Question 328753
During the first part of a trip, a canoeist travels 87 miles at a certain speed.
 The canoeist travels 9 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 the FIRST part of the trip?
 What was the speed on the SECOND part of the trip?
(type an integer or a decimal. Round to the nearest hundredth)
:
Let s = "a certain speed on the first part of the speed"
then
(s-5) = "speed on the second part of the trip"
:
:
Write a time equation: Time = dist/speed
:
1st part time + 2nd part time = 5 hrs
{{{87/s}}} + {{{9/((s-5))}}} = 5
:
Multiply equation by s(s-5), results:
87(s-5) + 9s = 5s(s-5)
:
87s - 435 + 9s = 5s^2 - 25s
:
96s - 435 = 5s^2 - 25s
Arrange as a quadratic equation
5s^2 - 25s - 96s + 435 = 0
:
5s^2 - 121s + 435 = 0
Use the quadratic formula to find s:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
in this problem: x=s; a=5; b=-121; c= 435
{{{s = (-(-121) +- sqrt(-121^2-4*5*435 ))/(2*5) }}} 
:
{{{s = (121 +- sqrt(14641 - 8700 ))/10 }}} 
:
{{{s = (121 +- sqrt(5941 ))/10 }}}
Two solutions, but this is the one that makes sense
{{{s = (121 + 77.08)/10 }}}
s = {{{198.08/10}}}
s = 19.81 mph on the first part of the trip
then 
19.81 - 5 = 14.81 mph on the 2nd part  
:
:
Check solution
{{{87/19.81}}} + {{{9/14.81}}} = 
4.39 + .61 = 5 hrs, confirms our solutions