📚 Kinematics Key Points Revision | 运动学考点精讲
Kinematics is the branch of mathematics that describes motion in terms of displacement, velocity, and acceleration without considering the forces that cause it. In both IB Mathematics and CIE A Level Mathematics, kinematics forms a crucial part of the calculus application syllabus, often appearing in Paper 2 or Paper 4 (Mechanics). Mastering the core definitions, graph interpretations, constant acceleration suvat equations, and the calculus link between displacement, velocity and acceleration is essential for success. This article covers all major topics tested in IB Analysis & Approaches, IB Applications & Interpretation, and CIE 9709 Mechanics, providing bilingual explanations with clear examples.
运动学是数学的一个分支，用位移、速度和加速度来描述运动，而不考虑产生运动的力。在 IB 数学和 CIE A Level 数学中，运动学是微积分应用的重要部分，常出现在试卷二或试卷四（力学）中。掌握核心定义、图像解读、匀加速 suvat 方程以及位移、速度、加速度之间的微积分关系是取得高分的关键。本文全面覆盖 IB 分析数学、IB 应用数学和 CIE 9709 力学中考查的所有重点，以中英双语提供清晰讲解。
1. Displacement, Velocity and Acceleration | 位移、速度和加速度
Displacement (s or x) is a vector quantity measuring the change in position from the origin. It can be positive or negative depending on direction. Distance, on the other hand, is a scalar total path length.
位移（s 或 x）是矢量，衡量从原点开始位置的变化量，有正负取决于方向。而路程是标量，表示运动轨迹的总长度。
Velocity v is the rate of change of displacement with respect to time t. It is the first derivative: v = ds/dt. Speed is the magnitude of velocity, always non-negative.
速度 v 是位移对时间 t 的变化率，即一阶导数：v = ds/dt。速率是速度的大小，始终为非负数。
Acceleration a is the rate of change of velocity with respect to time: a = dv/dt = d²s/dt². If acceleration has the same sign as velocity, the object speeds up; opposite signs indicate slowing down.
加速度 a 是速度对时间的变化率：a = dv/dt = d²s/dt²。若加速度与速度同号，物体加速；异号则表示减速。
2. Interpreting Displacement-Time Graphs | 位移-时间图像解析
The gradient of a displacement-time (s-t) graph gives the instantaneous velocity. A straight line represents constant velocity, while a curve indicates acceleration: the tangent at any point shows velocity at that instant.
位移-时间 (s-t) 图像的斜率给出瞬时速度。直线表示匀速运动，曲线表示有加速度；某点处切线的斜率代表该时刻的瞬时速度。
When the s-t graph is horizontal, the object is stationary. Turning points occur where the gradient changes sign – this corresponds to the object reversing direction.
当 s-t 图像为水平线时，物体静止。拐点出现在斜率改变符号的位置，这对应物体运动方向反转。
An upward-sloping curve getting steeper means positive velocity and positive acceleration. A curve that is levelling off shows velocity approaching zero.
上坡且越来越陡的曲线表示正速度和正加速度。曲线趋于平坦则速度趋近于零。
3. Interpreting Velocity-Time Graphs | 速度-时间图像解析
The gradient of a velocity-time (v-t) graph gives the instantaneous acceleration. A horizontal line represents constant velocity (zero acceleration).
速度-时间 (v-t) 图像的斜率给出瞬时加速度。水平线表示匀速运动（加速度为零）。
The area under a v-t graph between two times gives the displacement (or change in position). Be careful with areas below the time axis – they represent negative displacement and must be subtracted if computing net displacement, but added as positive if computing total distance travelled.
v-t 图像下方与时间轴围成的面积表示位移（位置变化量）。注意时间轴下方的面积表示负位移；计算净位移时要减去，计算总路程时要取绝对值相加。
In IB and CIE exams, students often need to sketch or interpret piecewise linear v-t graphs and calculate total distance from a velocity function.
