Afreight train and a passenger train share the same rail track. the freight train leaves station a at 8: 00 a. m. the train travels at a speed of 45 km/h for the first 10 minutes and then continues to travel at a speed of 60 km/h. at 8: 35 a. m., the passenger train leaves station a. the pas¬senger train travels first at a speed of 75 km/h for 5 minutes and then continues to travel at a speed of 105 km/h. determine the location of the siding where the freight train will have to be parked to allow the faster passenger train to pass through. as a safety precaution, it is determined that the time headway between the two trains should not be allowed to fall below 5 minutes.

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