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?

Question

bobpursley · Accepted Answer

distance train=1/2 a t^2
distance passenger=V*(t-6)

set them equal

1/2 a t^2=Vt-6V

.22t^2-Vt+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(12-6)=you do it

yoyo · Answer

no

Hannah Choi · Answer

The response by @bobpursley seems incorrect. I had a different method of solving for Vo for the passenger train.