📚 Constant Acceleration Formulae 1 | 匀加速运动公式（一）
Welcome to the first part of mastering constant acceleration, the cornerstone of kinematics in Edexcel A‑Level Mathematics. Here we will explore the four fundamental equations of motion — often called SUVAT — which describe uniformly accelerated motion in a straight line. From deriving each formula to applying them in real‑world contexts, you will learn to choose the right equation, avoid common pitfalls, and interpret displacement, velocity, acceleration, and time with confidence. Let’s get started.
欢迎来到匀加速运动的第一部分，这是 Edexcel 爱德思 A‑Level 数学中运动学的基石。在这里，我们将探讨四个基本的运动方程——通常称为 SUVAT——它们描述了直线上的匀加速运动。从推导每个公式到在实际情境中应用，你将学会选择合适的方程、避开常见错误，并自信地解读位移、速度、加速度和时间。让我们开始吧。
1. Understanding the SUVAT Variables | 认识 SUVAT 变量
In kinematics problems with constant acceleration, we work with five core quantities: s – displacement (m), u – initial velocity (m s⁻¹), v – final velocity (m s⁻¹), a – constant acceleration (m s⁻²), and t – time (s). The mnemonic SUVAT helps you remember them. It is essential to define a positive direction at the start of every problem and apply sign conventions consistently to all vector quantities.
在匀加速运动问题中，我们使用五个核心物理量：s – 位移（米），u – 初速度（米/秒²），v – 末速度（米/秒²），a – 恒定加速度（米/秒²），以及 t – 时间（秒）。助记符 SUVAT 有助于记忆。在每个问题开始时，必须定义正方向，并始终对所有矢量使用一致的符号规则。
2. The First Equation: v = u + at | 第一方程式：v = u + at
This formula emerges directly from the definition of constant acceleration: acceleration is the rate of change of velocity. When a is constant, Δv = a × t, so v − u = at, leading to v = u + at. It does not involve displacement, making it ideal for problems where s is unknown or irrelevant.
该公式直接来自恒定加速度的定义：加速度是速度的变化率。当 a 恒定时，Δv = a × t，因此 v − u = at，得到 v = u + at。它不涉及位移，因此非常适合 s 未知或不相关的问题。
rates from rest at 3 m s⁻² for 8 seconds. Its final velocity is v = 0 + (3)(8) = 24 m s⁻¹.
例如，一辆汽车从静止以 3 米/秒² 加速 8 秒。它的末速度为 v = 0 + (3)(8) = 24 米/秒。
3. The Second Equation: s = ut + ½ at² | 第二方程式：s = ut + ½ at²
sing average velocity. Starting from v = u + at and integrating with respect to time gives s = ut + ½at². This equation is powerful when time is known but final velocity is not needed.
匀加速下的位移可以通过对速度函数积分或使用平均速度求得。从 v = u + at 开始，对时间积分得到 s = ut + ½at²。当已知时间但不需要末速度时，该方程非常有用。
with u = 0 and a = 3 m s⁻² for t = 8 s, the displacement is s = (0)(8) + ½(3)(8²) = 96 m.
如果同一辆汽车以 u = 0、a = 3 米/秒² 行驶 t = 8 秒，位移为 s = (0)(8) + ½(3)(8²) = 96 米。
4. The Third Equation: s = ½(u + v)t | 第三方程式：s = ½(u + v)t
Since acceleration is constant, the average velocity is simply the arithmetic mean of initial and final velocities: (u + v)/2. Multiplying by time yields displacement: s = ½(u + v)t. This formula omits acceleration, which is convenient when a is not part of the question or when you have u, v, and t given.
由于加速度恒定，平均速度就是初速度和末速度的算术平均值：(u + v)/2。乘以时间即得位移：s = ½(u + v)t。这个公式不含加速度，当 a 不是问题的一部分或已知 u、v 和 t 时非常方便。
Using the earlier values, s = ½(0 + 24)(8) = 96 m, which agrees with the previous result.
使用前面的值，s = ½(0 + 24)(8) = 96 米，与之前的结果一致。
5. The Fourth Equation: v² = u² + 2as | 第四方程式：v² = u² + 2as
Eliminating time t from v = u + at and s = ut + ½at² gives v² = u² + 2as. This time‑independent relation is perfect for scenarios where the duration of motion is unknown, such as braking distances or final velocity after moving a known displacement.
从 v = u + at 和 s = ut + ½at² 中消去时间 t，得到 v² = u² + 2as。这个不含时间的关系式非常适合运动持续时间未知的情况，例如制动距离或经过已知位移后的末速度。
5 m s⁻¹ over a distance of 30 m, v² = 5² + 2(2)(30) = 25 + 120 = 145, so v ≈ 12.04 m s⁻¹.
如果一名骑自行车的人以 2 米/秒² 的加速度从初速度 5 米/秒 行驶 30 米，v² = 5² + 2(2)(30) = 25 + 120 = 145，因此 v ≈ 12.04 米/秒。
6. Derivation from a Velocity–Time Graph | 从速度–时间图推导
All four SUVAT equations can be visualised on a v–t graph. For constant acceleration, the graph is a straight line with gradient a and intercept u. The area under the graph between t = 0 and t = t equals the displacement s height times base gives s = ½(u + v)t, and geometric manipulation leads to v² = u² + 2as.
所有四个 SUVAT 方程都可以在 v–t 图上直观显示。对于恒定加速度，图形是一条斜率为 a、截距为 u 的直线。t = 0 到 t = t 之间图线下方的面积等于位移 s。该面积可分解为一个矩形（ut）和一个三角形（½ × t × at），得到 s = ut + ½at²。斜率给出 a = (v − u)/t，因此 v = u + at。平均高度乘以底边得到 s = ½(u + v)t，几何推导得出 v² = u² + 2as。
7. Choosing the Right Formula: SUVAT Selection Guide | 选择正确的公式：SUVAT 选择指南
When solving a problem, list the five SUVAT variables and identify which three are known and which one is required. The equation that connects these four quantities (and ignores the fifth) is the one to use. Avoid forcing a formula that includes an unknown you don't actually need to find.
