SOLUTION: Extreme hardship. Kim starts to walk 3 mi to school at 7:30 A.M. with a temperature of 0°F. Her brother Bryan starts at 7:45 A.M. on his bicycle, traveling 10 mph faster than Kim.
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-> SOLUTION: Extreme hardship. Kim starts to walk 3 mi to school at 7:30 A.M. with a temperature of 0°F. Her brother Bryan starts at 7:45 A.M. on his bicycle, traveling 10 mph faster than Kim.
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Question 161020: Extreme hardship. Kim starts to walk 3 mi to school at 7:30 A.M. with a temperature of 0°F. Her brother Bryan starts at 7:45 A.M. on his bicycle, traveling 10 mph faster than Kim. If they get to school at the same time, then how fast is each one traveling? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Kim starts to walk 3 mi to school at 7:30 A.M. with a temperature of 0°F. Her brother Bryan starts at 7:45 A.M. on his bicycle, traveling 10 mph faster than Kim. If they get to school at the same time, then how fast is each one traveling?
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Change 15 min to hrs: 15/60 = .25 hrs
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Let s = K's walking speed
then
(s+10) = B's biking speed
:
Write a time equation: Time =
:
K's travel time = B's travel time + .25 hr = + .25
:
Multiply equation by t(t+10) to get rid of the denominator, results
3(s+10) = 3s + .25(s(s+10)
:
3s + 30 = 3s + .25s^2 + 2.5s
Arrange as a quadratic equation on the right
0 = .25s^2 + 2.5s + 3s - 3s - 30
:
.25s^2 + 2.5s - 30 = 0
Get rid of those decimals, mult equation by 4, results
s^2 + 10s - 120 = 0
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Use the quadratic formula to solve for s:
In this equation a=1, b=10, c=-120
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solve this you should get a postive solution of:
s = 7.05 mph is K's walking speed
then
7.05 + 10 = 17.05 mph is B's biking speed
:
:
Confirm this by finding the time of each
3/7.05 = .425 hrs or 25.5 min
3/17.05 = .176 hrs or 10.6 min
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Difference ~ 15 min
