<PRE><B>Question 138837</B>
Debbie traveled by boat 5 miles upstream. Because of the 4-mph 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 of boat in still water
then
(s-4) = speed upstream
and
(s+4) = speed downstream
:
Change 40 min to hrs: 40/60 = {{{2/3}}}hr
:
Write a time equation: Time = Dist/speed:
Time upstream = time downstream + 2/3 hr
{{{5/((s-4))}}} = {{{5/((s+4))}}} + {{{2/3}}}
:
Multiply equation by 3(s+4(s-4) to get rid of the denominators:
3(s+4)(s-4)*{{{5/((s-4))}}} = 3(s+4)(s-4)*{{{5/((s+4))}}} + 3(s+4)(s-4)*{{{2/3}}}
Cancel out the denominators:
3(s+4)*5 = 3(s-4)*5 + 2(s+4)(s-4)
:
15(s+4) = 15(s-4) + 2(s^2 - 16)
:
15s + 60 = 15s - 60 + 2s^2 - 32
:
0 = 2s^2 + 15s - 15s - 60 - 60 - 32; combine on the right
:
2s^2 - 152 = 0
:
2s^2 = 152
s^2 ={{{152/2}}}
s = {{{sqrt(76)}}}
s = 8.7 mph speed in still water
:
:
Check the times of each trip, using decimals:
5/4.7 = 1.064 hrs
5/12.7 = .394 hrs
------------------
Differs .67 hrs; confirms our solution</PRE>