📚 IB Edexcel Maths: Kinematics Revision Essentials | IB Edexcel 数学：运动学 考点精讲
Kinematics is a core topic in IB Edexcel Mathematics that connects calculus with the physical description of motion. Understanding how to move between displacement, velocity and acceleration using differentiation and integration is crucial for top exam performance.
运动学是 IB Edexcel 数学的核心主题，它将微积分与运动的物理描述联系起来。理解如何通过求导和积分在位移、速度和加速度之间转换是取得高分的关键。
1. Kinematics: From Calculus to Motion | 运动学：从微积分到运动
In IB Maths, kinematics studies the motion of a particle along a straight line or in a plane using functions of time t. The subject relies heavily on the calculus tools of differentiation and integration to model changes in position and speed.
在 IB 数学中，运动学研究粒子沿直线或平面内随时间 t 变化的运动。这个主题高度依赖微积分中的求导和积分工具来模拟位置和速度的变化。
The central idea is that we can describe a particle’s location with a displacement function s(t), and then derive its velocity v(t) and acceleration a(t) as successive derivatives.
核心思想是我们可以用位移函数 s(t) 描述粒子的位置，然后将其速度 v(t) 和加速度 a(t) 作为逐次导数导出。
Conversely, if we start from acceleration, we can integrate to obtain velocity and then displacement, provided we know the initial conditions of the motion.
反之，如果我们从加速度出发，可以通过积分得到速度，再得到位移，前提是我们知道运动的初始条件。
2. Displacement, Velocity and Acceleration | 位移、速度和加速度
The displacement s(t) is the position of a particle relative to a fixed origin O at time t. It is measured in metres (m) and takes positive or negative values depending on direction.
位移 s(t) 是粒子在时间 t 相对于固定原点 O 的位置。它以米 (m) 为单位，根据方向取正值或负值。
Velocity v(t) is the rate of change of displacement with respect to time, i.e. the first derivative: v(t) = ds/dt. It indicates both the speed and the direction of motion.
速度 v(t) 是位移对时间的变化率，即一阶导数：v(t) = ds/dt。它同时表示运动的快慢和方向。
Acceleration a(t) is the rate of change of velocity, given by the second derivative of displacement or the first derivative of velocity: a(t) = dv/dt = d²s/dt².
加速度 a(t) 是速度的变化率，由位移的二阶导数或速度的一阶导数给出：a(t) = dv/dt = d²s/dt²。
These three quantities are intimately linked: s(t) -differentiate→ v(t) -differentiate→ a(t). The arrow can be reversed by integration.
这三个量紧密相连：s(t) ——求导→ v(t) ——求导→ a(t)。该过程可以通过积分反向进行。
3. Differentiation: From Displacement to Velocity and Acceleration | 求导：由位移求速度和加速度
When a displacement function s(t) is given, you obtain the velocity by differentiating once: v(t) = ds/dt. Repeating the process gives the acceleration: a(t) = dv/dt.
当给出位移函数 s(t) 时，通过一次求导得到速度：v(t) = ds/dt。重复这一过程得到加速度：a(t) = dv/dt。
For example, if s(t) = 3t³ − 2t² + 5t − 1 in metres, then v(t) = 9t² − 4t + 5 m s⁻¹ and a(t) = 18t − 4 m s⁻².
例如，若 s(t) = 3t³ − 2t² + 5t − 1 (米)，则 v(t) = 9t² − 4t + 5 m s⁻¹，a(t) = 18t − 4 m s⁻²。
You can then analyse the motion by substituting particular times, finding when the particle is at rest (v = 0), or determining the direction of motion from the sign of v.
然后你可以通过代入特定时刻、求出粒子何时静止 (v = 0) 或根据 v 的正负确定运动方向来分析运动。
In exam questions, you are often asked to find the velocity and acceleration at a given time, or to show that a particle changes direction when v = 0 and a ≠ 0.
在考试题中，经常要求你求某一时刻的速度和加速度，或者证明当 v = 0 且 a ≠ 0 时粒子改变方向。
4. Integration: From Acceleration to Velocity and Displacement | 积分：由加速度求速度和位移
Starting with acceleration a(t), integrate to find velocity: v(t) = ∫ a(t) dt + C₁. Use an initial condition (e.g. v(0) = u) to determine the constant of integration C₁.
从加速度 a(t) 出发，积分求速度：v(t) = ∫ a(t) dt + C₁。使用初始条件（例如 v(0) = u）确定积分常数 C₁。
Then integrate the velocity to obtain displacement: s(t) = ∫ v(t) dt + C₂. Again, use s(0) = s₀ (often zero) to find C₂.
然后再积分速度求位移：s(t) = ∫ v(t) dt + C₂。同样，用 s(0) = s₀（通常为零）求 C₂。
For instance, if a(t) = 6t − 2 and at t = 0, v = 3, s = 0, then v(t) = 3t² − 2t + 3 and s(t) = t³ − t² + 3t.
例如，若 a(t) = 6t − 2 且在 t = 0 时 v = 3，s = 0，则 v(t) = 3t² − 2t + 3，s(t) = t³ − t² + 3t。
Always remember to include the constants of integration, as forgetting them is one of the most common mistakes in kinematics problems.
永远记得加上积分常数，因为忘记常数是运动学问题中最常见的错误之一。
5. Graphical Interpretation | 图形解释
Kinematics graphs give a visual way to understand motion. A displacement–time graph (s-t) shows the position; its gradient at any point equals the velocity.
运动学图形提供了一种直观理解运动的方式。位移–时间图 (s-t) 显示位置；其上任意一点的斜率等于速度。
A velocity–time graph (v-t) has a gradient equal to acceleration, while the area under the graph between two times gives the change in displacement.
速度–时间图 (v-t) 的斜率等于加速度，而两时刻之间图形下的面积表示位移的变化量。
An acceleration–time graph (a-t) shows how acceleration varies; the area under the a-t graph gives the change in velocity.
加速度–时间图 (a-t) 显示加速度的变化情况；a-t 图下的面积表示速度的变化量。
IB exam questions may ask you to sketch these graphs or to interpret information such as maximum velocity, rest points, and return to the origin.
IB 考试题可能会要求你绘制这些图形，或者解读诸如最大速度、静止点以及返回原点等信息。
6. Constant Acceleration (SUVAT) Equations | 匀加速 (SUVAT) 方程
When acceleration is constant, the kinematic equations of motion (SUVAT) provide quick solutions. They relate five quantities: s (displacement), u (initial velocity), v (final velocity), a (acceleration), and t (time).
当加速度恒定时，运动学方程 (SUVAT) 提供快速的解。它们联系五个量：s (位移), u (初速度), v (末速度), a (加速度) 和 t (时间)。
The four standard equations are:
四个标准方程为：
- v = u + at
- s = ut + ½at²
- s = ½(u + v)t
- v² = u² + 2as
- v = u + at
- s = ut + ½at²
- s = ½(u + v)t
- v² = u² + 2as
You must choose the appropriate equation based on the given information. Always check that the signs of vectors are consistent with your chosen positive direction.
你必须根据已知信息选择合适的方程。始终检查向量的符号与所选正方向一致。
These equations can also be derived by integrating a constant acceleration, which reinforces the link between SUVAT and calculus.
这些方程也可以通过积分恒定的加速度来推导，这加强了 SUVAT 与微积分之间的联系。
7. Variable Acceleration Problems | 变加速问题
Many IB questions move beyond constant acceleration and require you to work with functions of time. Here, calculus is the only reliable method.
许多 IB 试题超出匀加速范围，要求你处理时间的函数。此时微积分是唯一可靠的方法。
For a given a(t), integrate to get v(t) and again to get s(t). Pay close attention to integrating piecewise if the function changes, though this is rare in standard problems.
对于给定的 a(t)，积分得到 v(t)，再积分得到 s(t)。如果函数分段变化，需注意分段积分，不过这在标准问题中较为罕见。
You may also be given v(t) or s(t) and be asked to find the maximum velocity or the farthest distance from the origin. This involves setting the derivative equal to zero and testing the second derivative.
你也可能被给出 v(t) 或 s(t)，并被要求求最大速度或离原点最远的距离。这涉及令导数为零并检验二阶导数。
Always identify which function is provided and whether you need to differentiate or integrate to answer the question.
始终识别给定了哪个函数，以及你需要求导还是积分来回答问题。
8. Turning Points and Maxima/Minima | 转折点与极值
In kinematics, the turning point corresponds to the moment when the particle changes direction, which happens when v(t) = 0 and a(t) ≠ 0.
在运动学中，转折点对应粒子改变方向的时刻，此时 v(t) = 0 且 a(t) ≠ 0。
To find the maximum displacement, set v(t) = ds/dt = 0 and solve for t. Then check that a(t) = d²s/dt² < 0 to confirm it is a maximum.
要求最大位移，令 v(t) = ds/dt = 0 并解出 t。然后验证 a(t) = d²s/dt² < 0 以确认其为极大值。
The maximum (or minimum) velocity can be found by setting a(t) = dv/dt = 0 and checking the sign of da/dt or using the second derivative test.
最大（或最小）速度可通过令 a(t) = dv/dt = 0 并检查 da/dt 的符号或使用二阶导数检验来找到。
Be careful: when v = 0, the particle might be at a maximum distance from the origin or merely at a turning point; always examine the full motion.
注意：当 v = 0 时，粒子可能处于离原点最远的距离，也可能只是处于一个转折点；务必考察完整的运动过程。
9. Displacement vs Distance Travelled | 位移与路程的区别
Displacement s(t) is the net change in position and can be positive, negative or zero. Distance travelled, however, is the total length of the path and is always non-negative.
位移 s(t) 是位置的净变化量，可为正、负或零。然而，路程是路径的总长度，始终非负。
If the particle changes direction, you must integrate the absolute value of velocity, |v(t)|, over time intervals, or split the motion into intervals where v has a constant sign.
若粒子改变方向，你必须对速度的绝对值 |v(t)| 在时间区间上积分，或者将运动分割成 v 符号不变的区间。
For example, if v(t) = t² − 4, the particle turns at t = 2. Distance from t = 0 to t = 3 is ∫₀² |v| dt + ∫₂³ |v| dt, not simply |s(3) − s(0)|.
例如，若 v(t) = t² − 4，粒子在 t = 2 时转向。从 t = 0 到 t = 3 的路程是 ∫₀² |v| dt + ∫₂³ |v| dt，而不仅仅是 |s(3) − s(0)|。
Many students lose marks by confusing displacement and distance; always check whether the question asks for ‘displacement’ or ‘total distance travelled’.
许多学生因混淆位移和路程而失分；一定要检查题目问的是“位移”还是“行驶的总路程”。
10. Vector Kinematics | 向量运动学
When motion occurs in two dimensions, the position is described by a vector r(t) = x(t) i + y(t) j. Velocity and acceleration vectors follow by differentiation: v(t) = dr/dt, a(t) = dv/dt.
当运动发生在二维时，位置由向量 r(t) = x(t) i + y(t) j 描述。速度和加速度向量通过求导得到：v(t) = dr/dt，a(t) = dv/dt。
Each component differentiates independently: v = (dx/dt) i + (dy/dt) j and a = (d²x/dt²) i + (d²y/dt²) j.
每个分量分别求导：v = (dx/dt) i + (dy/dt) j，a = (d²x/dt²) i + (d²y/dt²) j。
The speed of the particle is the magnitude of the velocity vector: |v| = √(vₓ² + vᵧ²). You may be asked to find the time when speed is minimum or when the particle moves parallel to a given vector.
粒子的速率是速度向量的大小：|v| = √(vₓ² + vᵧ²)。你可能需要求速率最小时的时间，或粒子何时平行于给定向量运动。
Vector integration follows the same principle: v(t) = ∫ a(t) dt + C₁, and r(t) = ∫ v(t) dt + C₂, where the constants are vector quantities determined by initial conditions.
向量积分遵循相同原则：v(t) = ∫ a(t) dt + C₁，r(t) = ∫ v(t) dt + C₂，其中常数是由初始条件确定的向量。
11. Common Errors and Exam Tips | 常见错误与备考技巧
Forgetting the integration constant when recovering velocity or displacement is the top mistake. Always write ‘+ C’ and use the given initial conditions (t = 0, v = u, s = s₀) to find its value.
在恢复速度或位移时忘记积分常数是头号错误。务必写出“+ C”并利用给定的初始条件 (t = 0, v = u, s = s₀) 求出其值。
Another frequent error is mismanaging signs. Choose a positive direction (usually to the right or upwards) and consistently apply it to displacement, velocity, and acceleration.
另一个常见错误是符号处理不当。选定一个正方向（通常向右或向上），并对位移、速度和加速度一致使用。
When asked for ‘distance’, do not simply evaluate |s(b) − s(a)|. Check for zeroes of velocity and break the motion into segments where the direction is constant.
当要求“路程”时，不要简单计算 |s(b) − s(a)|。检查速度是否为零，并将运动切分成方向恒定的若干段。
Read the question carefully to determine whether you are given s(t), v(t) or a(t), and decide whether to differentiate or integrate. Draw a quick sketch of the v-t graph if it helps visualise the motion.
仔细审题以确定给出的是 s(t)、v(t) 还是 a(t)，并决定是求导还是积分。如果有助于可视化运动，可快速绘制 v-t 草图。
In SUVAT problems, list the known symbols and choose the equation that contains them. Always check that the units are consistent (e.g. all in metres and seconds).
在 SUVAT 问题中，列出已知符号并选择包含它们的方程。始终检查单位是否一致（例如全部采用米和秒）。
Practice past paper questions where you must combine differentiation, integration, and graphical interpretation, as IB exams frequently test multiple concepts in a single kinematics problem.
练习历年真题，这些题目需要你结合求导、积分和图形解释，因为 IB 考试经常在一个运动学问题中测试多个概念。
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