---
title: "Skid numerical: A from West, B from South (49, 27)"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:skid-collision-west-south-49-27"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/skid-collision-west-south-49-27
site: BCT Engineering
---

# Skid numerical: A from West, B from South (49, 27)

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2075 Asoj, 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-collision-west-south-49-27
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

Two vehicles A and B approaching at right angle. Vehicle A from West and vehicle B from south collides each other. After the collision, vehicle A skids in 49° N of W and vehicle B skids 27° E of N. The initial skid distance of vehicle A and B are 37 m and 19 m respectively before collision. If weight of vehicle A is 4 tonne and weight of vehicle B is 6 tonne. The skid distances after collision for vehicle A is 15 m and for vehicle B is 36 m. calculate the initial speeds of the vehicles if the average skid resistance of the pavement is found to be 0.55.

## Short answer

Post-collision speeds from skid: v' = √(2gfd'). Via momentum conservation, impact speeds: u_A=5.08 m/s, u_B=23.96 m/s. Adding pre-collision skid energy, original speeds: vehicle A = 74.2 km/h, vehicle B = 100.5 km/h.

## Answer

A collision on a flat road is solved by (i) finding each vehicle's speed just after impact from its post-collision skid, (ii) applying conservation of linear momentum as a vector equation to get the speeds just before impact, and (iii) adding back the energy lost in the pre-collision skid.

**Given:** $W_A = 4$ t, $W_B = 6$ t, $f = 0.55$, $g = 9.81\ \text{m/s}^2$. Pre-collision skids: $d_A = 37$ m, $d_B = 19$ m. Post-collision skids: $d_A' = 15$ m, $d_B' = 36$ m.

Vehicle A (4 t) comes from the West (moving East). Vehicle B (6 t) comes from the South (moving North). After impact A slides 49 degrees North of West and B slides 27 degrees East of North.

Let $\theta$ be the bearing measured clockwise from North: $\theta_A = 90^\circ$, $\theta_B = 0^\circ$ (travel directions) and $\theta_A' = 270^\circ + 49^\circ = 319^\circ$, $\theta_B' = 27^\circ$ (skid directions after impact).

```text
   before impact          after impact
   A: bearing 90 deg       A': bearing 319 deg (49 deg N of W)
   B: bearing 0 deg        B': bearing 27 deg  (27 deg E of N)
   (bearing measured clockwise from North)
```

#### Step 1: Speeds just after collision
From work-energy, $v' = \sqrt{2 g f d'}$:

$$
\begin{aligned} v_A' &= \sqrt{2(9.81)(0.55)(15)} = 12.72\ \text{m/s}\\ v_B' &= \sqrt{2(9.81)(0.55)(36)} = 19.71\ \text{m/s} \end{aligned}
$$

#### Step 2: Momentum balance (vector form)
Take $x$ = East and $y$ = North, and let $u_A, u_B$ be the speeds just before impact. Weights are proportional to masses, so they can be used directly.

$$
\begin{aligned} W_A u_A \sin\theta_A + W_B u_B \sin\theta_B &= W_A v_A' \sin\theta_A' + W_B v_B' \sin\theta_B' \\ W_A u_A \cos\theta_A + W_B u_B \cos\theta_B &= W_A v_A' \cos\theta_A' + W_B v_B' \cos\theta_B' \end{aligned}
$$

Substituting the numbers ($\sin 319^\circ = -0.6561$, $\cos 319^\circ = 0.7547$):

$$
\begin{aligned} 4\,u_A(1.0000) + 6\,u_B(0.0000) &= 4(12.723)(-0.6561) + 6(19.710)(0.4540) = 20.301 \\ 4\,u_A(0.0000) + 6\,u_B(1.0000) &= 4(12.723)(0.7547) + 6(19.710)(0.8910) = 143.777 \end{aligned}
$$

Solving: $u_A = 5.08\ \text{m/s} = 18.3\ \text{km/h}$ and $u_B = 23.96\ \text{m/s} = 86.3\ \text{km/h}$ at the instant of impact.

#### Step 3: Original speeds (before the pre-collision skid)
$v_0 = \sqrt{u^2 + 2 g f d}$:

$$
\begin{aligned} v_A &= \sqrt{(5.08)^2 + 2(9.81)(0.55)(37)} = 20.62\ \text{m/s}\\ v_B &= \sqrt{(23.96)^2 + 2(9.81)(0.55)(19)} = 27.91\ \text{m/s} \end{aligned}
$$

**Answer:** original speed of vehicle $A$ = **74.2 km/h** and of vehicle $B$ = **100.5 km/h** (speeds at impact: 18.3 and 86.3 km/h).

## Check yourself

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

1. How is speed just after impact found from post-collision skid?
   A. v' = sqrt(2 g f d')
   B. v' = g f d'
   C. v' = sqrt(g d')
   D. v' = 2 f d'
   Answer: A. Work-energy gives v' = sqrt(2 g f d'), e.g. 12.72 m/s for A. (Key point: Post-collision speed)

2. How is the speed just before impact obtained?
   A. Energy conservation only
   B. Vector conservation of linear momentum in x and y
   C. Average of post-collision speeds
   D. Using weight only
   Answer: B. Momentum is conserved as a vector in two perpendicular directions; weights can replace masses. (Key point: Momentum principle)

3. How is the original speed obtained after finding speed at impact u?
   A. v0 = u + 2 g f d
   B. v0 = sqrt(u^2 + 2 g f d) using the pre-collision skid
   C. v0 = u
   D. v0 = sqrt(2 g f d)
   Answer: B. Energy lost in the pre-collision skid is added back: v0 = sqrt(u^2 + 2 g f d). (Key point: Original speed)

4. What are the original speeds of A and B?
   A. 13.8 and 83.3 km/h
   B. 73.2 and 97.9 km/h
   C. 20.3 and 27.2 km/h
   D. 45 and 60 km/h
   Answer: B. vA = 20.34 m/s = 73.2 km/h and vB = 27.20 m/s = 97.9 km/h; 13.8 and 83.3 are speeds at impact. (Key point: Final answer)

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