Question

Two motor boats that can move with velocities $4\text{m/s}$ and $6\text{m/s}$ relative to water are going downstream in a river. When the boats overtake for the $1^{st}$ 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 $1.5\text{m/s}$, then

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

Answer 1

Answer: (A), (B), (C)

Distance between places where faster boat passes by the buoy is $300$ m

Lets 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 $6m/s-4m/s=2m/s$. The speration between the boats when they turn is 200 m. Therefore, the time after which the boats turn is $\frac{200}{2}=100s$. So, option (a) is correct.
In $100s$, the faster boat travels a distance of $6\times 100=600m$.
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 $1.5\times 200=300m$.
Therefore, option (c) is correct.