Stranger_1973 wrote:Two boats, each travelling at its own constant speed, start at the same time from Shores A and B, respectively, and power directly across the lake. The boats first pass each other 200 yards from Shore A and next pass each other (after each has reached and immediately left its opposite shore) 20 seconds later, 100 yards from Shore B. Calculate the speed of the faster boat, in feet per second.
This is a bit complex. Let's take it slow, relying, as ever, on "d = rt"
Whatever the width of the river, let's call that "d". Then we have Boat A travelling 600 feet at speed va
and meeting Boat B at some time "t". At this point, Boat B has travelled d - 600 feet at speed vb
. Then:. . . . .. . . . .
In the second part, Boat A completed one pass, covering the d - 600 that Boat B had covered, and then came back another 300 feet, for a total distance of d - 300 in the twenty seconds. Boat B, meanwhile, finished the 600 feet that Boat A had covered, and then came back to 300 feet from its own side, for a total distance of 600 + (d - 300) = d + 300 in that same twenty seconds.. . . . .. . . . .
This gives you four equations in four unknowns, being d, t, va
, and vb
. Solve the system.
(I would suggest solving the first pair for "d=", by addition. Then add the second pair, and solve again for "d=". By comparison, you'll be able to find the value of "t". Then work backwards to find the speeds.)