Linear speed-density: free speed 90, jam 150

(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 $k_j$:

$$v = v_f\left(1-\frac{k}{k_j}\right) = 90\left(1-\frac{k}{150}\right) = 90 - 0.6\,k$$

Since $q = v\,k$:

$$q = v_f\,k\left(1-\frac{k}{k_j}\right) = 90\,k - 0.6\,k^2$$

(b) Diagrams and critical values

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

$$\begin{aligned} k_m &= 150/2 = 75\ \text{veh/km}\\ v_m &= 90/2 = 45\ \text{km/h}\\ q_{max} &= \frac{90 \times 150}{4} = 3375\ \text{veh/h} \end{aligned}$$

v-k diagram (straight line)
 v (km/h)
   90 |*
      |   *
   45 |- - - *   <- at q max
      |         *
      +-----------*--- k (veh/km)
      0    75     150

q-k diagram (parabola)
 q (veh/h)
 3375 |        _ * _   <- q max
      |     *         *
      |   *             *
      +--*-------|-------*-- k
      0        75       150

v-q diagram (half parabola)
 v (km/h)
   90 |*
      |  *
   45 |- - - *  <- q max (nose)
      |     *
      |   *   (congested branch)
      +--------|--------- q (veh/h)
      0       3375

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

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

$$0.6k^2 - 90k + 800 = 0 \quad\Rightarrow\quad k = \frac{90 \pm \sqrt{90^2 - 4(0.6)(800)}}{2(0.6)}$$ $$k_1 = 9.49\ \text{veh/km},\quad k_2 = 140.51\ \text{veh/km}$$

• Uncongested (stable) state: $k = 9.5$ veh/km and $v = 90(1-9.49/150) = 84.3$ km/h.
• Congested (forced) state: $k = 140.5$ veh/km and $v = 5.7$ km/h.

Answer: for 800 veh/h, speed = 84.3 km/h at density = 9.5 veh/km (stable flow); the congested alternative is 5.7 km/h at 140.5 veh/km.

(d) Headway, spacing, clearance and gap at maximum flow

At $q_{max} = 3375$ veh/h, $k_m = 75$ veh/km, $v_m = 45$ km/h:

$$\begin{aligned} \text{Average headway}\ h &= \frac{3600}{q_{max}} = \frac{3600}{3375} = 1.07\ \text{s}\\ \text{Average spacing}\ s &= \frac{1000}{k_m} = \frac{1000}{75} = 13.33\ \text{m}\\ \text{Clearance}\ &= s - L = 13.33 - 6.1 = 7.23\ \text{m}\\ \text{Gap (time)}\ &= h - \frac{L}{v_m} = 1.07 - \frac{6.1}{12.500} = 0.58\ \text{s} \end{aligned}$$

Answer: headway = 1.07 s, spacing = 13.33 m, clearance = 7.23 m, gap = 0.58 s.