Applying Vectors to Uniform Motion Problems | 向量在匀速运动问题中的应用

📚 Applying Vectors to Uniform Motion Problems | 向量在匀速运动问题中的应用

In IB Mathematics, vectors provide a natural language for describing motion. When an object moves in a straight line with constant velocity, its position vector changes linearly with time. This gives us a single vector equation that can answer questions about position, speed, direction, intersection, collision, and closest approach.

在IB数学中，向量为描述运动提供了一种自然语言。当一个物体沿直线做匀速运动时，其位置向量随时间线性变化。于是我们可以用一个向量方程，来解决关于位置、速率、方向、路径相交、碰撞以及最近距离等问题。

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1. Why Vectors Describe Motion | 为什么用向量描述运动

Uniform motion requires two pieces of information: a starting position and a constant velocity. A vector combines both direction and magnitude in one object, so it is ideal for modelling velocity. The position vector of a moving object is measured from a fixed origin, and the velocity vector shows how that position changes each unit of time.

匀速运动需要两个信息：起始位置和恒定速度。向量把方向和大小合并在一个数学对象中，因此非常适合表示速度。物体的位置向量以固定原点为基准，速度向量则表示位置在每个单位时间内的变化。

Suppose an object starts at position r₀ and moves with constant velocity v. After t seconds, the displacement from the starting point is t v, so the position vector is

假设物体从位置 r₀ 出发，以恒定速度 v 运动。经过 t 秒后，从起点出发的位移为 t v，因此位置向量为

r(t) = r₀ + t v, t ≥ 0

This one formula is the foundation for every problem in this article.

这一条公式就是本文所有问题的核心基础。

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2. Position, Displacement and Distance | 位置向量、位移与距离

The position vector r(t) gives the location of the object relative to the origin at time t. Its magnitude |r(t)| is the distance from the origin, not necessarily the distance the object has travelled.

位置向量 r(t) 表示在时刻 t 物体相对于原点的位置。其模 |r(t)| 是物体到原点的距离，但不一定是物体运动的总路程。

The displacement between two times t₁ and t₂ is the difference of the position vectors:

在 t₁ 与 t₂ 两个时刻之间的位移，是这两个位置向量的差：

Δr = r(t₂) − r(t₁) = (t₂ − t₁) v

Therefore the straight-line distance between these two positions is |Δr| = (t₂ − t₁)|v| eed multiplied by time when the path is straight and the motion does not change direction.

因此这两个位置之间的直线距离为 |Δr| = (t₂ − t₁)|v|。这说明在路径为直线且运动方向不变时，两时刻间经过的路程等于速率乘以时间。

osition is r(2) = (7, 10). The object travels 5 units of distance per unit of time because |v| = √3² + 4² = 5.

例如，若 r₀ = (1, 2)，v = (3, 4)，则 t = 2 时位置为 r(2) = (7, 10)。由于 |v| = √3² + 4² = 5，物体每单位时间前进 5 个单位距离。

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3. Velocity Vector and Speed | 速度向量与速率

he magnitude of the velocity vector. In two dimensions, if v = (v₁, v₂), then the speed is

速度是向量，速率是速度向量的模。在二维平面中，若 v = (v₁, v₂)，则速率为

speed = |v| = √v₁² + v₂²

The direction of motion is given by the unit vector

运动方向由单位向量给出

u = v / |v|

In three dimensions the idea is the same: add the third component v₃, and use |v| = √v₁² + v₂² + v₃². Because the motion is uniform, both speed and direction are constant.

三维情形完全相同：加入第三个分量 v₃，且 |v| = √v₁² + v₂² + v₃²。由于匀速运动的速度不变，速率和方向都不随时间变化。

In IB questions, be careful with units. If position is measured in kilometres and time in hours, then velocity is written in km h⁻¹. If position is in metres and time in seconds, then velocity is in m s⁻¹.

在IB题目中要注意单位。若位置用千米、时间用小时，则速度单位是 km h⁻¹；若位置用米、时间用秒，则速度单位是 m s⁻¹。

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4. The Vector Equation of a Straight-Line Path | 直线路径的向量方程

nt to the vector equation of a line, where v is a direction vector and t is the parameter.

对于匀速运动，路径本身就是一条直线。位置向量 r(t) = r₀ + t v 等价于直线的向量方程，其中 v 是方向向量，t 是参数。

In coordinate form, for a two-dimensional motion, write

在二维坐标形式下，可写成

x = x₀ + v₁t, y = y₀ + v₂t

If v₁ ≠ 0, we can eliminate t to obtain a Cartesian equation of the path:

若 v₁ ≠ 0，我们可以消去 t，得到路径的笛卡尔方程：

y = y₀ + (v₂ / v₁)(x − x₀)

This Cartesian equation is useful for checking whether a particular point lies on the path. It is also a good reminder that a particle in uniform motion always follows a straight line.

这个笛卡尔方程可用于判断某个点是否在路径上。它也提醒我们，做匀速运动的质点始终沿直线前进。

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5. Finding an Unknown Position or Time | 求未知位置与时刻

+ t v directly. More interesting problems give a coordinate and require solving for the time.

许多基础题给出 r₀ 和 v，然后要求某一时刻的位置。直接把该时刻代入 r(t) = r₀ + t v 即可。更有趣的题目会给出某个坐标，然后要求解出时间。

at (4, 7) with velocity v = (2, 1). If we need the time when its x-coordinate is 10, solve

例如，一质点从 (4, 7) 出发，速度 v = (2, 1)。若要求其 x 坐标为 10 的时刻，解

4 + 2t = 10

he position is (10, 10). If a problem gives both x and y, both equations must be satisfied by the same value of t.

得 t = 3。此时 y = 7 + 3 × 1 = 10，所以位置为 (10, 10)。若题目同时给出 x 和 y，则两个方程必须对同一个 t 同时成立。

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6. Does the Path Pass Through a Given Point? | 路径是否经过给定点

t v. This gives one equation for each coordinate. If all equations produce the same value of t, then the point is on the path. Otherwise it is not.

要判断位置向量为 p 的点是否在路径上，就解 p = r₀ + t v。每个坐标都给出一个方程。若所有方程都得到同一个 t，则点在路径上；否则不在。

For instance, consider the line r(t) = (1 + 2t, 3 − t). Is (7, 0) on the line? 1 + 2t = 7, so t = 3. From y = 0 we get 3 − t = 0, so t = 3. The same t works, so yes, (7

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