Fundamentals of Transportation/Shockwaves/Solution

Shockwave Problem: Rolling Roadblock

Flow on a road is $q_{1} = 1800 v e h / h r / l a n e$ , and the density of $k_{1} = 14.4 v e h / k m / l a n e$ . To reduce speeding on a section of highway, a police cruiser decides to implement a rolling roadblock, and to travel in the left lane at the speed limit ( $v_{2} = 88 k m / h r$ ) for 10 km. No one dares pass. After the police cruiser joins, the platoon density increases to 20 veh/km/lane and flow drops. How many vehicles (per lane) will be in the platoon when the police car leaves the highway?

Solution

Step 0

0. Solve for Unknowns Original speed

$v_{1} = \frac{q_{1}}{k_{1}} = \frac{1800}{14.4} = 125 k m / h$

Flow after police cruiser joins

$q_{2} = k_{2} v_{2} = 88 * 20 = 1760 v e h / h$

Step 1

1. Calculate the wave velocity:

$v_{w} = \frac{q_{2} - q_{1}}{k_{2} - k_{1}} = \frac{1760 - 1800}{20 - 14.4} = - 7.14 k m / h r$

Step 2

2. Determine the growth rate of the platoon (relative speed)

$v_{r 2} = v_{2} - v_{w} = 88 - (- 7.14) = 95.1 k m / h r$

Step 3

3. Determine the time spent by the police cruiser on the highway

$t = d / v = 10 k m / 88 k m / h r = 0.11 h r = 6.8 m i n u t e s$

Step 4

4. Calculate the Length of platoon (not a standing queue)

$L = v t = 95.1 k m / h r * 0.11 h r = 10.46 k m$

Step 5

5. What is the rate at which the queue grows, in units of vehicles per hour?

$Δ q = q_{1} - k_{1} v_{w} = q_{2} - k_{2} v_{w} = 1800 - (14.4 * - 7.14) = 1760 - (20 * - 7.14) = 1902 v e h / h r$

Step 6

6. The number of vehicles in platoon

$L_{p} k_{2} = 10.46 k m * 20 v e h / k m = 209.2 v e h i c l e s$

OR

$Δ q t = 1902 v e h / h r * 0.11 h r = 209.2 v e h$

Fundamentals of Transportation/Shockwaves/Solution

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Fundamentals of Transportation/Shockwaves/Solution

Solution

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