SOLUTION: At 6:00 a.m. a man left a town A and travelled toward a town B 19 miles away. At 6:30 another man started from B and travelled toward A. The two men met on the road at 8:30 a.m. If

Question 339383: At 6:00 a.m. a man left a town A and travelled toward a town B 19 miles away. At 6:30 another man started from B and travelled toward A. The two men met on the road at 8:30 a.m. If the rate of the first traveller is 1/2 mile per hour less than that of the second,find the rate of each.
Pls. answer.Thank you.
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
At 6:00 a.m. a man left a town A and traveled toward a town B 19 miles away. At 6:30 another man started from B and traveled toward A. The two men met on the road at 8:30 a.m. If the rate of the first traveler is 1/2 mile per hour less than that of the second, find the rate of each.
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6:00 man DATA:
time = 2.5 hrs ; rate = x mph ; distance = 2.5x miles
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6:30 man DATA:
time = 2 hrs ; rate = x+1/2 mph ; distance = 2(x+(1/2)) = 2x+1 miles
======================
Equation:
distance + distance = 19 miles
2.5x + 2x+1 = 19 miles
4.5x = 18
x = 4 mph (rate of the 6:00 man)
x+(1/2) = 4.5 mph (rate of the 6:30 man)
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Cheers,
Stan H.
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Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Write an equation for both men:

and

given:

---------------
I can rewrite the equations of motion

and

and, since both are equal to

The 1st man's rate is 4 mi/hr
The 2nd man's rate is 4.5 mi/hr
check:

mi
and

OK

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