在 IB 和 CIE 考试中，学生常需绘制或解读分段线性的 v-t 图，并根据速度函数计算总路程。
4. Constant Acceleration Formulae (SUVAT) | 匀加速运动公式 (SUVAT)
For motion in a straight line with constant acceleration a, initial velocity u, final velocity v, displacement s, and time t, the following five equations (often called suvat) are used. Only four are independent; s = ½(u+v)t follows from the others.
对于加速度 a 恒定的直线运动，设初速度为 u，末速度为 v，位移为 s，时间为 t，可使用以下五个方程（常称 SUVAT）。其中只有四个独立，s = ½(u+v)t 可由其他推导。
v = u + at s = ut + ½at² v² = u² + 2as s = ½(u + v)t s = vt − ½at²
Choose the equation that contains the known three quantities and the one unknown. Always take care with signs: decide a positive direction first and stick to it. In vertical motion under gravity, a = ±g (often 9.8 m/s² or 9.81 m/s²).
选择包含三个已知量和一个未知量的方程。务必注意符号：先规定正方向并保持一致。在重力下的垂直运动中，a = ±g（通常取 9.8 m/s² 或 9.81 m/s²）。
5. Calculus in Kinematics: Variable Acceleration | 运动学中的微积分：变加速运动
When acceleration is not constant, we use differentiation and integration to connect displacement, velocity and acceleration. Given displacement s(t), v(t)=s'(t) and a(t)=v'(t)=s”(t).
当加速度不恒定时，我们用微分和积分联系位移、速度和加速度。已知位移 s(t)，则 v(t)=s'(t)，a(t)=v'(t)=s”(t)。
Conversely, given acceleration a(t) and initial conditions, velocity v(t) = ∫a(t)dt + C, and displacement s(t) = ∫v(t)dt + D. Boundary conditions are used to find the constants of integration.
反之，已知加速度 a(t) 及初始条件，则速度 v(t) = ∫a(t)dt + C，位移 s(t) = ∫v(t)dt + D。利用边界条件确定积分常数。
Typical exam problems give a = f(t), a = f(v), or a = f(s). For a function of v or s, use the chain rule: a = dv/dt = v dv/ds. This allows solving problems where acceleration depends on position.
典型考题会给出 a = f(t)、a = f(v) 或 a = f(s)。若为 v 或 s 的函数，用链式法则：a = dv/dt = v dv/ds，从而求解加速度依赖位置的问题。
6. Vertical Motion under Gravity | 重力作用下的垂直运动
When an object moves vertically near the Earth’s surface with negligible air resistance, it experiences constant downward acceleration g (≈ 9.8 m/s² or 9.81 m/s²). All suvat equations apply directly if the sign of a is set to −g when upward is positive.
当物体在忽略空气阻力的情况下于地表附近垂直运动时，受恒定向下的重力加速度 g（≈ 9.8 m/s² 或 9.81 m/s²）。若规定向上为正，取 a = −g，所有 suvat 公式可直接使用。
Common examples include: free fall from rest (u=0), object thrown vertically upward (velocity zero at highest point), and object projected downward with initial speed. Time to maximum height is found from v=0, and total flight time is often twice the time to the peak when launch and landing levels are the same.
常见例子包括：自由落体 (u=0)、竖直上抛（最高点速度为零）和竖直下抛。利用 v=0 可求出到达最高点的时间；若起落点在同一水平面，总飞行时间常为上升时间的两倍。
In CIE Mechanics questions, you may need to combine the constant acceleration equations for two parts of motion, e.g. upwards and then downwards.
在 CIE 力学考题中，可能需要将运动分为上升和下降两段分别应用匀加速方程。
7. Relative Velocity in One Dimension | 一维相对速度
Relative velocity describes the motion of one object from the perspective of another moving observer. The relative velocity of A with respect to B is vA/B = vA − vB (as vectors, considering sign).
相对速度描述从另一个运动观察者眼中看某个物体的运动。A 相对于 B 的速度为 vA/B = vA − vB（矢量，考虑正负号）。
This concept is used to determine overtaking times, closing speeds, or when two objects meet. In one dimension, assign proper signs for direction and solve linear equations.
