SOLUTION: Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4 mph current, it took her 20 minutes longer to get there than to return. How fast will her bo

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4 mph current, it took her 20 minutes longer to get there than to return. How fast will her bo      Log On


   

Question 99494: Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4 mph current, it took her 20 minutes longer to get there than to return. How fast will her boat go in still water?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Debbie traveled by boat 5 miles upstream. Because of the 4 mph current, it took her 20 minutes longer to get there than to return. How fast will her boat go in still water?
:
Let s = speed of the boat in still water
Then
(s+4) = speed down-stream
and
(s-4) = speed up-stream
:
Since we are dealing in mph, change 20 min to hrs: 20/60 = 1/3 hr
:
Write a time equation: Time = distance/speed
:
Time up-stream - Time down-stream = 1/3 hr
5%2F%28%28s-4%29%29 - 5%2F%28%28s%2B4%29%29 = 1%2F3
:
Multiply equation by 3(s-4)(s+4), eliminates the denominator, resulting in:
3(s+4)*5 - 3(s-4)*5 = (s+4)(s-4)
:
15(s+4) - 15(s-4) = s^2 - 16
:
15s + 60 - 15s + 60 = s^2 - 16
:
120 = s^2 -16
or
s^2 - 16 = 120
:
s^2 = 120 + 16
:
s^2 = 136
:
s = sqrt%28136%29
:
s = 11.66 mph in still water
:
:
Check solution by finding the times:
5/(11.66-4) = .6527 hrs
5/(11.66+4) = .3193
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difference = .3334 hrs which ~ 20 min
:
:
How about this? Did it make sense to you?