📚 A - Kinematics | 运动学
Kinematics is the branch of mathematics that describes the motion of objects using displacement, velocity and acceleration as functions of time. In IB Mathematics, students learn to apply differentiation and integration to analyse both linear and two‑dimensional motion, linking algebraic, graphical and vector representations.
运动学是用位移、速度和加速度作为时间函数来描述物体运动的数学分支。在 IB 数学中，学生需要应用微分和积分来分析直线运动和二维运动，并将代数、图像和向量表示联系起来。
1. Introduction to Kinematics | 运动学简介
Kinematics deals with the 'how' of motion without asking 'why' an object moves. It provides the mathematical tools to model the path of a particle, calculate its speed at any instant, and predict where it will be after a given time.
运动学研究物体“如何”运动，而不追究“为什么”运动。它提供数学工具来建模质点的轨迹、计算任意时刻的速率，并预测给定时间后的位置。
In IB syllabi, kinematics appears in the context of calculus. By treating displacement, velocity and acceleration as related by derivatives and integrals, you can move seamlessly between these quantities.
在 IB 大纲中，运动学出现在微积分内容里。把位移、速度和加速度看作用导数和积分关联的量，你就能够在这几个量之间自由转换。
2. Displacement, Velocity and Acceleration | 位移、速度与加速度
Displacement (often denoted by s) is a vector quantity measuring the change in position of a particle relative to an origin. It is not the same as distance, which is a scalar measuring the total length of the path travelled.
位移（常用 s 表示）是矢量，测量质点相对于原点的位置变化。它与距离不同，距离是标量，测量经过路径的总长度。
Velocity is the rate of change of displacement with respect to time: v = ds/dt. Acceleration is the rate of change of velocity: a = dv/dt or equivalently a = d²s/dt². The sign of these quantities indicates direction.
速度是位移对时间的变化率：v = ds/dt。加速度是速度对时间的变化率：a = dv/dt，也等价于 a = d²s/dt²。这些量的正负号表示方向。
If the velocity is positive and the acceleration is negative, the object is slowing down. Conversely, both positive or both negative means the object is speeding up.
如果速度为正而加速度为负，物体在减速。反之，两者同号表示物体在加速。
3. Differentiation and Integration in Motion | 运动中的微分与积分
Given a displacement function s(t), you obtain velocity by differentiation: v(t) = s'(t). Differentiate again to get acceleration: a(t) = v'(t) = s''(t).
若已知位移函数 s(t)，通过微分得到速度：v(t) = s'(t)。再次微分得到加速度：a(t) = v'(t) = s''(t)。
When starting from acceleration, integration recovers velocity: v(t) = ∫ a(t) dt + C₁. A second integration gives displacement: s(t) = ∫ v(t) dt + C₂. The constants of integration are determined by initial conditions, such as velocity at t = 0.
如果从加速度出发，积分可以求出速度：v(t) = ∫ a(t) dt + C₁。二次积分得到位移：s(t) = ∫ v(t) dt + C₂。积分常数由初始条件决定，例如 t = 0 时的速度。
For example, if a(t) = 2t + 3 and the initial velocity is v(0) = 5, then v(t) = ∫ (2t + 3) dt = t² + 3t + 5. Further, if s(0) = 2, then s(t) = ∫ (t² + 3t + 5) dt = (1/3)t³ + (3/2)t² + 5t + 2.
例如，若 a(t) = 2t + 3 且初速度 v(0) = 5，则 v(t) = ∫ (2t + 3) dt = t² + 3t + 5。进一步，若 s(0) = 2，则 s(t) = ∫ (t² + 3t + 5) dt = (1/3)t³ + (3/2)t² + 5t + 2。
4. Constant Acceleration Equations (SUVAT) | 匀加速运动方程
When acceleration a is constant, five key equations link the variables s (displacement), u (initial velocity), v (final velocity), a (acceleration) and t (time).
当加速度 a 恒定时，五个关键方程将变量 s（位移）、u（初速度）、v（末速度）、a（加速度）和 t（时间）联系起来。
| Variable | Meaning |
| s | displacement (m) |
| u | initial velocity (m s⁻¹) |
| v | final velocity (m s⁻¹) |
| a | acceleration (m s⁻²) |
| t | time (s) |
The standard SUVAT equations are derived from the definitions of velocity and acceleration or from the area under a velocity–time graph.
标准的 SUVAT 方程源于速度和加速度的定义，也可由速度‑时间图下的面积导出。
v = u + at
s = ut + ½at²
s = ½(u + v)t
v² = u² + 2as
s = vt − ½at²
To choose the right equation, list the known quantities and the one you need, then pick the formula that includes all of them. Watch the sign of acceleration when the object is slowing down.
