📚 Deriving the Equations of Motion for Constant Acceleration (SUVAT) | 恒定加速度运动方程（SUVAT）推导
In the OxfordAQA AS Physics Unit 1 (PH01) examination, a solid grasp of kinematics is essential. One of the most examined skills is the ability to derive and apply the equations of motion for constant acceleration, often referred to as the SUVAT equations. This article presents a clear, step-by-step derivation of these four fundamental formulas from basic definitions, aligning perfectly with the requirements of the 9630 specification. Understanding where these equations come from not only deepens your conceptual knowledge but also prepares you for the derivation questions frequently featured in PH01 past papers such as the June 2023 series.
在 OxfordAQA AS 物理单元一（PH01）考试中，牢固掌握运动学至关重要。考查最多的技能之一就是推导和应用匀加速直线运动的公式，通常称为 SUVAT 方程。本文从基本定义出发，逐步清晰地推导这四个基本公式，完全符合 9630 大纲的要求。理解这些公式的来源不仅能加深你的概念认知，还能帮你应对 PH01 真题中经常出现的推导题，比如 2023 年 6 月系列试卷中的题目。
1. What Are the SUVAT Equations? | 什么是 SUVAT 方程？
The SUVAT equations are a set of four kinematic formulas that relate five key quantities: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t). They are only valid when acceleration is constant in both magnitude and direction. The acronym SUVAT comes from the standard symbols used in A-Level physics.
SUVAT 方程组是由四个运动学公式组成的，它们联系着五个关键量：位移（s）、初速度（u）、末速度（v）、加速度（a）和时间（t）。它们只在加速度的大小和方向都恒定时才成立。这个首字母缩略词 SUVAT 来源于 A-Level 物理中使用的标准符号。
The four equations are:
这四个方程是：
- v = u + at
- s = ½(u + v)t
- s = ut + ½at²
- v² = u² + 2as
Each equation omits one of the five quantities, making it convenient to select the right one based on the information given in a problem.
每个方程都会省略五个量中的一个，因此可以根据题目给出的信息方便地选用合适的方程。
2. Basic Definition: Acceleration | 基本定义：加速度
Acceleration is defined as the rate of change of velocity. If an object moves from an initial velocity u to a final velocity v in a time interval t, the constant acceleration a is given by:
加速度定义为速度的变化率。如果一个物体在时间间隔 t 内从初速度 u 变化到末速度 v，则恒定加速度 a 由下式给出：
a = (v – u) / t
Rearranging this definition immediately yields the first SUVAT equation. This simple relationship is the cornerstone of all subsequent derivations.
重新整理这个定义，马上就可以得到第一个 SUVAT 方程。这个简单的关系是所有后续推导的基石。
3. Deriving the First Equation: v = u + at | 推导第一个方程：v = u + at
Starting from a = (v – u) / t, multiply both sides by t to obtain:
从 a = (v – u) / t 出发，两边同乘 t 得到：
a × t = v – u
Then add u to both sides:
然后两边加上 u：
v = u + at
This equation expresses final velocity as the sum of initial velocity and the product of acceleration and time. It is used when time is known but displacement is not required.
这个方程表示末速度等于初速度加上加速度与时间的乘积。当时间已知但不需要位移时，可使用这个方程。
4. Area Under a Velocity-Time Graph | 速度–时间图下的面积
For constant acceleration, a velocity-time graph is a straight line. The displacement s of the object is equal to the area under the v-t graph. The shape is a trapezium, and its area can be computed as the average velocity multiplied by time.
对于恒定加速度，速度–时间图是一条直线。物体的位移 s 等于 v-t 图下的面积。这个形状是一个梯形，其面积可以用平均速度乘以时间来计算。
Average velocity = (initial velocity + final velocity) / 2 = (u + v)/2. Therefore:
平均速度 = （初速度 + 末速度）/ 2 = (u + v)/2。因此：
s = average velocity × t = ½(u + v)t
This is the second SUVAT equation. It is particularly useful when acceleration is not given.
