Posted by Chip on .
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?

physics 
bobpursley,
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 
physics 
yoyo,
no