Question

A train travels between two stations 1/2 mile apart in a minimum time of 41 sec. If the train accelerates and decelerates at 8 ft/sec^2, starting from the first station and coming to a stop at the second station, what is its maximum speed in mph? How long does it travel at this top speed?

Answer 1

Answe677.38 feets

Explanation:

Answer 2

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

distance at full speed

x = v*t
x = 145.36*4.66 = 677.38 feets