Modelling with Vectors | 向量建模

📚 Modelling with Vectors | 向量建模

Vectors are the natural language for describing motion and forces in two dimensions. In Edexcel A-Level Mathematics, modelling with vectors means translating a physical situation into i-j notation or column vectors, then using algebra, calculus or constant-acceleration formulae to predict positions, velocities and forces.

向量是描述二维运动和力的自然语言。在 Edexcel A-Level 数学中，向量建模意味着将物理情境转化为 i-j 表示或列向量，然后运用代数、微积分或匀加速公式来预测位置、速度和力。

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1. What is a vector model? | 什么是向量模型

A vector model represents quantities that have both magnitude and direction. In mechanics, displacement, velocity, acceleration and force are all vectors, whereas speed, distance and mass are scalars.

向量模型表示既有大小又有方向的量。在力学中，位移、速度、加速度和力都是向量，而速率、路程和质量是标量。

In two dimensions, any vector can be written as r = xi + yj or as a column vector. The unit vectors i and j are perpendicular, usually taken as east and north.

在二维空间中，任何向量都可以写成 r = xi + yj 或列向量。单位向量 i 和 j 相互垂直，通常取为东向和北向。

• Vector examples: displacement, velocity, acceleration, force, momentum.

向量示例：位移、速度、加速度、力、动量。

• Scalar examples: speed, distance, mass, time, energy.

标量示例：速率、路程、质量、时间、能量。

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2. Position and displacement vectors | 位置向量与位移向量

A position vector r = xi + yj gives the location of a point relative to a fixed origin. The displacement from point A to point B is the change in position vector.

位置向量 r = xi + yj 表示点相对于固定原点的位置。从点 A 到点 B 的位移是位置向量的变化量。

AB = r_B − r_A = (x_B − x_A)i + (y_B − y_A)j

The distance between A and B is the magnitude of the displacement vector, not the sum of component differences.

A 与 B 之间的距离是位移向量的模，而不是坐标差之和。

|AB| = √[(x_B − x_A)² + (y_B − y_A)²]

on vector 2i + 3j and B has position vector 5i + 7j, then AB = 3i + 4j and the distance is √3² + 4² = 5 units.

示例：如果 A 的位置向量为 2i + 3j，B 的位置向量为 5i + 7j，则 AB = 3i + 4j，距离为 √3² + 4² = 5 单位。

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3. Velocity and acceleration as vectors | 速度与加速度向量

Velocity is the rate of change of displacement, so it has both direction and magnitude. Speed is the magnitude of the velocity vector.

速度是位移的变化率，因此它既有方向又有大小。速率是速度向量的模。

v = Δr / Δt

Acceleration is the rate of change of velocity. A particle can change direction while keeping the same speed, so acceleration is not always about changing speed.

加速度是速度的变化率。粒子可以在保持速率不变的情况下改变方向，因此加速度并不总是与速率变化有关。

a = Δv / Δt

If v = (3i + 4j) m s⁻¹, then speed = |v| = √3² + 4² = 5 m s⁻¹. The direction is the direction of the vector 3i + 4j.

如果 v = (3i + 4j) m s⁻¹，则速率 = |v| = √3² + 4² = 5 m s⁻¹。方向即向量 3i + 4j 的方向。

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4. Constant acceleration formulae in vector form | 向量形式的匀加速公式

When acceleration is constant, the standard SUVAT equations can be used directly with vector quantities. Each equation applies separately to the i and j components.

当加速度恒定时，标准 SUVAT 方程可直接用于向量。每个方程分别适用于 i 分量和 j 分量。

v = u + a t

r = r₀ + u t + ½ a t²

r = r₀ + ½ (u + v) t

Here u is the initial velocity, v is the final velocity, a is the constant acceleration, r₀ is the initial position vector and r is the final position vector.

这里 u 是初速度，v 是末速度，a 是恒定加速度，r₀ 是初始位置向量，r 是最终位置向量。

city u = (2i + 3j) m s⁻¹ and constant acceleration a = (0.5i − 0.2j) m s⁻². After t = 4 s:

例题：粒子从原点出发，初速度 u = (2i + 3j) m s⁻¹，恒定加速度 a = (0.5i − 0.2j) m s⁻²。经过 t = 4 s 后：

v = (2 + 0.5×4)i + (3 − 0.2×4)j = 4i + 2.2j m s⁻¹

r = (2×4 + ½×0.5×4²)i + (3×4 + ½×(−0.2)×4²)j = 12i + 10.4j m

Speed after 4 s = √4² + 2.2² ≈ 4.57 m s⁻¹.

4 秒后的速率 = √4² + 2.2² ≈ 4.57 m s⁻¹。

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5. Resultant forces and equilibrium | 合力与平衡

Newton's second law can be written in vector form. The resultant force acting on a particle equals its mass multiplied by its acceleration.

牛顿第二定律可以写成向量形式。作用在粒子上的合力等于其质量乘以加速度。

F = m a

If several forces act on an object, they must be added as vectors to find the resultant force. The i components and j components are added separately.

