Linear speed-density: free speed 80, jam 120

(Jam density is a number of vehicles per kilometre, so "veh/hr" in the question is read as veh/km.)

(a) Speed-density and flow-density equations

For a linear (Greenshields) relation, speed falls from the free-flow speed $v_f$ at zero density to zero at jam density $k_j$:

$$v = v_f\left(1-\frac{k}{k_j}\right) = 80\left(1-\frac{k}{120}\right) = 80 - 0.6667\,k$$

Since $q = v\,k$:

$$q = v_f\,k\left(1-\frac{k}{k_j}\right) = 80\,k - 0.6667\,k^2$$

(b) Diagrams and critical values

Setting $dq/dk = 0$ gives $k_m = k_j/2$, $v_m = v_f/2$ and $q_{max} = v_f k_j/4$:

$$\begin{aligned} k_m &= 120/2 = 60\ \text{veh/km}\\ v_m &= 80/2 = 40\ \text{km/h}\\ q_{max} &= \frac{80 \times 120}{4} = 2400\ \text{veh/h} \end{aligned}$$

v-k diagram (straight line)
 v (km/h)
   80 |*
      |   *
   40 |- - - *   <- at q max
      |         *
      +-----------*--- k (veh/km)
      0    60     120

q-k diagram (parabola)
 q (veh/h)
 2400 |        _ * _   <- q max
      |     *         *
      |   *             *
      +--*-------|-------*-- k
      0        60       120

v-q diagram (half parabola)
 v (km/h)
   80 |*
      |  *
   40 |- - - *  <- q max (nose)
      |     *
      |   *   (congested branch)
      +--------|--------- q (veh/h)
      0       2400

(c) Speed and density for a flow of 800 veh/h

Substitute in $q = v_f k - (v_f/k_j)k^2$:

$$0.6667k^2 - 80k + 800 = 0 \quad\Rightarrow\quad k = \frac{80 \pm \sqrt{80^2 - 4(0.6667)(800)}}{2(0.6667)}$$ $$k_1 = 11.01\ \text{veh/km},\quad k_2 = 108.99\ \text{veh/km}$$

• Uncongested (stable) state: $k = 11.0$ veh/km and $v = 80(1-11.01/120) = 72.7$ km/h.
• Congested (forced) state: $k = 109.0$ veh/km and $v = 7.3$ km/h.

Answer: for 800 veh/h, speed = 72.7 km/h at density = 11.0 veh/km (stable flow); the congested alternative is 7.3 km/h at 109.0 veh/km.

(d) Headway, spacing at maximum flow

At $q_{max} = 2400$ veh/h, $k_m = 60$ veh/km, $v_m = 40$ km/h:

$$\begin{aligned} \text{Average headway}\ h &= \frac{3600}{q_{max}} = \frac{3600}{2400} = 1.50\ \text{s}\\ \text{Average spacing}\ s &= \frac{1000}{k_m} = \frac{1000}{60} = 16.67\ \text{m} \end{aligned}$$

Answer: headway = 1.50 s, spacing = 16.67 m.