Question

can you help me with this one

A train travels between two stations ½ mile apart in a minimum time of 41 sec. If the train accelerates and decelerates at 8 ft/〖sec〗^2, starting from rest at the first station and coming to a stop at the end of the station, what is its maximum speed in mph? how long does it travel at this top speed?

Answer 1

very simple question....
just use equations of motions..

Answer 2

let it accelerates for x units and then deaccelerates the rest diatance (1/2 -x)mile
use the formule third equation of motion
speed is maximum when it has completely accelerated for x miles from starting

Answer 3

@mayank13589 you're ignoring the min 41 seconds

Answer 4

ddd

Answer 5

the velocity-time graph for this
the train is moving with maximum velocity when the graph is parallel to the x-axis.thearea under this graph is the distance covered.
let oa=bc=x sec and ab=41-x
v.m is the maximum velocity
area of the graph=distance covered (area of oae=area of bcd, as the train accelerates and decelerates for same amount of time)
2.(area of oae)+area of abde=2640 ft
2.(1/2).x.(v.m)+(v.m).(41-2x)=2640 -------> its the first equation.

for the initial time duration x,applying equation of motion-->s=ut+(at^2)/2
x.(v.m)/2=(8.x^2)/2
v.m=8x ------>now putting the value of v.m in first equation u will get the value of x
and (41-2x) will be the time for which the train is travelling at its maximum speed