Combining Displacements | 位移的合成

📚 Combining Displacements | 位移的合成

When an object moves from one point to another, its displacement is the straight-line change in position. Combining displacements matters whenever two or more movements happen in succession. In CIE A-Level Physics, you need to add and subtract displacement vectors using diagrams and components, not just ordinary arithmetic.

当物体从一点运动到另一点时，它的位移是位置的直线变化。只要两个或更多的运动连续发生，就需要合成位移。在 CIE A-Level 物理中，你需要用图形和分量法来加减位移矢量，而不是只做普通的算术运算。

1. Displacement as a Vector | 位移是矢量

Displacement is a vector quantity: it has both magnitude and direction. The magnitude is the straight-line distance from the starting point to the finishing point, and the direction is usually given as an angle or compass bearing.

位移是矢量：既有大小又有方向。大小是从起点到终点的直线距离，方向通常用角度或罗盘方位表示。

Do not confuse displacement with distance. Distance is a scalar equal to the total path length travelled, while displacement depends only on the initial and final positions. For example, walking 3 m east and then 4 m north gives a distance of 7 m but a displacement of 5 m northeast.

不要把位移与路程混淆。路程是标量，等于实际经过的路径总长度；位移只取决于初、末位置。例如，先向东走 3 m，再向北走 4 m，路程是 7 m，但位移是 5 m，方向为东北。

2. Representing Displacements | 位移的表示方法

A displacement is represented by a directed line segment. The arrow length is proportional to the magnitude, and the arrowhead shows the direction. A common label is s or d with an arrow above it in printed text.

位移用有向线段表示。线段的长度与位移大小成比例，箭头指向表示方向。常用符号 s 或 d，在印刷体中字母上方加箭头。

Always state the reference direction when quoting an angle. For example, ‘5.0 m at 30° north of east’ means the angle is measured from east towards north. Bearings are measured clockwise from north, so the same direction can be written as a bearing of 060°.

在给出角度时必须说明参考方向。例如，“5.0 m，东偏北 30°”表示从正东向北量 30°。罗盘方位从正北按顺时针量起，因此同一方向可写为方位角 060°。

3. The Triangle Rule for Addition | 三角形法则

To add displacement A followed by displacement B, draw A first, then draw B starting at the arrowhead of A. The resultant R is the single arrow drawn from the start of A to the tip of B. This is called the head-to-tail or triangle rule.

若物体先发生位移 A，再发生位移 B，则先画 A，再从 A 的箭头末端起画 B。合位移 R 是从 A 的起点画到 B 的末端的单个箭头。这种方法叫首尾相接法或三角形法则。

Vector addition is commutative, so A + B gives the same resultant as B + A. The triangle rule also shows that the resultant is independent of the order in which displacements are drawn.

矢量加法满足交换律，因此 A + B 与 B + A 得到的合位移相同。三角形法则也表明，合位移与画矢量的先后顺序无关。

4. The Parallelogram Rule | 平行四边形法则

When two displacements act from the same starting point, draw both arrows from that common origin. Complete the parallelogram by drawing parallel copies, and the diagonal from the origin to the opposite corner is the resultant.

当两个位移从同一起点出发时，从共同原点画出两个箭头。作平行线补成平行四边形，从原点到对角的对角线就是合位移。

The parallelogram rule is equivalent to the triangle rule. It is especially useful when you need to show the resultant of two simultaneous displacement vectors rather than sequential motion.

平行四边形法则与三角形法则等价。当需要表示两个同时发生的位移矢量而不是先后运动时，平行四边形法尤其有用。

5. Adding Perpendicular Displacements | 垂直位移的合成

If two displacements are perpendicular, the resultant magnitude is found from Pythagoras’ theorem:

R = √(A² + B²)

如果两个位移互相垂直，合位移的大小由勾股定理求得：

R = √(A² + B²)

The direction of the resultant is found from tan θ = opposite/adjacent:

θ = tan⁻¹(B / A)

合位移的方向由正切函数求得：

θ = tan⁻¹(B / A)

Here θ is measured from the direction of A. Example: an object travels 3.0 m east and then 4.0 m north. R = √(3.0² + 4.0²) = 5.0 m, and θ = tan⁻¹(4.0/3.0) ≈ 53° north of east.