这一概念用于确定超车时间、接近速度或两物体相遇的时刻。在一维问题中，正确分配方向符号并求解线性方程即可。
CIE P4 frequently includes a relative motion problem, while IB may ask for overtaking scenarios using motion equations.
CIE 试卷四经常包含相对运动问题，IB 则可能通过运动方程考查超车场景。
8. Projectile Motion (2D Kinematics) | 抛体运动（二维运动学）
Projectile motion with initial speed u at angle θ to the horizontal can be modelled by treating horizontal and vertical components independently. Horizontal velocity ux = u cosθ is constant; vertical acceleration is ay = −g.
初速度大小为 u、与水平面夹角为 θ 的抛体运动，可将水平和竖直分量独立处理。水平速度 ux = u cosθ 恒定；竖直加速度 ay = −g。
Position equations: x = u cosθ · t, y = u sinθ · t − ½gt². Eliminating t gives the trajectory equation: y = x tanθ − (g/(2u²cos²θ)) x², which is a parabola.
位置方程：x = u cosθ · t，y = u sinθ · t − ½gt²。消去 t 得到轨迹方程：y = x tanθ − (g/(2u²cos²θ)) x²，为抛物线。
Time of flight, maximum height, and horizontal range are derived from vertical motion. Range R = (u² sin2θ)/g, max height H = (u² sin²θ)/(2g). These formulae are directly tested in IB AI HL and CIE Mechanics.
飞行时间、最大高度和水平射程均由竖直运动推导。射程 R = (u² sin2θ)/g，最大高度 H = (u² sin²θ)/(2g)。这些公式在 IB AI HL 和 CIE 力学中直接考查。
9. Connected Rates and Differential Equations | 关联变化率与微分方程
In some motion problems, the velocity or acceleration is given as a function of displacement, leading to a differential equation. For instance, a = k√v or a = −ω²x. These are solved by separating variables or recognising simple harmonic motion (SHM).
在某些运动问题中，速度或加速度作为位移的函数给出，从而形成微分方程。例如 a = k√v 或 a = −ω²x。可通过分离变量或识别简谐运动 (SHM) 来求解。
Using a = v dv/dx, we can set up an integral to find v(x) directly. IB Analysis & Approaches HL often includes questions requiring the use of dv/dx to link velocity and position.
利用 a = v dv/dx，可直接建立积分求得 v(x)。IB 分析数学 HL 常包含需要使用 dv/dx 联系速度与位置的问题。
When solving, always include the constants of integration and use initial conditions to match the specific scenario described.
求解时务必添加积分常数，并利用初始条件代入特定情境。
10. Exam Tips and Common Mistakes | 考试技巧与常见错误
Always define your positive direction clearly before writing equations. Mixing signs is the most frequent error in kinematics.
在列出方程前务必明确正方向。正负号混淆是运动学中最常见的错误。
For graph questions, remember: gradient of s-t gives v; gradient of v-t gives a; area under v-t gives displacement. Do not confuse area with gradient.
对于图像题，牢记：s-t 斜率得 v；v-t 斜率得 a；v-t 下方面积得位移。不要把面积和斜率弄混。
When using suvat, check that acceleration truly is constant. If acceleration varies, you must apply calculus or the chain rule a = v dv/ds.
使用 suvat 时，确保加速度确实恒定。若加速度变化，必须使用微积分或链式法则 a = v dv/ds。
In projectile problems, write two independent sets of suvat equations for horizontal and vertical motion. Horizontal motion is constant speed; vertical motion is constant acceleration. Time t is the common link.
在抛体问题中，分别写出水平与竖直方向的两组独立 suvat 方程。水平方向为匀速；竖直方向为匀加速。时间 t 是连接两方向的纽带。
For CIE Mechanics paper, always include units in final answers; give exact forms or three significant figures unless otherwise instructed.
在 CIE 力学试卷中，最终答案始终包含单位；除非特别说明，答案保留精确形式或三位有效数字。
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