SOLUTION: PLEASE HELP! A cyclist traveled 60 miles at a constant before reducing the speed by 2 mph. Another 40 miles was traveled at the reduced speed. The total time for the 100-mile trip

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Question 446986: PLEASE HELP! A cyclist traveled 60 miles at a constant before reducing the speed by 2 mph. Another 40 miles was traveled at the reduced speed. The total time for the 100-mile trip was 9 hours. Find the rate during the first 60 miles.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A cyclist traveled 60 miles at a constant before reducing the speed by 2 mph.
Another 40 miles was traveled at the reduced speed.
The total time for the 100-mile trip was 9 hours.
Find the rate during the first 60 miles.
:
Let s = rate during the 1st 60 mi
then
(s-2) = reduced rate for the last 40 mi
:
Write a time equation; time = dist/speed
:
Normal rate time + reduced rate time = 9 hrs
60%2Fs + 40%2F%28%28s-2%29%29 = 9
multiply by s(s-2), results
60(s-2) + 40s = 9s(s-2)
60s - 120 + 40s = 9s^2 - 18s
100s - 120 = 9s^2 - 18s
Arrange as a quadratic equation on the right
0 = 9s^2 - 18s - 100s + 120
9s^2 - 118s + 120 = 0
You can solve for s using the quadratic formula, but this will factor to
(9s-10)(s-12) = 0
Two solutions, but only one is reasonable
9s = 10
and
s = 12 mph, is the speed on the 1st 60 mi
:
:
Confirm this by finding the times
60/12 + 40/10 =
5 + 4 = 9 hrs