Question 781803: Two cars enter the Florida Turnpike at commercial Boulevard at 8 AM, each heading for Wildwood. One car's average speed is 10 mph more than the other's. The faster car arrives at Wildwood at 11 AM, 1/2 hour before the other car. What is the average speed of each car? How far did each travel?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Two cars enter the Florida Turnpike at commercial Boulevard at 8 AM, each heading for Wildwood. One car's average speed is 10 mph more than the other's. The faster car arrives at Wildwood at 11 AM, 1/2 hour before the other car. What is the average speed of each car? How far did each travel?
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Faster Car DATA:
rate = x + 10 mph ; time = 3 hrs ; distance = 3x+30 miles
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Slower Car DATA:
rate = x mph ; time = 3.5 hrs ; distance = 3.5x miles
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Equation:
distance = distance
3.5x = 3x+30
(1/2)x = 30
x = 60 mph (slower car rate)
x+10 = 70 mph (faster car rate)
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Faster car distance = 3x+30 = 180 + 30 = 210 miles
Slower car distance: = 3.5*60 = 210 miles
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Cheers,
Stan H.
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