📚 Constant Velocity Problems | 匀速运动问题
Constant velocity problems are a fundamental part of kinematics, requiring you to model motion where speed and direction remain unchanged. In these scenarios, displacement changes at a steady rate, and the relationship between distance, time, and velocity becomes linear, making the mathematics both accessible and powerful for describing real-world movement.
匀速运动问题是运动学的基础部分，要求你建立速度和方向保持不变的模型。在这些情形中，位移以恒定速率变化，距离、时间和速度之间的关系是线性的，这使得相关的数学运算既简单又能够有力地描述真实世界的运动。
1. Definition of Constant Velocity | 匀速运动的定义
Constant velocity means an object moves in a fixed direction with unchanging speed. Both the magnitude and direction of the velocity vector remain constant. This implies zero acceleration, as any change in speed or direction would require a net force according to Newton's first law.
匀速运动意味着物体以恒定的速度沿固定方向运动。速度向量的大小和方向都保持不变。这意味着加速度为零，因为根据牛顿第一定律，任何速度大小或方向的改变都需要有净外力作用。
Mathematically, velocity v is defined as the rate of change of displacement s with respect to time t: v = Δs / Δt. When v is constant, this simplifies to s = v t (assuming initial displacement is zero). In IB Mathematics, you will often work with vector forms such as r = r₀ + v t, where r is position vector and r₀ is initial position.
在数学上，速度 v 定义为位移 s 对时间 t 的变化率：v = Δs / Δt。当 v 恒定时，可简化为 s = v t（假设初始位移为零）。在 IB 数学中，你经常会使用向量形式，如 r = r₀ + v t，其中 r 是位置向量，r₀ 是初始位置。
2. Displacement-Time Graphs | 位移-时间图像
nward, velocity is negative. A horizontal line indicates the object is stationary (v = 0).
匀速运动的位移-时间图像是一条直线。直线的斜率等于速度。如果直线向上倾斜，速度为正值；如果向下倾斜，速度为负值。水平线则表示物体静止（v = 0）。
t, but for constant velocity, the graph is a horizontal line, so the area is simply v × t.
在 IB 问题中，你可能会被要求从图像中求速度、根据文字描述画出图像，或解释两条线的交点即为两物体相遇点。记住，速度-时间图像下的面积表示位移，但对于匀速运动，图像是一条水平线，因此面积就是简单的 v × t。
3. Using s = vt | 使用公式 s = vt
The scalar equation s = v t applies when an object moves at constant velocity v for a time interval t, covering displacement s in the direction of motion. If initial displacement s₀ is not zero, the equation becomes s = s₀ + v t. This is the one-dimensional form useful for problems along a straight line.
当物体沿运动方向以恒定速度 v 运动一段时间 t，产生的位移为 s 时，可以使用标量方程 s = v t。如果初始位移 s₀ 不为零，则方程变为 s = s₀ + v t。这是一维形式，适用于直线上的问题。
Units must be consistent: when velocity is in m s⁻¹, time in seconds, displacement is in metres. Problems may involve unit conversions, such as km h⁻¹ to m s⁻¹ (multiply by 1000/3600, i.e. ÷3.6). Always check that the numbers make physical sense.
单位必须保持一致：当速度以 m s⁻¹ 表示，时间以秒为单位时，位移以米为单位。题目可能涉及单位换算，例如将 km h⁻¹ 换算为 m s⁻¹（乘以 1000/3600，即除以 3.6）。始终要检查数值在物理上是否合理。
4. Vector Representation | 向量表示
In two or three dimensions, constant velocity is expressed using vectors. The position vector r of an object at time t is given by r(t) = r₀ + v t vector equation separates into independent components, e.g. x = x₀ + vₓ t, y = y₀ + vᵧ t.
在二维或三维空间中，匀速运动用向量来表示。物体在时刻 t 的位置向量 r 由 r(t) = r₀ + v t 给出，其中 r₀ 是初始位置向量，v 是恒定的速度向量。这个向量方程可分解为独立的分量，例如 x = x₀ + vₓ t，y = y₀ + vᵧ t。
xam questions frequently ask for the speed, which is the magnitude of the velocity vector: |v| = √vₓ² + vᵧ². You might also need to find the distance between two moving objects as a function of time, or determine when they are closest together.
IB 考试题目经常要求计算速率，即速度向量的大小：|v| = √vₓ² + vᵧ²。你可能还需要求两个运动物体之间距离关于时间的函数，或者确定它们何时相距最近。
5. Relative Velocity | 相对速度
Relative velocity is the velocity of one object as observed from another moving object. If object A has velocity vA and object B has velocity vB, then the velocity of A relative to B is vAB = vA − vB. This vector subtraction helps determine how one object perceives the motion of the other.
相对速度是从一个运动的物体上观察另一个物体的速度。如果物体 A 的速度为 vA，物体 B 的速度为 vB，则 A 相对于 B 的速度是 vAB = vA − vB。这种向量减法有助于判断一个物体如何感知另一个物体的运动。
Constant velocity problems often involve two moving objects. By using relative velocity, you can reduce a two-body problem to one body moving relative to a stationary frame. Then the meeting time or collision condition can be found with simple geometry or algebra.
匀速运动问题常涉及两个运动物体。通过使用相对速度，你可以将二体问题简化为一个物体相对于静止参考系的运动。然后就可以用简单的几何或代数方法求出相遇时间或碰撞条件。
6. Problem-Solving Steps | 解题步骤
A systematic approach is essential for solving constant velocity problems. Start by defining a clear coordinate system and recording known quantities (initial positions, velocities, time). Translate the wording into vector equations. Identify what is being asked: meeting time, position, closest distance, etc.
