Two cyclists start twenty miles apart and head toward each other each going at a steady rate of 10 m.p.h. At the same time, a fly that travels at a steady 15 m.p.h. starts from the front wheel of the southbound bicycle and flies to the front wheel of the northbound one, then turns around and flies to the front wheel of the southbound bicycle again, and continues in this manner till he is crushed between the two front wheels. What total distance will the fly cover before being squashed between the bike tires.
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I vaguely remember my undergrad director of studies in applied maths mentioning this problem. It was all fine until he asked, "and just before the cyclists meet, which way will the fly be facing?"
Both travel at 10 mph so in one hour the both will reach the center and collide. This means the fly has one hour for its travels. In that time it travels 15 miles.
Total distance = 2 0 miles
Speed of Cycle 1 = 1 0 mph
Speed of Cycle 2 = 1 0 mph
Since the cycles are moving in opposite directions,
Relative speed = 1 0 mph + 1 0 mph = 2 0 mph
Relative speed = 2 0 mph
Total distance = 2 0 miles
Time taken = Total distance ÷ Relative speed = 1 hour
The fly runs around for 1 hour.
Speed of fly = 1 5 mph
Hence, it travels 1 5 miles.
Nice Vinayak try this other problem I made https://brilliant.org/problems/real-algebra/
Can you solve it?
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The two cyclists are headed to each other until collision. The time frame until collision is also the time frame the bird travels. The time frame until collision is
d = Vt
10 miles = (10 mph)(t)
t = 1 hour
The total travel distance of the fly is
d f l y = V f l y t = ( 1 5 m p h ) ( 1 h o u r ) = 15 miles