It is important to notice firs of all that the train with no stop acceleration and them no stop acceleration would have a faster travel between stops. given this is safe to assume that there is a period during while the train is not accelerating nor decelerating. To find the time that is speed is constant let's assume non-stop accelerating and decelerating and the time lost (subtraction of both) is the time the train had constant speed:
assuming same time accelerating a decelerating we have that the distance is
x = voT + 1/2aT^2 (initiial speed = 0) (0.25 miles = 1320 feets) 1320 = 1/2 (8)(T/2)^2 T/2 = 18.17 seg T = 36.34 seg time of constant speed T = 41 -36.34 = 4.66 seg
maximum speed
v = v0 + aT (initial speed = 0) v = 8(18.17) (time accelerating) v = 145.36 feets/sec
