📚 Constant Velocity Problems | 匀速运动问题
Constant velocity problems form a crucial part of the IB Mathematics curriculum, bridging vector algebra with kinematics. When an object moves with constant velocity, its acceleration is zero and its position vector changes linearly with time. These problems often involve two objects moving along straight lines, requiring us to determine whether and when they meet, find the shortest distance between them, or analyse relative motion. Mastering this topic relies on a solid understanding of vector equations, parametric forms, and quadratic optimisation.
匀速运动问题是 IB 数学课程中的一个关键部分，它将向量代数与运动学联系在一起。当物体以恒定速度运动时，加速度为零，其位置向量随时间线性变化。这类问题常常涉及两个沿直线运动的物体，要求我们判断它们是否相遇、何时相遇，计算它们之间的最短距离，或者分析相对运动。掌握该主题依赖于对向量方程、参数形式以及二次函数最优化有扎实的理解。
1. Definition of Constant Velocity | 匀速运动的定义
An object is said to move with constant velocity if its velocity vector v does not change in magnitude or direction over time. This implies the acceleration vector a = 0 and the displacement after time t is exactly v t. In one dimension, this reduces to uniform motion along a line: s = s₀ + v t.
如果物体的速度向量 v 的大小和方向都不随时间变化，那么我们就说它做匀速运动。这意味着加速度向量 a = 0，并且经过时间 t 后的位移恰好为 v t。在一维情况下，这简化为直线上的匀速运动：s = s₀ + v t。
In two or three dimensions, constant velocity means the object travels in a straight line at a steady speed. The path is linear, and equal distances are covered in equal time intervals. This is the starting point for all the vector-based motion problems in IB Mathematics: Analysis and Approaches.
在二维或三维空间中，匀速运动意味着物体沿一条直线以恒定速率运动。运动路径是线性的，并且在相同的时间间隔内移动的距离相等。这是 IB 数学分析与解释课程中所有基于向量的运动问题的出发点。
2. Position Vector and the Equation of Motion | 位置向量与运动方程
The fundamental equation for constant velocity motion in vector form is r(t) = r₀ + v t, where r₀ is the initial position vector at time t = 0 and v is the constant velocity vector. Time t is a scalar parameter, usually measured in seconds.
匀速运动的向量基本方程为 r(t) = r₀ + v t，其中 r₀ 是 t = 0 时的初始位置向量，v 是恒定的速度向量。时间 t 是一个标量参数，通常以秒为单位。
If r₀ = (x₀ i + y₀ j) and v = (a i + b j), then the position at time t is (x₀ + a t) i + (y₀ + b t) j. This expression allows us to track both components of the object’s position separately.
如果 r₀ = (x₀ i + y₀ j)，v = (a i + b j)，那么时刻 t 的位置就是 (x₀ + a t) i + (y₀ + b t) j。这个表达式使我们能够分别追踪物体位置的两个分量。
3. Parametric Equations from Vectors | 从向量到参数方程
From the vector equation we can immediately write the parametric equations for the x and y coordinates: x(t) = x₀ + a t, y(t) = y₀ + b t. These describe the path of the object in the plane. If we eliminate t between the two equations, we obtain the Cartesian equation of the line, confirming that constant velocity motion always produces a straight line.
根据向量方程，我们可以直接写出 x 坐标与 y 坐标的参数方程：x(t) = x₀ + a t，y(t) = y₀ + b t。它们描述了物体在平面内的轨迹。如果在这两个方程之间消去 t，我们就得到直线的笛卡尔方程，从而确认匀速运动总是沿直线进行。
x = x₀ + a t, y = y₀ + b t
For example, an object starting at (2, 5) with velocity (3i – j) would have parametric equations x = 2 + 3t and y = 5 – t. Its Cartesian path is obtained by substituting t = x – 2 over 3 into y, giving y = 5 – (x – 2)/3.
