Problema Solution
I went canoeing on the Winding River last Saturday. I traveled 15 miles downstream and then turned around and traveled the same 15 miles upstream. The trip took me a total of 4 hours. If my canoe travels 8mph (miles per hour) in still water, what is te speed of the current river?
Answer provided by our tutors
let
d = 15 miles
t1 = the time of the trip downstream
t2 = the time of the trip upstream
v = 8 mph the speed of the canoe is still water
w = the speed of the current, w>0
the trip took me a total of 4 hours
t1 + t2 = 4 => t2 = 4 - t1
we will use the formula speed = distance/time
traveling downstream the speed is v + w
(v + w)*t1 = d
(8 + w)*t1 = 15 => t1 = 15/(8 + w)
traveling upstream the speed is v - w
(v - w)*t2 = d
(8 - w)*t2 = 15
plug t2 = 4 - t1 into the previous equation
(8 - w)*(4 - t1) = 15
plug t1 = 15/(8 + w) into the last equation
(8 - w)*(4 - 15/(8 + w)) = 15
by solving we find and consider the positive roots
w = 2 mph
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the speed of the current is 2 miles per hour.