要选正确的方程，先列出已知量和待求量，然后选择包含它们全部的那个公式。当物体在减速时，要注意加速度的正负。
5. Motion with Variable Acceleration | 变加速运动
When acceleration is not constant, you must use calculus to link the motion quantities. If acceleration is given as a function of time, a(t), the velocity is the definite integral v(t) = v₀ + ∫₀ᵗ a(τ) dτ, and displacement is s(t) = s₀ + ∫₀ᵗ v(τ) dτ.
当加速度不恒定时，必须使用微积分联系各运动量。若加速度以时间函数 a(t) 给出，速度就是定积分 v(t) = v₀ + ∫₀ᵗ a(τ) dτ，位移则是 s(t) = s₀ + ∫₀ᵗ v(τ) dτ。
Alternatively, you may be given a = f(v) or a = f(s). In such cases, you use the chain rule and separate variables: a = dv/dt = v(dv/ds) allows you to write v dv = a ds. Integrating with appropriate limits then yields the velocity–displacement relation.
也可能给定的是 a = f(v) 或 a = f(s)。此时利用链式法则和分离变量：a = dv/dt = v(dv/ds)，从而得到 v dv = a ds。代入合适的上下限积分，就能找出速度‑位移关系。
Worked example: given a = 6t, v = 0 when t = 0, find s after 2 seconds. First, v(t) = ∫₀ᵗ 6t dt = 3t². Then, s(t) = ∫₀² 3t² dt = [t³]₀² = 8 m.
例题：已知 a = 6t，t = 0 时 v = 0，求 2 秒后的位移。首先 v(t) = ∫₀ᵗ 6t dt = 3t²，然后 s(t) = ∫₀² 3t² dt = [t³]₀² = 8 m。
6. Graphical Analysis of Motion | 运动图解分析
Displacement–time graphs show the position of a particle over time; their gradient gives velocity. Velocity–time graphs have a gradient equal to acceleration and the area under the curve represents displacement.
位移‑时间图展示质点随时间的位置；其斜率给出速度。速度‑时间图的斜率等于加速度，曲线下方的面积代表位移。
Acceleration–time graphs are less common, but their area between two times gives the change in velocity. Being able to switch between s‑t, v‑t and a‑t graphs is a core skill tested in IB exams.
加速度‑时间图较少见，但两个时间之间的面积表示速度变化量。能够在 s‑t、v‑t 和 a‑t 图之间转换是 IB 考试考查的核心技能。
For example, a horizontal line on a v‑t graph indicates constant velocity; a sloping line indicates constant acceleration. Curved segments imply variable acceleration, which often requires integral calculus for precise analysis.
例如，v‑t 图上的水平线表示匀速；倾斜直线表示匀加速。曲线段意味着变加速，通常需要积分计算才能精确分析。
7. Vector Kinematics in Two Dimensions | 二维向量运动学
When motion occurs in a plane, position, velocity and acceleration are expressed as vectors. A typical representation uses the unit vectors i and j: r(t) = x(t)i + y(t)j.
当运动发生在平面上时，位置、速度和加速度都用向量表示。典型的表达方式是使用单位向量 i 和 j：r(t) = x(t)i + y(t)j。
The velocity vector is the derivative of the position vector: v(t) = r'(t) = (dx/dt)i + (dy/dt)j. Its magnitude gives the speed: |v| = √((dx/dt)² + (dy/dt)²). Acceleration is the second derivative: a(t) = (d²x/dt²)i + (d²y/dt²)j.
速度向量是位置向量的导数：v(t) = r'(t) = (dx/dt)i + (dy/dt)j。其大小给出速率：|v| = √((dx/dt)² + (dy/dt)²)。加速度是二阶导数：a(t) = (d²x/dt²)i + (d²y/dt²)j。
For a particle with r(t) = (t² − 3t)i + (2t + 1)j, the velocity is v = (2t − 3)i + 2j, acceleration is a = 2i. You can then find, for instance, the speed at t = 2: |v| = √((1)² + 2²) = √5.
对于质点 r(t) = (t² − 3t)i + (2t + 1)j，速度为 v = (2t − 3)i + 2j，加速度为 a = 2i。接着你可以求出例如 t = 2 时的速率：|v| = √((1)² + 2²) = √5。
8. Projectile Motion | 抛体运动
Projectile motion with an initial speed u at an angle θ to the horizontal can be modelled by treating the horizontal and vertical components separately. Horizontally, acceleration is zero, so x = u cos θ t. Vertically, acceleration is −g, giving y = u sin θ t − ½ g t².
以初速率 u、与水平面夹角 θ 进行的抛体运动，可以通过将水平和竖直分量分开来建模。水平方向加速度为零，因此 x = u cos θ t。竖直方向加速度为 −g，得到 y = u sin θ t − ½ g t²。
Eliminating t yields the Cartesian equation of the trajectory:
y = x tan θ − (g x²) / (2 u² cos² θ)
消去 t 得到轨迹的直角坐标方程：
y = x tan θ − (g x²) / (2 u² cos² θ)
Key results include the time of flight T
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