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?
distance train=1/2 a t^2
set them equal
1/2 a t^2=Vt-6V
Lets see if it works.
distance train: 1/2 .44*144=you do it
distance passenger= 5.20(12-6)=you do it