解匀速运动问题时，采用系统的方法至关重要。首先定义清晰的坐标系，并记录已知量（初始位置、速度、时间）。将文字转化为向量方程。明确题目所求：相遇时间、位置、最近距离等。
Next, set up the relevant equations for each object. If an object changes direction or speed, the problem is no longer a constant velocity case and may require piecewise analysis. Solve the equations, often simultaneously, and interpret the solutions in context, discarding negative times or unrealistic results.
接下来，为每个物体建立相关方程。如果物体改变了方向或速度，问题就不再是匀速运动的情况，可能需要进行分段分析。解这些方程，通常是联立求解，并在实际情境中解释所得解，舍弃负的时间值或不合理的结果。
7. Example 1: One-Dimensional Motion | 例1：一维运动
t a constant speed of 20 m s⁻¹ along a straight road. How far does it travel in 45 seconds? d car starts 500 m ahead and travels at 15 m s⁻¹ in the same direction, when does the first car catch up?
一辆汽车以 20 m s⁻¹ 的恒定速度沿直线道路行驶。在 45 秒内它行驶多远？使用 s = v t：s = 20 × 45 = 900 m。如果第二辆车在前方 500 m 处以 15 m s⁻¹ 的速度同向匀速行驶，第一辆车何时追上前车？
Set up equations: Car 1: s₁ = 20 t = 500 + 15t → 5t = 500 → t = 100 s. The meeting position is s = 20×100 = 2000 m from Car 1' . This demonstrates solving simple linear equations derived from constant velocity models.
建立方程：车 1：s₁ = 20 t 车 2：s₂ = 500 + 15 t。当 s₁ = s₂ 时相遇 → 20t = 500 + 15t → 5t = 500 → t = 100 s。相遇位置距车 1 起点为 s = 20×100 = 2000 m。这展示了如何求解由匀速模型导出的简单线性方程。
8. Example 2: Two-Dimensional Vector Problem | 例2：二维向量问题
A boat leaves a dock at position vector (2i + 3j) km and moves with constant velocity (–i + 4j) km h⁻¹. Find its position after 2 hours. r(t) = r₀ + v t = (2i + 3j) + 2(–i + 4j) = (2–2)i + (3+8)j = 0i + 11j, so 11 km north of the origin on the y-axis.
一艘船从位置向量 (2i + 3j) km 的码头出发，以恒定的速度 (–i + 4j) km h⁻¹ 移动。求 2 小时后的位置。r(t) = r₀ + v t = (2i + 3j) + 2(–i + 4j) = (2–2)i + (3+8)j = 0i + 11j，即位于原点正北 11 km 处。
To find the distance from the origin: |r| = √0² + 11² e between them after t hours is the magnitude of the difference of their position vectors: |r₁ – r₂| = |(2–t – 3t)i + (3+4t – 2t)j| be minimised using calculus or completing the square to find the time of closest approach.
要计算到原点的距离：|r| = √0² + 11² = 11 km。如果第二艘船同时从原点出发，速度为 (3i + 2j) km h⁻¹，则 t 小时后两船之间的距离为它们位置向量之差的模：|r₁ – r₂| = |(2–t – 3t)i + (3+4t – 2t)j|。可以通过微积分或配方法来最小化这个表达式，从而找到最近接近的时间。
9. Common Mistakes | 常见错误
One frequent error is confusing speed with velocity. Speed is scalar and cannot be negative, while velocity includes direction. Another is misinterpreting displacement-time graphs: a steep slope means high velocity, not necessarily fast speed if the direction is negative. Also, forgetting to convert units leads to wrong answers.
一个常见错误是混淆速率和速度。速率是标量，不能为负值，而速度包含方向。另一个错误是误解位移-时间图像：陡峭的斜率意味着高速度，但如果方向是负的，不一定意味着速率快。另外，忘记换算单位会导致错误答案。
he meeting occurred before t = 0, which is invalid unless the problem specifies otherwise.
在向量问题中，学生有时会错误地相加位置向量，或者在比较距离时忘记对各分量分别处理。务必检查最终答案是否合理：负时间通常意味着相遇发生在 t = 0 之前，除非题目另有说明，否则就是无效的。
10. Applications | 实际应用
Constant velocity models, though simplified, describe many real-world situations sufficiently: conveyor belts, escalators, steady walking or driving on a straight road, and aircraft cruising at constant altitude and heading. Even planetary motion can be approximated over very short time intervals where the path is nearly straight and speed nearly constant.
匀速模型虽然简化了情况，但足以描述许多现实世界的情境：传送带、自动扶梯、匀速行走或在笔直道路上稳定驾驶、飞机以恒定高度和航向巡航。即使是行星运动，在极短的时间间隔内，路径接近直线且速率几乎恒定，也可以近似为匀速运动。
In IB Mathematics, these problems build foundational skills for more complex kinematics, such as constant acceleration and calculus-based motion. Mastery of constant velocity scenarios ensures you can handle vectors, parametric equations, and relative motion with confidence in later topics.
在 IB 数学中，这些问题为更复杂的运动学内容（如匀加速运动和基于微积分的运动）打下了基础。掌握匀速运动情境能确保你在后续课题中自信地处理向量、参数方程和相对运动。
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