其中 θ 从 A 的方向量起。例：物体先向东运动 3.0 m，再向北运动 4.0 m。R = √(3.0² + 4.0²) = 5.0 m，θ = tan⁻¹(4.0/3.0) ≈ 53°，即东偏北 53°。

6. Resolving into Components | 位移的正交分解

Any displacement can be resolved into two perpendicular components, usually chosen as x (east-west) and y (north-south). If the displacement has magnitude s and direction θ from the x-axis, then the components are:

x = s cos θ, y = s sin θ

任何一个位移都可以分解为两个互相垂直的分量，通常取 x（东西方向）和 y（南北方向）。如果位移大小为 s，方向与 x 轴成 θ 角，则分量为：

x = s cos θ, y = s sin θ

Resolving is the reverse of vector addition. It allows you to replace one vector with two perpendicular vectors that have exactly the same combined effect.

分解是矢量合成的逆运算。它允许你用一个矢量替换为两个互相垂直、合效果完全相同的矢量。

7. General Component Method | 一般分量法

For two or more displacements that are not perpendicular, resolve every vector into its x- and y-components. Add all the x-components algebraically to obtain the total X component, and add all the y-components to obtain the total Y component.

当两个或更多位移不互相垂直时，把每个矢量都分解为 x 分量和 y 分量。将所有 x 分量代数相加得到总的 X 分量，将所有 y 分量代数相加得到总的 Y 分量。

X = x₁ + x₂ + …, Y = y₁ + y₂ + …

Then combine the total components using Pythagoras and trigonometry:

R = √(X² + Y²), θ = tan⁻¹(Y / X)

然后用勾股定理和三角函数合成总分量：

R = √(X² + Y²), θ = tan⁻¹(Y / X)

This method works for any number of displacements and avoids scale-drawing errors.

这种方法适用于任意数量的位移，并能避免比例绘图带来的误差。

8. Subtracting Displacements | 位移的减法

Subtracting a displacement B from A means adding the opposite of B: A − B = A + (−B). The vector −B has the same magnitude as B but points in the opposite direction.

从位移 A 中减去位移 B，等于加上 B 的反矢量：A − B = A + (−B)。矢量 −B 与 B 大小相同，但方向相反。

Subtraction is used when finding a relative displacement or a change in displacement. For example, if an object moves from P to Q and then partially back, the resultant can be found by adding the negative of the return vector.

在求相对位移或位移变化时会用到减法。例如，物体从 P 运动到 Q，然后又返回一部分路程，其合位移可以通过加上返回矢量的负矢量来求得。

9. Resultant of Multiple Displacements | 多个位移的合成

For three or more displacements, place the vectors head-to-tail in sequence. The resultant is the arrow from the start of the first vector to the tip of the last vector. This polygon method is a direct extension of the triangle rule.

对于三个或更多位移，按顺序将矢量首尾相接。合位移就是从第一个矢量的起点指向最后一个矢量末端的箭头。这种多边形法是三角形法则的直接推广。

Algebraically, add all x-components and all y-components separately. The order of addition does not affect the final resultant.

用代数方法处理时，分别把所有 x 分量和所有 y 分量相加。相加的顺序不影响最终的合位移。

Example: 2.0 m east, 3.0 m north and 4.0 m west. Total east-west component X = 2.0 − 4.0 = −2.0 m (west). Total north-south component Y = 3.0 m. Hence R = √(2.0² + 3.0²) ≈ 3.6 m, and θ = tan⁻¹(3.0/2.0) ≈ 56° north of west.

例：向东 2.0 m，向北 3.0 m，再向西 4.0 m。东西方向总分量 X = 2.0 − 4.0 = −2.0 m（向西）。南北方向总分量 Y = 3

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Combining Displacements | 位移的合成