Physics
posted by Kim on .
A subway train starts from rest at a station and accelerates at a rate of 1.50m/s^2 for 13.3s . It runs at constant speed for 71.0s and slows down at a rate of 3.48m/s^2 until it stops at the next station.
Find the total distance covered.
Express your final answer in kilometers

d1 = Vo*t + 0.5a*t^2,
di = 0 + 0.75*(13.3)^2 = 132.7 m.
Vf = at = 1.5m/s^2 * 13.3s = 19.95 m/s.
Vo = 19.95 m/s.
Vf^2 = Vo^2 + 2a*d,
d2 = (Vf^2Vo^2) / 2a,
d2 = (0(19.95)^2 /3.48 = 114.4 m.
D = d1 + d2 = 132.7 + 114.4 = 247 m. =
0.247 km. = Tot. dist. 
Correction: d2 = (0(19.95)^2 / 6.96 =
57.2 m.
D = di + d2 = 132.7 + 57.2 = 190 m. =
0.19 km. = Tot. dist. 
Since the train ran at constant speed for 71 seconds, we must add d3:
d3=Vo*t = 19.95m/s * 71s. = 1416.45 m.
D = d1 + d2 + d3,
D = 132.7 + 57.2 + 1416.45 = 1606.4 m. = 1.61 km = Tot. dist.