physics
posted by Chip .
A train pulls away from a station with a constant acceleration of 0.44 m/s2. A passenger arrives at a point next to the track 6.2 s 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?

distance train=1/2 a t^2
distance passenger=V*(t6)
set them equal
1/2 a t^2=Vt6V
.22t^2Vt+6V=0
t=(V+sqrt(V^2.88*6V)/.44
Well, V^2>.88*6V
V>5.20m/s
Lets see if it works.
t=(5.20+sqrt(0))/.44=12 seconds
distance train: 1/2 .44*144=you do it
distance passenger= 5.20(126)=you do it 
no