Oh, thank you. I needed an exercise in applied math.
[...] So if they take around 700 feet to slow down 30mph, it'd take about 2100 feet to slow from 90? And brakes need to kick in so.... 2500-2700?
Is my math right or am I totally off?
As Trogdor noted, "Stopping distance is not linear with speed." I wonder if deceleration is constant, because braking is changing kinetic energy to some other form of energy (often heat energy), and kinetic energy varies with the square of speed. Something going 90 mph has nine times the kinetic energy of something going 30 mph. But a varying deceleration makes computation more difficult, so let's assume a constant deceleration.
[...] some trains have an emergency braking deceleration of 1.4 m/s2, which translates, very roughly, to 3 mph/second. [...]
I have no better number, so let's use 3 mph/second as deceleration.
The time dependent location of an object (train) moving on a straight line (track, with no curve) in the direction of travel at constant acceleration is the starting point (milepost, assumed to be zero) plus the change due to the initial speed plus the change due to the acceleration. In other words,
s(t) = s0 + v0*t + 0.5*a0*t^2 where s0 is the initial location, v0 is the initial speed, and a0 is the initial acceleration (assumed to be constant).
Consistency in units is important. Let's use miles and hours. As I said, I assume the initial location is at zero. I'll assume the initial speed is 60 mph, although it was probably higher, based on
TinCan782's comment. (Plug in a different number if you want to assume a different initial speed.) I'll assume constant acceleration of -3 mph/second, or -(3*3600) mph/hour, based on
Trogdor's comment.
The location of interest is where the speed is zero. If the initial speed is 60 mph and it drops by 10,800 mph every hour, then it will be zero after 60/10800 hours, or 0.00556 hours. (Three digits should provide sufficient accuracy, but feel free to use more digits if you like.) Plugging that value into the formula,
s(0.00556) = 0 + 60*0.00556 - 0.5*10800*0.00556*0.00556
Evaluating the expression, we find the train comes to a stop after 0.1667 miles, one sixth of a mile.
Whenever conducting a thought experiment like this, it is useful to ask, "Is that answer reasonable?" With an initial speed of 60 mph and a deceleration of 3 mph/second, it would take 20 seconds to slow to zero. Happily, 0.00556 hours is 20.016 seconds; the difference is a rounding error because I didn't carry more digits. A train going 60 mph will cover 1/3 of a mile in 20 seconds, so certainly the answer should be less than 1/3 of a mile. Since the assumed deceleration is constant, it seem reasonable that the distance covered would be half the distance covered with no deceleration. Happily, 0.1667 miles is half of 1/3 of a mile. Whatever mistakes I made, I seem to have made them consistently. Corrections and clarifications are welcome.
[...] I say "ish" because the actual distance covered is more of a curve than linear. In reality, you're covering a higher distance because of the time you spend at a higher speed while you slow down. Even using these rough numbers, a train slowing down from 90 mph would require 9 times the distance of a train slowing down from 30 mph. [...]
I'm gratified to see that a 90 mph train takes 9 times longer to stop than a 30 mph train, exactly in line with the change in kinetic energy.
Because the deceleration is constant, you needn't say "ish". If you plot the speed versus time, you get a triangle. The distance covered is the area under the triangle. You are correct that "the distance covered is more of a curve" but the change in the rate of covering the distance is constant, as indicated by assuming a deceleration of 3 mph/second.
However, I think you slipped up somewhere in your calculations. If the train is going 90 mph, then it travels 1.5 miles every minute or 0.75 miles every 30 seconds. If it takes the train 30 seconds to slow to a stop, it cannot go more than 0.75 miles as comes to that stop.
