📚 A-Level Physics: Kinematic Equations Explained | A-Level 物理：运动学方程详解
Kinematics is the study of motion without considering the forces that cause it. In CIE A-Level Physics, the constant-acceleration equations, often called the suvat equations, are essential tools for solving motion problems in one dimension. This article explains each equation, how it is derived, and how to apply it correctly in exam situations.
运动学研究物体的运动，而不涉及引起运动的力。在 CIE A-Level 物理中，匀加速运动方程通常被称为 suvat 方程，是解决一维运动问题的核心工具。本文将逐一讲解每个方程的推导过程以及如何在考试中正确运用。
1. Displacement, Velocity and Acceleration | 位移、速度与加速度
Displacement is a vector quantity measured in metres (m). It describes the straight-line distance from a starting point in a specific direction. Distance, by contrast, is a scalar and has no direction. Velocity is the rate of change of displacement and is measured in m/s. Acceleration is the rate of change of velocity and is measured in m/s².
位移是矢量，单位是米（m），表示物体从起点沿特定方向到终点的直线距离。相比之下，距离是标量，没有方向。速度是位移的变化率，单位是 m/s。加速度是速度的变化率，单位是 m/s²。
Average acceleration can be written as the change in velocity divided by the time taken:
平均加速度可以写成速度变化量除以所用时间：
a = (v – u) / t
This relationship is the starting point for deriving the first equation of motion.
这一关系是推导第一个运动方程的起点。
2. When Can We Use These Equations? | 这些方程的适用条件
The suvat equations are only valid when acceleration is constant, or uniform. This means the acceleration must not change during the time interval being considered. The equations also assume straight-line motion in one dimension.
suvat 方程仅在加速度恒定（即匀加速）时才成立。这意味着在所研究的时间间隔内，加速度不能发生变化。这些方程还假设物体沿直线做一维运动。
If an object moves with changing acceleration, you must either split the motion into smaller intervals with approximately constant acceleration or use graphical and calculus methods. In examination questions, look for phrases such as “uniform acceleration”, “constant acceleration” or “assuming air resistance is negligible”.
如果物体的加速度发生变化，则必须将运动分割为若干个加速度近似恒定的阶段，或者使用图象法和微积分方法。在考试题目中，请注意“匀加速”“恒定加速度”或“忽略空气阻力”等关键词。
3. The First Equation: v = u + at | 第一方程：v = u + at
Starting from the definition of average acceleration, a = (v – u) / t, we can rearrange to make v the subject. Multiplying both sides by t and adding u gives the first equation of motion.
从平均加速度的定义 a = (v – u) / t 出发，将 v 变为公式的主项。两边乘以 t，然后加上 u，就得到第一个运动方程。
v = u + at
Here, u is the initial velocity, v is the final velocity, a is the constant acceleration and t is the time taken. This equation is most useful when you need to find the final velocity after a known time.
其中，u 是初速度，v 是末速度，a 是恒定加速度，t 是经过的时间。当已知时间而要求末速度时，这个方程最常使用。
4. The Second Equation: s = ut + ½at² | 第二方程：s = ut + ½at²
The displacement s can be found from the area under a velocity-time graph. For constant acceleration, the area is made up of a rectangle of area u × t and a triangle of area ½ × t × (v – u). Since v – u = at, the triangle has area ½ × t × at = ½at².
位移 s 可以通过速度-时间图象下的面积求得。在匀加速运动中，该面积由一个矩形和一个三角形组成：矩形的面积为 u × t，三角形的面积为 ½ × t × (v – u)。因为 v – u = at，所以三角形的面积为 ½ × t × at = ½at²。
s = ut + ½at²
This equation is ideal when the time t is known and the final velocity is not needed.
当已知时间 t 且不需要末速度时，这个方程非常适用。
5. The Third Equation: v² = u² + 2as | 第三方程：v² = u² + 2as
This equation is derived by eliminating t from the first two equations. From v = u + at, we obtain t = (v – u) / a. Substituting this into s = ut + ½at² and simplifying gives a time-independent equation.
这个方程通过消去前两个方程中的时间 t 推导而来。由 v = u + at 可得 t = (v – u) / a。将其代入 s = ut + ½at² 并化简，就得到一个与时间无关的方程。
v² = u² + 2as
This is particularly useful when the value of t is neither given nor required, as in many vertical projection problems.
当题目没有给出时间 t，也不需要求时间 t 时，这个方程尤其好用，例如许多竖直抛体问题。
6. Average Velocity and Graphical Meaning | 平均速度与图象意义
When acceleration is constant, the average velocity over a time interval is exactly halfway between the initial and final velocities. Therefore, displacement can be written as the average velocity multiplied by the time taken.
当加速度恒定时，一段时间内的平均速度正好等于初速度和末速度的平均值。因此，位移可以写成平均速度乘以时间。
s = ½(u + v)t
This equation is also equal to the area under the velocity-time graph. In a velocity-time graph, the gradient represents acceleration, and the area under the graph represents displacement. A displacement-time graph has a gradient that represents velocity.
这个方程也等于速度-时间图象下的面积。在速度-时间图中，斜率表示加速度，图象下方的面积表示位移。在位移-时间图中，斜率表示速度。
7. The Five Key Symbols and Units | 五个关键符号与单位
| Quantity / 物理量 | Symbol / 符号 | Unit / 单位 | Notes / 说明 |
| Displacement / 位移 | s | m | Vector quantity / 矢量 |
| Initial velocity / 初速度 | u | m/s | Velocity at t = 0 / 在 t = 0 时的速度 |
| Final velocity / 末速度 | v | m/s | Velocity at time t / 在时刻 t 的速度 |
| Acceleration / 加速度 | a | m
Published by TutorHao | A-Level Physics Revision Series | aleveler.com 更多咨询请联系16621398022（同微信） CommentsMore posts |
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导