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Answers
The time taken by the car to cover 20 km before the turn is = 20/40 = 0.5 h
(a)
The fly moves at a constant speed of 100 km/h during this time. Hence, the total distance covered by it is = (100)(0.5) = 50 km
(b)
Suppose the car is at A which is x km away when the fly is at the wall, O. The fly hits the glass pane at B, taking a time t we have,
AB = 40t
OB = 100t
Therefore, x = AB + OB = 140t
=> t = x/140
And, OB = (5/7)x
The fly returns to the wall and during this period the car moves through distance BC. The time taken by the fly in this return path is = (5/7)x/100 = x/140
Thus, BC = (time)(speed of car) = (x/140)(40) = (2/7)x
And, OC = OB – BC = (5/7)x – (2/7)x = (3/7)x
If at the beginning of the round trip (wall to car and back) the car is at a distance x away when the next trip again begins.
Distance of the car at the beginning of the 1st trip is = 20 km
Distance of the car at the beginning of the 2nd trip is = (3/7)(20) km
Distance of the car at the beginning of the 3rd trip is = (3/7)2(20) km
Distance of the car at the beginning of the 4th trip is = (3/7)3(20) km
Distance of the car at the beginning of the nth trip is = (3/7)n-1(20) km
The trips will continue till the car reaches the turn that is the distance reduces to zero. This will be the case when n becomes infinity. Thus the fly makes an infinite number of trips before the car takes the turn.
Answer:
Refer to the attachment:
