SOLUTION: A truck traveled the first 100 miles of a trip at one speed and the last 135 miles at an average speed of 5 miles per hour less. The entire trip took 5 hours. What was the average

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Question 1058176: A truck traveled the first 100 miles of a trip at one speed and the last 135 miles at an average speed of 5 miles per hour less. The entire trip took 5 hours. What was the average speed for the first part of the trip?
Found 2 solutions by stanbon, Boreal:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A truck traveled the first 100 miles of a trip at one speed and the last 135 miles at an average speed of 5 miles per hour less. The entire trip took 5 hours. What was the average speed for the first part of the trip?
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1st leg DATA:
dist = 100 miles ; rate = x mph ; time = d/r = 100/x hrs
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2nd leg DATA:
dist = 135 miles ; rate = x-5 mph ; time = 135/(x-5) hrs
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Equation:
100/x + 135/(x-5) = 5
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100(x-5) + 135x = 5x(x-5)
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235x - 500 = 5x^2 - 25
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5x^2 - 235x + 475 = 0
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x^2 - 47x + 95 = 0
x = 44.68 mph
x-5 = 39.68 mph
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Cheers,
Stan H.
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Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
100 miles at x mph, and the time was 100/x hours. Speed*time=distance.
135 miles at (x-5) mph, and the time was 135/(x-5).
Those two add to 5.
(100/x)+(135/(x-5)=5
multiply everything by x*(x-5)
100(x-5)+135x=5x^2-25x
100x-500+135x=5x^2-25x
5x^2-260x+500=0
x^2-52x+100=0, dividing everything by 5
(x-2)(x-50)=0
x=2,50, only 50 makes sense
Truck went 50 mph for the first two hours ANSWER
It went 45 mph for the next 3 hours, or 135 miles.