<PRE><B>Question 405959</B>
A boat traveled downstream a distance of 95 miles and then came right back.
 If the speed of the current was 8 mph and the total trip took 6 hours and 
20 min, find the average speed of the boat relative to the water
:
Let s = boat speed in still water
then
(s+8) = speed (relative to land) downstream
and
(s-8) = speed upstream
:
Change 6 hr 20 min to 6{{{1/3}}} hrs
:
Write a time equation: time = dist/speed
:
{{{95/(s+8)}}} + {{{95/(s-8)}}} = 6{{{1/3}}}
change to an improper fraction
{{{95/(s+8)}}} + {{{95/(s-8)}}} = {{{19/3}}}
:
Multiply by 3(s+8)(s-8), results:
95*3(s-8) + 95*3(s+8) = 19(s+8)(s-8)
:
95(3s-24) + 95(3s+24) = 19(s^2-64)
:
285s - 2280 + 285s + 2280 = 19s^2 - 1216
:
570s = 19s^2 - 1216
:
A quadratic equation
19s^2 - 570s - 1216 = 0
divide by 19
s^2 - 30s - 64 = 0
Factors to
(s+2)(s-32) = 0
The positive solution
s = 32 mph is the boat speed relative to the water
:
:
Confirm this by finding the time for each half of the trip
95/40 = 2.375 hrs
95/24 = 3.958 hrs
------------------
total = 6.333 hrs which is 6 hr 20 min
</PRE>