I'm having trouble with this physics problem. Please help!?

Question

A train pulls away from a station witha constant acceleration of .4m/s^2. A passenger arrives at at point next to the track 6s after the end of the train has passed the very same point. What is the slowest constant speed at which she can run and still catch the train?

Answer 1

The train is at a distance 0.5*0.4*6^2 = 7.2m from the station when the passenger arrives. The train's distance from the station after that time will be 7.2 + 0.5*0.4*t^2. The passenger's distance from the station will be v*t, where v is her speed. She will catch the train at a time T when these distances are equal

7.2+0.2*T^2 = v*T

v = 7.2/T+0.2*T

v is minimum when dv/dT = 0

-7.2/T^2 +0.2 = 0

T = 36

The train at this time is 7.2 + 0.2*36^2 = 266.4 meters away.

She must run that 266.4 m in 36 sec, or 7.4 m/sec