Yes I realize many factors affect the answer, including coefficient of static friction, train mass and speed, track incline, and whatnot. I’m simply looking for a rough order-of-magnitude answer to the question, for a loaded freight train of, say, 100 cars on a flat track and going at 50 MPH, when max emergency braking is applied, how many miles before that train stops?

Does the number of cars in the train make a difference? All of the wheels have brakes, don’t they; so if each car is capable of stopping itself does coupling a hundred of them together change the stopping distance?

Does it matter if the train is fully loaded or not ? I mean the force of friction is directly proportional to the weight, so effectively the weight will cancel out when looking at deceleration (?)

It might be up to the brakes rather than wheel traction on the rails. If the brakes have limited clamping force, or limited heat dissipation capacity, then the total weight of the train may affect stopping distance.

So to explain… no, there is too much, let me sum up:

To answer the OP, depending on many factors, it can take up to approximately a mile, maybe a bit more, for a fully loaded freight train of, say, 100 cars on a flat track and going at 50 MPH, when max emergency braking is applied. It would definitely be stopped before 2 miles.

There’s YouTube clip of a freight train going 28 MPH when the emergency brake is applied. It takes 36 seconds to come to a halt. Using extrapolation, at 50 MPH (73.3 feet per second) it would take 64 seconds to come to a halt. 64 seconds at an average speed ot 36.6 feet per second, that’s 2360 feet, or 0.45 miles.

Wouldn’t the possibility of derailing and jackknifing the trailer cars factor into braking force? The brake force of all 100 cars is unlikely to equal so I would think there would have to be sort of anti-lock pumping of the brakes.

Looking at your math, it seems you’ve assumed a linear relationship between speed and stopping distance. It turns out that stopping distance isn’t linear with speed but proportional to the square of v_{0 }.

No. My linear relationship is between speed and stopping time (not stopping distance). If you double the speed, you double the time to stop. You also double the average speed so the distance is quadrupled.

Not sure if this is any help, but there was a story in the Kansas City Star this morning about a BNSF freight train colliding with a car at a railroad crossing. While the story doesn’t mention how long the train was or how fast it was going, it did mention that the car was pushed for about half a mile before the train came to a stop.

(I’d post a link but they don’t seem to copy/paste very well from the newspaper’s phone app.)

Google says a train can weigh any where from 3000-18,000 tons. A mid-size car is about 2 tons, so you can assume the frictional drag of pushing one down the tracks is (at worst) about 2 tons. So the added deceleration, measured in G, would range from 2/18,000 to 2/3000, almost nothing compared to what the train’s own brakes are doing.