---
title: "Skid numerical: bridge abutment impact at 20 kmph"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:skid-bridge-abutment-gravel-shoulder"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/skid-bridge-abutment-gravel-shoulder
site: BCT Engineering
---

# Skid numerical: bridge abutment impact at 20 kmph

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2073 Shrawan, 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/skid-bridge-abutment-gravel-shoulder
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

A vehicle hits a bridge abutment at a speed estimated by investigations as 20kmph. Skid marks of 30 m on the pavement (f=0.35) followed by skid marks of 60 m on the gravel shoulder approaching the abutment (f=0.50). What was the initial speed of vehicle?

## Short answer

Using the skid-distance equation in km/h with v_a=20 km/h, f1=0.35, d1=30 m, f2=0.50, d2=60 m gives initial speed = sqrt(400 + 254(10.5+30)) = 103.4 km/h (28.73 m/s).

## Answer

The vehicle skidded 30 m on the pavement, left the pavement and skidded another 60 m on the gravel shoulder, and then hit the abutment at 20 km/h. Working backwards by the work-energy equation $v_1^2 = v_2^2 + 2gfd$ over each stretch:

```text
 start -> [ 30 m pavement f=0.35 ] -> [ 60 m gravel f=0.50 ] -> abutment
 v0                                  v1                        20 km/h
```

**Data:** $v_{abutment} = 20$ km/h $= 5.556$ m/s.

#### Step 1: Speed at the start of the gravel shoulder
$$
v_1 = \sqrt{v_a^2 + 2 g f_2 d_2} = \sqrt{(5.556)^2 + 2(9.81)(0.50)(60)} = 24.889\ \text{m/s} = 89.6\ \text{km/h}
$$

#### Step 2: Initial speed at the start of the pavement skid
$$
v_0 = \sqrt{v_1^2 + 2 g f_1 d_1} = \sqrt{(24.889)^2 + 2(9.81)(0.35)(30)} = 28.731\ \text{m/s}
$$

**Answer: initial speed of the vehicle = 28.73 m/s = 103.4 km/h.**

(Check with the IRC/Indian field formula $v = \sqrt{254(f_1d_1+f_2d_2)+v_a^2} = \sqrt{254(0.35\times30 + 0.5\times60) + 20^2} = 103.4$ km/h.)

## Check yourself

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

1. Why is the calculation done starting from the abutment speed and working backwards?
   A. Because the final speed (20 km/h) is known and each skid stretch adds to the speed squared
   B. Because friction is lower near the abutment
   C. Because v^2 = u^2 - 2gfd cannot be used forwards
   D. Because the pavement skid happens last
   Answer: A. The known 20 km/h at impact is the end of the gravel skid; adding 2gfd for each stretch backwards gives the earlier speeds. (Key point: Direction of calculation)

2. Which equation is used over each skid stretch?
   A. v1 = v2 + gfd
   B. v1^2 = v2^2 + 2gfd
   C. v1^2 = v2^2 - 2gfd
   D. v1 = v2 x fd
   Answer: B. Energy lost to friction is gfd per unit mass, so v1^2 = v2^2 + 2gfd with v2 the final speed. (Key point: Work-energy equation)

3. What is the speed at the start of the 60 m gravel skid (f = 0.50)?
   A. 20 km/h
   B. 55.4 km/h
   C. 76.2 km/h
   D. 89.6 km/h
   Answer: D. sqrt(5.556^2 + 2 x 9.81 x 0.5 x 60) = 24.89 m/s = 89.6 km/h. (Key point: Speed at the start of gravel)

4. What is the initial speed of the vehicle?
   A. 89.6 km/h
   B. 103.4 km/h
   C. 110.0 km/h
   D. 20 km/h
   Answer: B. Adding the pavement stretch gives sqrt(24.889^2 + 2 x 9.81 x 0.35 x 30) = 28.73 m/s = 103.4 km/h. (Key point: Initial speed)

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