Two boats on opposite shores of a river start moving towards each other. When they pass each other, they are 750 yards from one of the shorelines. They each continue to the opposite shore, and once they reach it, immediately turn around and start back. When they pass again, they are 250 yards from the other shoreline. Each boat maintains a constant speed throughout (i.e. quick sharp turns). How wide is the river?
Source: R & S (WWU students)
Hint: Draw a diagram....what formula expresses useful information relative to the problem?
Solution Commentary: Variation of soution commentary from R&S: Let d be the river's width, and rA and rB represent the rates of the two boats A and B. Then, meeting on the first trip across, t1 = 750/rA =(d - 750)/rB. Massaging this with some algebra, we get rA = (750 rB)/(d - 750).
And meeting on the trip back, things are a little more complicated. We have t2 = (d - 750 + 250)/rA =(750 + d - 250)/rB. Massaging this with some algebra, we get rA = [(d - 500)( rB)]/(d + 500).
Setting the two rA equal, the rB cancels and we get d = 2000.
Do you agree....What other solution approaches can you find?