如果有多个力作用在物体上，必须按向量加法求合力。i 分量与 j 分量分别相加。

A particle is in equilibrium when the resultant force is zero. In that case acceleration is zero, so velocity is constant.

当合力为零时，粒子处于平衡状态。此时加速度为零，因此速度恒定。

eleration 3i − 4j m s⁻². The resultant force is F = 2(3i − 4j) = 6i − 8j N, with magnitude √6² + 8² = 10 N.

示例：质量为 2 kg 的物体加速度为 3i − 4j m s⁻²。合力为 F = 2(3i − 4j) = 6i − 8j N，模为 √6² + 8² = 10 N。

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6. Using i-j notation and column vectors | 使用 i-j 表示法与列向量

The vector xi + yj can also be written as a column vector. Addition, subtraction and scalar multiplication work in the same way.

向量 xi + yj 也可以写成列向量。加法、减法和数乘的运算方式相同。

i-j notation	Column vector
3i + 4j	(3, 4)ᵀ
−2i + 5j	(−2, 5)ᵀ

The unit vector i has components (1, 0) and the unit vector j has components (0, 1). When a vector has magnitude r and makes an angle θ with the i direction, its components are r cos θ and r sin θ.

单位向量 i 的分量为 (1, 0)，单位向量 j 的分量为 (0, 1)。当向量的模为 r 且与 i 方向成 θ 角时，其分量为 r cos θ 和 r sin θ。

r = r cos θ i + r sin θ j

This is often called resolving a vector into horizontal and vertical components. It is essential for modelling forces and velocities in real problems.

这通常称为将向量分解为水平分量和竖直分量。对于实际问题中的力和速度建模，这是必不可少的。

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7. Magnitude, unit vectors and direction | 模、单位向量与方向

r = xi + yj is the length of the line segment representing it. It is always non-negative.

向量 r = xi + yj 的模是表示该向量的线段长度。它始终为非负值。

|r| = √x² + y²

A unit vector has magnitude 1. To find the unit vector in the direction of r, divide r by its magnitude.

单位向量的模为 1。要找到 r 方向上的单位向量，将 r 除以其模。

unit vector = r / |r|

navigation problems, direction is usually given as a bearing measured clockwise from north.

向量的方向可以用相对于 i 正方向的夹角来表示，即 θ = arctan(y / x)。在导航问题中，方向通常用从北向顺时针测量的方位角来表示。

s⁻¹. Its direction is 67.4° below the positive i direction, which corresponds to a bearing of 157.4°.

示例：v = 5i − 12j m s⁻¹ 的速率为 13 m s⁻¹。其方向为 i 正方向下方 67.4°，对应的方位角为 157.4°。

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8. Calculus with vectors | 向量的微积分

If position vector r is a function of time, velocity is obtained by differentiating r with respect to t, and acceleration is obtained by differentiating velocity.

如果位置向量 r 是时间的函数，则速度可通过 r 对 t 求导得到，加速度可通过速度求导得到。

v = dr/dt

a = dv/dt = d²r/dt²

Integration works in the reverse direction. Given acceleration as a vector function, integrate to find velocity, then integrate velocity to find position. Initial conditions are used to find the constant vectors of integration.

积分是导数的逆运算。给定加速度向量函数，先积分得到速度，再积分速度得到位置。初始条件用于确定积分常数向量。

v = ∫ a dt + c

r = ∫ v dt + d

f a = 6t i + 2 j and at t = 0, v = 0 and r = 0, then v = 3t² i + 2t j and r = t³ i + t² j.

示例：如果 a = 6t i + 2 j，且在 t = 0 时 v = 0、r = 0，则 v = 3t² i + 2t j，r = t³ i + t² j。

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9. Bearings and relative motion | 方位角与相对运动

Relative motion problems often ask for the velocity of one object as seen by another. The relative velocity of A with respect to B is found by subtracting B's velocity from A's velocity.

相对运动问题通常会问一个物体相对于另一个物体的速度。A 相对于 B 的相对速度等于 A 的速度减去 B 的速度。

v_{A relative to B} = v_A − v_B

To describe a vector direction as a bearing, draw a right-angled triangle with the i and j components. Bearings are always measured clockwise from north, so be careful when converting from the angle given by arctan(y/x).

要将向量方向表示为方位角，请用 i 和 j 分量画出直角三角形。方位角始终从北向顺时针测量，因此从 arctan(y/x) 给出的角度转换时要小心。

For a vector with positive east component x and positive north component y, the bearing is arctan(x/y) measured from north. If the north component is negative, add 180°.

对于东分量为正 x、北分量为正 y 的向量，方位角为从北向测量的 arctan(x/y)。如果北分量为负，则加 180°。

In closest-approach problems, write the relative position vector and relative velocity. The shortest distance occurs when the relative position is perpendicular to the relative velocity.

在最近距离问题中，写出相对位置向量和相对速度。当相对位置与相对速度垂直时，距离最短。

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10. Modelling real-world motion: an example | 实际运动建模示例

A boat

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