trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 40 mph and train B at 44 mph. Train A passes a station at 1:2-am. If train B passes the same station at 1:50am, at what time will train B catch up to train A?

Question

anonymous · Answer

that's supposed to be 1:20am not 1:2-am

anonymous · Answer

Using the line format: $$y=mx+b$$ You solve for the two equations: $$y_{a} = 0, x_{a} = 1.333$$$$y_{a} = 40x_{a} + b$$$$0 = 40(1.333) + b$$$$-53.333 = b$$$$y_{a} = 40x - 53.333$$
$$y_{b} = 0, x_{b} = 1.8333$$$$y_{b} = 44x_{b} + b$$$$0= 44(1.8333) + b$$$$-73.333 = b$$$$y_{b} = 44x -73.333$$You then solve the two equations:
$$40x - 53.333 = 44x - 73.333$$$$20 = 4x$$$$x = 5$$If I didn't mess up horribly somewhere, that means they should meet at 5:00 am.

anonymous · Answer

You can calculate the distance train A has moved since it passed the station untill train B passes that station, that will be 20 miles, as the time diffenrence is a half hour. 
If we say the train B catches up with 4 mph, then divide the distance with the speed, 20miles / 4 mph =  5 hours,  5 hours plus 1:50 am = 6:50 am

anonymous · Answer

r t = d
Solve 40 (t + 1/2 ) = 44 t, for t
t = 5 hours
time of day = 1:50AM +5 hours or 6:50AM