9514 1404 393
Answer:
210 km
Explanation:
The position of the first train as a function of the time after it left is ...
f(t) = 30t
The position of the second train as a function of time after the first train left is ...
s(t) = 105(t -5)
The positions are equal when ...
30t = 105(t -5)
525 = 75t . . . . . . add 525-30t to both sides
7 = t . . . . . . . . . divide by 75
f(7) = s(7) = 30(7) = 210 = 105(7 -5)
The trains will meet 210 km from the station.
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Alternate solution
The first train has a 5-hour head start, amounting to 5×30 = 150 km. The second train catches up at the rate of (105 -30) = 75 km/h. So, the second train covers the 150 km gap in 150/75 = 2 hours. It will have traveled 105×2 = 210 km in that time. (Effectively, we're using the start time of the second train as the time reference. 30(t+5) = 105t)