Question 4 of 7: Greenshields model and shock‑wave analysis of a stalled vehicle

Question 4: Greenshields model and shock‑wave analysis of a stalled vehicle (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. One‑lane road; Greenshields linear speed–density model.

Find. Jam density and density at capacity (a); platoon length at the instant of release (b); time for the platoon to clear (c).

Approach. Fit Greenshields to get the jam density from the capacity, identify the three traffic states (approach, jam, capacity discharge), then use the shock‑wave speed $u_w=\Delta q/\Delta k$ for the back‑of‑queue and release waves.

1. Jam density and density at capacity (a). For Greenshields $q=u_f\,k(1-k/k_j)$ the capacity occurs at $k=k_j/2$ with $q_{\max}=u_f k_j/4$. Solving for $k_j$:
   $$k_j=\frac{4\,q_{\max}}{u_f}=\frac{4(1500)}{75}=\boxed{80\ \text{veh/km}},\qquad k_m=\frac{k_j}{2}=\boxed{40\ \text{veh/km}}.$$
2. Identify the traffic states. The three states that bound the shock waves are:
   $$A:\ k_A=20,\ q_A=k_Au_A=1200;\quad B:\ k_B=k_j=80,\ q_B=0;\quad C:\ k_C=40,\ q_C=1500.$$
   State $C$ is the capacity discharge that leaves the front of the queue once the vehicle moves.
3. Back‑of‑queue shock wave (approach → jam).
   $$u_{AB}=\frac{q_B-q_A}{k_B-k_A}=\frac{0-1200}{80-20}=-20\ \text{km/h}$$
   (negative = travelling upstream). During the 4‑min stall the back of the stopped platoon moves upstream by
   $$L=|u_{AB}|\,\Delta t=20\left(\tfrac{4}{60}\right)=\boxed{1.33\ \text{km}}\ \ (\approx 107\ \text{stopped veh at }k_j).$$
   This is the platoon length the instant the vehicle regains power (b).
4. Release (starting) shock wave (jam → capacity). When the lead vehicle moves, vehicles discharge at capacity (state $C$):
   $$u_{BC}=\frac{q_C-q_B}{k_C-k_B}=\frac{1500-0}{40-80}=-37.5\ \text{km/h}$$
   (the release front also moves upstream, but faster than the back of the queue).
5. Time to dissipate (c). After release the back of the queue keeps retreating at 20 km/h while the release front advances into the queue at 37.5 km/h; the platoon clears when the release wave overtakes the back‑of‑queue wave. Their closing speed is $37.5-20=17.5$ km/h across the 1.33 km platoon:
   $$\tau=\frac{L}{|u_{BC}|-|u_{AB}|}=\frac{1.333}{17.5}=0.0762\ \text{h}=\boxed{4.57\ \text{min}.}$$