Question

On a specific westbound section of highway, studies show that the speed-density relationship is: u = uf[1-(k/kj)3.5]. It is known that the capacity is 4200 veh/h and the jam density is 210 veh/mi. What is the space-mean speed of the traffic at capacity, and what is the free-flow speed?

Answer 1

The space-mean speed of traffic at capacity is equal to the free-flow speed.

Step-by-step explanation:

The speed-density relationship for traffic on a specific westbound section of highway is given by the equation u = uf[1-(k/kj)3.5]. In this equation, u represents the space-mean speed of traffic, uf represents the free-flow speed, k represents the traffic density, and kj represents the jam density. The capacity of the highway is 4200 veh/h and the jam density is 210 veh/mi. To find the space-mean speed at capacity, we substitute k = kj into the equation: u = uf[1-(kj/kj)3.5]. Simplifying this gives u = uf[1-1] = uf. Therefore, the space-mean speed of traffic at capacity is equal to the free-flow speed uf.