这就是第二个 SUVAT 方程。当未给出加速度时，它特别有用。
5. Deriving the Displacement Equation: s = ut + ½at² | 推导位移方程：s = ut + ½at²
We can substitute the expression for v from the first equation (v = u + at) into the area-based equation s = ½(u + v)t. Replace v:
我们可以把第一个方程中 v 的表达式 (v = u + at) 代入基于面积得到的方程 s = ½(u + v)t 中。替换 v：
s = ½(u + [u + at])t = ½(2u + at)t
Simplify the bracket:
化简括号：
s = ½(2u + at)t = (u + ½at)t
Finally, distribute t:
最后，分配 t：
s = ut + ½at²
This equation gives displacement directly from initial velocity, acceleration, and time. It is the workhorse for problems where final velocity is not mentioned.
这个方程直接从初速度、加速度和时间给出位移。对于未提及末速度的问题，它是常用公式。
6. Deriving the Velocity-Displacement Equation: v² = u² + 2as | 推导速度–位移方程：v² = u² + 2as
To eliminate time t, start from the first equation v = u + at and solve for t:
为了消去时间 t，从第一个方程 v = u + at 解出 t：
t = (v – u) / a
Now substitute this expression for t into s = ½(u + v)t:
现在把这个 t 的表达式代入 s = ½(u + v)t：
s = ½(u + v) × (v – u) / a
Notice that (u + v)(v – u) is the difference of two squares, which equals v² – u².
注意到 (u + v)(v – u) 是平方差，等于 v² – u²。
s = (v² – u²) / (2a)
Multiply both sides by 2a and rearrange to obtain the familiar form:
两边同乘 2a 并重新整理，得到熟悉的形式：
v² = u² + 2as
This equation is used when time is unknown, for example in stopping distance calculations.
当时间未知时，例如在计算刹车距离时，使用这个方程。
7. Deriving the Average Velocity Equation: s = ½(u + v)t (Recap) | 推导平均速度方程：s = ½(u + v)t（回顾）
Although we derived it using the area under a graph, it is worth reiterating that this equation assumes uniform acceleration. The average velocity is only equal to (u+v)/2 under constant acceleration. If acceleration changes, the area method must be adapted to a general integral, which is beyond the scope of PH01.
虽然我们已经用图下面积的方法推导了这个方程，但有必要重申，这个方程假设加速度均匀。只有在恒定加速度下，平均速度才等于 (u+v)/2。如果加速度变化，面积法必须改用一般的积分，这超出了 PH01 的范畴。
In exam derivations, you may be asked to start from basic principles and show the steps. Always explicitly state the assumption a = constant.
在考试推导题中，可能会要求你从基本原理出发并展示步骤。一定要明确写出 a = 常数的假设。
8. Checking the Derivations | 验证推导
A quick dimensional analysis can verify each equation. For instance, in s = ut + ½at², the term ut has dimensions of [LT⁻¹]×[T] = [L], and at² has dimensions of [LT⁻²]×[T²] = [L]. Both terms are lengths, so the equation is dimensionally consistent.
快速进行量纲分析可以验证每个方程。例如，在 s = ut + ½at² 中，项 ut 的量纲是 [LT⁻¹]×[T] = [L]，而 at² 的量纲是 [LT⁻²]×[T²] = [L]。两项都是长度，因此方程在量纲上是一致的。
You can also test the equations with a simple numerical scenario. Suppose u=0, a=2 m/s², t=3 s. From v=u+at, v=0+2×3=6 m/s. From s=ut+½at², s=0+½×2×9=9 m. From v²=u²+2as, 36=0+2×2×9=36, which holds true. This cross-check builds confidence.
你也可以用一个简单的数值场景来检验方程。假设 u=0，a=2 m/s²，t=3 s。由 v=u+at 得 v=0+2×3=6 m/s。由 s=ut+½at² 得 s=0+½×2×9=9 m。由 v²=u²+2as 得 36=0+2×2×9=36，成立。这种交叉检验可以增强信心。
9. Worked Example: Free Fall | 应用示例：自由落体
Consider an object dropped from rest near the Earth’s surface, where a = g = 9.81 m/s² downward. After 2.0 seconds, what is its velocity and how far has it fallen?
