Question 187033
 The Ohio River flows at a rate of 5 miles per hour. 
A patrol boat travels 40 miles upriver and returns in a total of 6 hours.
 What is the speed of the boat in still water? 
:
Let s = boat speed in still water
then
(s+5) = boat speed down-stream
and
(s-5) = boat speed up-stream
:
Write a time equation; Time = {{{dist/speed}}}
;
Upstream time + downstream time = 6 hrs
{{{40/((s-5))}}} + {{{40/((s+5))}}} = 6
:
Multiply equation by (s-5)(s+5) and you have:
40(s+5) + 40(s-5) = 6(s+5)(s-5)
:
40s + 200 + 40s - 200 = 6(s^2 - 25)
:
80s = 6s^2 - 150
:
0 = 6s^2 - 80s - 150; a quadratic equation
Simplify, divide by 2
3s^2 - 40s - 75 = 0
Factors to:
(3s + 5) (s - 15) = 0
The positive solution is what we want
s = 15 mph is the speed in still water
;
;
Check solution in our original equation
40/20 = 2 hr
40/10 = 4 hr
;
:
Did this reduce the confusion somewhat? Any questions?