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

Algebra ->  Square-cubic-other-roots -> SOLUTION: Debbi traveled by boat by 5 miles upstream to fish in her favorite spot. Because of the 4-mhp current, it took her 20 minutes longer to get there than to return. How fast will her       Log On


   

Question 123331: Debbi traveled by boat by 5 miles upstream to fish in her favorite spot. Because of the 4-mhp 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!
Debbi traveled by boat by 5 miles upstream to fish in her favorite spot. Because of the 4-mhp current, it took her 20 minutes longer to get there than to return. How fast will her boat go in still water?
:
Let s = boat speed in still water
then
(s-4) = speed upstream
and
(s+4) = speed downstream
:
Change 20 min to hrs: 20/60 = 1%2F3 hr
:
Write a time equation: Time = Distance%2Fspeed
:
Time upstream - time downstream = 20 min (1/3 hr)
5%2F%28%28s-4%29%29 - 5%2F%28%28s%2B4%29%29 = 1%2F3
Multiply equation by the common denominator 3(s-4)(s+4) and you have:
:
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
:
120 + 16 = s^2
:
s^2 = 136
s = sqrt%28136%29
s = 11.66 mph boat speed in still water
:
:
Check solution in original equation using decimals:
speed upstream = 7.66
speed downstream = 15.66
5%2F%287.66%29 - 5%2F%2815.66%29 =
.653 - .319 = .333; confirms our solution of s = 11.66