The figure 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 the figure 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?
Solution:
Let us represent the speed a person travels at as v . The time it takes for each person to move a distance of L is \triangle t=\frac{L}{v}. Next, we see that with each additional person, the distance increases with the depth of the body, which is d .
a) Let us calculate the average rate of increase of the layer of people.
R= \frac{d}{\triangle t} = \frac{d}{\frac{L}{v}} = \frac{dv}{L}
=\frac{(0.25 m)(3.5 m/s)}{(1.75 m)}
=0.50 m/s
b) To calculate the time it takes for the depth layer to increase to 5.0m is:
t = \frac{D}{R} = \frac{5.0 m}{0.5 m/s} = 10 s.
This is question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 8.
