 Q

Upload: ain-syahira

Post on 21-Jul-2016

1 views

Category:

Documents


0 download

DESCRIPTION

kertas kerja addmath

TRANSCRIPT

A car travels along a road and its velocity-time function is illustrated in Diagram 1. The straight line

PQ is parallel to the straight line RS.

80 .

20

A B C

D E F

-80 .

V = 60t + 20

Let y = 60x + 20

If x = 0

y = 60x + 20

y = 60(0) + 20

y = 20

If x = 1

y = 60x + 20

y = 60(1) + 20

y = 80

(a) From the graph, find

(i) the acceleration of the car in the first hour,

Acceleration = area under graph

Equation given ; v = 60t + 20

Let y = 60x + 20

Gradient = 60

Acceleration = 60kmh-2

1.0 1.5 2.0 2.5 3.0 3.5 4.0 0

v (km/h)

t (h)

P

R

S Diagram 1

Q

(ii) the average speed of the car in the first two hours.

Average speed = total distance

total time

Distance travelled = area under graph

V = 60t + 20

Let y = 60x + 20

Area A

Area trapezium

=

( a + b) (h)

=

( 20 + 80 ) (1)

= 50

Area B

= 0.5 x 80

= 40

Area C

=

( 80) (0.5)

= 20

Total area = 110

Average speed = 110 / 2

= 55 kmh-1

b) What is the significance of the position of the graph

(i) above the t-axis,

The car travells from a place to other place

(ii) below the t-axis ?

Negative velocity indicates that an object

is travelling in a direction opposite to its

original direction

The car is reverse or return or toward the the start position

(c) Using two different methods, find the total distance travelled by the car.

Method I

Distance travelled = area under graph

Area A = 50

Area B = 40

Area C = 20

Area D =

( 80) (0.5)

= 20

Area E = 0.5 x 80

= 40

Area F =

( 80) (0.5)

= 20

Total area = Distance travelled

Total area = 190

Distance travelled = 190 km

Method II

-tak tahu- TT.TT

(d) Based on the above graph, write an interesting story of the journey in not more than 100

words.

Abu is mechanic. One day, Abu is driving to his workshop. Suddenly, his phone rings. It is

his father, Ahmad. Ahmad tells him that his car broke down. So Ahmad wants Abu to come

and fix the car. Then Abu immediately starts the journey to the location. He passes by his

workshop. He wants to arrive quickly so he accelerates his car at the acceleration of 60kmh-2.

After an hour, he starts driving at constant speed, 80 kmh-1 for 30 minutes. Then when he

enters the rural area, he decelerates his car at the acceleration of -160 kmh-1 and arrives at

the location after half an hour. He fix his father’s for 30 minutes before starts his journey

back. He arrives at his workshop after one and half hour. Then Abu starts his work.