---
title: "Floating car method on 3.2 km road C-D"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:floating-car-3-2km-cd-trips"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/floating-car-3-2km-cd-trips
site: BCT Engineering
---

# Floating car method on 3.2 km road C-D

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2070 Chaitra, 8 marks
- Video: [Average flow, Journey speed, Running Speed | stop & delay study | Numerical | Transportation |](https://www.youtube.com/watch?v=ZhQcd2cbUB4) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/floating-car-3-2km-cd-trips
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

The data collected after speed and delay studies by floating car method on a stretch of road 3.2 km long are given below. Determine the average values of volume, journey speed and running speed of the traffic stream along either direction.

| Trip | Direction of trip | Journey time (min) | Total stopped delay (min) | Overtaking | Overtaken | From opposite direction |
|---|---|---|---|---|---|---|
| 1 | C-D | 6.50 | 1.58 | 4 | 7 | 270 |
| 2 | D-C | 7.48 | 1.72 | 5 | 4 | 250 |
| 3 | C-D | 6.92 | 1.62 | 5 | 4 | 300 |
| 4 | D-C | 7.82 | 1.82 | 3 | 3 | 275 |
| 5 | C-D | 6.33 | 1.40 | 3 | 2 | 295 |
| 6 | D-C | 8.13 | 2.10 | 2 | 1 | 280 |
| 7 | C-D | 6.71 | 1.73 | 4 | 4 | 300 |
| 8 | D-C | 7.40 | 1.85 | 3 | 3 | 230 |
| 9 | C-D | 6.23 | 1.60 | 4 | 2 | 275 |
| 10 | D-C | 6.98 | 1.78 | 2 | 1 | 242 |

## Short answer

Floating car method: flow from opposing counts and overtaking, journey time adjusted. C-D: 1088 veh/h, journey 29.4 km/h, running 38.9 km/h. D-C: 1228 veh/h, 25.5 km/h, 33.8 km/h. Average: 1158 veh/h, 27.5 km/h, 36.3 km/h.

## Answer

For each direction, the floating car (moving observer) method gives the flow $q = \dfrac{n_a + y}{t_a + t_w}$ where $n_a$ = vehicles counted from the opposite direction while running against the stream (these are the vehicles of the direction being studied), $y$ = overtaking minus overtaken during the run with the stream, $t_w$ = journey time with the stream and $t_a$ = journey time against it. Odd trips are C-D and even trips are D-C; $L = 3.2$ km.

#### Direction C-D (stream moving C to D)
Averages of 5 trips: $t_w = 6.538$ min (this direction), $t_a = 7.562$ min (opposite direction), opposing vehicles $n_a = 255.4$, overtaking 4.0, overtaken 3.8 so $y = 0.2$, mean stopped delay 1.586 min.

$$
\begin{aligned} q &= \frac{n_a + y}{t_a + t_w} \times 60 = \frac{255.4 + (0.2)}{7.562 + 6.538} \times 60 = 1088\ \text{veh/h}\\ \bar t &= t_w - \frac{y}{q} = 6.538 - \frac{0.2}{18.13} = 6.527\ \text{min}\\ V_j &= \frac{L}{\bar t} = \frac{3.2 \times 60}{6.527} = 29.4\ \text{km/h}\\ V_r &= \frac{L}{\bar t - \text{delay}} = \frac{3.2 \times 60}{6.527 - 1.586} = 38.9\ \text{km/h} \end{aligned}
$$

#### Direction D-C (stream moving D to C)
Averages of 5 trips: $t_w = 7.562$ min (this direction), $t_a = 6.538$ min (opposite direction), opposing vehicles $n_a = 288.0$, overtaking 3.0, overtaken 2.4 so $y = 0.6$, mean stopped delay 1.854 min.

$$
\begin{aligned} q &= \frac{n_a + y}{t_a + t_w} \times 60 = \frac{288.0 + (0.6)}{6.538 + 7.562} \times 60 = 1228\ \text{veh/h}\\ \bar t &= t_w - \frac{y}{q} = 7.562 - \frac{0.6}{20.47} = 7.533\ \text{min}\\ V_j &= \frac{L}{\bar t} = \frac{3.2 \times 60}{7.533} = 25.5\ \text{km/h}\\ V_r &= \frac{L}{\bar t - \text{delay}} = \frac{3.2 \times 60}{7.533 - 1.854} = 33.8\ \text{km/h} \end{aligned}
$$

#### Average of the two directions
| Direction | Volume (veh/h) | Journey speed (km/h) | Running speed (km/h) |
|---|---|---|---|
| C-D | 1088 | 29.4 | 38.9 |
| D-C | 1228 | 25.5 | 33.8 |
| Average | 1158 | 27.5 | 36.3 |

**Answer: C-D: volume 1088 veh/h, journey speed 29.4 km/h, running speed 38.9 km/h; D-C: volume 1228 veh/h, journey speed 25.5 km/h, running speed 33.8 km/h. Average of both: 1158 veh/h, 27.5 km/h and 36.3 km/h.**

## Check yourself

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

1. In q = (n_a + y)/(t_a + t_w), what is n_a?
   A. Vehicles overtaking the test car
   B. Vehicles counted from the opposite direction while running against the stream
   C. Vehicles overtaken by the test car
   D. Number of trips made
   Answer: B. The opposing vehicles counted during the run against the stream are the vehicles of the direction studied. (Key point: Meaning of n_a)

2. How is y defined?
   A. Overtaking plus overtaken
   B. Overtaken minus overtaking
   C. Overtaking minus overtaken during the run with the stream
   D. Opposing vehicles minus overtaking
   Answer: C. y = vehicles overtaking the test car minus vehicles it overtakes, in the run with the stream. (Key point: Overtaking correction y)

3. Why is running speed (38.9 km/h) higher than journey speed (29.4 km/h) for C-D?
   A. Running speed excludes stopped delay from the time
   B. Running speed uses a longer distance
   C. Running speed uses opposing traffic
   D. Journey speed ignores the length of the stretch
   Answer: A. V_r = L/(t - delay) uses a shorter time, so it is larger than V_j = L/t. (Key point: Journey vs running speed)

4. What is the flow for direction C-D?
   A. 1228 veh/h
   B. 1158 veh/h
   C. 255 veh/h
   D. 1088 veh/h
   Answer: D. q = (255.4 + 0.2)/(7.562 + 6.538) x 60 = 1088 veh/h. (Key point: Result for C-D)

## Related in Transportation Engineering II

- [Floating car method: flow, journey and running speed](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Afloating-car-25-5km-flow-speed) — Transportation Engineering II, chapter 1
- [Moving observer: car counts on 800 m road](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Amoving-observer-car-800m-20kmph) — Transportation Engineering II, chapter 1