Question 376588
During the first part of a trip, a canoeist travels 71 miles at a certain speed.
 The canoeist travels 12 miles on the second part of the trip at a speed 5 mph slower.
 The total time for the trip is 5 hours. What was the speed on EACH part of the trip?
:
Let s = speed for the 1st part of the trip
then
(s-5) = speed on the 2nd part of the trip
:
Write time equation; time = dist/speed
:
1st speed time + 2nd speed time = 5 hrs
{{{71/s}}} + {{{12/((s-5))}}} = 5
Multiply by s(s-5) to clear the denominators, results:
71(s-5) + 12s = 5s(s-5)
71s - 355 + 12s = 5s^2 - 25s
83s - 355 = 5s^2 - 25s
0 = 5s^2 - 25s - 83s + 355
5s^2 - 108s + 355 = 0
Use the quadratic formula to find s
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
Where x = s; a=5; b=-108; c=355
{{{s = (-(-108) +- sqrt(-108^2-4*5*355 ))/(2*5) }}}
:
{{{s = (108 +- sqrt(11664-7100 ))/10 }}}
:
{{{s = (108 +- sqrt(4564 ))/10 }}}
two solutions
{{{s = (108 + 67.557)/10 }}}
s = {{{175.557/10}}}
s = 17.5557
and
{{{s = (108 - 67.557)/10 }}}
s = {{{40.443/10}}}
s = 4.0443
:
s = 17.5557 mph is the only reasonable speed for the 1st part of the trip
and
17.5557 - 5 = 12.5557 mph is the speed on the 2nd part of the trip
:
:
See if this is true, find the time at each speed
71/17.5557 = 4.044 hrs
12/12.5557 = .9557
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total time: 4.9997 ~ 5 hrs