---
title: "Importance of vision; acceleration numerical"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:vision-importance-acceleration-numerical"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/vision-importance-acceleration-numerical
site: BCT Engineering
---

# Importance of vision; acceleration numerical

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2080 Baisakh, 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/vision-importance-acceleration-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

Explain the importance of vision to Road users. The acceleration of a vehicle is given by the equation below: dv/dt = 2 − 0.04v. Compute the time when the acceleration of the vehicle will be 1.0 m/s² if the vehicle was initially travelling at 100 kph.

## Short answer

About 90 percent of driver information is visual; acuity, field, colour, glare, depth, and night vision affect safe driving. Given dv/dt=2-0.04v at 100 km/h, initial acceleration is 0.889 m/s2 and decreases, so 1.0 m/s2 is never reached afterward; it occurred 2.94 s earlier at 90 km/h.

## Answer

#### Importance of vision to road users
About 90 percent of the information a road user needs (road alignment, signs, signals, other vehicles, pedestrians and hazards) is received through the eyes, so vision is the most important sense in driving. Defective vision causes late detection, wrong judgement of speed and distance, and crashes.

- **Visual acuity** decides the distance at which signs and objects can be read; sign letter heights are fixed to suit it.
- **Field of vision** (clear cone about 3-5 degrees, fairly clear up to 10-12 degrees, peripheral up to about 120-160 degrees) fixes where signs are placed; it narrows with speed.
- **Colour vision** is needed to read red, amber and green signals.
- **Glare vision and recovery** affect night driving; they decide anti-glare design and lighting.
- **Depth perception** is needed for judging gaps when overtaking and crossing.
- **Night vision** needs street lighting, retro-reflective signs and markings.

Therefore design provides adequate sight distance (IRC:66, IRC:SP:23), legible signs (IRC:67), lighting and markings (IRC:35).

#### Numerical
Given $\dfrac{dv}{dt} = 2 - 0.04v$ (v in m/s), initial speed $v_0 = 100$ km/h $= 27.78$ m/s.

Since $a = 2 - 0.04v$, differentiating gives $\dfrac{da}{dt} = -0.04\,\dfrac{dv}{dt} = -0.04a$, so

$$
a(t) = a_0 e^{-0.04t},\qquad a_0 = 2 - 0.04(27.78) = 0.889\ \text{m/s}^2
$$

For $a = 1.0$ m/s$^2$ the speed would have to be $v = (2-1)/0.04 = 25$ m/s $= 90$ km/h:

$$
t = \frac{1}{0.04}\ln\frac{a_0}{1.0} = 25\ln(0.889) = -2.94\ \text{s}
$$

**Answer:** the acceleration at the start (100 km/h) is only 0.889 m/s$^2$, and it keeps falling as the vehicle speeds up, so it never reaches 1.0 m/s$^2$ afterwards. The acceleration was 1.0 m/s$^2$ about **2.94 s before** the 100 km/h instant (t = -2.94 s, when the speed was 90 km/h). The value 1.0 m/s$^2$ is reached after time zero only if the vehicle starts below 90 km/h (25 m/s); for example, from rest $t = 25\ln 2 = 17.33$ s.

## Check yourself

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

1. About how much of the information a road user needs is received through the eyes?
   A. About 10 percent
   B. About 50 percent
   C. About 90 percent
   D. 100 percent
   Answer: C. Around 90 percent of driving information is visual, which makes vision the most important sense. (Key point: Importance of vision)

2. Why is colour vision needed by road users?
   A. To judge gaps in overtaking
   B. To read red, amber and green signals
   C. To see at night
   D. To recover from glare
   Answer: B. Colour vision is needed for signals; depth perception is for gap judgement and glare recovery for night driving. (Key point: Colour vision and glare)

3. For dv/dt = 2 - 0.04v and v0 = 100 km/h (27.78 m/s), the initial acceleration is:
   A. 2.00 m/s^2
   B. 1.00 m/s^2
   C. 0.889 m/s^2
   D. 1.11 m/s^2
   Answer: C. a0 = 2 - 0.04(27.78) = 0.889 m/s^2. (Key point: Initial acceleration)

4. Since da/dt = -0.04a, acceleration varies with time as:
   A. a = a0 + 0.04t
   B. a = a0 e^(-0.04t)
   C. a = a0 - 0.04t
   D. a = a0 e^(0.04t)
   Answer: B. Differentiating the equation gives da/dt = -0.04a, so a decays exponentially. (Key point: Acceleration decay)

5. Why is the time to reach a = 1.0 m/s^2 negative (-2.94 s)?
   A. The vehicle already passed that acceleration; at 100 km/h a is 0.889 and keeps falling as speed rises
   B. A calculation error, the time should be +2.94 s
   C. Because the vehicle is decelerating
   D. Because 1.0 m/s^2 is the maximum acceleration
   Answer: A. a = 1.0 corresponds to v = 25 m/s (90 km/h), which occurred 2.94 s before the 100 km/h instant. (Key point: Interpretation of result)

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