How far apart are the ferryboats when they start their return trips?

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Discussion Overview

The discussion revolves around a problem involving two ferryboats traveling across a river, focusing on determining the width of the river based on their travel distances and meeting points. The scope includes mathematical reasoning and problem-solving related to geometry and relative motion.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the width of the river can be calculated using the distances the ferryboats travel before and after they meet, proposing a relationship between these distances.
  • Another participant confirms the calculated width of the river as 1760 yards, indicating agreement with the previous calculation.
  • A later reply acknowledges the correctness of the previous contributions without introducing new calculations or perspectives.

Areas of Agreement / Disagreement

Participants generally agree on the calculated width of the river being 1760 yards, with no significant disagreement presented in the discussion.

Contextual Notes

The discussion relies on the assumptions regarding the constant speeds of the ferryboats and the geometry of their paths across the river. There may be additional considerations regarding the time spent in slips that are not fully explored.

moe darklight
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post your answers in white :biggrin: . no pen or paper allowed.

Two ferryboats start at the same instant from opposite sides of a river, traveling across the water on routes at right angles to the shores. Each travels at a constant speed, but one is faster than the other. They pass at a point 720 yards from the nearest shore. both boats remain in their slips for 20 minutes before starting back. On the return trips they meet 400 yards from the other shore.

How wide is the river?I didn't get the right answer and then I felt like an idiot because the answer is really easy :frown: .
 
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The answer is:
1760[/color]
yards
 
moe darklight said:
How wide is the river?

Well,
at the two moments the ferryboats meet, the relation between the distances covered by them is the same:
(w - a) / a = (2w - b) / (w + b)
So,
w = 3a - b = 3*720 - 400
w = 1760 yards

:smile:
 
oops, totally forgot about this thread— yep, both right :biggrin: .
 

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