---
title: "Linear speed-density: free speed 80, jam 120"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:linear-speed-density-free-80-jam"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/linear-speed-density-free-80-jam
site: BCT Engineering
---

# Linear speed-density: free speed 80, jam 120

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2081 Bhadra, 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-jam
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

Assuming the linear speed-density relationship, the mean free speed is observed to be 80 kmph near zero density and the corresponding jam density is 120 veh/hr.

a) Write down the speed-density and flow density equation.
b) Draw v-k, v-q and q-v diagram indicating the critical values.
c) Compute speed and density corresponding to flow of 800 veh/hr.
d) Compute the avg headways, spacing when the flow is maximum.

## Short answer

Speed-density relation v=80(1−k/120) km/h, flow-density q=80k−0.6667k². Maximum flow 2400 veh/h at density 60 veh/km, speed 40 km/h. At flow 800 veh/h, speeds 72.7 and 7.3 km/h at densities 11.0 and109.0 veh/km; average headway1.50 s, spacing16.67 m at capacity.

## Answer

(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 $q_{max} = v_f k_j/4$:

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

```text
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.**

## 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 = 120 veh/km, which equation is correct?
   A. v = 80(1 + k/120)
   B. v = 80(1 - k/120)
   C. v = 120(1 - k/80)
   D. v = 80 k/120
   Answer: B. Greenshields: v = vf(1 - k/kj) = 80 - 0.6667k. (Key point: Speed-density equation)

2. What are the speed and density at maximum flow?
   A. 80 km/h and 120 veh/km
   B. 40 km/h and 60 veh/km
   C. 20 km/h and 30 veh/km
   D. 60 km/h and 40 veh/km
   Answer: B. At capacity vm = vf/2 = 40 and km = kj/2 = 60; qmax = 80 x 120/4 = 2400 veh/h. (Key point: Critical values)

3. Why does a flow of 800 veh/h give two densities (11.0 and 109.0 veh/km)?
   A. The equation is quadratic in k: one uncongested and one congested state
   B. Data are wrong
   C. Jam density changes
   D. Speed is constant
   Answer: A. q is a parabola in k, so a flow below capacity occurs in a stable and a forced state. (Key point: Two states for given flow)

4. At qmax = 2400 veh/h and km = 60 veh/km, what are headway and spacing?
   A. 1.5 s and 16.67 m
   B. 2.4 s and 60 m
   C. 1.5 s and 60 m
   D. 3.0 s and 16.67 m
   Answer: A. Headway = 3600/2400 = 1.5 s; spacing = 1000/60 = 16.67 m. (Key point: Headway and spacing at capacity)

## Close variants set in other exams

- [Linear speed-density: free speed 90, jam 150](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Alinear-speed-density-free-90-jam) — Transportation Engineering II, chapter 1, asked 2×

## Related in Transportation Engineering II

- [Linear speed-density: free speed 80, jam 130](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Alinear-speed-density-free-80-130) — Transportation Engineering II, chapter 1
- [Linear speed-density V = 60 - 0.43K](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Alinear-speed-density-v-60-minus) — Transportation Engineering II, chapter 1