---
title: "Moving observer: car counts on 800 m road"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:moving-observer-car-800m-20kmph"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/moving-observer-car-800m-20kmph
site: BCT Engineering
---

# Moving observer: car counts on 800 m road

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2082 Baisakh, 8 marks
- Video: [Engineering Traffic Chapter 4.(Moving Observer Method)](https://www.youtube.com/watch?v=yc8GKUbbKU8) — Eng.ibrahim.hevall
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/moving-observer-car-800m-20kmph
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

A car driver was driving along 800 m road. She counted the number of cars that she meets while driving southwards (Ns), the number of cars that overtake her while driving northward (No), and the numbers of cars that she overtakes while driving northward (Np) is shown in table below. Find the average traffic flow and journey speed. Given that she travels at a constant speed of 20 kmph.

| Vehicle met while driving southward | Vehicle overtaking while driving north | Vehicle overtaken while driving north |
|---|---|---|
| 107 | 10 | 74 |
| 113 | 25 | 41 |
| 30 | 15 | 5 |
| 79 | 18 | 9 |

## Short answer

Average counts give Ns=82.25, No=17, Np=32.25, y=-15.25. Flow q = (82.25-15.25)/(0.08) = 837.5 veh/h; journey time 0.0582 h, so journey speed = 0.8/0.0582 = 13.7 km/h.

## Answer

The moving observer method gives the flow $q$ and mean stream speed from counts made by a test car running with and against the stream.

**Reading of the table:** the vehicles she meets while driving southward are the northbound stream she is measuring ($N_s$); while driving northward, $N_o$ = cars that overtake her and $N_p$ = cars she overtakes. Section length 800 m, test speed 20 km/h on both runs.

#### Average counts (4 runs)
$$
\begin{aligned} \bar N_s &= \frac{107+113+30+79}{4} = 82.25\\ \bar N_o &= \frac{10+25+15+18}{4} = 17.00, \quad \bar N_p = \frac{74+41+5+9}{4} = 32.25\\ y &= \bar N_o - \bar N_p = -15.25 \end{aligned}
$$

#### Travel times
$$
t_w = t_a = \frac{0.8}{20} = 0.04\ \text{h} = 2.4\ \text{min}
$$

#### Traffic flow
$$
q = \frac{N_s + y}{t_w + t_a} = \frac{82.25 + (-15.25)}{0.04 + 0.04} = 837.5\ \text{veh/h}
$$

#### Journey time and speed of the stream
$$
\bar t = t_w - \frac{y}{q} = 0.04 - \frac{(-15.25)}{837.5} = 0.0582\ \text{h} = 3.49\ \text{min}
$$

$$
V_j = \frac{L}{\bar t} = \frac{0.8}{0.0582} = 13.74\ \text{km/h}
$$

**Answer: average flow = 837.5 veh/h; journey speed = 13.7 km/h** (journey time 3.49 min). The stream is slower than the test car because more vehicles were overtaken by her than overtook her.

## Check yourself

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

1. In the moving observer method, what is Ns here?
   A. Cars overtaking the observer
   B. Cars the observer overtakes
   C. Vehicles met while driving against the stream (the stream measured)
   D. Total cars in the lane
   Answer: C. The vehicles met while driving southward are the northbound stream being measured. (Key point: Meaning of counts)

2. How is the correction y calculated?
   A. y = No - Np (overtaking minus overtaken)
   B. y = No + Np
   C. y = Ns - No
   D. y = Np only
   Answer: A. y = 17.00 - 32.25 = -15.25. (Key point: Correction y)

3. Which formula gives flow q?
   A. q = (Ns + y)/(tw + ta)
   B. q = Ns x (tw + ta)
   C. q = y/(tw + ta)
   D. q = (Ns - y) x tw
   Answer: A. q = (82.25 - 15.25)/(0.04 + 0.04) = 837.5 veh/h. (Key point: Flow formula)

4. Why is the stream's journey speed (13.7 km/h) lower than the test car's 20 km/h?
   A. Because y is negative: the observer overtook more cars than overtook her
   B. Because the road is long
   C. Because Ns is large
   D. Because the flow is high
   Answer: A. A negative y raises mean journey time (t = tw - y/q), so the stream is slower than the test car. (Key point: Journey speed meaning)

## Related in Transportation Engineering II

- [Moving observer: bicycle racer on 800 m bikeway](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Amoving-observer-bicycle-800m) — Transportation Engineering II, chapter 1

## Related in other subjects

- [Power of a car moving uphill](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Acar-power-uphill-efficiency) — Engineering Mechanics, chapter 3