Average velocity and average speed problem.

[afcwestwarrior]

Homework Statement

The people move toward the door at a speed V s=3.50m/s,are each d=.25m in depth, and are separated by L=1.75 m. The arrangement in Fig.2=22 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 5.0m? (the answers reveal how quickly such a situation becomes dangerous.
In the picture there are 3 people running towards a door


v= change in x/ change in t

 

[tiny-tim]

Homework Statement

The people move toward the door at a speed V s=3.50m/s,are each d=.25m in depth, and are separated by L=1.75 m. The arrangement in Fig.2=22 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 5.0m? (the answers reveal how quickly such a situation becomes dangerous.
In the picture there are 3 people running towards a door


v= change in x/ change in t

Hi afcwestwarrior!

I suspect everyone's as confused as I am …

are the 3 people running side-by-side or front-to-back?

where is everyone else?

 

[baba89]

= (1.75m)/0.25m = 7m/s

a) The average rate at which the layer of people at the door increases is 7m/s. This means that for every second that passes, the layer of people at the door increases by 7 meters.

b) The layer's depth will reach 5.0m when t = 5/7 seconds. This is calculated by dividing the desired depth (5.0m) by the average rate of increase (7m/s). This means that after 5/7 seconds, the layer of people at the door will have reached a depth of 5.0m.

These calculations show that this situation can quickly become dangerous, as the layer of people at the door is increasing at a high average rate and can reach a dangerous depth in just a few seconds. It is important to monitor and manage crowd control in situations like this to ensure the safety of everyone involved.