Do you know that subway, especially underground parts, are often built so that way lowers when leaving the station and rises when coming to the next? This both helps keeping water sipping through walls off the stations and simplifies braking on arrival!
The train travels distance S between two stations in T minutes. Leaving the first station, it applies constant acceleration until maximal speed Vmax is
reached - and then keeps this speed, until nearing the second station. At appropriate point driver applies constant deceleration (i.e. acceleration
directed backwards), until coming to full stop. It is experienced driver, so this stop happens exactly at the station.
We, on behalf of government comission, want to know the amount of acceleration / deceleration experienced by passengers, supposing that they are equal in magnitude (just opposite in direction).
Numeric Example:
T = 2.33333 min
S = 2 km
Vmax = 60 km/h
Result should be 0.83333 m/s^2 Note that speed is in kilometers per hour, time in minutes and acceleration should be in meters per second squared.
source: Sakharov