We can find the final velocity v at the end of the station. v^2 = (v0)^2 + 2ax v = sqrt{ (v0)^2 + 2ax } v = sqrt{ (22.2 m/s)^2 - (2)(0.142 m/s^2)(206 m) } v = 20.84 m/s We can find the time to decelerate to this velocity. v = v0 + at t = (v - v0) / a t = (20.84 m/s - 22.2 m/s) / -0.142 m/s^2 t = 9.58 seconds The nose of the train is in the station for 9.58 seconds.