📚 The Equations of Motion | 运动学方程
The equations of motion, often called SUVAT equations, are the bedrock of kinematics in A-Level Physics. They describe how an object moves in a straight line under constant acceleration, linking displacement, initial velocity, final velocity, acceleration, and time. Mastering these four equations allows you to solve a vast range of problems, from free-falling objects to braking vehicles.
运动学方程（通常称为 SUVAT 方程）是 A-Level 物理运动学的基石。它们描述物体在恒定加速度下如何沿直线运动，将位移、初速度、末速度、加速度和时间联系起来。掌握这四个方程，你就能解决从自由落体到车辆刹车等大量问题。
1. Introduction to the Equations of Motion | 运动学方程简介
Kinematics is the branch of mechanics that deals with the description of motion without considering its causes. The equations of motion provide a mathematical framework for uniformly accelerated linear motion. In CIE A-Level Physics, you are expected to recall and apply the four standard equations, often referred to by the variables they contain: s (displacement), u (initial velocity), v (final velocity), a (acceleration), and t (time).
运动学是力学中一个分支，仅描述运动而不考虑其成因。运动学方程为匀加速直线运动提供了数学框架。在 CIE A-Level 物理考试中，你需要记住并应用四个标准方程，通常以它们包含的变量命名：s（位移）、u（初速度）、v（末速度）、a（加速度）和 t（时间）。
2. Key Assumptions in Kinematics | 运动学关键假设
All four equations are valid only under two strict conditions: the acceleration is constant in magnitude and direction, and the motion occurs along a straight line. This means they apply to situations such as free fall near the Earth's surface (ignoring air resistance), a car accelerating uniformly, or a train braking with constant deceleration. If acceleration changes with time, these equations cannot be used directly.
这四个方程仅在两个严格条件下成立：加速度的大小和方向恒定，且运动沿直线进行。这意味着它们适用于诸如地表附近的自由落体（忽略空气阻力）、汽车匀加速或列车匀减速制动等情况。如果加速度随时间变化，就不能直接使用这些方程。
3. Defining the Five Fundamental Variables | 定义五个基本变量
Before applying the equations, it is essential to understand each symbol precisely. In the CIE syllabus, vector quantities are approached with sign conventions to indicate direction. The five variables are:
在应用方程之前，必须准确理解每个符号。在 CIE 大纲中，矢量量使用符号法则来表示方向。这五个变量是：
s – displacement (m): the overall change in position from start to finish, a vector that may be positive or negative depending on the chosen direction.
s – 位移（米）：从起点到终点的位置总变化量，是矢量，根据所选方向可为正或负。
u – initial velocity (m s⁻¹): the velocity at time t = 0.
u – 初速度（米/秒）：在时间 t=0 时的速度。
v – final velocity (m s⁻¹): the velocity at some later time t.
v – 末速度（米/秒）：在稍后时间 t 时的速度。
a – acceleration (m s⁻²): the constant rate of change of velocity. It can be positive (speeding up in the positive direction) or negative (slowing down or reversing).
a – 加速度（米/秒²）：速度恒定的变化率。可为正（沿正方向加速）或为负（减速或反向）。
t – time interval (s): the time elapsed between the initial and final state.
t – 时间间隔（秒）：从初状态到末状态所经过的时间。
4. The First Equation: v = u + at | 第一方程：v = u + at
This equation is derived directly from the definition of constant acceleration. It states that the final velocity equals the initial velocity plus the acceleration multiplied by time. It is the only equation that does not involve displacement s.
这个方程直接来自恒定加速度的定义。它表明末速度等于初速度加上加速度与时间的乘积。它是唯一不涉及位移 s 的方程。
v = u + at
You would use this equation when time is known, or when you need to find the velocity after a given time, provided displacement is not required. For example, a car accelerating at 2 m s⁻² from 5 m s⁻¹ will reach a speed of 13 m s⁻¹ after 4 seconds.
当时间已知，或者需要求某段时间后的速度但不需要位移时，可使用此方程。例如，一辆汽车以 2 m s⁻² 的加速度从 5 m s⁻¹ 加速，4 秒后速度将达到 13 m s⁻¹。
5. The Second Equation: s = ut + ½at² | 第二方程：s = ut + ½at²
The second equation gives the displacement as a function of initial velocity, time, and acceleration. It arises from the area under a velocity-time graph for constant acceleration and does not contain final velocity v.
第二个方程将位移表示为初速度、时间和加速度的函数。它源自匀加速情况下速度-时间图下的面积，不含末速度 v。
s = ut + ½at²
This is the go-to equation when calculating how far an object travels during an acceleration phase. Note the ½ factor, which reminds us that the additional displacement due to acceleration grows quadratically with time.
这是计算物体加速期间行驶距离的首选方程。注意因子 ½，它提醒我们由加速度引起的额外位移随时间呈二次关系增长。
6. The Third Equation: s = ½(u + v)t | 第三方程：s = ½(u + v)t
This equation shows that displacement is the product of average velocity and time when acceleration is constant. The expression ½(u + v) is simply the arithmetic mean of the initial and final velocities, which equals the average velocity for uniform acceleration. It does not involve acceleration a.
此方程表明，当加速度恒定时，位移等于平均速度与时间的乘积。表达式 ½(u + v) 就是初速度与末速度的算术平均值，对于匀加速而言，这就是平均速度。该方程不涉及加速度 a。
s = ½(u + v) t
Use this equation when acceleration is unknown but you have the initial and final velocities and the time. It is particularly handy for problems where the duration is given and the motion is symmetrical, such as an object thrown upwards returning to its launch height.
当加速度未知，但已知初速度、末速度和时间时，使用此方程。对于给定的时长且运动对称的问题（如抛体上升后返回到起始高度），这个方程特别方便。
7. The Fourth Equation: v² = u² + 2as | 第四方程：v² = u² + 2as
The fourth equation links the squares of velocities, acceleration, and displacement, completely omitting time t. It is extremely useful when the time interval is not given or not relevant, such as determining the speed of a car after travelling a certain distance while braking.
第四个方程将速度的平方、加速度和位移联系起来，完全不包含时间 t。当时间间隔未给出或不相关时，如求车辆刹车经过一定距离后的速度，该方程极为有用。
v² = u² + 2as
Always take care with signs: if a is negative (deceleration), the term 2as reduces the value of v². This equation also explains why the stopping distance of a car is proportional to the square of its initial speed.
务必注意符号：如果 a 为负（减速），项 2as 会减小 v² 的值。该方程也解释了为什么汽车的制动距离与其初速度的平方成正比。
| Equation | Missing Variable |
|---|---|
| v = u + at | s |
| s = ut + ½at² | v |
| s = ½(u + v)t | a |
| v² = u² + 2as | t |
8. Deriving the Equations from a Velocity-Time Graph | 从速度-时间图推导方程
ual to a, ending at (t, v). The area under the graph gives displacement s. The area can be split into a rectangle (u × t) and a triangle (½ × t × (v - u)). Since v - u = at, the triangle' cal understanding reinforces the deep connection between gradient (acceleration) and area (displacement).
所有四个方程都可以用一条直线的速度-时间图来可视化和推导。该直线始于 (0, u)，恒定梯度等于 a，终于 (t, v)。图下面积给出位移 s。面积可分为矩形 (u × t) 和三角形 (½ × t × (v - u))。由于 v - u = at，三角形面积变为 ½at²，得到 s = ut + ½at²。对 u 和 v 取平均可得 s = ½(u + v)t，结合 v = u + at 进行整理，就能推出其余方程。这种图形理解巩固了梯度（加速度）与面积（位移）之间的深层联系。
9. Sign Conventions and Vector Nature | 符号约定与矢量性质
ally, you choose a reference direction, for example upward positive for a thrown ball. If upward is positive, then acceleration due to gravity is g = -9.81 m s⁻². A negative acceleration can mean either slowing down in the positive direction or speeding up in the negative direction. Confusing sign conventions is one of the most common errors in using the equations of motion. Always define your positive direction at the start and apply signs consistently.
