📚 IB Mathematics: Relative Velocity Vector Analysis | IB数学：相对速度的向量分析
Relative velocity describes the motion of one object as seen from another moving object. In IB Mathematics, this topic appears in the vectors section, where all velocities are represented as vectors and calculations are performed using vector addition and subtraction.
相对速度描述的是一个物体在另一个运动物体参考系中的运动。在IB数学中，这一考点出现在向量章节，我们需要用向量来表示速度，并通过向量加减完成计算。
1. The Definition of Relative Velocity | 相对速度的定义
Suppose two cars drive along the same straight road. If car A moves at 20 m s⁻¹ and car B moves at 15 m s⁻¹ in the same direction, then the velocity of A relative to B is 5 m s⁻¹. If B moves in the opposite direction, the relative speed becomes 35 m s⁻¹.
假设两辆车在同一条直路上行驶。若车A速度为20 m s⁻¹，车B速度为15 m s⁻¹且方向相同，那么A相对B的速度为5 m s⁻¹。若B反向行驶，相对速度则为35 m s⁻¹。
In two or three dimensions, direction matters, so relative velocity cannot be treated as a simple scalar difference. It must be calculated by subtracting velocity vectors.
在二维或三维运动中，方向非常重要，因此相对速度不能简单地作标量差。我们必须用向量减法来计算。
2. The Vector Equation vₐᵦ = vₐ − v_b | 向量方程 vₐᵦ = vₐ − v_b
For two objects A and B with ground-frame velocities vA and vB, the velocity of A relative to B is defined as the vector difference vA − vB.
设物体A和B相对地面的速度分别为 vA 与 vB，则A相对B的速度定义为向量差 vA − vB。
vAB = vA − vB
This equation says that to observe A from B’s moving frame, we subtract B’s velocity vector. Equivalently, the velocity of B relative to A is vBA = vB − vA = −vAB.
这个方程说明：为了在B的运动参考系中观察A，我们要减去B的速度向量。反过来，B相对A的速度为 vBA = vB − vA = −vAB。
3. Position Vectors and Relative Position | 位置向量与相对位置
The relative position vector from B to A is the vector difference of their ground position vectors:
从B指向A的相对位置向量等于它们地面位置向量的差：
rAB = rA − rB
If A and B move with constant velocities vA and vB, then at time t their positions are rA(t) = rA0 + t vA and rB(t) = rB0 + t vB. Therefore the relative position vector changes as:
若A、B以恒定速度 vA、vB运动，则t时刻位置为 rA(t) = rA0 + t vA，rB(t) = rB0 + t vB。因此相对位置向量随时间变化为：
rAB(t) = (rA0 − rB0) + t (vA − vB) = rAB0 + t vAB
This single equation is the foundation of most IB relative velocity problems.
这个方程是解决IB相对速度问题的基础。
4. Coordinate Systems and Component Vectors | 坐标系与分量向量
IB questions often use unit vectors i (east) and j (north). A velocity vector can be written as v = vx i + vy j, or as an ordered pair (vx, vy).
IB题目常使用单位向量 i（东）和 j（北）。速度向量可写成 v = vx i + vy j，也可写成有序数对 (vx, vy)。
When the velocity is given as a magnitude and a bearing, resolve it into components using sine and cosine. For example, a speed s on a bearing of θ° from north has components:
当速度以大小和方位角给出时，我们需要用正弦和余弦分解分量。例如，大小为s、方位角为θ°（自北顺时针）的速度分量为：
v = (s sin θ) i + (s cos θ) j
Always draw a vector diagram before substituting into equations; this minimizes sign errors.
在代入公式前一定要画向量图，这样可以减少符号错误。
5. River Crossing: Combining Velocities | 渡河问题：速度合成
A classic relative motion problem involves a boat crossing a river. Let vBW be the boat’s velocity relative to the water, and vWG be the water’s velocity relative to the ground. The boat’s velocity relative to the ground is:
经典的相对运动问题是船渡河。设 vBW 为船相对水的速度，vWG 为水流相对地面的速度，则船相对地面的速度为：
vBG = vBW + vWG
Example: a river is 60 m wide and the current is 3 m s⁻¹ downstream. The boat’s speed in still water is 4 m s⁻¹. To land directly opposite the starting point, the boat must head upstream at an angle such that its upstream component equals the current speed.
