Figure shows a general situation in which a stream of people attempts to escape through an exit door that turns out to be locked. The people move toward the door at speed $V_s=3.5m/s$, are each d=0.25 m in depth, and are separated by L=1.75 m. The arrangement in figure 2-24 occurs at time t=0. (a) At what average rate does the layer of people at door increase? (b) At what time does the layer’s depth reach 5 m? (The answers reveal how quickly such a situation becomes dangerous)

Figure 2-24Problem 8

Speed of the people toward the door ,$v_s=3.5\mathrm{m}/\mathrm{s}$ .

The initial depth of layer, $d=0.25\mathrm{m}$.

The separation between the layers, $L=1.75\mathrm{m}$.

Speed may be defined as the time rate of change in distance.The time taken for each person to move a distance L with speed is used to find the rate of increase of layer of people.

The expression for the speed is given as follows:

$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$

The amount of time it takes for each person to move a distance L with speed $V_s$is calculated as follows:

$\begin{array}{c}\mathrm{\Delta }t=\frac{L}{v_s}\\ =\frac{1.75\mathrm{m}}{3.5\mathrm{m}/\mathrm{s}}\\ =0.5\mathrm{s}\end{array}$

With each additional person, the depth increases by one body depth d.

Therefore, the rate of increase of the layer of people is calculated as follows:

$\begin{array}{c}R=\frac{d}{\mathrm{\Delta }t}\\ =\frac{0.25}{0.5}\\ =0.50\mathrm{m}/\mathrm{s}\end{array}$

Hence, the average rate of increase in the layer of people is .$0.50m/s$

The amount of time required to reach a depth of D=5.0 m is calculated as follows:

$\begin{array}{c}\Delta t=\frac{D}{R}\\ =\frac{5.0\mathrm{m}}{0.5\mathrm{m}/\mathrm{s}}\\ =10\mathrm{s}\end{array}$

Hence, the time required when the layer depth reaches 5 m is 10 s.