📚 A-Level Maths: Constant Acceleration Formulae | A-Level数学：匀加速运动公式
In mechanics, one of the first mathematical models we meet is the motion of a particle moving in a straight line with constant acceleration. This simple assumption lets us describe the motion completely using five key quantities: displacement, initial velocity, final velocity, acceleration and time. These are commonly known as the SUVAT equations.
在力学中，我们最先接触的数学模型之一，就是质点在直线上做匀加速运动的情况。通过“加速度恒定”这一假设，我们可以用五个关键量完整描述运动：位移、初速度、末速度、加速度和时间。这些量之间的关系常被称为SUVAT公式。
The SUVAT equations are central to the A-Level mathematics mechanics modules and to many examination questions. Once you understand where they come from and how to apply them safely, most constant acceleration problems become straightforward.
SUVAT公式是A-Level数学中力学模块的核心内容，也是许多考试题目的基础。一旦你理解了它们的来源，并且知道如何正确使用，大多数匀加速运动问题都会变得非常简单。
1. Introduction | 引言
Uniform acceleration means that the velocity of the object changes by the same amount in every equal interval of time. For example, in free fall near the Earth’s surface, if we ignore air resistance, the acceleration is approximately constant.
匀加速运动是指物体在任意相等的时间间隔内，速度的变化量都相同。例如，在地球表面附近的自由落体运动中，如果忽略空气阻力，加速度可以近似看作恒定值。
Because acceleration is constant, the velocity-time graph is a straight line. This straight-line graph gives us a powerful visual tool: its gradient is the acceleration, and the area under it is the displacement.
因为加速度恒定，速度-时间图像是一条直线。这条直线图像是一个非常有力的工具：它的斜率就是加速度，而图线下方的面积就是位移。
In this article, we will define the five variables, present the five SUVAT equations, show how to derive them, and work through realistic examples in the A-Level style.
在本文中，我们将定义五个变量，列出五条SUVAT公式，展示它们的推导过程，并按照A-Level考试的典型风格逐步解答实例。
2. The Five Suvat Variables | 五个基本变量
Before using any formula, you must know what each symbol represents. In most A-Level materials, the symbols below are used.
在使用任何公式之前，你必须知道每个符号代表什么。在大多数A-Level教材中，通常使用以下符号。
| Symbol 符号 | Quantity 物理量 | SI Unit 国际单位 |
| s | displacement 位移 | metres (m) 米 |
| u | initial velocity 初速度 | metres per second (m/s) 米每秒 |
| v | final velocity 末速度 | metres per second (m/s) 米每秒 |
| a | constant acceleration 匀加速度 | metres per second squared (m/s²) 米每二次方秒 |
| t | time interval 时间间隔 | seconds (s) 秒 |
The symbol s, not d, is used for displacement because it is a standard notation in mechanics and avoids confusion with the word “distance”.
位移不使用d而是使用s，这是因为s是力学中的标准记号，同时也能避免与英文单词“distance”混淆。
3. The Formulae | 公式汇总
The five SUVAT equations are all equivalent, and knowing them gives you a complete toolkit for constant acceleration problems.
这五条SUVAT公式彼此等价，掌握了它们，你就拥有了解决匀加速运动问题的完整工具包。
v = u + at
This equation is simply the definition of constant acceleration rearranged. It connects final velocity, initial velocity, acceleration and time.
这条公式其实就是匀加速度定义的变形，它联系了末速度、初速度、加速度和时间。
s = ½(u + v)t
Since acceleration is constant, average velocity is the midpoint of u and v. Displacement equals average velocity multiplied by time.
因为加速度恒定，平均速度等于u和v的中点值。位移等于平均速度乘以时间。
s = ut + ½at²
This version is useful when we do not know the final velocity v and need displacement directly.
当末速度v未知，而我们需要直接求位移时，这个形式非常有用。
s = vt – ½at²
This version is useful when the initial velocity u is unknown. It can be obtained from s = ut + ½at² by replacing u with v – at.
当初速度u未知时，这个形式很有用。它由 s = ut + ½at² 将u替换为 v – at 后得到。
v² = u² + 2as
This formula is ideal when time t is not given and not required.
当时间t没有给出，也无需计算时，这条公式是最佳选择。
4. Deriving the Equations | 公式推导
Understanding the derivation helps you remember the formulas and see why they are true.
理解推导过程有助于你记忆公式，也能让你明白它们为什么成立。
Start from the definition of acceleration. For constant acceleration,
首先从加速度的定义出发。对于匀加速运动，
a = (v – u) / t
Multiplying both sides by t gives the first equation.
两边同时乘以t，就得到第一条公式。
v = u + at
Next, because the velocity changes linearly, the average velocity is (u + v)/2. Therefore displacement is
接着，因为速度随时间线性变化，平均速度等于 (u + v)/2。因此位移为
s = ½(u + v)t
Now substitute v = u + at into this equation:
现在将 v = u + at 代入上式：
s = ½(u + u + at)t = ut + ½at²
Finally, eliminate t from the first two equations. From v = u + at, we get t = (v – u)/a. Substituting into s = ½(u + v)t gives
最后，从前两个方程中消去t。由 v = u + at 可得 t = (v – u)/a，将其代入 s = ½(u + v)t，得到
v² = u² + 2as
Thus all five equations are connected; you really only need two or three of them, but it is safer to remember all five.
