Kinematics Problem and stream of people

[Bashyboy]

Homework Statement

Figure 2-21 shows a general situation
in which a stream of people attempt to escape through an exit door
that turns out to be locked. The people move toward the door at
speed vs
3.50 m/s, are each d
0.25 m in depth, and are separated
by L
1.75 m.The arrangement in Fig. 2-21 occurs at time t
0. (a)
At what average rate does the layer of people at the door increase?
(b) At what time does the layer’s depth reach 5.0 m? (The answers
reveal how quickly such a situation becomes dangerous.)

Homework Equations

The Attempt at a Solution

I am not certain how to solve this.

 

[gneill]

Homework Statement

Figure 2-21 shows a general situation
in which a stream of people attempt to escape through an exit door
that turns out to be locked. The people move toward the door at
speed vs
3.50 m/s, are each d
0.25 m in depth, and are separated
by L
1.75 m.The arrangement in Fig. 2-21 occurs at time t
0. (a)
At what average rate does the layer of people at the door increase?
(b) At what time does the layer’s depth reach 5.0 m? (The answers
reveal how quickly such a situation becomes dangerous.)

Homework Equations

The Attempt at a Solution

I am not certain how to solve this.

What equations in general do you think are applicable?

How long do you think the time interval will be between successive impacts?

 

[Bashyboy]

Perhaps t = x/v would be one of the equations. I know the first person impacts the wall in .5 seconds.

 

[gneill]

Perhaps t = x/v would be one of the equations. I know the first person impacts the wall in .5 seconds.

Okay, so suppose the first person has hit the door and come to a halt. How long till the next impact occurs?

 

[Bashyboy]

Would it be another .5 seconds?

 

[gneill]

Would it be another .5 seconds?

Is that a guess, or do you have some reasoning behind that value?

 

[Bashyboy]

Well, I believe it would be another .5 seconds because they all have to travel the same distance--that is, L.

 

[gneill]

Well, I believe it would be another .5 seconds because they all have to travel the same distance--that is, L.

Okay, so if the guy in front stops, the guy following has distance L to travel before impacting. Good.

So, what can you say about the rate of impacts? Can you turn that into an average rate that the layer of people will increase?

 

[Bashyboy]

So, in 1 second the first person collides into the door, and then the second person collides into the first person. Since each person has a depth of .25 m, that would mean the depth at the door would increase .5 m/s? That doesn't seem right. Isn't the depth at the door increases exponentially?

 

[gneill]

So, in 1 second the first person collides into the door, and then the second person collides into the first person. Since each person has a depth of .25 m, that would mean the depth at the door would increase .5 m/s? That doesn't seem right. Isn't the depth at the door increases exponentially?

As you stated earlier, the first person hits the door in 0.5 seconds, not 1 second. After that they "pile on" at a rate of one every 0.5 seconds. At the moment you're interested in this "after that" scenario, since it represents the ongoing rate of accumulation of people on the pile.

Note that there is no exponential increase in the rate that people arrive --- they arrive at a constant rate, so the depth is increasing at an average constant rate, too: At 0.5m/sec in fact, as you've discovered.

You're now in a position to answer part (b).