A naughty bird is sitting on top of a car.It sees another car approaching it at a distance of 12 kms.The speed of 2 cars is 60 kmph each.The bird starts flying from 1st car and moves towards 2nd car, reaches the 2nd car and comes back to 1st car and so on. If the speed at which bird flies is 120 kmph then answer the following questions.Assume the 2 cars have a crash.
1) The total distance travelled by the bird before the crash is
2)The total distance travelled by the bird before it reaches the second car for the 2nd time
a. 10.55 km
c. 12.33 km
1) There is a very quick solution.
Cars A and B are 12 km farther between each other. They get this distance in 6 minutes. (60minutes*12km/(60kmph+60kmph)=6 minutes). It means after 6 minutes they will crash. Bird is flying at 120 kmph for this 6 minutes. So the bird flyes 120*6/60=12 km. B.
2)This is a scheduling problem.
When bird flyes there is 12 km from bird to Car B. Bird flyes at 120 kmph and car B is approaching at 60 kmph. So they are approaching each other at 180 kmph. But car A and B are approaching each other at only 120 kmph. So bird reach Car B in 12*60/180=4 minutes. In 4 minutes distance between Car A and Car B becomes 4 km (12-120*4/60=4). In 4 minutes bird flyes 8
km's and Car B approaches 4 km. After that the same logic works. Bird and Car A approaches each other at 180 kmph total. 4 km's are covered in 1 minutes 20 seconds. In that time bird covers 2,66 kms and the distance between car A and car B becomes 1,33 km's. In third round bird flyes 2,66/3 km's. In fourth round bird flyes 2,66/(3*3) km's. So all sum to 8 + 2,66 + 2,66/3 + 2,66/9 = 11,8 km's. I think the choice is B.