Question 4 of 7: Greenshields Model and Shock Waves Behind a Slow Truck

Question 4: Greenshields Model and Shock Waves Behind a Slow Truck (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 length of the platoon that has built up behind the truck at the moment it leaves the highway; the speed at which the front of that platoon moves once the obstruction is removed; and the time the platoon then takes to disappear.

Approach. Calibrate the Greenshields curve from $u_f$ and $q_{\max}$, identify the three traffic states involved — the approach flow, the platoon travelling at the truck's speed, and the discharge state after the truck exits — and locate each on the flow–density diagram. Every boundary between two states is a shock wave whose speed is the chord slope $\omega = \Delta q/\Delta k$ between them, and the whole question is then bookkeeping on where those boundaries are at each instant.

1. (a) Calibrate the Greenshields model. The Greenshields speed–density relation is linear, $$u = u_f\left(1 - \frac{k}{k_j}\right)\qquad\text{so}\qquad q = ku = u_f\left(k - \frac{k^{2}}{k_j}\right)$$ Setting $dq/dk = 0$ gives the density at capacity at exactly half the jam density, $k_c = k_j/2$, and therefore $q_{\max} = u_f k_j/4$. Inverting for the jam density: $$k_j = \frac{4\,q_{\max}}{u_f} = \frac{4(1800)}{120}$$ $$\boxed{k_j = 60\ \text{veh/km}\qquad k_c = \frac{k_j}{2} = 30\ \text{veh/km}}$$ The corresponding speed at capacity is $u_c = u_f/2 = 60$ km/h, and a check confirms the calibration: $q = 60 \times 30 \times (1 - 30/60) \times 2 = 1800$ veh/h. A jam density of 60 veh/km implies an average space headway of 16.7 m per vehicle at a standstill, which is realistic for mixed traffic and confirms the calibration is physically sensible.
2. Identify state A, the approach flow. At the prevailing density, $$u_A = u_f\left(1 - \frac{k_A}{k_j}\right) = 120\left(1 - \frac{10}{60}\right) = 100\ \text{km/h},\qquad q_A = k_A u_A = 10(100) = 1000\ \text{veh/h}$$ The highway is operating well below capacity on the uncongested branch of the curve.
3. Identify state B, the platoon behind the truck. The vehicles trapped behind the truck all travel at the truck's speed, so the Greenshields relation inverted for density gives $$k_B = k_j\left(1 - \frac{u_B}{u_f}\right) = 60\left(1 - \frac{40}{120}\right) = 40\ \text{veh/km},\qquad q_B = k_B u_B = 40(40) = 1600\ \text{veh/h}$$ State B lies on the congested branch, at four times the approach density. Note that $q_B = 1600$ veh/h exceeds $q_A = 1000$ veh/h even though the platoon is travelling more slowly; that is not a contradiction, because flow is the product of speed and density and the density has risen by more than the speed has fallen. It is also the reason the back of the platoon moves forward in Step 4 rather than backward.
4. (b) Locate the rear of the platoon — the A∣B shock. The boundary between the approach state and the platoon state travels at the chord slope joining the two states on the flow–density diagram: $$\omega_{AB} = \frac{q_B - q_A}{k_B - k_A} = \frac{1600 - 1000}{40 - 10} = \frac{600}{30} = +20\ \text{km/h}$$ The positive sign means the rear of the platoon advances downstream at 20 km/h. Physically, the approach flow of 1000 veh/h is light enough that vehicles do not join the back of the queue as fast as the queue discharges into it, so the congested region does not grow backward against the traffic.
5. Compute the platoon length at the moment the truck exits. The front of the platoon is the truck itself, moving at 40 km/h; the rear is the shock, moving at 20 km/h. The platoon therefore lengthens at the difference of the two, $40 - 20 = 20$ km/h. The truck occupies the highway for $$t_b = \frac{L_T}{u_T} = \frac{2}{40} = 0.05\ \text{h} = 3.0\ \text{min}$$ so at the instant it exits, $$\boxed{L_{\text{platoon}} = \left(u_T - \omega_{AB}\right)t_b = (40 - 20)(0.05) = 1.0\ \text{km}}$$ Two independent counts of the vehicles stored in that platoon must agree, and they do — this is the cheapest available check that the shock speed is right. Counting by density within the platoon, $k_B L = 40(1.0) = 40$ vehicles. Counting the vehicles the shock has overtaken, $\left(u_A - \omega_{AB}\right)k_A t_b = (100 - 20)(10)(0.05) = 40$ vehicles. Note that this is not the same as the 50 vehicles that would pass a fixed point in three minutes at 1000 veh/h; the shock is itself moving, so the relative speed and not the absolute speed governs how many vehicles it absorbs.
6. (c) Locate the front of the platoon after the truck exits. Once the truck leaves, nothing downstream restrains the lead vehicles, and the question states there is no congestion further downstream, so the platoon discharges at the highest rate the highway can sustain — capacity. That is state C, with $k_C = k_c = 30$ veh/km, $q_C = q_{\max} = 1800$ veh/h and $u_C = 60$ km/h. The boundary between the still-congested platoon and this discharging state is $$\omega_{BC} = \frac{q_C - q_B}{k_C - k_B} = \frac{1800 - 1600}{30 - 40} = \frac{200}{-10}$$ $$\boxed{\omega_{BC} = -20\ \text{km/h, i.e. 20 km/h travelling upstream}}$$ The negative sign means the front boundary of the platoon eats backward into the platoon at 20 km/h: successive vehicles are released and accelerate, so the congested region shrinks from the front. The phrase "speed of the front of the platoon" admits a second reading, and both are worth one line because the examiner may intend either. As a boundary, the front of the platoon retreats upstream at 20 km/h; as a vehicle, the lead car of the platoon accelerates to the discharge speed $u_C = 60$ km/h and moves downstream at that speed. The 20 km/h figure is the one that is used in part (d).
7. (d) Dissipation time. The platoon is now bounded at the rear by $\omega_{AB} = +20$ km/h and at the front by $\omega_{BC} = -20$ km/h. The two boundaries approach each other at the sum of their magnitudes, 40 km/h, and the platoon disappears when they meet: $$t_{\text{diss}} = \frac{L_{\text{platoon}}}{\omega_{AB} - \omega_{BC}} = \frac{1.0}{20 - (-20)} = \frac{1.0}{40} = 0.025\ \text{h}$$ $$\boxed{t_{\text{diss}} = 0.025\ \text{h} = 1.5\ \text{min} = 90\ \text{s}}$$ The whole disturbance therefore lasts 4.5 minutes from the moment the truck enters, and its effect is confined to a stretch of road 1.5 km long. Once the two shocks meet, states A and C are in contact and a new shock forms between them at $\omega_{AC} = (1800-1000)/(30-10) = +40$ km/h, which simply sweeps the last of the disturbance downstream.