<PRE><B>Question 387547</B>
During the first part of a trip a canoeist travels 65 miles at a certain speed.
 The canoeist travels 11 miles on the second part of the trip at a speed of 5 mph slower.
 The total time for the trip is 5hr.
 What was the speed for each part of the trip?
:
Let s = speed on the 1st part of the trip
then
(s-5) = speed the last 11 miles
:
 write a time equation; time = dist/speed
:
65 mi time + 11 mi time = 5 hrs
{{{65/s}}} + {{{11/((s-5))}}} = 5
multiply by s(s-5), results
65(s-5) + 11s = 5s(s-5)
:
65s - 325 + 11s = 5s^2 - 25s
:
76s - 325 = 5s^2 - 25s
Combine as a quadratic equation on the right:
0 = 5s^2 - 25s - 76s + 325
:
5s^2 - 101s + 325 = 0
Use the quadratic formula to find s:
{{{s = (-(-101) +- sqrt(-101^2-4*5*325 ))/(2*5) }}}
Do the math, two solutions, only one makes sense:
:
s = 16.18 mph for the 1st 65 mi
and, obviously
11.18 mph for the last 11 mi
:
:
Check solution by finding the times
65/16.18 = 4.02 hrs
11/11.18 = .98 hrs
------------------
total time: 5 hrs
 

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