Posted by **Kim** on Tuesday, January 17, 2012 at 3:37pm.

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

- Physics -
**Henry**, Thursday, January 19, 2012 at 5:01pm
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^2-Vo^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.

- Physics -
**Henry**, Thursday, January 19, 2012 at 5:17pm
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.

- Physics -
**Henry**, Thursday, January 19, 2012 at 7:07pm
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.

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