Two subway stops are separated by 1.1 km. if a subway train accelerates at +1.2 m/s2 from rest
through the first half of the distance and decelerates at −1.2 m/s2 through the second half, what are
(a) its travel time and (b) its maximum speed? (c) graph x, v and a versus t for the trip.

a) Travel time = 60.56 s

b) Maximum speed = 36.33 m/s

Explanation:

a)  Distance = 1.1 km = 1100 m

 A subway train accelerates at +1.2 m/s² from rest  through the first half of the distance and decelerates at −1.2 m/s² through the second half.

 So half the distance is traveled at an acceleration of +1.2 m/s².

 We have equation of motion

 Substituting

 Travel time = 2 x 30.28 = 60.56 s

b) We have equation of motion v = u+at

   Substituting t = 30.28 s and a = 1.2 m/s²

   v = 0 + 1.2 x 30.28 = 36.33 m/s

   Maximum speed = 36.33 m/s

c) Photos of graphs are given

i think if it were me i whould divide 300 dide 15

step-by-step explanation: