---
title: "Skid numerical: car 430 kg and truck 1299 kg"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:skid-collision-car-430kg-truck-1299kg"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/skid-collision-car-430kg-truck-1299kg
site: BCT Engineering
---

# Skid numerical: car 430 kg and truck 1299 kg

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2081 Baisakh, 8 marks
- Video: [FIND INITIAL SPEED OF VEHICLE | TRANSPORTATION 2 NUMERICAL | 2019 PU](https://www.youtube.com/watch?v=dJrbm8txSsU) — 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-collision-car-430kg-truck-1299kg
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

A passenger car of mass 430 kg moving east collides with a truck of mass 1299 kg that is moving south. Initial skid distances of the car and the truck are 15 m and 10 m respectively. After collision they lock together and skid through a distance equal to 8 m in direction 65 degrees south of east before stopping. Compute the initial speeds of the car and the truck. Assume average coefficient of friction as 0.5.

## Short answer

Post-impact speed from skid:8.86 m/s; momentum conservation gives impact speeds15.05 m/s (car) and10.69 m/s (truck). Initial speeds: car69.6 km/h (19.33 m/s), truck52.5 km/h (14.57 m/s).

## Answer

After the collision the two vehicles move together, so the common post-impact speed comes from the skid distance, the impact speeds from momentum conservation in two directions, and the original speeds from the pre-collision skids.

**Given:** car $m_1 = 430$ kg moving East, truck $m_2 = 1299$ kg moving South, $f = 0.5$, pre-collision skids 15 m (car) and 10 m (truck), post-collision skid $d' = 8$ m at $65^\circ$ South of East.

```text
        N
        ^
        |      car (East) ->
  W ----+----> E
        |\ 65 deg
        | \  locked pair
        v  v (S of E)
        S   truck comes moving South
```

#### Step 1: Speed of the locked vehicles after impact
$$
v' = \sqrt{2gfd'} = \sqrt{2(9.81)(0.5)(8)} = 8.859\ \text{m/s}
$$

#### Step 2: Momentum conservation
Total mass $M = 430 + 1299 = 1729$ kg. East component (only the car had eastward momentum) and South component (only the truck had southward momentum):

$$
\begin{aligned} m_1 u_1 &= M v'\cos 65^\circ \Rightarrow u_1 = \frac{1729(8.859)(0.4226)}{430} = 15.054\ \text{m/s}\\ m_2 u_2 &= M v'\sin 65^\circ \Rightarrow u_2 = \frac{1729(8.859)(0.9063)}{1299} = 10.687\ \text{m/s} \end{aligned}
$$

#### Step 3: Original speeds
$$
\begin{aligned} v_1 &= \sqrt{u_1^2 + 2gf(15)} = \sqrt{(15.054)^2 + 2(9.81)(0.5)(15)} = 19.33\ \text{m/s}\\ v_2 &= \sqrt{u_2^2 + 2gf(10)} = \sqrt{(10.687)^2 + 2(9.81)(0.5)(10)} = 14.57\ \text{m/s} \end{aligned}
$$

**Answer: initial speed of the car = 69.6 km/h and of the truck = 52.5 km/h** (impact speeds 54.2 and 38.5 km/h).

## Check yourself

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

1. How is the common speed of the locked vehicles found?
   A. v' = sqrt(2 g f d') with d' = 8 m
   B. v' = f d'
   C. Average of car and truck speeds
   D. v' = d'/t
   Answer: A. Work-energy gives v' = sqrt(2 x 9.81 x 0.5 x 8) = 8.859 m/s. (Key point: Speed after impact)

2. Why does only the car appear in the East equation?
   A. The truck is heavier
   B. Only the car had eastward momentum before impact
   C. The truck had stopped
   D. East is the skid direction
   Answer: B. The car moved East and the truck South, so m1 u1 = M v' cos65 and m2 u2 = M v' sin65. (Key point: Momentum components)

3. The locked pair moves 65 degrees south of east. What does this mean for the components?
   A. East component uses cos 65 and South component uses sin 65
   B. East uses sin 65 and South uses cos 65
   C. Both use cos 65
   D. Both use tan 65
   Answer: A. The angle is measured from East, so East gets cos 65 and South gets sin 65. (Key point: Direction after impact)

4. How is the car's original speed found from its impact speed u1 = 15.054 m/s?
   A. v1 = u1 - 2 g f (15)
   B. v1 = sqrt(u1^2 + 2 g f (15)) = 19.33 m/s
   C. v1 = u1 + 15
   D. v1 = u1 / f
   Answer: B. The pre-collision skid of 15 m dissipated energy, so v1^2 = u1^2 + 2 g f d. (Key point: Original speeds)

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