Answer:
74.2 km from station A.
Explanation:
We set a frame of reference with the station at the origin and the X axis pointing in the direction the trains run.
The freight train leaves at 8 AM and travels 10 minutes at 45 km/h.
For this problem it is better to convert the speeds to km/min
45 km/h = 0.75 km/min
The equation for position under constant speed is:
X(t - t0) = X0 + v0 * (t - t0)
Since we know the time it will stop moving at this speed:
X(10 - 0) = 0 + 0.75 * (10 - 0) = 7.5 km
After it ran those 7.5 km it will keep running at 60 km/h.
60 km/h = 1 km/min
The position equation for it is now:
X(t - 10) = 7.5 + 1 * (t - 10)
The passenger train leaves the station at 8:35 AM. It travels at 75 km/h for 5 minutes.
75 km/h = 1.25 km/min
After those 5 minutes it will have traveled:
X(40 - 35) = 0 + 1.25 * (40 - 35) = 6.25 km
Then it travels at 105 km/h
105 km/h = 1.75 km/min
Its position equation is now:
X(t - 40) = 6.25 + 1.75 * (t - 40)
Equating both positions we find the time at which they would meet:
7.5 + 1 * (t - 10) = 6.25 + 1.75 * (t - 40)
7.5 + t - 10 = 6.25 + 1.75*t - 70
t - 1.75*t = 6.25 - 70 +10 - 7.5
-0.75*t = -61.25
t = 61.25 / 0.75
t = 81.7 minutes
The freight train will have to be parked 5 minutes before this at t = 76.7 minutes.
At that moment the freight train will be at:
X(76.7 - 10) = 7.5 + 1 * (76.7 - 10) = 74.2 km
