Chapter 5, Problem 1P

To determine

The flow, the average speed and the density of the traffic in the given lane.

Answer to Problem 1P

The traffic flow is $1.125\text{veh}/\text{hr}$ , the speed is $35.15\text{mi}/\text{hr}$ and the density is $\text{32}\text{veh}/\text{mi}$ .

Explanation of Solution

Given:

The average headway is $3.2\sec$ .

The average spacing of passing is $165\text{ft}$ .

Formula Used:

Write the expression for the traffic flow.

  $q=\frac{1}{\overline{h}}$....................(I)

Here, $q$ is the traffic flow and $\overline{h}$ is the time headway.

Write the expression for the average traffic speed.

  $u=\frac{\overline{s}}{\overline{h}}$....................(II)

Here, $u$ is the average traffic speed, $\overline{s}$ is the average spacing.

Write the expression for the density.

  $k=\frac{q}{u}$....................(III)

Here, $k$ is the traffic density.

Calculation:

Substitute $3.2\sec$ for $\overline{h}$ in equation (I).

  $\begin{array}{c}q=\frac{1}{3.2\sec }\\ =0.3125\text{veh}/\text{sec}\times \frac{3600\text{veh}/\text{hr}}{1\text{veh}/\text{sec}}\\ =1125\text{veh}/\text{hr}\end{array}$

Substitute $3.2\sec$ for $\overline{h}$ and $165\text{ft}$ for $\overline{s}$ in equation (II).

  $\begin{array}{c}u=\frac{\left(165\text{ft}\times \frac{\left(1/5280\right)\text{mi}}{1\text{ft}}\right)}{\left(3.2\sec \times \frac{\left(1/3600\right)\text{hr}}{1\text{sec}}\right)}\\ =\left(\frac{0.03125}{8.89\times {10}^{-4}}\right)\text{mi}/\text{hr}\\ =35.15\text{mi}/\text{hr}\end{array}$

Substitute $35.15\text{mi}/\text{hr}$ for $u$ and $1.125\text{veh}/\text{hr}$ for $q$ in equation (III).

  $\begin{array}{c}k=\frac{1,125\text{veh}/\text{hr}}{35.15\text{mi}/\text{hr}}\\ =32\text{veh}/\text{mi}\end{array}$

Conclusion:

Thus, the traffic flow is $1.125\text{veh}/\text{hr}$ , the speed is $35.15\text{mi}/\text{hr}$ and the density is $\text{32}\text{veh}/\text{mi}$ .