# Fundamentals of Transportation/Shockwaves/Solution

Problem:

Flow on a road is ${\displaystyle q_{1}=1800veh/hr/lane}$, and the density of ${\displaystyle k_{1}=14.4veh/km/lane}$. 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 (${\displaystyle v_{2}=88km/hr}$) 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?

How long will it take for the queue to dissipate?

Solution:

Step 0

Solve for Unknowns:

Original speed

${\displaystyle v_{1}={\frac {q_{1}}{k_{1}}}={\frac {1800}{14.4}}=125km/h\,\!}$

Flow after police cruiser joins

${\displaystyle q_{2}=k_{2}v_{2}=88*20=1760veh/h\,\!}$

Step 1

Calculate the wave velocity:

${\displaystyle v_{w}={\frac {q_{2}-q_{1}}{k_{2}-k_{1}}}={\frac {1760-1800}{20-14.4}}=-7.14km/hr\,\!}$

Step 2

Determine the growth rate of the platoon (relative speed)

${\displaystyle v_{r2}=v_{2}-v_{w}=88-(-7.14)=95.1km/hr\,\!}$

Step 3

Determine the time spent by the police cruiser on the highway

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

Step 4

Calculate the Length of platoon (not a standing queue)

${\displaystyle L=vt=95.1km/hr*0.11hr=10.46km\,\!}$

Step 5

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

${\displaystyle \Delta {q}=q_{1}-k_{1}v_{w}=q_{2}-k_{2}v_{w}=1800-(14.4*-7.14)=1760-(20*-7.14)=1902veh/hr\,\!}$

Step 6

The number of vehicles in platoon

${\displaystyle L_{p}k_{2}=10.46km*20veh/km=209.2vehicles\,\!}$

OR

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

How long will it take for the queue to dissipate?

(a) Where does the shockwave go after the police cruiser leaves? Just reverse everything?

${\displaystyle {{v}_{w}}={\frac {{{q}_{2}}-{{q}_{1}}}{{{k}_{2}}-{{k}_{1}}}}={\frac {1800-1760}{14.4-20}}=-7.14km/hr}$

Second wave never catches first.

(b) Return speed to 125, keep density @ 20? ${\displaystyle q_{2}=20*125=2500\,\!}$

${\displaystyle {{v}_{w}}={\frac {{{q}_{2}}-{{q}_{1}}}{{{k}_{2}}-{{k}_{1}}}}={\frac {2500-1760}{20-20}}=\infty km/hr}$

Second wave instantaneously catches first

(c) Return speed to 125, keep density @ halfway between 14.4 and 20? ${\displaystyle q_{2}=20*125=2500\,\!}$

${\displaystyle {{v}_{w}}={\frac {{{q}_{2}}-{{q}_{1}}}{{{k}_{2}}-{{k}_{1}}}}={\frac {2150-1760}{17.2-20}}=139km/hr}$

Second wave quickly catches first

{\displaystyle {\begin{aligned}&139*t=7.14*\left(.11+t\right)\\&\left(139-7.14\right)t=7.14*11\\&t=0.006h=.35\min =21\sec \\\end{aligned}}}

(d) Drop upstream q,k to slow down formation curve.

${\displaystyle {{v}_{w}}={\frac {{{q}_{2}}-{{q}_{1}}}{{{k}_{2}}-{{k}_{1}}}}={\frac {1780-1760}{14.24-20}}=-3.47km/hr}$

Second wave eventually catches first

(e) If ${\displaystyle q_{upstream}}$ falls below downstream capacity, wave dissipates

${\displaystyle {{v}_{w}}={\frac {{{q}_{2}}-{{q}_{1}}}{{{k}_{2}}-{{k}_{1}}}}={\frac {1250-1760}{10-20}}=+51km/hr}$

Forward moving wave.