Question
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?
Answers
bobpursley
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
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
no
Hannah Choi
The response by @bobpursley seems incorrect. I had a different method of solving for Vo for the passenger train.