    Question

# Two motor boats that can move with velocities 4 m/s and 6 m/s relative to water are going downstream in a river. When the boats overtake for the 1st time a buoy(a navigation mark) is dropped from one of the boats. After sometime, both boats turn back simltaneously and move upstream. If maximum separation between boats is 200 meters after buoy is dropped and river flow velocity is 1.5 m/s, then

A
Time interval after which boats turn is 100 sec.
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B
Seperation between fast boat and buoy when boats turn is 600 m
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C
Distance between places where faster boat passes by the buoy is 300 m
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D
Distance between places where faster boat passes by the buoy is 450 m
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Solution

## The correct option is C Distance between places where faster boat passes by the buoy is 300 mLets look at the problem with respect to the river. This means, the speeds of the two boats when they first meet are 4 m/s and 6 m/s respectively. The difference between their velocities is 6 m/s−4 m/s=2 m/s. The speration between the boats when they turn is 200 m. Therefore, the time after which the boats turn is 2002=100 s. So, option (a) is correct. In 100 s, the faster boat travels a distance of 6×100=600 m. So, option (b) is correct. They got seperated for 100s, so it takes again 100 s for them to meet, so they will meet after 200s. In the mean time, the distance travelled by the river is 1.5×200= 300m. Therefore, option (c) is correct.  Suggest Corrections  0      Related Videos   Speed and Velocity
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