例：河宽60 m，水流向下游速度为3 m s⁻¹。船在静水中速度为4 m s⁻¹。若要在正对岸登陆，船必须向上游偏一个角度，使上游分量恰好抵消水流速度。
Let θ be the angle made with the line perpendicular to the bank. Then:
设θ为船头与垂直于河岸方向的夹角，则：
sin θ = 3 / 4 ⇒ θ ≈ 48.6°
The component across the river is 4 cos θ = √7 ≈ 2.65 m s⁻¹, so the crossing time is 60 / √7 ≈ 22.7 s.
垂直河岸的分量为4 cos θ = √7 ≈ 2.65 m s⁻¹，因此渡河时间为60 / √7 ≈ 22.7 s。
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If the boat points straight across, it drifts downstream but crosses faster.
若船头正对河岸，船会被水流冲向下游，但渡河更快。
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Time straight across = 60 / 4 = 15 s; drift distance = 3 × 15 = 45 m.
正对河岸时渡河时间为60 / 4 = 15 s，漂流距离为3 × 15 = 45 m。
6. Collision and Interception Problems | 碰撞与拦截问题
Two moving objects collide if their position vectors are equal at some time t > 0. Using relative vectors, this condition is rAB(t) = 0.
两个运动物体在某一时刻t > 0位置向量相等时发生碰撞。用相对向量表示，即 rAB(t) = 0。
From rAB0 + t vAB = 0, we obtain:
由 rAB0 + t vAB = 0，得：
vAB = − rAB0 / t
This is useful when one object wants to intercept another: the required relative velocity points directly from B to A along the initial line of sight. In an exam, write the x and y components as linear equations and solve for t or the unknown velocity.
当某个物体需要拦截另一物体时，这个公式尤其有用：所需相对速度的方向必须沿初时B指向A的方向。考试中，可以把x和y分量写成线性方程，求解t或未知速度。
Example: A is at (10,0) with velocity (2,1) m s⁻¹, and B is at (0,0) with velocity (4,3) m s⁻¹. Here rAB0 = (10,0), vAB = (−2,−2). Solving rAB(t) = (10 − 2t, −2t) = (0,0) gives contradictory t = 5 and t = 0, so they do not collide.
例：A位于(10,0)，速度为(2,1) m s⁻¹；B位于(0,0)，速度为(4,3) m s⁻¹。这里rAB0 = (10,0)，vAB = (−2,−2)。令rAB(t) = (10 − 2t, −2t) = (0,0)，得到t = 5和t = 0矛盾，因此它们不会相撞。
7. Closest Distance Between Two Moving Objects | 两运动物体之间的最近距离
Even if two objects never collide, we may need to find the closest distance between them. At the instant of closest approach, the relative position vector rAB(t) is perpendicular to the relative velocity vAB.
即使两个物体不会相撞，我们也可能需要求它们之间的最近距离。在最近距离对应的时刻，相对位置向量 rAB(t) 与相对速度 vAB 垂直。
rAB(t) · vAB = 0
Worked example: At noon, ship A is at (0,0) with velocity (4,3) km h⁻¹, and ship B is at (20,0) with velocity (−2,1) km h⁻¹. Find the time and distance of closest approach.
例题：中午12点，船A位于(0,0)，速度为(4,3) km h⁻¹；船B位于(20,0)，速度为(−2,1) km h⁻¹。求最近距离发生的时刻和距离。
Solution: For A relative to B,
解：对A相对B，
rAB0 = (−20,0), vAB = (6,2)
So rAB(t) = (−20 + 6t, 2t). Dot with vAB:
因此 rAB(t) = (−20 + 6t, 2t)。与 vAB
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