SOLUTION: two cars leave a city on the same road, one driving 12 mph faster the the other. After 4 hours , the car traveling faster stops for lunch. after 4 hours and 30 minutes the car trav
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Question 462377: two cars leave a city on the same road, one driving 12 mph faster the the other. After 4 hours , the car traveling faster stops for lunch. after 4 hours and 30 minutes the car traveling slower stops for lunch. assuming that the person in the faster car is still eating lunch, the cars are now 24 miles apart. How fast is each driving?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
two cars leave a city on the same road, one driving 12 mph faster then the other.
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Slower car rate = "x" mph.
Faster car rate = "x+12" mph
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After 4 hours, the car traveling faster stops for lunch.
Faster car distance = 4*(x+12) = 4x+48 miles
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After 4 hours and 30 minutes the car traveling slower stops for lunch.
Slower car distance = 4.5x miles
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Assuming that the person in the faster car is still eating lunch, the cars are now 24 miles apart.
How fast is each driving?
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faster car distance - slower car distance = 24 miles
4x+48 - 4.5x = 24
-0.5 = -24
x = 48 mph (speed of the slower car)
x+12 = 60 mph (speed of the faster car)
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Cheers,
Stan H.
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