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Two motorboats, which can move with velocities 4.0 m/s and 6.0 m/s relative to water are going up-stream. When the faster one overtakes the slower one, a buoy (floating device) is dropped from the slower one. After sometime both the boats turn back simultaneously and move at the same speeds relative to the water as before. Their engines are switched off when they reach the buoy again. If the maximum separation between the boats is 200 m after the buoy is dropped and water flow velocity is 1.5 m/s, find the distance( in metres) travelled by the buoy from the dropping point till the point where the boats meet the buoy again.


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Solution

Let after time t they turn back at C and D respectively.


d = 200 m
x = (4 - 15) t
x + 200 = (6 - 1.5) t
t = 100 s, x = 250 m
After turning, both boats take another 100 s to meet the buoy. So in total 200 sec, the distance moved by buoy = u × 200 = 1.5 × 200 = 300 m


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