Understanding Traffic Flow Parameters

This question involves calculating the maximum traffic flow and the speed at which this maximum occurs on a highway segment. We are given that there is a linear relationship between speed and density. The key information provided is the free flow speed and the jam density.

Given Information:

• Free Flow Speed ($v_f$): The maximum speed vehicles can travel when there is no congestion. Given as 100 km/hour.
• Jam Density ($k_j$): The maximum number of vehicles that can occupy a given length of road, typically when traffic is stopped or moving very slowly. Given as 120 vehicles/km.

The Linear Speed-Density Model

A linear relationship between speed ($v$) and density ($k$) can be represented by the following equation:

$v = v_f \left(1 - \frac{k}{k_j}\right)$

Substituting the given values ($v_f = 100$ km/hr and $k_j = 120$ vehicles/km):

$v = 100 \left(1 - \frac{k}{120}\right)$

This model assumes speed decreases linearly from the free flow speed to zero as density increases from zero to jam density.

Calculating Flow and Finding the Maximum

Traffic flow ($q$) is the rate at which vehicles pass a point on the highway, measured in vehicles per hour. It is calculated as the product of speed and density:

$q = v \times k$

To find the flow as a function of density, we substitute the speed-density relationship into the flow equation:

$q(k) = \left[ 100 \left(1 - \frac{k}{120}\right) \right] \times k$

Simplifying this expression gives us the flow equation:

$q(k) = 100k - \frac{100}{120}k^2$

$q(k) = 100k - \frac{5}{6}k^2$

Finding the Density for Maximum Flow:

The maximum flow occurs at a specific density, often called the optimal density ($k_{opt}$). To find this, we take the derivative of the flow equation with respect to density and set it to zero:

$\frac{dq}{dk} = \frac{d}{dk} \left(100k - \frac{5}{6}k^2\right)$

Calculating the derivative:

$\frac{dq}{dk} = 100 - 2 \times \frac{5}{6}k$

$\frac{dq}{dk} = 100 - \frac{5}{3}k$

Setting the derivative to zero to find the maximum:

$100 - \frac{5}{3}k_{opt} = 0$

Solving for $k_{opt}$:

$\frac{5}{3}k_{opt} = 100$

$k_{opt} = 100 \times \frac{3}{5}$

$k_{opt} = 60 \text{ vehicles/km}$

So, the maximum flow occurs when the density is 60 vehicles per kilometer.

Calculating the Maximum Flow ($q_{max}$):

Now, we plug this optimal density ($k_{opt} = 60$) back into the flow equation $q(k)$:

$q_{max} = 100(60) - \frac{5}{6}(60)^2$

$q_{max} = 6000 - \frac{5}{6}(3600)$

$q_{max} = 6000 - (5 \times 600)$

$q_{max} = 6000 - 3000$

$q_{max} = 3000 \text{ vehicles/hour}$

The maximum flow expected is 3000 vehicles per hour.

Determining the Speed at Maximum Flow

To find the speed at which the maximum flow occurs, we use the optimal density ($k_{opt} = 60$ vehicles/km) in the speed-density equation:

$v_{opt} = 100 \left(1 - \frac{k_{opt}}{120}\right)$

$v_{opt} = 100 \left(1 - \frac{60}{120}\right)$

$v_{opt} = 100 \left(1 - 0.5\right)$

$v_{opt} = 100 \times 0.5$

$v_{opt} = 50 \text{ km/hour}$

The speed corresponding to the maximum flow is 50 km per hour.

Conclusion

Based on the linear speed-density relationship, the maximum flow expected on this highway section is 3000 vehicles per hour, and this occurs at a speed of 50 km per hour.