例如，一个物体从 (2, 5) 出发，以速度 (3i – j) 运动，其参数方程为 x = 2 + 3t，y = 5 – t。将 t = (x – 2)/3 代入 y 便可求得笛卡尔方程 y = 5 – (x – 2)/3。
4. Velocity Vector and Speed | 速度向量与速率
The velocity vector v is often given in component form, such as (p i + q j). Its magnitude, the speed, is calculated using Pythagoras: speed = |v| = √(p² + q²). While velocity gives the direction of motion, speed is a scalar that tells us how fast the object is moving.
速度向量 v 通常以分量形式给出，例如 (p i + q j)。其大小，即速率，可用勾股定理计算：速率 = |v| = √(p² + q²)。速度决定了运动方向，而速率则是标量，告诉我们物体运动的快慢。
speed = √(a² + b²)
In IB problems, the velocity vector may be expressed using unit vectors i and j, or as a column vector. The key point is that even when two objects have the same speed, their velocity vectors can be different if their directions differ, leading to very different paths.
在 IB 考题中，速度向量可以用单位向量 i 和 j 表示，也可以写成列向量。关键点在于，即使两个物体的速率相同，如果方向不同，速度向量也就不同，从而导致完全不同的路径。
5. Setting Up Problems with Two Moving Objects | 两个运动物体的情景建模
Most examination questions feature two particles, boats, or aircraft moving with constant velocities. We define separate position vectors for A and B: r_A(t) = r_A0 + v_A t and r_B(t) = r_B0 + v_B t. It is essential to use the same time parameter t and to check that the units are consistent.
大多数考题都会涉及两个以恒定速度运动的质点、船只或飞机。我们为 A 和 B 分别定义位置向量：r_A(t) = r_A0 + v_A t 以及 r_B(t) = r_B0 + v_B t。使用相同的时间参数 t 并检查单位是否一致，这一点至关重要。
Often these moving objects start at different positions and travel in different directions. The vector equations serve as the foundation for answering all subsequent questions: Do they collide? When are they closest? What is the distance between them after a certain time?
这些运动物体通常从不同的位置出发，并沿不同方向运动。向量方程是回答所有后续问题的基础：它们会相撞吗？它们何时距离最近？经过一段时间后它们之间的距离是多少？
6. Meeting and Collision Conditions | 相遇与碰撞条件
Two objects collide if they occupy exactly the same position at the same time. Mathematically, this means there exists a t ≥ 0 such that r_A(t) = r_B(t). We equate the i and j components separately to obtain two equations in t. If the same value of t satisfies both, a collision occurs.
如果两个物体在同一时间恰好占据同一位置，它们就会相撞。从数学上讲，这意味着存在 t ≥ 0 使得 r_A(t) = r_B(t)。我们分别令 i 和 j 分量相等，得到两个关于 t 的方程。如果同一个 t 值同时满足两者，就会发生碰撞。
r_A0 + v_A t = r_B0 + v_B t
If the two equations yield different t values, the objects do not meet. In some problems one object sets off later; then we must use a shifted time variable, e.g. for a delay of 2 seconds, replace t by (t – 2) in that object’s equation for t ≥ 2.
如果两个方程得出的 t 值不同，物体就不会相遇。在某些问题中，一个物体出发较晚；此时我们必须使用平移后的时间变量，例如延迟 2 秒出发，就在该物体的方程中将 t 替换为 (t – 2)（当 t ≥ 2 时）。
7. Closest Distance Between Two Moving Objects | 两个运动物体之间的最短距离
Even when objects do not collide, we are often asked to find the minimum distance between them. We first form the relative position vector: d(t) = r_B(t) – r_A(t) = d₀ + u t, where d₀ = r_B0 – r_A0 and u = v_B – v_A.
即便物体不相撞，我们也常被要求求出它们之间的最小距离。我们首先写出相对位置向量：d(t) = r_B(t) – r_A(t) = d₀ + u t，其中 d₀ = r_B0 – r_A0，u = v_B – v_A。
The square of the distance is a quadratic function of t: D² = |d(t)|² = (d₀ + u t)·(d₀ + u t). We find the time t_min that minimises this quadratic by differentiating or, more simply, by using t_min = – (d₀·u) / |u|². If t_min is negative, the minimum distance occurs at t = 0.
距离的平方是关于 t 的二次函数：D² = |d(t)|² = (d₀ + u t)·(d₀ + u t)。我们通过求导，或者更简单地，利用 t_min = – (d₀·u) / |u|² 来找到使该二次式取最小值的时刻 t_min。如果 t_min 为负，最短距离出现在 t = 0 处。
t_min = – (d₀·u) / |u|²
Substituting t_min back into |d(t)| gives the minimum distance. This technique avoids solving messy square-root equations and is extremely efficient in exam settings.
将 t_min 代回 |d(t)| 即可得到最短距离。这种方法能避免求解繁琐的根式方程，在考试中非常高效。
8. Relative Velocity Approach | 相对速度法
An elegant geometric interpretation uses the concept of relative velocity. Imagine object A as stationary; B then moves with velocity u = v_B – v_A. The initial separation vector is d₀. The minimum distance is simply the perpendicular distance from the fixed point (A) to the line along which B appears to travel.
一种优美的几何解释使用了相对速度的概念。想象物体 A 静止不动；那么 B 就以速度 u = v_B – v_A 运动。初始分离向量为 d₀。最短距离就是从固定点 (A) 到 B 看上去所沿直线的最短距离，即垂直距离。
This minimum distance can also be expressed as |d₀| sin θ, where θ is the angle between d₀ and u. Using the dot product formula, we obtain the same result as before, reaffirming the connection between the algebraic and geometric methods.
这个最短距离也可以表示为 |d₀| sin θ，其中 θ 是 d₀ 与 u 之间的夹角。利用点积公式我们可以得到与之前相同的结果，这再次印证了代数方法与几何方法之间的联系。
9. Constant Velocity with Wind or Current | 风或水流影响下的匀速运动
In navigation problems, an aircraft or boat moves with a velocity relative to the air or water, while the medium itself has a drift velocity due to wind or current. The resultant velocity is the vector sum: v_ground = v_air + v_wind. The same constant velocity equations then apply using this resultant vector.
在导航问题中，飞机或船相对于空气或水流有一个速度，而介质本身因风或水流存在漂移速度。合速度是向量和：v_ground = v_air + v_wind。然后用这个合速度向量套用同样的匀速运动方程即可。
To find the true course or the required heading to reach a destination, we set up vector triangles. For example, if a swimmer wants to cross a river directly, they must aim upstream so that the cross-stream component of their velocity cancels the current. This is solved by equating components.
为了找出实际航向或为到达目的地所需的航行方向，我们建立向量三角形。例如，如果游泳者想垂直横渡河流，就必须朝上游方向游，使其速度的横向分量抵消水流的影响。这可以通过令分量相等来求解。
10. Worked Example: Two Boats | 典型例题：两艘船
Problem: At 12:00, boat A is at position (2i + 5j) km and moves with velocity (3i – j) km/h. Boat B is at (8i + 1j) km and moves with velocity (-i + 2j) km/h. Find (a) the time they would collide if they are on course, and (b) the minimum distance between them if they continue moving.
题目：在 12:00，船 A 位于 (2i + 5j) 千米处，以 (3i – j) 千米/小时的速度行驶。船 B 位于 (8i + 1j) 千米处，以 (-i + 2j) 千米/小时的速度行驶。(a) 若它们航线相交，求碰撞时间；(b) 如果它们继续行驶，求两船间的最短距离。
Solution (a): Set r_A = (2+3t)i + (5-t)j and r_B = (8 – t)i + (1+2t)j. Equate i: 2+3t = 8 – t → 4t = 6 → t = 1.5. Equate j: 5 – t = 1 + 2t → 3t = 4 → t = 4/3. Times differ, so no collision.
