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PART 2

A car travels along a road and its velocity-time function is illustrated in Diagram 1. The straightline PQ is parallel to the straight line RS.

a) From the graph, findI) The acceleration of the car in the first hour,

(1,80)

(0,20)

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

Finding average speed, use concept of area under graph.

20 80 + 80 + 80

1.0 0.5 0.5

Distance = (20+80)(1.0) + 0.5(80) + (0.5)(80)

= 50km + 40km+ 20kmd = 110kmt = 2 h

= 55 km/h

b) What is the significance of the graph

i) significance of the position of the graph above the t-axis, the car moving to the right.ii) significance of the position of the graph below the t-axis, the car is moving to the left.

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c) Using 2 different method, find the total distance travelled by the car.

METHOD 1 : AREA UNDER GRAPH

From Question a(ii) area above t-axis is 110km. Now, area below t-axis

0.5 0.5 0.5

80 + 80 + 80

AREA = 2 ( ) = 40km + 40km

= 80km

Distance to the right is 110km, distance traveled to the left is 80km

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METHOD 1 : INTEGRATION

Integrate above t-axis,

From t=0 to t= 1,

I = = = [30+20]-0= 50

From t=1 to t= 1.5ii =

= = [80(1.5)- 80(1)]= 120-80= 40

From t=1.5 to t=2iii =

= = [-80(2)+320(2) (-80(1.5) + 320(1.5))]= 320-300= 20

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Since , the gradient is same

, R(2.5,0)

, (2.5,0)

( ),

When S(3,y)

( ), (

Since, we integrate below t-axis, modulus is used.

From t=2.5 to t=3,

IV = |

= = [ ( ) ( ) ]-[-80 ( ) ( ) = 480-500 = -20 = 20km

From t=3 to t= 3.5

V =

= = [-80(3.5)- 80(3)]

= -280+240

= -40 =40km

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Find equation of the point (3.5, -80) and (4,0)

, R(2.5,0)

( ) , (4,0)

( ),

, ( From t=3.5 to t=4

VI = = = [80(4)-640(4) (80(3.5) -640(3.5))] = -1280-(-1260) = -20 = 20km

Distance travel below t-axis is 80km

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d) In a Sunday, we are friends of four, plan a journey from Bangi to Perak town. We take a proton saga for the journey. We started our journey at 9am.

At start, the car moves 20km/h in a positive direction. Then it gradually increase thespeed (constant acceleration) until reaches 80km/h in 1 hour. The speed 80km/h remain

constant (0 acceleration) for half an hour. Then car had a constant deceleration andstopped in t=2hr.

The car comes to stop and rest for half an hour. We had a delicious Nasi Kandar inRestoran Vision. Then we decide to go back home. So, we drive the car, car moves innegative direction. The car accelerates until reaches the speed of 80km/h and remainsconstant for another half an hour. Then, car decelerates and finally stops. Total journeytook 4 hours.