SOLUTION: At 10:00 A.M. a car leaves a house at a rate of 60mi/h. At the same time another car leaves the same house at a rate of 50mi/h in the opposite direction. At what time will the car

Question 108296: At 10:00 A.M. a car leaves a house at a rate of 60mi/h. At the same time another car leaves the same house at a rate of 50mi/h in the opposite direction. At what time will the cars be 330 miles apart?
Found 2 solutions by stanbon, wgunther:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
At 10:00 A.M. a car leaves a house at a rate of 60mi/h. At the same time another car leaves the same house at a rate of 50mi/h in the opposite direction. At what time will the cars be 330 miles apart?
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1st car DATA:
rate = 60 mph ; time = x hrs ; distance = rt = 60x miles
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2nd car DATA:
rate = 50 mph ; time = x hrs ; distance = rt = 50x miles
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EQUATION:
distance + distance = 330 miles
60x + 50x = 330
110x = 330
x = 3 hrs
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10:00 AM + 3 hrs = 1:PM
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Cheers,
Stan H.
Answer by wgunther(43) (Show Source): You can put this solution on YOUR website!
if you draw a picture you get a better idea. Call the first car Car A and the second Car B. Car A is moving 60mi/h from the house, and Car B is moving 50mi/h from the house the other direction.
So, we can come up with 2 equations, one to describe each Car. Define Car A's motion to be positive, which means Car B's is negative as it's in the opposite direction.

so, we want to find when the distance between the two is 330. Distance is the absolute value of the difference.

The absolute value drop as t must be positive (it's time). Then, we're looking when this distance is 330

So the answer is 3 hours.
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