Atrain 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?

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

 

For a 0.25 pfor b o.3pp (0.25+0.3)=0.28

Ineed on quadratic equations to solve problems