<PRE><B>Question 526139</B>
During the first part of a trip, a canoeist travels 61 miles at a certain speed.
 The canoeist travels 2 miles on the second part of the trip at a speed 5 mph slower.
 The total time for the trip is 3 hours. What was the speed on each part of the trip?
:
Let s = the speed for the 1st 61 mi
then
(s-5) = the speed for the last 2 mi
:
Write a time equation, time = dist/speed
:
faster time + slower time = 3 hrs
{{{61/s}}} + {{{2/(s-5)}}} = 3
:
multiply by s(s-5), results:
61(s-5) + 2s = 3s(s-5)
:
61s - 305 + 2s = 3s^2 - 15s
63s - 305 = 3s^2 - 15s
Arrange as a quadratic equation on the right
0 = 3s^2 - 15s - 63s + 305
3s^2 - 78s + 305 = 0
Use the quadratic formula to find s:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this problem, a=3, b=-78, c=305, x=s
{{{s = (-(-78) +- sqrt(-78^2-4*3*305 ))/(2*3) }}}
:
{{{s = (78 +- sqrt(6084-3660 ))/6 }}}
:
{{{s = (78 +- sqrt(2424))/6 }}}
Two solutions
{{{s = (78 + 49.234)/6 }}}
s = {{{127.234/6}}}
s = 21.206  
and
{{{s = (78 - 49.234)/6 }}}
s = {{{28.766/6}}}
s = 4.794
:
Obviously the only reasonable solution:
s = 21.206 mph on the first 61 mi
and
 16.206 mph on the last two miles 
:
:
See if that check out, find the actual time of each
61/21.206 = 2.876 hrs
2/16.206  =  .123 hrs
----------------------
total time: 3.0 hrs </PRE>