<PRE><B>Question 806891</B>
During the first part of a trip a canoeist travels 18 miles at a certain speed.
 The canoeist travels 4 miles on the second part of the trip at a speed of 5 mph slower.
 The total time for the trip is 3 hrs.
 What was the speed on each part of the trip?
:
Let s = speed on the first part of the trip
then
(s-5) = speed on the last part
:
Write a time equation; time = dist/speed:
:
Time for the 1st 18 mi + time for the last 4 mi = 3 hrs
{{{18/s}}} + {{{4/(s-5)}}} = 3
multiply by s(s-5), cancel the denominators, resulting in:
18(s-5) + 4s = 3s(s-5)
18s - 90 + 4s = 3s^2 - 15s
22s - 90 = 3s^2 - 15s
a quadratic equation
0 = 3s^2 - 15s - 22s + 90
3s^2 - 37s + 90 = 0
We can us the quadratic formula, but this will factor to
(3s-10)(s-9) = 0
s = 9 mph on the 1st part of the trip is the reasonable solution
then
9-5 = 4 mph is the speed on the last part
:
:
Check this, find the time at each speed
18/9 = 2 hrs
4/4 = 1 hr
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total 3 hrs</PRE>