Question

Based on Greenshield's model, a speed-density relationship is developed on the data of a traffic stream. This relationship is represented as \(v = 75 - 0.03k\), where \(v\) (in km/hr) is the mean speed at density \(k\) (in vehicle/km). The jam density (in vehicle/km) of this traffic stream is (in integer).

Hint:

Jam density is the density at which speed becomes zero. Put \(v=0\) in the given equation and solve for \(k\).

Answer 1

Correct Answer: 2500

Step 1: Understand what jam density means.
In Greenshield's linear speed-density model, the mean speed of a traffic stream falls as the density rises. Jam density \(k_j\) is the density at which vehicles are packed bumper to bumper and no gap is left to move, so the mean speed drops to zero at this density.

Step 2: Write the given relation in Greenshield's standard form.
The problem gives \(v = 75 - 0.03k\), where \(v\) is the mean speed in km/hr and \(k\) is the density in vehicle/km. Greenshield's model is written in general as \(v = v_f - \left(\dfrac{v_f}{k_j}\right)k\), where \(v_f\) is the free flow speed, the speed a vehicle would get on an empty road.

Step 3: Apply the zero speed condition.
By definition, at jam density the stream has come to a complete stop, so \(v = 0\). Put this into the given equation:
\[ 0 = 75 - 0.03k_j \]

Step 4: Solve for the jam density.
\[ 0.03k_j = 75 \]
\[ k_j = \frac{75}{0.03} \]
\[ k_j = 2500 \]

Final Answer:
The jam density of this traffic stream works out to 2500 vehicle/km.
\[ \boxed{k_j = 2500 \ \text{vehicle/km}} \]