📚 Combining Displacements: Vector Addition in Kinematics | 位移的合成：运动学中的矢量加法
In kinematics, understanding how to combine displacements is fundamental to describing motion. Whether an object moves in a straight line or follows a multi-step path, the net effect of its movements can be represented by a single displacement vector. This article explains the principles of vector addition for displacements, covering one-dimensional and two-dimensional cases, graphical methods, and component calculations.
在运动学中，理解如何合成位移是描述运动的基础。无论物体沿直线运动还是经历多段路径，其运动的净效果都可以用一个合位移矢量来表示。本文讲解位移矢量加法的原理，涵盖一维和二维情形、作图法以及分量计算法。
1. What is Displacement? | 什么是位移？
Displacement is a vector quantity that describes the change in position of an object. It is defined as the shortest straight-line distance from the initial position to the final position, together with the direction. In contrast, distance is a scalar quantity that measures the total length of the path travelled, regardless of direction. In A-Level Physics, displacement is usually denoted by s or d and its SI unit is the metre (m).
位移是描述物体位置变化的矢量。它定义为从起点到终点的最短直线距离，并带有方向。相对地，路程是标量，测量的是所经过路径的总长度，不考虑方向。在 A-Level 物理中，位移通常用 s 或 d 表示，其 SI 单位是米 (m)。
Understanding that displacement is a vector means that when we combine several movements, we must account for both magnitude and direction; simple arithmetic addition is only valid if all motions occur along the same straight line.
理解位移是一个矢量意味着，当组合多个运动时，我们必须同时考虑大小和方向；只有所有运动发生在同一直线上时，简单的算术加法才有效。
2. Scalar and Vector Quantities | 标量与矢量
Before combining displacements, it is vital to distinguish between scalar and vector quantities. A scalar has magnitude only, while a vector has both magnitude and direction. Examples of scalars include distance, speed, mass and time. Examples of vectors include displacement, velocity, acceleration and force.
在合成位移之前，区分标量和矢量至关重要。标量只有大小，而矢量既有大小又有方向。标量的例子包括路程、速率、质量和时间。矢量的例子包括位移、速度、加速度和力。
When we say an object has been displaced by 5 m, the information is incomplete without a direction. A displacement of 5 m east is very different from a displacement of 5 m north. Vector addition rules therefore apply whenever we combine two or more displacements.
当我们说一个物体移动了 5 m 时，如果没有方向，这个信息是不完整的。向东 5 m 的位移与向北 5 m 的位移截然不同。因此，每当组合两个或更多位移时，都必须应用矢量加法规则。
3. Representing Displacement Vectors | 位移矢量的表示
splacement, and the arrowhead indicates its direction. For example, a scale of 1 cm = 10 m allows a 50 m displacement to be drawn as a 5 cm arrow.
位移矢量通常在图上用箭头表示。箭头的长度按选定的比例尺绘制，表示位移的大小，箭头指向表示其方向。例如，比例尺 1 cm = 10 m 可将 50 m 的位移画成一个 5 cm 的箭头。
The starting point of the arrow is the initial position, and the tip is the final position. When several successive displacements occur, each is drawn with its tail at the head of the previous arrow. This head-to-tail method is the basis of graphical vector addition.
箭头的起点是初位置，尖端是末位置。当发生多个相继的位移时，每一个位移的箭尾都画在前一个箭头的尖端。这种首尾相接的方法是矢量加法作图法的基础。
4. One-Dimensional Displacement Addition | 一维位移相加
instance, take east as positive. A displacement of 8 m east is +8 m, and a subsequent displacement of 3 m west is -3 m. The resultant displacement is (+8) + (-3) = +5 m, meaning 5 m east.
如果所有位移发生在同一直线上，在规定了正方向之后，矢量加法就简化为代数加法。例如，设向东为正。向东 8 m 的位移为 +8 m，接着再向西 3 m 的位移为 -3 m。合位移为 (+8) + (-3) = +5 m，即向东 5 m。
This method works for any number of one-dimensional movements. The sign of the resultant indicates its direction relative to the chosen positive sense. However, if an object retraces its path, the total distance travelled can be much larger than the magnitude of the displacement. For example, walking 5 m east and then 5 m west gives a total distance of 10 m but a net displacement of zero.
这种方法适用于任意数量的一维运动。合位移的符号表明方向与所选正方向的关系。但是，如果物体沿原路返回，所经过的总路程可能比位移的大小大得多。例如，向东走 5 m 然后向西走 5 m，总路程为 10 m，而合位移为零。
5. Vector Addition in Two Dimensions: The Triangle Law | 二维矢量相加：三角形法则
When displacements are not along the same line, we use the triangle law of vector addition. To add two displacements A and B, draw vector A first. Then draw vector B so that its tail starts at the head of A. The resultant displacement R is the vector drawn from the tail of A to the head of B. This head-to-tail method can be extended to any number of vectors, forming a displacement polygon.
