Question 517975
if one car goes thirty miles per hour, and another goes 60 miles per hour but leaves two hours later, when will the meet up
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slower car DATA:
rate = 30 mph ; time = x hrs ; distance = rt = 30x miles
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faster car DATA:
rate = 60 mph ; time = x-2 hrs; distance = rt = 60(x-2) miles
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Equation:
distance = distance
30x = 60(x-2)
30x = 60x - 60*2
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30x = 60*2
x = 4 hrs (time for faster to catch up to slower)
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Cheers,
Stan H.
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