---
title: "Skid numerical: moving vehicle hits parked vehicle"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:skid-moving-vehicle-hits-parked-40m"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/skid-moving-vehicle-hits-parked-40m
site: BCT Engineering
---

# Skid numerical: moving vehicle hits parked vehicle

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2068 Baisakh, 8 marks
- Video: [Find initial Speed | Transportation numerical | Collision of 2 vehicle | TU.PU.KU.POU](https://www.youtube.com/watch?v=ICzAWfNu_Ks) — 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/skid-moving-vehicle-hits-parked-40m
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

A vehicle skids through a distance of 40m before colliding with another parked vehicle, the weight of which is 75 percent of the former. After collision both the vehicles skid through 14m before stopping. Compute the initial speed of moving vehicle. Assume coefficient of friction of 0.62.

## Short answer

Post-collision speed from 14 m joint skid: 13.05 m/s. Impact speed from momentum conservation (mass ratio 1:0.75): 22.84 m/s. Initial speed from 40 m pre-impact skid: 31.75 m/s (114.3 km/h).

## Answer

The moving vehicle (weight $W$) skids 40 m, hits the parked vehicle (weight $0.75W$) and both skid together 14 m. Work backwards: speed after impact, then momentum, then the 40 m skid.

#### Step 1: Speed after the collision (both vehicles together)
$$
v' = \sqrt{2gfd'} = \sqrt{2(9.81)(0.62)(14)} = 13.050\ \text{m/s}
$$

#### Step 2: Momentum conservation
The parked vehicle has zero speed before impact. With impact speed $u$ of the moving vehicle:

$$
W\,u + 0.75W\,(0) = (W + 0.75W)\,v' \Rightarrow u = 1.75\,v' = 1.75(13.050) = 22.837\ \text{m/s}
$$

#### Step 3: Initial speed before the 40 m skid
$$
v_0 = \sqrt{u^2 + 2gfd} = \sqrt{(22.837)^2 + 2(9.81)(0.62)(40)} = 31.751\ \text{m/s}
$$

**Answer: initial speed of the moving vehicle = 31.75 m/s = 114.3 km/h** (speed at impact 82.2 km/h).

## Check yourself

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

1. What is the correct order of working?
   A. Initial speed, impact speed, post-impact speed
   B. Post-impact speed, then momentum for impact speed, then the 40 m skid for initial speed
   C. Only the 40 m skid
   D. Energy conservation across collision
   Answer: B. Work backwards: v' from the 14 m skid, u by momentum, v0 by adding the 40 m skid. (Key point: Method order)

2. What is the speed v' of the two vehicles right after impact?
   A. 13.05 m/s
   B. 22.8 m/s
   C. 31.75 m/s
   D. 7.46 m/s
   Answer: A. v' = sqrt(2 x 9.81 x 0.62 x 14) = 13.05 m/s. (Key point: Post-collision speed)

3. How is the impact speed u related to v'?
   A. u = v'/1.75
   B. u = 0.75 v'
   C. u = 1.75 v' since W u = 1.75W v'
   D. u = v'
   Answer: C. W u + 0.75W x 0 = 1.75W v', so u = 22.84 m/s. (Key point: Momentum conservation)

4. What is the initial speed of the moving vehicle?
   A. 82.2 km/h
   B. 22.8 km/h
   C. 50 km/h
   D. 114.3 km/h
   Answer: D. v0 = sqrt(22.84^2 + 2x9.81x0.62x40) = 31.75 m/s = 114.3 km/h. (Key point: Initial speed)

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