当位移不在同一直线上时，我们使用矢量加法的三角形法则。要相加两个位移 A 和 B，先画出矢量 A，然后画矢量 B，使其箭尾始于 A 的尖端。合位移 R 是从 A 的尾端画到 B 的尖端的矢量。这种首尾相接法可推广到任意多个矢量，从而形成一个位移多边形。
east, then from its tip draw a 4 cm arrow pointing north. The resultant, from start to finish, is a 5 cm arrow at an angle of about 53° north of east. The magnitude is found by Pythagoras' theorem or direct measurement.
考虑一个人先向东走 3.0 m，然后向北走 4.0 m。用 1 cm = 1 m 的比例尺，画一条指向东的 3 cm 箭头，然后从其尖端画一条指向北的 4 cm 箭头。合位移从起点到终点是一条 5 cm 的箭头，方向约为东偏北 53°。大小可用勾股定理求出，也可直接测量。
R = √( (3.0 m)² + (4.0 m)² ) = 5.0 m
θ = tan⁻¹(4.0/3.0) ≈ 53° north of east
6. The Parallelogram Law of Vector Addition | 矢量加法的平行四边形法则
The parallelogram law is an equivalent graphical method. To add two displacements A and B, draw them both starting from the same point. Complete the parallelogram using A and B as adjacent sides. The diagonal from the common starting point represents the resultant R in both magnitude and direction.
平行四边形法则是一种等效的作图方法。要相加两个位移 A 和 B，将它们都从同一点出发画出。以 A 和 B 为邻边完成平行四边形。从共同起点画出的对角线即表示合位移 R 的大小和方向。
This method is particularly useful when two vectors act at the same point, such as a boat being carried by a river current while its engine propels it in a certain direction. The boat's resultant displacement in a given time interval is the vector sum of the displacement due to its own motion and the displacement caused by the current.
当有两个矢量共同作用于同一点时，此法特别有用，例如船被水流带动，同时自身的发动机使其朝某一方向运动。在给定时间间隔内，船的合位移是自身运动引起的位移与水流引起的位移的矢量和。
7. Resolving Displacements into Components | 分解位移为分量
For analytical calculations, it is often easier to resolve each displacement into perpendicular components, typically horizontal (x) and vertical (y). If a displacement s makes an angle θ with the positive x-axis (measured anticlockwise), its components are:
进行解析计算时，通常将每个位移分解为相互垂直的分量，一般是水平 (x) 分量和竖直 (y) 分量。若一位移 s 与正 x 轴的夹角为 θ（逆时针测量），其分量为：
sₓ = s cosθ
sy = s sinθ
take care of the direction. For example, a displacement of 10 m at 120° to the positive x‑axis has sₓ = 10 cos120° = −5 m, and sy = 10 sin120° ≈ 8.66 m.
这些分量方程对任意角度都成立；cosθ 和 sinθ 的符号会自动处理方向。例如，一位移大小为 10 m，与正 x 轴成 120°，则 sₓ = 10 cos120° = −5 m，sy = 10 sin120° ≈ 8.66 m。
Resolving vectors is a powerful technique because once all displacements are expressed in terms of their x- and y-components, the net effect can be found by simple algebraic addition of the components.
分解矢量是一种强大的方法，因为一旦所有位移都用 x 和 y 分量表示，就可以通过分量的简单代数相加来求得净效果。
8. Calculating Resultant Displacement from Components | 由分量计算合位移
first calculate the sum of all x-components, Rₓ = Σ sₓ, and the sum of all y-components, Ry = Σ sy. The magnitude of the resultant displacement is then given by:
要求出几个位移的合位移，首先计算所有 x 分量的总和 Rₓ = Σ sₓ，以及所有 y 分量的总和 Ry = Σ sy。合位移的大小则由下式给出：
R = √ Rₓ² + Ry²
The direction of the resultant is found using the inverse tangent function. However, care must be taken to identify the correct quadrant. The angle θ measured from the positive x‑axis satisfies:
合位移的方向用反正切函数求得。但必须注意识别正确的象限。从正 x 轴算起的角度 θ 满足：
θ = tan⁻¹( Ry / Rₓ )
If Rₓ is negative and Ry positive, the displacement lies in the second quadrant, and 180° should be added to the calculator value. Similarly, for the third quadrant (both negative) add 180°, and for the fourth quadrant (Rₓ positive, Ry negative) add 360° to obtain a positive angle.
如果 Rₓ 为负而 Ry 为正，位移位于第二象限，应在计算器值上加 180°。同样，对于第三象限（两者皆负）加 180°，对于第四象限（Rₓ 为正，Ry 为负）加 360° 以得到正角度。
9. Worked Example: Perpendicular Paths | 实例：垂直路径
A student walks 40.0 m due east and then 30.0 m due south. Calculate the resultant displacement.
一个学生向正东走了 40.0 m，然后向正南走了 30.0 m。计算合位移。
isplacement is 30.0 m south, which is −30.0 m in the y-direction: s2ₓ = 0, s2_y = −30.0 m.
