---
title: "Linear speed-density: free speed 80, jam 130"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:linear-speed-density-free-80-130"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/linear-speed-density-free-80-130
site: BCT Engineering
---

# Linear speed-density: free speed 80, jam 130

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2072 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/linear-speed-density-free-80-130
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

Assuming a linear speed-density relationship, the mean free speed is observed to be 80 km/h near zero density and the corresponding jam density is 130 veh/km. Assume that the average length of vehicles is 6 m.
i) Write down the speed-density and flow-density equations
ii) Compute speed and density corresponding to flow of 1000 veh/hr.
iii) Compute the average headways, spacing, clearance and gaps when the flow is maximum

## Short answer

Equations: v=80(1-k/130), q=80k-0.6154k^2. For q=1000 veh/h, k=14.0 veh/km and v=71.4 km/h (stable) or k=116.0 and v=8.6 km/h (congested). At qmax=2600 veh/h, km=65, vm=40; headway 1.38 s, spacing 15.38 m, clearance 9.38 m, gap 0.84 s.

## Answer

#### (i) 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 $k_j$:

$$
v = v_f\left(1-\frac{k}{k_j}\right) = 80\left(1-\frac{k}{130}\right) = 80 - 0.6154\,k
$$

Since $q = v\,k$:

$$
q = v_f\,k\left(1-\frac{k}{k_j}\right) = 80\,k - 0.6154\,k^2
$$

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

$$
\begin{aligned} k_m &= 130/2 = 65\ \text{veh/km}\\ v_m &= 80/2 = 40\ \text{km/h}\\ q_{max} &= \frac{80 \times 130}{4} = 2600\ \text{veh/h} \end{aligned}
$$

```text
v-k diagram (straight line)
 v (km/h)
   80 |*
      |   *
   40 |- - - *   <- at q max
      |         *
      +-----------*--- k (veh/km)
      0    65     130

q-k diagram (parabola)
 q (veh/h)
 2600 |        _ * _   <- q max
      |     *         *
      |   *             *
      +--*-------|-------*-- k
      0        65       130

v-q diagram (half parabola)
 v (km/h)
   80 |*
      |  *
   40 |- - - *  <- q max (nose)
      |     *
      |   *   (congested branch)
      +--------|--------- q (veh/h)
      0       2600
```

#### (ii) Speed and density for a flow of 1000 veh/h
Substitute in $q = v_f k - (v_f/k_j)k^2$:

$$
0.6154k^2 - 80k + 1000 = 0 \quad\Rightarrow\quad k = \frac{80 \pm \sqrt{80^2 - 4(0.6154)(1000)}}{2(0.6154)}
$$

$$
k_1 = 14.01\ \text{veh/km},\quad k_2 = 115.99\ \text{veh/km}
$$

- Uncongested (stable) state: $k = 14.0$ veh/km and $v = 80(1-14.01/130) = 71.4$ km/h.
- Congested (forced) state: $k = 116.0$ veh/km and $v = 8.6$ km/h.

**Answer: for 1000 veh/h, speed = 71.4 km/h at density = 14.0 veh/km (stable flow); the congested alternative is 8.6 km/h at 116.0 veh/km.**

#### (iii) Headway, spacing, clearance and gap at maximum flow
At $q_{max} = 2600$ veh/h, $k_m = 65$ veh/km, $v_m = 40$ km/h:

$$
\begin{aligned} \text{Average headway}\ h &= \frac{3600}{q_{max}} = \frac{3600}{2600} = 1.38\ \text{s}\\ \text{Average spacing}\ s &= \frac{1000}{k_m} = \frac{1000}{65} = 15.38\ \text{m}\\ \text{Clearance}\ &= s - L = 15.38 - 6 = 9.38\ \text{m}\\ \text{Gap (time)}\ &= h - \frac{L}{v_m} = 1.38 - \frac{6}{11.111} = 0.84\ \text{s}
\end{aligned}
$$

**Answer: headway = 1.38 s, spacing = 15.38 m, clearance = 9.38 m, gap = 0.84 s.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With vf = 80 km/h and kj = 130 veh/km, which is the linear speed-density equation?
   A. v = 130 - 1.625k
   B. v = 80 - 0.6154k
   C. v = 80 + 0.6154k
   D. v = 80k - 0.6154k^2
   Answer: B. v = vf(1 - k/kj) = 80 - (80/130)k = 80 - 0.6154k. (Key point: Speed-density equation)

2. What are the maximum flow and the corresponding speed and density?
   A. 2600 veh/h at 40 km/h and 65 veh/km
   B. 5200 veh/h at 80 km/h and 130 veh/km
   C. 2600 veh/h at 80 km/h and 65 veh/km
   D. 1300 veh/h at 40 km/h and 32.5 veh/km
   Answer: A. qmax = vf kj/4 = 2600 veh/h at km = 65 veh/km and vm = 40 km/h. (Key point: Maximum flow)

3. At 1000 veh/h, what are the speed and density in stable (uncongested) flow?
   A. 8.6 km/h and 116 veh/km
   B. 40 km/h and 65 veh/km
   C. 71.4 km/h and 14.0 veh/km
   D. 80 km/h and 12.5 veh/km
   Answer: C. The smaller root of 0.6154k^2 - 80k + 1000 = 0 is k = 14.0, giving v = 80(1 - 14/130) = 71.4 km/h. (Key point: Flow of 1000 veh/h)

4. At maximum flow with vehicle length 6 m, what are the spacing and clearance?
   A. Spacing 9.38 m, clearance 15.38 m
   B. Spacing 15.38 m, clearance 9.38 m
   C. Spacing 15.38 m, clearance 21.38 m
   D. Spacing 6 m, clearance 9.38 m
   Answer: B. Spacing = 1000/65 = 15.38 m; clearance = spacing - length = 9.38 m. (Key point: Spacing and clearance at capacity)

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