# 1 dimensional motion escape from a locked door

1. Aug 6, 2010

1. The problem statement, all variables and given/known data

[PLAIN]http://img443.imageshack.us/img443/3598/physicsproblem.png [Broken]
2. Relevant equations

Ok so I am first of I don't know when they said layer if they meant the number of people or the thickness? of the people I don't understand the question

3. The attempt at a solution
what I noticed is that at first the first guy will have to cross L distance but the second guy must cross (L-d) but then it get so complicated I mean how can I solve this? what I noticed is that for all 1 seconds they cross L distances so at the first 1/2 second the first guy will be at the door but then the second guy will be at his place but he will have to cross 1.5 m not 1.75 which will take him .42 seconds but that wont matter for the ones before him right? they will still have to cross .5 seconds to reach the before last place pfff it's too complicated to me that I can't even describe it rather than solve it
ok so what I can make out is that they move L distance for all .5 seconds
and if I call the door is where x = 0 and the left side is positive
then when they reach x=L they must cross (L-nd) distance while the first guy is where n = 0 .but that is not useful at all at finding how the layers increases.... but I tried lol
EDIT: I know this is stupid but is the book assuming the people are transperant because it would be easier if they can go through each other

Last edited by a moderator: May 4, 2017
2. Aug 6, 2010

### sronen71

madah12, you started to think in the right direction, but went off track.
After time t_1=L/v the first guy got stuck at the door. What about the 2'nd guy?
look at the figure provided. The distance to the first guy is still L, not L-d. That's because the figure shows that the distance L between the guys does not include their thickness d.
Therefore, 2'nd guy bumps into the first at time t_2=2*L/v; It's simpler now, right? Try again to solve it now, good luck!

3. Aug 7, 2010

yes but when the first guy reaches the door he will occupy d distance to the left of the door so when the second guy comes his finishing point wont be the door but it will be that distance minus the thickness of the first guy so it will be reduced to L-d .

4. Aug 7, 2010

Are you saying I should neglect the fact that the people will occupy space next to the door?

Last edited: Aug 7, 2010
5. Aug 7, 2010

I don't know about the rules of bumping but I hope it's allowed.

6. Aug 7, 2010

Just to be clear on what I want , I do know how to solve it if we were to neglect their depth it would be d / t where t= L/v because that would be the time to add another layer
L/v=1.75m/3.50(m/s) =.50 d/t = .25m/.50s =.50 m/s and for b it's 5m/.50(m/s) =10seconds
but in the question it does not say to ignore the thickness of the men so the time can't be L/d right?

7. Aug 8, 2010

### sronen71

yes but when the first guy reaches the door he will occupy d distance to the left of the door so when the second guy comes his finishing point wont be the door but it will be that distance minus the thickness of the first guy so it will be reduced to L-d .

->
Alright, if it is L you show you know to solve it, and that's good! Now let's see why it is L and not L-d as you thought. The figure shows that:

1)The distance between the *back* of each guy to the *front* body of the guy after him is L,
as shown in the figure.
2) The distance between the front of a guy and the back of same guy is d (that is a guy's thickness).
So when first guy stops at the door, his back is at distance d from the door.
Exactly at the moment of his stopping, the distance from his back to the front of the 2'nd guy, is L - as it was before (given data no. 1), not L-d. Because the distance L was measured also before from his back side, not from his front side.
This is why the 2'nd guy willl now take time L/v to bump into the first guy, and so on.

your original argument of L-d would have been correct if the distance L were between front of one person to front of the person after him. But this is not the case.

Is this clearer now?

8. Aug 8, 2010

Yeah I got it now I guess I didn't look at the picture thoroughly.

I got it in a sense that I picked the point x = 0 at the front of the second guy initially
and x1,x2 are the front of the second , back of the first
so initially x(2)=0
x(1)=L
delta x =L

after half a second
x(2)=L
x(1)=2 L delta x =L
so I got that second guy needs to also move a distance L to finish
is that correct?

Last edited: Aug 8, 2010
9. Aug 8, 2010