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Q.

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 is dropped from the slower one. After lapse of 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 between the two places where the boats meet the buoy is found to be 100×n meters, the value of ‘n’ is ______________.

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answer is 3.

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Detailed Solution

Let after time t they turn back at C and D respectively.d=200mx=(4−1.5)tx+200=(6−1.5)tt=100s,x=250mAfter turning, both boats take another 100 s to meet the buoy.So in total 200 s, the distance moved by buoy - u x 200 = 1.5 x 200 = 300 m
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