Displacement Composition and Analysis Methods | 位移合成分析方法

📚 Displacement Composition and Analysis Methods | 位移合成分析方法

Displacement is a fundamental vector quantity in physics, representing the change in position of an object. Understanding how to combine and analyse displacements is essential for solving problems in kinematics, mechanics, and even electricity and magnetism. This article provides a systematic guide to displacement composition methods for CIE A-Level Physics.

位移是物理学中最基本的矢量物理量之一，它表示物体位置的变化。理解如何合成与分析位移，对于解决运动学、力学乃至电磁学问题都至关重要。本文为 CIE A-Level 物理考生提供一套系统的位移合成分析方法指南。

1. Scalar vs Vector Quantities | 标量与矢量

A scalar quantity has magnitude only, while a vector quantity has both magnitude and direction. Distance is a scalar, whereas displacement is a vector. For example, walking 3 km north then 4 km east results in a total distance of 7 km, but a displacement of only 5 km at an angle of 53.1° east of north.

标量只有大小，而矢量既有大小又有方向。路程是标量，而位移是矢量。例如：向北走 3 km 再向东走 4 km，总路程为 7 km，但位移仅为 5 km，方向为北偏东 53.1°。

The magnitude of displacement is the shortest straight-line distance between the starting and ending points.

位移的大小是起点到终点之间的最短直线距离。
Displacement is path-independent; the route taken does not affect the final displacement vector.

位移与路径无关；所走的路线不影响最终的位移矢量。

2. Vector Representation and Notation | 矢量的表示与符号

In printed text, a vector is often denoted by a bold symbol such as s or an arrow above the symbol. In handwritten work, you should always place an arrow above the letter, e.g., s⃗. The magnitude of a vector is written as |s⃗| or simply s.

在印刷文本中，矢量通常用粗体符号如 s 或字母上方加箭头表示。在手写作业中，务必在字母上方加上箭头，例如 s⃗。矢量的大小写作 |s⃗| 或简写为 s。

When dealing with displacement, we commonly use the symbol s or d. In two dimensions, a displacement vector can be expressed in Cartesian component form:

在处理位移时，我们常用符号 s 或 d。在二维空间中，位移矢量可以用笛卡尔分量形式表示：

s⃗ = sₓ·î + sᵧ·ĵ

where sₓ = s·cos θ and sᵧ = s·sin θ are the horizontal and vertical components respectively, and î and ĵ are unit vectors along the x- and y-axes.

其中 sₓ = s·cos θ 和 sᵧ = s·sin θ 分别为水平分量和竖直分量，î 和 ĵ 是沿 x 轴和 y 轴的单位矢量。

3. Graphical Methods: The Parallelogram Law | 图解法：平行四边形定则

The parallelogram law states that when two displacement vectors are drawn from the same starting point, their resultant displacement is given by the diagonal of the parallelogram formed by these two vectors. This method is particularly useful when the two vectors are at an angle to one another.

平行四边形定则指出：当两个位移矢量从同一起点画出时，由这两个矢量构成的平行四边形的对角线即为它们的合位移。当两个矢量之间存在夹角时，此方法尤为实用。

To apply this method: draw both vectors to scale from the same origin, complete the parallelogram, and the diagonal from the common origin to the opposite corner represents the resultant displacement.

应用步骤：按比例从同一原点画出两个矢量，补全平行四边形，从共同原点到对角顶点的对角线就是合位移。

|R⃗| = √(a² + b² + 2ab·cos θ)

Here, a and b are the magnitudes of the two displacement vectors, θ is the angle between them, and R is the resultant vector. The direction of the resultant can be found using the sine rule or by resolving components.

其中 a 和 b 是两个位移矢量的大小，θ 是它们之间的夹角，R 是合矢量。合矢量的方向可用正弦定理或分量分解求得。

4. The Triangle (Head-to-Tail) Method | 三角形法（首尾相接法）

The triangle method involves drawing the first displacement vector, then placing the tail of the second vector at the head of the first. The resultant vector is drawn from the tail of the first vector to the head of the second. This is also called the nose-to-tail rule.

三角形法的步骤是：先画出第一个位移矢量，然后将第二个矢量的尾部置于第一个矢量的头部。合矢量是从第一个矢量的尾部指向第二个矢量的头部。这也称为首尾相接法则。

This method extends naturally to three or more vectors, forming a vector polygon.

此方法可自然推广到三个或更多矢量，形成矢量多边形。
If the head of the final vector meets the tail of the first, the resultant displacement is zero, indicating a closed path.

如果最后一个矢量的头部恰好与第一个矢量的尾部重合，则合位移为零，表示路径闭合。

For three or more displacements, a vector polygon works best. The resultant is the single vector that closes the polygon from the start point to the end point.