位移、速度和加速度都是矢量。在一维运动中，我们用正/负号表示方向。通常，你要选定一个参考方向，例如向上抛球时取向上为正。如果向上为正，重力加速度就是 g = -9.81 m s⁻²。负加速度可能意味着沿着正方向减速，或者是沿着负方向加速。混淆符号约定是使用运动学方程时最常见的错误之一。务必从一开始就确定正方向，并始终一致地使用符号。
10. Problem-Solving Framework | 解题框架
A systematic approach will help you avoid mistakes. First, list the five SUVAT quantities and write down the values you know, adding a question mark for the unknown. Second, choose the positive direction and assign signs to all vector quantities. Third, determine which equation contains the unknown and uses only the known variables. Fourth, rearrange the equation algebraically before inserting numbers. Fifth, substitute the values with units and calculate the unknown. Finally, check that your answer is physically reasonable and expressed with the correct units and direction.
系统的方法能帮助你避免错误。首先，列出五个 SUVAT 量，写下已知的值，对未知量标上问号。其次，选择正方向并为所有矢量赋上符号。第三，判断哪个方程既包含未知量又仅使用已知变量。第四，先对方程进行代数变形，再代入数字。第五，代入带单位的数值，计算未知量。最后，检查答案在物理上是否合理，是否用正确的单位和方向表达。
11. Worked Examples | 典型例题
Example 1: A car accelerates from rest at 3.0 m s⁻² for 8.0 s. Find the distance travelled.
例题1：一辆汽车从静止以 3.0 m s⁻² 加速 8.0 秒。求行驶距离。
Known: u = 0, a = 3.0 m s⁻², t = 8.0 s, s = ?. Missing v, so use s = ut + ½at². s = 0 + ½ × 3.0 × (8.0)² = 96 m. The car covers 96 m.
已知：u = 0，a = 3.0 m s⁻²，t = 8.0 s，s = ?。缺少 v，因此使用 s = ut + ½at²。s = 0 + ½ × 3.0 × (8.0)² = 96 m。汽车行驶了 96 m。
Example 2: vertically upwards with 20 m s⁻¹. Take g = 9.81 m s⁻² downwards. Find the maximum height.
例题2：一个球以 20 m s⁻¹ 竖直向上抛出。取向下为 g = 9.81 m s⁻²。求最大高度。
e upwards as positive. Then u = +20 m s⁻¹, a = -9.81 m s⁻², v = 0 (at maximum height), s = ? = (20)² + 2 × (-9.81) × s → 0 = 400 - 19.62 s → s = 20.4 m. The maximum height is 20.4 m.
选向上为正。则 u = +20 m s⁻¹，a = -9.81 m s⁻²，v = 0（在最高点），s = ?。使用 v² = u² + 2as。0 = 20² + 2 × (-9.81) × s → 0 = 400 - 19.62 s → s = 20.4 m。最大高度为 20.4 m。
Example 3: A train travelling at 30 m s⁻¹ decelerates uniformly at 1.5 m s⁻². How far does it travel before stopping?
例题3：一列火车以 30 m s⁻¹ 行驶，以 1.5 m s⁻² 均匀减速。停止前行驶多远？
Take direction of motion as positive: u = 30 m s⁻¹, v = 0, a = -1.5 m s⁻², s = ?. Missing t, so v² = u² + 2as → 0 = 900 + 2(-1.5)s → s = 300 m. The train travels 300 m.
取运动方向为正：u = 30 m s⁻¹，v = 0，a = -1.5 m s⁻²，s = ?。缺少 t，因此用 v² = u² + 2as → 0 = 900 + 2(-1.5)s → s = 300 m。火车行驶 300 m。
12. Common Mistakes and How to Avoid Them | 常见错误及避免方法
tive direction and stick to it. Another common slip is using the wrong equation when acceleration is zero— ½ factor correctly. Finally, check that you have converted all units to SI (metres, seconds, m s⁻¹, m s⁻²) before substituting. Practising with a consistent list of SUVAT quantities will steadily build your confidence.
一个常见错误是在抛体问题中将位移和速度的符号弄混。务必选定正方向并坚持到底。另一个常见疏忽是当加速度为零时——运动是匀速的，应使用 s = vt 或 s = ut——却选错方程。学生还经常在 s = ut + ½at² 中忘记将时间平方，或者没有正确使用 ½ 因子。最后，代数值前要检查是否已将所有单位转换为国际单位（米、秒、m s⁻¹、m s⁻²）。按照 SUVAT 量表格进行稳定练习，能逐步建立你的信心。
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