考虑一个在近地表从静止下落的物体，此处 a = g = 9.81 m/s² 向下。2.0 秒后，它的速度是多少，下落了多远？
Use u=0, a=9.81 m/s², t=2.0 s. Then v = 0 + 9.81×2.0 = 19.62 m/s. Displacement s = 0 + ½×9.81×(2.0)² = 19.62 m. Notice that the numbers for v and s are numerically equal because of the specific inputs; such coincidences can be used to check algebraic derivations.
取 u=0，a=9.81 m/s²，t=2.0 s。那么 v = 0 + 9.81×2.0 = 19.62 m/s。位移 s = 0 + ½×9.81×(2.0)² = 19.62 m。注意 v 和 s 的数值相等是因为特定的输入值；这种巧合可以用来检验代数推导。
10. Worked Example: Car Braking | 应用示例：汽车刹车
A car travelling at 20 m/s brakes with a constant deceleration of 4.0 m/s². How far does it travel before stopping?
一辆汽车以 20 m/s 的速度行驶，以 4.0 m/s² 的恒定减速度刹车。它在停下前行驶了多远？
Here v=0, u=20 m/s, a = –4.0 m/s² (negative because opposite to motion). Use v² = u² + 2as → 0 = 400 + 2×(–4.0)×s → 0 = 400 – 8s → s = 50 m. You could also find t first and then use s = ut + ½at², giving the same result.
这里 v=0，u=20 m/s，a = –4.0 m/s² （负号表示与运动方向相反）。使用 v² = u² + 2as 得到 0 = 400 + 2×(–4.0)×s → 0 = 400 – 8s → s = 50 m。你也可以先求 t，然后用 s = ut + ½at²，结果相同。
11. Common Mistakes | 常见错误
One frequent error is forgetting to use consistent sign conventions. Always define a positive direction and assign signs to u, v, a, and s accordingly. Also, remember that the SUVAT equations only hold for constant acceleration; they cannot be applied directly to scenarios with varying acceleration, such as a spring-mass system.
一个常见错误是忘记使用一致的符号规定。务必先定义正方向，并相应地给 u、v、a 和 s 赋予符号。另外，记住 SUVAT 方程只对恒定加速度成立；它们不能直接用于加速度变化的情景，比如弹簧–质量系统。
Another pitfall is confusing the displacement equation s = ut + ½at² with simply distance travelled. If the object changes direction, s is the net displacement, not the total path length.
另一个易错点是混淆位移方程 s = ut + ½at² 与简单的路程。如果物体改变方向，s 是净位移，而不是总路程。
Finally, ensure you can derive these equations from the definitions, as the PH01 paper may ask you to show the derivation of v² = u² + 2as starting from a = (v-u)/t and s = ½(u+v)t.
最后，要确保你能从定义出发推导这些方程，因为 PH01 试卷可能会要求你从 a = (v-u)/t 和 s = ½(u+v)t 出发，展示 v² = u² + 2as 的推导过程。
12. Conclusion | 结语
Mastering the derivation of the SUVAT equations equips you with a powerful set of tools for solving kinematics problems in the OxfordAQA Unit 1 exam. By internalising the logical flow — from the definition of acceleration, through the velocity-time graph area, to algebraic elimination of variables — you not only prepare for derivation questions but also become more adept at spotting which equation to use. Regular practice with these derivations will reinforce your understanding and accuracy, giving you a distinct advantage on the PH01 paper, including the June 2023 variant examined under code 9630.
掌握 SUVAT 方程的推导为你在 OxfordAQA 单元一考试中解决运动学问题提供了一套强大的工具。通过内化推导的逻辑流程——从加速度的定义，到速度–时间图面积，再到代数消元——你不仅为推导题做好了准备，而且能更熟练地判断应该使用哪个方程。经常练习这些推导将强化你的理解和准确率，让你在 PH01 试卷（包括编号为 9630 的 2023 年 6 月试题）中占据明显优势。
Published by TutorHao | Physics Revision Series | aleveler.com
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