对于三个或更多位移，矢量多边形法最为合适。合矢量是使多边形从起点到终点闭合的唯一矢量。

5. Resolution of Displacement Vectors into Components | 位移矢量的正交分解

Resolving a displacement vector into perpendicular components is the single most powerful analytical technique in A-Level physics. Any displacement vector s⃗ of magnitude s making an angle θ with the x-axis can be resolved as:

将位移矢量分解为相互垂直的分量，是 A-Level 物理中最强有力的分析技术。任何大小为 s、与 x 轴夹角为 θ 的位移矢量 s⃗ 都可以分解为：

sₓ = s·cos θ（水平分量） sᵧ = s·sin θ（竖直分量）

When the angle is measured from the vertical axis, the sine and cosine functions are interchanged. Always draw a clear right-angled triangle diagram before performing trigonometric calculations.

当角度从竖直轴测量时，正弦和余弦函数会互换。在进行三角函数计算之前，务必画清晰的直角三角形示意图。

The choice of axes is arbitrary but should align with the geometry of the problem for convenience. For example, on an inclined plane, choose axes parallel and perpendicular to the plane surface.

坐标轴的选择是任意的，但应与问题的几何形状对齐以方便计算。例如在斜面上，应选择平行和垂直于斜面的坐标轴。

6. Analytical Method: Adding Components | 解析法：分量相加

The analytical method is the most accurate and preferred method in examinations. The procedure is:

解析法是最精确且在考试中最受青睐的方法。步骤如下：

Step 1: Resolve each displacement vector into its x- and y-components.

第一步：将每个位移矢量分解为 x 分量和 y 分量。
Step 2: Sum all x-components to obtain the total x-component Rₓ = Σsₓᵢ.

第二步：将所有 x 分量相加得到总 x 分量 Rₓ = Σsₓᵢ。
Step 3: Sum all y-components to obtain the total y-component Rᵧ = Σsᵧᵢ.

第三步：将所有 y 分量相加得到总 y 分量 Rᵧ = Σsᵧᵢ。
Step 4: Combine the components to find the resultant magnitude and direction.

第四步：合成分量求合矢量的大小和方向。

|R⃗| = √(Rₓ² + Rᵧ²) tan θ = Rᵧ / Rₓ

Note that the quadrant of θ must be determined from the signs of Rₓ and Rᵧ. A calculator only gives the principal value, so always check the diagram to ensure the angle is in the correct direction.

注意 θ 所在象限必须由 Rₓ 和 Rᵧ 的正负号确定。计算器只给出主值，因此务必对照图形检查角度方向是否正确。

7. Special Cases: Perpendicular and Collinear Vectors | 特殊情况：垂直与共线矢量

Perpendicular vectors: When two displacement vectors are perpendicular to each other, the resultant magnitude is simply given by the Pythagorean theorem:

垂直矢量：当两个位移矢量相互垂直时，合矢量大小直接用勾股定理给出：

|R⃗| = √(a² + b²) tan θ = b / a

Collinear vectors: When vectors act in the same straight line, they are added algebraically. A positive sign is assigned to one direction (e.g., east or north) and a negative sign to the opposite direction (west or south).

共线矢量：当矢量在同一直线上时，直接代数相加。规定一个方向（如东或北）为正，相反方向（西或南）为负。

Same direction: add magnitudes; resultant direction remains unchanged.

同向：大小相加；合矢量方向不变。
Opposite direction: subtract magnitudes; resultant direction follows the larger vector.

反向：大小相减；合矢量方向与较大的矢量相同。

8. Worked Example 1: Year 12 Standard | 典型例题 1：AS 水平

Problem: A student walks 40 m due east, then 30 m due north. Determine the resultant displacement.

题目：一名学生先向东走 40 m，再向北走 30 m。求合位移。

Solution: The two displacements are perpendicular:

解答：两个位移相互垂直：

s = √(40² + 30²) = √(1600 + 900) = √2500 = 50 m

θ = tan⁻¹(30/40) = tan⁻¹(0.75) ≈ 36.87°

Therefore the resultant displacement is 50 m at an angle of approximately 36.9° north of east. This can also be written as 50 m at 53.1° east of north.

因此合位移为 50 m，方向为东偏北约 36.9°。也可表示为北偏东 53.1°。

9. Worked Example 2: Component Method | 典型例题 2：分量法

Problem: Three displacement vectors are given: A⃗ = 5 m at 30°, B⃗ = 8 m at 120°, and C⃗ = 6 m at 210° (angles measured counterclockwise from the +x axis). Find the resultant displacement.

题目：三个位移矢量：A⃗ = 5 m，方向 30°；B⃗ = 8 m，方向 120°；C⃗ = 6 m，方向 210°（角度均从 +x 轴逆时针测量）。求合位移。

Solution: Resolve each vector into components:

解答：将每个矢量分解为分量：

Vector	x-component	y-component
A⃗ = 5 m, 30°	5·cos 30° = 4.33 m	5·sin 30° = 2.50 m
B⃗ = 8 m, 120°	8·cos 120° = −4.00 m	8·sin 120° = 6.93 m
C⃗ = 6 m, 210°	6·cos 210° = −5.20 m	6·sin 210° = −3.00 m

Summing: Rₓ = 4.33 − 4.00 − 5.20 = −4.87 m; Rᵧ = 2.50 + 6.93 − 3.00 = 6.43 m.