解 (a)：令 r_A = (2+3t)i + (5-t)j，r_B = (8 – t)i + (1+2t)j。令 i 分量相等：2+3t = 8 – t → 4t = 6 → t = 1.5。令 j 分量相等：5 – t = 1 + 2t → 3t = 4 → t = 4/3。时间不一致，因此不会相撞。
Solution (b): d₀ = (8-2)i + (1-5)j = (6i – 4j). u = (-1-3)i + (2 – (-1))j = (-4i + 3j). Then t_min = – (d₀·u) / |u|² = – (6×(-4) + (-4)×3) / (16+9) = – (-24 -12)/25 = 36/25 = 1.44 h. Distance = |d₀ + u × 1.44| = … (simplifies to 0? Check: relative path passes through origin? The minimum distance is indeed zero? Wait, vectors d₀ and u might be proportional? d₀ = (6, -4), u = (-4,3) are perpendicular? 6×(-4)+(-4)×3=-24-12=-36 not zero, so not perpendicular. The minimum distance is |d₀| |sin θ|, compute: |d₀|=√52, |u|=5, dot=-36, cos θ = -36/(5√52), sin θ = √(1 – 1296/(25×52)) = … actually simpler: distance = |d₀ × u|/|u|. Cross product magnitude: |6×3 – (-4)×(-4)| = |18 – 16| = 2. So distance = 2/5 = 0.4 km.) Okay, we can present this calculation.)
解 (b)：d₀ = (6i – 4j)。u = (-4i + 3j)。最短距离可用叉积法求得：|d₀ × u| / |u| = |6×3 – (-4)×(-4)| / √(16+9) = |18 – 16|/5 = 2/5 = 0.4 千米。因此最短距离为 0.4 千米。
11. Common Pitfalls and Exam Tips | 常见陷阱与考试技巧
A frequent mistake is confusing distance with displacement or using the wrong time variable for objects that start at different times. Always read the problem carefully: does it ask for the shortest distance or the time of closest approach? Also, check if the question expects a distance or a position vector.
常见的错误包括混淆距离与位移，或在物体出发时间不同的情况下使用了错误的时间变量。务必仔细审题：题目要求的是最短距离还是最接近的时刻？同时还要看清问题要求的是距离还是位置向量。
Another oversight is forgetting to verify that a calculated collision time is non‑negative. In some cases, a solution for t might be positive but correspond to a point behind the starting position – always check the context.
另一个容易疏忽的地方是忘记验证计算出的碰撞时间是否为非负数。有时 t 的解虽然是正的，但可能对应于出发位置背后的点——一定要结合具体情景进行检查。
For closest approach, many students mistakenly set the velocity vectors equal or try to solve r_A = r_B. Instead, focus on minimising |r_B – r_A|². Mastering the quadratic method or the perpendicular distance method gives a huge time advantage.
针对最短距离问题，许多学生错误地令速度向量相等，或试图求解 r_A = r_B。正确做法应专注于最小化 |r_B – r_A|²。熟练掌握二次函数法或垂直距离法可以节省大量时间。
Finally, always draw a quick sketch. Visualising the initial positions and velocity directions helps detect sign errors and builds intuition about whether the objects will approach or separate.
最后，一定要迅速画一个草图。将初始位置和速度方向画出来有助于发现符号错误，并培养对物体是相互靠近还是远离的直观判断。
12. Summary and Key Formulas | 总结与关键公式
Constant velocity problems rest on three foundations: the position vector equation r(t) = r₀ + v t, the collision condition resulting in equal position vectors at the same t, and the minimum distance technique minimising |d₀ + u t|².
匀速运动问题建立在三个基石之上：位置向量方程 r(t) = r₀ + v t，碰撞条件即同一时刻位置向量相等，以及最短距离方法即最小化 |d₀ + u t|²。
- Velocity magnitude: speed = √(a² + b²)
- Collision: solve r_A(t) = r_B(t) for t
- Minimum distance: t_min = – (d₀·u) / |u|², then substitute
- Relative velocity: u = v_B – v_A
- 速度大小：速率 = √(a² + b²)
- 碰撞：解方程 r_A(t) = r_B(t) 求 t
- 最短距离：t_min = – (d₀·u) / |u|²，然后代回
- 相对速度：u = v_B – v_A
With practice, these structured approaches turn seemingly complex motion scenarios into a straightforward application of vector algebra and quadratic optimisation.
通过练习，这些结构化的方法能将看似复杂的运动情景转化为对向量代数和二次函数最优化的直接应用。
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