Question

For a given traffic stream, the speed-density relationship is given as:

\[ v = v_o \ln\left(\frac{k_j}{k}\right) \]

where \(v\) is the mean speed (in km/h), and \(k\) is the density (in veh/km).

Considering \(v_o\) as 45 km/h, and \(k_j\) as 200 veh/km, the maximum flow (in veh/h) for the given stream is (rounded off to the nearest integer).

Hint:

Maximize \(q = kv_o\ln(k_j/k)\) by differentiating with respect to \(k\); the peak occurs at \(k = k_j/e\), where \(v = v_o\).

Answer 1

Correct Answer: 3311

Step 1: Identify the traffic flow model.
The relationship \(v = v_o \ln(k_j/k)\) is Greenberg's logarithmic speed-density model, where \(v_o\) is a model constant and \(k_j\) is the jam density (density at which speed drops to zero).

Step 2: Write flow as a function of density.
Traffic flow is the product of speed and density:
\[ q = kv = k v_o \ln\left(\frac{k_j}{k}\right) \]

Step 3: Maximize flow with respect to density.
Differentiate \(q\) with respect to \(k\) and set it to zero. Using the product rule, with \(\frac{d}{dk}\left[\ln(k_j/k)\right] = -\frac{1}{k}\):
\[ \frac{dq}{dk} = v_o\ln\left(\frac{k_j}{k}\right) + k v_o\left(-\frac{1}{k}\right) = v_o\left[\ln\left(\frac{k_j}{k}\right) - 1\right] \]
Setting \(dq/dk = 0\):
\[ \ln\left(\frac{k_j}{k}\right) = 1 \implies \frac{k_j}{k} = e \implies k_m = \frac{k_j}{e} \]

Step 4: Find the speed at this density.
At \(k = k_m\), \(\ln(k_j/k_m) = 1\), so
\[ v_m = v_o \ln\left(\frac{k_j}{k_m}\right) = v_o(1) = v_o = 45 \text{ km/h} \]

Step 5: Compute the maximum density and flow.
\[ k_m = \frac{k_j}{e} = \frac{200}{2.71828} = 73.576 \text{ veh/km} \]
\[ q_{max} = k_m v_m = 73.576 \times 45 = 3310.9 \text{ veh/h} \]

Final Answer:
Rounded to the nearest integer,
\[ \boxed{q_{max} \approx 3311 \text{ veh/h}} \]