SOLUTION: I am attempting to solve an applied problem and having trouble setting it up.
Train A is traveling at 60 mph and train B is traveling at 80 mph, train A passes a bust stop at 7:25
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Train A is traveling at 60 mph and train B is traveling at 80 mph, train A passes a bust stop at 7:25
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Question 147707: I am attempting to solve an applied problem and having trouble setting it up.
Train A is traveling at 60 mph and train B is traveling at 80 mph, train A passes a bust stop at 7:25 AM and train B passes the bus stop at 7:40 AM, what time will B catch up with A?
Is this a situation where I find the slope
80-60/7:40 – 7:25 20/15
y-20= 20/15(x – 15)
y= 20/15x -20 +20
y=20/15x
I do not believe this is correct. Can you help me please?
You can put this solution on YOUR website! Train A is traveling at 60 mph and train B is traveling at 80 mph, train A passes a bus stop at 7:25 AM and train B passes the bus stop at 7:40 AM, what time will B catch up with A?
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Train A DATA:
rate = 60 mph ; time = x hrs ; distance = 60x miles
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Train B DATA:
rate = 80 mph ; time = (x-(1/4)) hr; distance = 80(x-(1/4) = (80x - 20) miles
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EQUATION:
distance for A = distance for B
60x = 80x - 20
20x = 20
x = 1 hr
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B will catch A at 8:25
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Cheers,
Stan H.
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