Question 4 of 7: Greenshields' Model and Shock Waves behind a Slow Vehicle

Question 4: Greenshields' Model and Shock Waves behind a Slow 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.

Find. The jam density and the density at capacity; the platoon length at the instant the tractor leaves; the speed of the front of the platoon thereafter; and the time the platoon takes to disappear.

Approach. Fit the Greenshields parabola from the free-flow speed and the capacity, place the three traffic states A, B and C on it, and take each shock-wave speed as the slope of the chord between the two states it separates.

1. Part (a) — fit the Greenshields curve and read the two densities. Greenshields assumes a linear speed–density relation u = u_f(1 − k/k_j), so q = ku is a parabola whose maximum is q_max = u_fk_j/4. Inverting for the jam density, $$k_j=\frac{4\,q_{max}}{u_f}=\frac{4(2500)}{100}=\boxed{100\ \text{veh/km}}$$ and because the parabola is symmetric the capacity occurs at half the jam density, $$\begin{aligned}k_c&=\frac{k_j}{2}=\boxed{50\ \text{veh/km}} \\ \text{at which} \\ u_c&=\frac{u_f}{2}=50\ \text{km/h}\end{aligned}$$
2. Locate the two traffic states and confirm the data are consistent. The undisturbed upstream state A travels at 80 km/h, so k_A = k_j(1 − u_A/u_f) = 100(1 − 0.80) = 20 veh/km and q_A = 80(20) = 1,600 veh/h. The platoon state B is given directly as k_B = 85 veh/km and q_B = 1,275 veh/h, which implies u_B = 1275/85 = 15 km/h — the tractor's speed, as stated. The fitted curve returns the same density, 100(1 − 15/100) = 85 veh/km, so the four supplied data are mutually consistent and no competing reading of the problem arises. It is worth performing this one-line check before solving: if the printed operating point contradicted the fitted parabola, the two readings would have to be reconciled.
3. Part (b) — the time the tractor spends on the highway. It covers 1.0 km at 15 km/h, so $$t_b=\frac{1.0}{15}=0.06667\ \text{h}=4.0\ \text{minutes}$$
4. Find the speed of the shock wave at the back of the platoon. The boundary between state A behind and state B ahead moves at the chord slope $$\omega_{AB}=\frac{q_B-q_A}{k_B-k_A}=\frac{1275-1600}{85-20}=\frac{-325}{65}=\boxed{-5.0\ \text{km/h}}$$ The negative sign means the back of the queue advances upstream, against the direction of travel, at 5 km/h — the platoon is growing backwards while the tractor drags it forward.
5. Assemble the platoon length. Both boundaries start at the point where the tractor entered. In the 4.0 minutes the tractor is on the road, its front travels 15(0.06667) = 1.0 km downstream while the rear shock travels −5(0.06667) = −0.333 km upstream, so $$L=\bigl(u_B-\omega_{AB}\bigr)\,t_b=\bigl(15-(-5)\bigr)(0.06667)=\boxed{1.333\ \text{km}}$$
6. Cross-check the platoon by counting vehicles two ways. The platoon holds k_BL = 85(1.333) = 113.3 vehicles. Independently, the number of vehicles that the advancing rear shock overtakes in time t_b is (u_A − ω_AB)k_At_b = (80 + 5)(20)(0.06667) = 113.3 vehicles. The two agree, which confirms the wave speed. Note that this is not the 1600(0.06667) = 107 vehicles that would pass a fixed point in the same four minutes; the shock runs upstream to meet the arriving traffic, so it sweeps up more vehicles than a stationary observer would count.
7. Part (c) — the speed of the front of the platoon. Once the tractor leaves, the vehicles at the head of the platoon are released into an uncongested road and accelerate to the capacity discharge state C (k_c = 50 veh/km, q_max = 2,500 veh/h). The boundary between the platoon and that discharging state moves at $$\omega_{BC}=\frac{q_{max}-q_B}{k_c-k_B}=\frac{2500-1275}{50-85}=\frac{1225}{-35}=\boxed{-35\ \text{km/h}}$$ The phrase "speed of the front of the platoon" admits two readings, and both are worth one line. As a boundary, the release wave travels upstream at 35 km/h, eating into the platoon from the front. As a vehicle, the lead car accelerates to the discharge speed u_c = u_f/2 = 50 km/h and moves downstream. The first is what shock-wave theory means by the front of the platoon; the second is what a driver would observe.
8. Part (d) — the dissipation time. The platoon disappears when the release wave at the front catches the stopping wave at the back. Crucially the rear shock keeps running upstream at 5 km/h after the tractor leaves, because state-A traffic is still arriving and joining the tail, so the two boundaries close on each other at the difference of their speeds, not at the release speed alone: $$t_{dis}=\frac{L}{|\omega_{BC}|-|\omega_{AB}|}=\frac{1.333}{35-5}=\frac{1.333}{30} =0.04444\ \text{h}=\boxed{2.67\ \text{minutes}\;(160\ \text{s})}$$ The platoon therefore vanishes 4.0 + 2.67 = 6.67 minutes after the tractor entered, at a point 1.0 − 35(0.04444) = −0.556 km, that is 556 m upstream of where the tractor joined the highway.

Dividing the platoon length by the release-wave speed alone would give 1.333/35 = 2.29 min, about 14 per cent short; that omission — forgetting that the tail is still growing — is the commonest error on this question.