解题时，列出五个 SUVAT 变量，确定已知的三个和需要求的一个。连接这四个量（并忽略第五个）的方程就是要用的。不要强行使用包含你实际上不需要求的未知量的公式。
| Missing variable | Equation to use |
| s | v = u + at |
| u | s = vt − ½at² |
| v | s = ut + ½at² |
| a | s = ½(u + v)t |
| t | v² = u² + 2as |
如果缺少 s，使用 v = u + at；缺少 u 时用 s = vt − ½at²；等等。务必确保代数值时单位一致，并在使用前将所有数值转换为 SI 单位。
8. Common Sign Errors and How to Avoid Them | 常见符号错误及如何避免
Always draw a clear diagram with a defined positive direction. Quantities acting in the positive direction are positive; those opposite are negative. Deceleration is acceleration in the negative direction, so insert a as negative. For vertical motion, taking up as positive makes gravitational acceleration g = −9.8 m s⁻². Ignoring signs is the most frequent mistake in SUVAT problems.
始终画一个清晰的图，标明正方向。沿正方向的量为正；相反方向的量为负。减速是负方向的加速度，因此将 a 代入负值。对于竖直运动，取向上为正，重力加速度 g = −9.8 米/秒²。忽略符号是 SUVAT 问题中最常见的错误。
point when v = 0. Using v² = u² + 2as with a = −9.8 gives 0 = 400 − 19.6s, so s ≈ 20.4 m.
例如，一个球以 u = 20 米/秒 向上抛，在最高点时 v = 0。使用 v² = u² + 2as，取 a = −9.8，得到 0 = 400 − 19.6s，因此 s ≈ 20.4 米。
9. Motion Under Gravity: Vertical SUVAT | 重力作用下的运动：竖直 SUVAT
When a particle moves freely under gravity, the acceleration is g positive direction is up, then a = −g. The equations remain exactly the same; just replace a with −g. The symmetry of projectile motion means time up equals time down to the same vertical level, and speed at a given height is the same on the way up and down (ignoring air resistance).
当物体在重力作用下自由运动时，加速度为向下的 g。如果取向上为正，则 a = −g。方程完全相同，只需将 a 替换为 −g。抛体运动的对称性意味着上升到某一高度和落回同一高度的时间相等，且在同一高度的速率在上抛和下落时相同（忽略空气阻力）。
t falls for 3 seconds. With u = 0, a = 9.8 m s⁻² (down positive), s = ½(9.8)(3²) = 44.1 m.
一块石头从静止落下 3 秒。取向下为正，u = 0，a = 9.8 米/秒²，s = ½(9.8)(3²) = 44.1 米。
10. Using SUVAT in Multi‑Stage Problems | 多阶段问题中的 SUVAT 应用
Many exam questions involve a journey split into stages where acceleration changes (e.g., acceleration, constant velocity, deceleration). Treat each stage separately: the final velocity of one stage becomes the initial velocity of the next. Keep a consistent sign convention throughout. Draw a velocity–time graph often to check consistency and visualise the motion.
许多考题涉及将运动分成加速度变化的阶段（例如，加速、匀速、减速）。分别处理每个阶段：一个阶段的末速度成为下一阶段的初速度。全程保持一致的符号规则。常画速度–时间图以检查一致性并直观呈现运动过程。
For a train that accelerates at 1.2 m s⁻² for 20 s from rest, maintains constant speed for 45 s, and decelerates at 0.8 m s⁻² to rest, the total distance can be found by summing the displacements of three segments.
对于一辆火车，从静止以 1.2 米/秒² 加速 20 秒，匀速 45 秒，再以 0.8 米/秒² 减速至静止，总距离可以通过三个阶段的位移求和得到。
11. Deriving the Fourth Equation from Work–Energy Principles | 从功能原理推导第四方程
in kinetic energy provides a satisfying derivation of v² = u² + 2as. If constant net force F = ma acts over displacement s ves a s = ½(v² − u²), rearranged to v² = u² + 2as. This connects kinematics with dynamics.
将净力做的功等同于动能的变化，可以漂亮地推导出 v² = u² + 2as。如果恒定净力 F = ma 在位移 s 上做功，则 F s = ½mv² − ½mu²。代入 F = ma 并消去 m，得到 a s = ½(v² − u²)，整理得 v² = u² + 2as。这建立了运动学与动力学的联系。
12. Common Exam Pitfalls and How to Master SUVAT | 常见考试陷阱与如何精通 SUVAT
Beware of questions that provide redundant data — ng s = ut + ½at² for return journeys, pay attention to sign of displacement. Practise writing down your SUVAT list before attempting a solution; it vastly reduces errors.
谨防提供多余数据的题目——只使用与你选择的方程匹配的量。记住，除非方向不变，SUVAT 给出的是位移，而非总路程。在往返运动中使用 s = ut + ½at² 时，注意位移的符号。在尝试解答之前，养成列出 SUVAT 列表的习惯；这会大大减少错误。
Master the four constant acceleration equations by doing a variety of problems: horizontal, vertical, and multi‑stage. Familiarity leads to fluency, and the selection table will become second nature. Practice, and you will handle any kinematics challenge with ease.
通过做各种题目（水平、竖直和多阶段）来掌握四个匀加速运动方程。熟练会带来流畅，选择表格将成为第二天性。多加练习，你就能轻松应对任何运动学挑战。
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