Question 161270
Cathy traveled by boat 4 miles upstream to fish. Because of the 5mph current it took her 40 minutes longer to get there than to return. How fast will her boat go in still water?
:
Let s = speed in still water
then
(s+5) = speed downstream
and
(x-5) = speed upstream
:
Change 40 min to hrs: 40 min = {{{2/3}}} hr
:
Write time equation; time = {{{dist/speed}}}
:
Downstream time + 2/3 hr = upstream time
{{{4/((s+5))}}} + {{{2/3}}} = {{{4/((s-5))}}}
:
Multiply equation by 3(s-5)(s+5) to get rid of the denominators
:
4(3(s-5)) + 2(s-5)(s+5) = 4(3(s+5))
:
12s - 60 + 2(s^2 - 25) = 12s + 60
:
12s - 60 + 2s^2 - 50 = 12s + 60
Combine like terms on the left
2s^2 + 12s - 12s - 60 - 50 - 60 = 0
:
2s^2 - 170 = 0
:
2s^2 = +170
s^2 = {{{170/2}}}
s^2 = 85
s = {{{sqrt(85)}}}
s = 9.22 mph in still water
:
:
Check solution on a calc:
4/14.22 + .67 = .95
4/4.22 = .95