Asked 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?
Answers
Answered by
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
Answered by
yoyo
no
Answered by
Hannah Choi
The response by @bobpursley seems incorrect. I had a different method of solving for Vo for the passenger train.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.