<PRE><B>Question 565547</B>
The Special leaves campus at 10:00AM and is travelling at a constant rate of 40 miles per hour. A car full of fans leaves campus at 11:30 AM and travels the same route at a constant rate of 70 miles per hour. At what time will the car catch up with the Special?
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Special DATA:
rate = 40 mph ; time = x hrs ; distance = 40x miles
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Fan DATA:
rate = 70 mph ; time = (x-(3/2)) hrs ; distance = 70(x-(3/2)) miles
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Equation:
distance = distance
40x = 70(x-(3/2))
40x = 70x - 105
30x = 105
x = 3.5 hrs
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Catch-up time would be 10AM + 3 1/2 hrs = 1:30 PM
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Cheers,
Stan H.
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