求总和：Rₓ = 4.33 − 4.00 − 5.20 = −4.87 m；Rᵧ = 2.50 + 6.93 − 3.00 = 6.43 m。

|R⃗| = √((−4.87)² + (6.43)²) = √(23.72 + 41.34) = √65.06 ≈ 8.07 m

θ = tan⁻¹(6.43/−4.87) ≈ −52.9°（即 127.1°）

Since Rₓ is negative and Rᵧ is positive, the resultant lies in the second quadrant at approximately 127° from the +x axis.

由于 Rₓ 为负、Rᵧ 为正，合矢量位于第二象限，距离 +x 轴约 127°。

10. Application: Relative Displacement | 应用：相对位移

Relative displacement describes the position of one object as observed from another moving object. If object A has displacement s⃗ₐ from the origin and object B has displacement s⃗ᵦ, then the displacement of B relative to A is:

相对位移描述的是一个物体相对于另一个运动物体的位置。若物体 A 相对原点的位移为 s⃗ₐ，物体 B 的位移为 s⃗ᵦ，则 B 相对于 A 的位移为：

s⃗ᵦₐ = s⃗ᵦ − s⃗ₐ

This vector subtraction is equivalent to adding s⃗ₐ reversed (i.e., −s⃗ₐ) to s⃗ᵦ using the head-to-tail method. Relative displacement is crucial in river-crossing problems and collision avoidance calculations.

这种矢量减法等价于将 s⃗ₐ 反向（即 −s⃗ₐ）与 s⃗ᵦ 用首尾相接法相加。相对位移在渡河问题和避碰计算中至关重要。

For example, if a boat aims straight across a river while the current carries it downstream, the resultant displacement relative to the bank is found by vector addition of the boat’s velocity displacement and the current’s displacement over the same time interval.

例如，若船垂直向河对岸行驶，同时水流将其带向下游，则船相对河岸的合位移就需要将船的位移和水流的位移在同一时间间隔内进行矢量合成。

11. Common Mistakes and Exam Tips | 常见错误与考试技巧

Students frequently lose marks on displacement composition questions due to a few recurring errors. Being aware of these pitfalls can significantly improve your score.

学生在位移合成题目中常因一些反复出现的错误而失分。了解这些陷阱可以显著提高你的分数。

Mistake 1: Adding vectors as scalars without considering direction.

错误一：不考虑方向，直接将矢量当标量相加。
Mistake 2: Using the parallelogram law incorrectly when vectors form a triangle rather than sharing a common origin.

错误二：当矢量构成三角形而非共起点时，错误地使用平行四边形定则。
Mistake 3: Forgetting to place the angle in the correct quadrant when using tan⁻¹.

错误三：使用 tan⁻¹ 时忘记判断角度所在的象限。
Mistake 4: Confusing the angle measured from the horizontal with the angle measured from the vertical.

错误四：把从水平方向测量的角度和从竖直方向测量的角度混淆。

Exam tip: Always draw a labelled diagram before starting calculations. State whether your final angle is “bearing” (measured clockwise from north) or “angle from the x-axis” — CIE examiners award marks for unambiguous direction statements.

考试技巧：开始计算前务必画带标注的示意图。在最终答案中明确说明角度是用”方位角”（从正北顺时针测量）还是”与 x 轴的夹角”——CIE 考官对明确无歧义的方向表述会给分。

12. Summary and Formula Sheet | 总结与公式速查表

The table below summarises the essential formulas for displacement composition. When solving any vector problem, follow the systematic approach: resolve → add components → reconstruct resultant.

下表总结了位移合成的必备公式。解决任何矢量问题时，遵循系统流程：分解 → 分量相加 → 重建合矢量。

Quantity / 量	Formula / 公式
Components / 分量	sₓ = s·cos θ; sᵧ = s·sin θ
Resultant magnitude / 合位移大小	\|R⃗\| = √(Rₓ² + Rᵧ²)
Direction / 方向	tan θ = Rᵧ / Rₓ (check quadrant)
Two vectors, angle θ / 两矢量夹角 θ	\|R⃗\| = √(a² + b² + 2ab·cos θ)
Perpendicular vectors / 垂直矢量	\|R⃗\| = √(a² + b²)
Relative displacement / 相对位移	s⃗ᵦₐ = s⃗ᵦ − s⃗ₐ

Mastering displacement composition provides the foundation for understanding velocity composition, force analysis, and projectile motion — all of which are heavily tested in both AS and A2 papers. Practice drawing vector diagrams under timed conditions to build speed and confidence.

掌握位移合成为理解速度合成、受力分析和抛体运动奠定基础——这些内容在 AS 和 A2 试卷中都是考查重点。在限时条件下多练习画矢量图，以提升速度和信心。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Displacement Composition and Analysis Methods | 位移合成分析方法