取东为 x 轴正方向，北为 y 轴正方向。第一个位移向东 40.0 m：s1ₓ = 40.0 m，s1_y = 0。第二个位移向南 30.0 m，在 y 方向上为 −30.0 m：s2ₓ = 0，s2_y = −30.0 m。
Summing components: Rₓ = 40.0 m, Ry rom the positive x-axis, meaning 36.9° south of east, or equivalently a bearing of 126.9°.
合成分量：Rₓ = 40.0 m，Ry = −30.0 m。大小为 R = √(40.0² + (−30.0)²) = 50.0 m。方向：tan⁻¹(−30.0/40.0) = −36.9°，从正 x 轴起算，即东偏南 36.9°，或等价方位角 126.9°。
10. Worked Example: Multiple Non‑Perpendicular Displacements | 实例：多个非垂直位移
A hiker undergoes three successive displacements: A: 5.0 km in a direction 30° north of east, B: 4.0 km due east, and C: 6.0 km in a direction 60° south of west. Find the resultant displacement.
一名徒步者经历了三段相继的位移：A：5.0 km，东偏北 30°；B：4.0 km，正东；C：6.0 km，西偏南 60°。求合位移。
Again set east as +x, north as +y. Resolve each vector:
同样设东为 +x，北为 +y。分解每个矢量：
| Vector | Angle θ from +x | sₓ / km | sy / km |
|---|---|---|---|
| A: 5.0 km | 30° | 5.0 cos30° ≈ 4.33 | 5.0 sin30° = 2.50 |
| B: 4.0 km | 0° | 4.0 | 0 |
| C: 6.0 km | 180°+60°=240° | 6.0 cos240° = −3.00 | 6.0 sin240° ≈ −5.20 |
Summing: Rₓ = 4.33 + 4.0 + (−3.00) = 5.33 km, Ry = 2.50 + 0 + (−5.20) = −2.70 km.
求和：Rₓ = 4.33 + 4.0 + (−3.00) = 5.33 km，Ry = 2.50 + 0 + (−5.20) = −2.70 km。
R = √( (5.33 km)² + (−2.70 km)² ) ≈ 5.97 km
θ = tan⁻¹(−2.70 / 5.33) ≈ −26.9°
The negative angle indicates 26.9° below the positive x-axis, i.e. 26.9° south of east, or a bearing of about 116.9°. The resultant displacement is therefore 5.97 km at an angle of 26.9° south of east.
负角度表明在正 x 轴下方 26.9°，即东偏南 26.9°，或方位角约 116.9°。因此合位移为 5.97 km，方向东偏南 26.9°。
11. Displacement as a Free Vector and Path Independence | 位移作为自由矢量与路径无关
Displacement is a free vector: its effect depends only on its magnitude and direction, not on the particular initial position. This means two displacement vectors can be added regardless of where they occur. In combining displacements, the resultant is independent of the order of addition; vector addition is commutative.
位移是一个自由矢量：其效应仅取决于大小和方向，而与具体的初始位置无关。这意味着两个位移矢量无论发生在哪里都可以相加。在合成位移时，合位移与相加的顺序无关；矢量加法满足交换律。
This property reflects the physical reality that the net change in position between two points is independent of the path taken. Whether a particle travels directly from A to B or goes via C, D and E, the overall displacement is identical. This is a central concept in kinematics and underpins the independence of displacement from the actual trajectory.
这一性质反映了一个物理事实：两点之间位置的净变化与经过的路径无关。粒子无论从 A 直接到 B，还是经 C、D、E 绕行，总位移完全相同。这是运动学的一个核心概念，说明了位移与实际轨迹无关。
12. Common Mistakes and Exam Tips | 常见错误与考试技巧
Students often confuse displacement with distance; remember that displacement is a straight-line vector from start to finish, and its magnitude can be zero even after considerable travel. Never forget to include the direction when stating a displacement. In calculations, always draw a sketch to visualise the vectors and check that your angle is in the correct quadrant.
学生经常混淆位移和路程；请记住，位移是从起点到终点的直线矢量，即使经过了很长的运动，位移的大小也可能为零。叙述位移时切勿忘记标明方向。在计算中，始终画一个草图来直观显示矢量，并检查角度是否处于正确的象限。
When using the component method, double-check the signs of trigonometric functions for angles beyond 90°. Write down the components in a neat table for easy summation. In an exam, present your final resultant with both magnitude and an unambiguous direction, such as '5.0 m at 53° north of east' or 'bearing 053°'.
使用分量法时，对于超过 90° 的角，要仔细检查三角函数的符号。将分量整齐地列表以便求和。在考试中，给出最终的合位移时，既要给出大小，也要有明确的方向，例如“5.0 m，东偏北 53°”或“方位角 053°”。
For graphical problems, use a sharp pencil and a ruler, and choose a scale that makes your diagram large enough to minimise measurement errors. Record the scale clearly and indicate how you obtained your resultant.
对于作图题，请使用削尖的铅笔和直尺，并选择使图足够大的比例尺，以减小测量误差。明确标出比例尺，并说明你是如何得到合位移的。
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