SOLUTION: In a 10 mile race, Janet covered the first 2 miles at a constant rate. She then sped up and rode her bike the last 8 miles at a rate that was 0.5 miles per minute faster. Janet's o

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Question 843391: In a 10 mile race, Janet covered the first 2 miles at a constant rate. She then sped up and rode her bike the last 8 miles at a rate that was 0.5 miles per minute faster. Janet's overall time would have been 2 minutes faster had she ridden her bike the whole race at the faster pace. What was Janet's average speed (in miles per minute) for the whole race?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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In a 10 mile race, Janet covered the first 2 miles at a constant rate.
She then sped up and rode her bike the last 8 miles at a rate that was 0.5 miles per minute faster.
Janet's overall time would have been 2 minutes faster had she ridden her bike the whole race at the faster pace.
What was Janet's average speed (in miles per minute) for the whole race?
:
Let s = the starting speed for the 1st two miles
then
(s+.5) = the speed for the last 8 mi of the race
:
Actual time = 2%2Fs + 8%2F%28%28s%2B.5%29%29
faster time = 10%2F%28%28s%2B.5%29%29
:
Actual time - faster time = 2 min
2%2Fs + 8%2F%28%28s%2B.5%29%29 - 10%2F%28%28s%2B.5%29%29 = 2
multiply equation by s(s+.5), cancel the denominators:
2(s+.5) + 8s - 10s = 2s(s+.5)
2s + 1 - 2s = 2s^2 + s
combine on right to form a quadratic equation
0 = 2s^2 + s - 1
Factors to
(2s-1)(s+1) = 0
the positive solution
2s = 1
s = .5 mi/min, starting speed, then adding .5, 1 mi/min is the ending speed
:
" What was Janet's average speed (in miles per minute) for the whole race?"
let a = the av speed
2%2F.5 + 8%2F1 = 10%2Fa
4 + 8 = 10/a
12a = 10
a = 10/12
a = .83 mi/min av speed for the race
:
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You can check this for yourself, find the actual times for each scenario