Fundamentals of Transportation/Shockwaves/Solution

Problem:

Flow on a road is q₁ = 1800veh / hr / lane, and the density of k₁ = 14.4veh / km / lane. 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 (v₂ = 88km / hr) 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

Solve for Unknowns:

Original speed

$v_1=\frac{q_1}{k_1}=\frac{1800}{14.4} = 125 km/h \,\!$

Flow after police cruiser joins

$q_2= k_2 v_2 = 88*20 = 1760 veh/h \,\!$

Step 1

Calculate the wave velocity:

$v_w = \frac{{q_2 - q_1 }}{{k_2 - k_1 }} = \frac{{1760 - 1800}}{{20 - 14.4}} = - 7.14km/hr \,\!$

Step 2

Determine the growth rate of the platoon (relative speed)

$v_{r2} = v_2 - v_w = 88 - ( - 7.14) = 95.1km/hr \,\!$

Step 3

Determine the time spent by the police cruiser on the highway

$t = d/v = 10 km / 88 km/hr = 0.11 hr = 6.8 minutes \,\!$

Step 4

Calculate the Length of platoon (not a standing queue)

$L = v t = 95.1 km/hr * 0.11 hr = 10.46 km \,\!$

Step 5

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

$\Delta{q} = q_1 - k_1 v_w = q_2 - k_2 v_w = 1800 - (14.4*-7.14) = 1760- (20*-7.14) = 1902 veh/hr \,\!$

Step 6

The number of vehicles in platoon

$L_p k_2 = 10.46 km * 20 veh/km = 209.2 vehicles \,\!$

OR

$\Delta{q} t = 1902 veh/hr*0.11hr=209.2veh \,\!$