Linear speed-density V = 60 - 0.43K

For the linear model $V = V_f - bK$ with $V_f = 60$ km/h and $b = 0.43$, the jam density is where $V = 0$:

$$k_j = \frac{60}{0.43} = 139.5\ \text{veh/km}$$

The flow is $Q = VK = 60K - 0.43K^2$, so the critical (maximum-flow) values are at $dQ/dK = 0$:

$$\begin{aligned} K_m &= \frac{k_j}{2} = 69.8\ \text{veh/km}\\ V_m &= \frac{V_f}{2} = 30\ \text{km/h}\\ Q_{max} &= \frac{V_f k_j}{4} = \frac{60 \times 139.5}{4} = 2093\ \text{veh/h} \end{aligned}$$

(a) Diagrams with critical values

v-k diagram (straight line)
 v (km/h)
   60 |*
      |   *
   30 |- - - *   <- at q max
      |         *
      +-----------*--- k (veh/km)
      0    69.8     139.5

q-k diagram (parabola)
 q (veh/h)
2093.0 |        _ * _   <- q max
      |     *         *
      |   *             *
      +--*-------|-------*-- k
      0        69.8       139.5

v-q diagram (half parabola)
 v (km/h)
   60 |*
      |  *
   30 |- - - *  <- q max (nose)
      |     *
      |   *   (congested branch)
      +--------|--------- q (veh/h)
      0       2093.0

(b) Saturation flow

The saturation flow is the maximum flow (capacity) at the critical density:

Saturation flow = $Q_{max}$ = 2093 veh/h

(c) Speed and density at a flow of 1000 veh/h

$$0.43K^2 - 60K + 1000 = 0 \Rightarrow K = \frac{60 \pm \sqrt{60^2 - 4(0.43)(1000)}}{2(0.43)} = \frac{60 \pm 43.36}{0.86}$$ $$K_1 = 19.35\ \text{veh/km},\quad K_2 = 120.18\ \text{veh/km}$$

• Uncongested: $K = 19.35$ veh/km, $V = 60 - 0.43(19.35) = 51.68$ km/h.
• Congested: $K = 120.18$ veh/km, $V = 8.32$ km/h.

Answer: at 1000 veh/h the speed is 51.7 km/h at a density of 19.4 veh/km (stable flow); in congested flow it is 8.3 km/h at 120.2 veh/km.