Question 445754
During the first part of a trip, a canoeist travels 50 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 3hrs.
 What was the speed on each part of the trip?
:
Let s = speed on the 1st part of the trip
then
(s-5) = speed on the 2nd part
:
Write a time equation, Time = dist/speed
:
1st part time + 2nd part time = 3
{{{50/s}}} + {{{2/((s-5))}}} = 3
Multiply by s(s-5); results:
50(s-5) + 2s = 3s(s-5)
50s - 250 + 2s = 3s^2 - 15s
52s - 250 = 3s^2 - 15s
A quadratic equation
3s^2 - 15s - 52s + 250 = 0
3s^2 - 67s + 250 = 0
Solve this using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this problem x=s; a=3; b=-67; c=250
{{{s = (-(-67) +- sqrt(-67^2-4*3*250 ))/(2*3) }}} 
:
{{{s = (67 +- sqrt(4489-3000))/6 }}}
:
{{{s = (67 +- sqrt(1489))/6 }}}
Two solutions. only this one is reasonable
{{{s = (67 + 38.58756)/6 }}}
s = {{{105.58765/6}}}
s = 17.598 mph for the 1st 50 miles
then
17.598 - 5 = 12.598 mph for the last 2 mi
;
:
Check this by finding the actual times
50/17.598 = 2.84 hrs
2/12.598  = .16 hrs
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total time: 3 hrs