Question 1168865: Two boats ply back and forth across a river with different velocity, without
any loss in time. They leave the opposite shores at the same instant. They
meet the first time 900 meters from one shore and 500 meters the second
time they meet from the opposite shore. What is the width of the river?
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Okay, I'll give this one a try:
Let d = the width of the river
The first time they meet, boat 1 will have traveled 900 m and boat 2 will
have traveled d - 900 m
The travel time for boat 1 = 900/v1, and for boat 2 = (d-900)/v2
Setting these times equal we have 900/v1 = (d-900)/v2 -> v1/v2 = 900/(d-900)
For the 2nd meeting, boat 1 travels over to the far shore and back 500 m
and boat 2 travels the 900 m back to the near shore and then the width
of the river - 500 m.
The travel times for the 2nd meeting are:
t = ((d-900)+500)/v1 = (d-400)/v1
t = (900 + (d - 500)/v2 = (d+400)/v2
This gives v1/v2 = (d-400)/(d+400)
Using the relation above we can equate ratios of the speeds:
900/(d-900) = (d-400)/(d+400)
Solving for d, I get d = 2200 m
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