SOLUTION: If a car had increased its average speed for a 18-mile journey by 5 mph, the journey would have been completed in 30 minutes less. What was the car's original average speed?
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Question 163342: If a car had increased its average speed for a 18-mile journey by 5 mph, the journey would have been completed in 30 minutes less. What was the car's original average speed? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! If a car had increased its average speed for a 18-mile journey by 5 mph, the journey would have been completed in 30 minutes less. What was the car's original average speed?
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Just looking at this we have to assume the car's speed will be awfully slow!
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Let s = car's original speed
Then
(s+5) = cars increased speed
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The distances at both speeds is given as 18 mi
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Change 30 min to .5 hrs
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Write a time equation: Time =
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Original speed time - half hour = increased speed time - .5 =
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Multiply equation by s(s+5):
s(s+5)* - s(s+5)*.5 = s(s+5)*
cancel the denominators and you have:
18(s+5) - .5(s^2 + 5s) = 18s
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Multiply what's in the brackets
18s + 90 - .5s^2 - 2.5s = 18s
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Arrange as a quadratic equation:
-.5s^2 + 18s - 18s - 2.5s + 90 = 0
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-.5s^2 - 2.5s + 90 = 0
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Multiply equation by -2 to get rid of the decimals and change the signs;
s^2 + 5s - 180
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This will not factor, use the quadratic formula:
In this problem a=1; b=5; c=-180
Find the positive solution here:
s =
s = 11.15 mph is the original speed
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Check solution by finding the times
11.15 + 56 = 16.15 is the faster speed
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18/11.15 = 1.6 hrs
18/16.15 = 1.1 hrs
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differ by .5 hrs as indicated