由此可见，五条公式相互关联；实际上你只需要两到三条就足够了，但记住全部五条会更稳妥。
5. When Can We Use Suvat? | 适用条件
The SUVAT equations are not universal. They may only be used when the following conditions are satisfied.
SUVAT公式并非万能的，只有在满足以下条件时才能使用。
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Acceleration must be constant.
加速度必须是恒定值。
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Motion must take place along a straight line, or along a single fixed direction with signs.
运动必须沿直线发生，或者沿单一固定方向并配合正负号处理。
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The object is treated as a particle, so rotation and internal shape are ignored.
物体被看作质点，因此不考虑转动和物体本身的形状。
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All quantities are measured over the same time interval, and in consistent units.
所有物理量必须在同一个时间间隔内测量，并且使用一致的单位。
If acceleration changes during the motion, for example when a rocket’s thrust changes, you must split the motion into intervals where acceleration is constant, or use calculus instead.
如果运动过程中加速度发生变化，例如火箭推力改变，你就必须把运动分成若干加速度恒定的阶段，或者改用微积分方法。
6. Choosing the Right Equation | 如何选择正确的公式
In an exam problem, you will normally be given three quantities and asked to find a fourth. The key is to identify the unknown, then choose the equation that contains the unknown and does not contain the extra fifth quantity.
在考试题目中，通常会给三个量，要求你求第四个量。关键是先明确未知量，然后选择包含该未知量、同时不包含另一个多余量的公式。
A useful decision table is shown below.
下面是一个很有用的选择表。
| Known 已知量 | Want 待求量 | Equation to use 应使用的公式 |
| u, a, t | v | v = u + at |
| u, v, t | s | s = ½(u + v)t |
| u, a, t | s | s = ut + ½at² |
| v, a, t | s | s = vt – ½at² |
| u, v, a | s | v² = u² + 2as |
| u, v, a | t | v = u + at |
Always start by writing down every quantity with its value and sign. This small habit prevents many careless errors.
做题时一定要先把每个量连同正负号写出来。这个小小的习惯能避免许多粗心错误。
7. Worked Example 1: Free Fall | 例题1：自由落体
A ball is dropped from rest from a high building. Find the distance it falls in the first 3 seconds and its speed after 3 seconds. Take g = 9.8 m/s² and downwards as positive.
一个小球从高楼顶端由静止开始下落。求它在前3秒内下落的距离，以及3秒后的速度。取 g = 9.8 m/s²，以下方向为正方向。
List the known values:
列出已知量：
u = 0, a = 9.8 m/s², t = 3 s
To find displacement s, use the equation without v.
要求位移s，需要使用不含v的公式。
s = ut + ½at²
s = 0 × 3 + ½ × 9.8 × 3² = 44.1 m
To find the final speed, use v = u + at.
要求末速度，使用 v = u + at。
v = 0 + 9.8 × 3 = 29.4 m/s
So the ball falls 44.1 metres and reaches a speed of 29.4 m/s.
因此小球下落了44.1米，末速度达到29.4米/秒。
8. Worked Example 2: Braking Distance | 例题2：刹车距离
A car is travelling at 20 m/s when the driver applies the brakes. The car decelerates uniformly at 4 m/s². Find the distance travelled before it stops, and the time taken.
一辆汽车以20米/秒的速度行驶，司机踩下刹车。汽车以4米/秒²的加速度均匀减速。求汽车停止前滑行的距离，以及所需时间。
Since the car is slowing down, the acceleration is negative if we take the direction of motion as positive.
因为汽车在减速，如果我们取运动方向为正，那么加速度应为负值。
u = 20 m/s, v = 0, a = -4 m/s²
Time t is not mentioned, so use v² = u² + 2as to find s.
题目没有提到时间t，所以用 v² = u² + 2as 来求s。
0² = 20² + 2 × (-4) × s
0 = 400 – 8s ⇒ s = 50 m
Now use v = u + at to find the stopping time.
再用 v = u + at 求停止所需时间。
0 = 20 + (-4)t ⇒ t = 5 s
The braking distance is 50 metres and it takes 5 seconds to stop.
刹车距离为50米，需要5秒才能停止。
9. Worked Example 3: Vertical Projection | 例题3：竖直上抛
A ball is thrown vertically upwards with an initial speed of 14.7 m/s. Find the maximum height reached and the total time before it returns to the throwing point. Take g = 9.8 m/s² and upwards as positive.
一个小球以14.7米/秒的初速度竖直向上抛出。求它达到的最大高度，以及返回抛出点所需的总时间。取 g = 9.8 m/s²，向上为正方向。
At maximum height, the ball is momentarily at rest, so v = 0. We know:
在最高点，小球瞬间静止，所以 v = 0。已知：
u = 14.7 m/s, v = 0, a = -9.8 m/s²
Use v² = u² + 2as to find the maximum height s.
用 v² = u² + 2as 求最大高度s。
0² = 14.7² + 2 × (-9.8) × s
s = 14.7² / (2 × 9.8) = 11.025 m
To find total time of flight, use s = ut + ½at² with s = 0.
要求总飞行时间，使用 s = ut + ½at²，并令 s = 0。
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