SOLUTION: During the first part of a trip a conoeist travels 65 miles at a certain speed. The conoeist travesl 11 miles on the second part of the trip at a speed of 5 mph slower. The total t

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Question 387547: During the first part of a trip a conoeist travels 65 miles at a certain speed. The conoeist travesl 11 miles on the second part of the trip at a speed of 5 mph slower. The total time for the trip is 5hr. What was the speed for each part of the trip?
Simplify your answer. Type an integer or a decimal. Round to the nearest hundredth.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
During the first part of a trip a canoeist travels 65 miles at a certain speed.
The canoeist travels 11 miles on the second part of the trip at a speed of 5 mph slower.
The total time for the trip is 5hr.
What was the speed for each part of the trip?
:
Let s = speed on the 1st part of the trip
then
(s-5) = speed the last 11 miles
:
write a time equation; time = dist/speed
:
65 mi time + 11 mi time = 5 hrs
65%2Fs + 11%2F%28%28s-5%29%29 = 5
multiply by s(s-5), results
65(s-5) + 11s = 5s(s-5)
:
65s - 325 + 11s = 5s^2 - 25s
:
76s - 325 = 5s^2 - 25s
Combine as a quadratic equation on the right:
0 = 5s^2 - 25s - 76s + 325
:
5s^2 - 101s + 325 = 0
Use the quadratic formula to find s:
s+=+%28-%28-101%29+%2B-+sqrt%28-101%5E2-4%2A5%2A325+%29%29%2F%282%2A5%29+
Do the math, two solutions, only one makes sense:
:
s = 16.18 mph for the 1st 65 mi
and, obviously
11.18 mph for the last 11 mi
:
:
Check solution by finding the times
65/16.18 = 4.02 hrs
11/11.18 = .98 hrs
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total time: 5 hrs