Deriving the Equations of Motion | 运动学方程的推导

📚 Deriving the Equations of Motion | 运动学方程的推导

In A-Level Physics, the equations of motion describe the relationship between displacement, initial velocity, final velocity, acceleration and time for an object moving in a straight line with constant acceleration. Often called the SUVAT equations, they are incredibly useful for solving a wide range of kinematics problems. In this article, we will derive each of these equations step by step from first principles, using both algebraic and graphical methods. This thorough understanding is essential for tackling exam questions confidently.

在 A-Level 物理中，运动学方程描述了物体在恒加速度下沿直线运动时位移、初速度、末速度、加速度和时间之间的关系。这些方程常被称为 SUVAT 方程，对于求解大量运动学问题极为有用。在这篇文章中，我们将从基本原理出发，通过代数和图像两种方法逐步推导每一个方程。透彻理解这些推导对于自信应对考试题目至关重要。

1. Introduction to SUVAT Equations | SUVAT方程简介

The term SUVAT is an acronym derived from the five variables used in the equations: s (displacement), u (initial velocity), v (final velocity), a (acceleration), and t (time). These equations are only valid when the acceleration is constant in magnitude and direction, and the motion is along a straight line. Knowing how to derive them deepens your grasp of kinematics beyond simply memorising the formulas.

SUVAT 一词是由方程中使用的五个变量首字母组成的缩写：s（位移）、u（初速度）、v（末速度）、a（加速度）和 t（时间）。这些方程仅在加速度大小和方向恒定、且运动沿直线进行时才成立。懂得如何推导它们能让你对运动学的理解超越单纯的公式记忆。

2. Basic Assumptions for the Derivations | 推导的基本假设

We assume the object moves along a straight line with uniform acceleration. This means the velocity changes at a constant rate. The direction of motion is taken as positive, so vector quantities like velocity and acceleration will carry signs accordingly. Air resistance and other complicating factors are neglected. Under these ideal conditions, the relationships between the kinematic variables are exactly described by the SUVAT equations.

我们假设物体沿直线做匀加速运动，这意味着速度以恒定速率变化。运动方向取为正方向，因此速度和加速度等矢量将相应地带有正负号。空气阻力和其他复杂因素忽略不计。在这些理想条件下，运动学变量之间的关系可由 SUVAT 方程精确描述。

3. Symbol Definitions and Units | 符号定义与单位

Before we start deriving, let’s define each symbol clearly: s is the displacement (measured in metres, m); u is the initial velocity (m s⁻¹); v is the final velocity (m s⁻¹); a is the constant acceleration (m s⁻²); and t is the time interval (s). Displacement is a vector quantity, so it can be positive or negative depending on direction relative to the chosen positive axis.

在开始推导之前，我们先明确定义每个符号：s 是位移（单位：米，m）；u 是初速度（m s⁻¹）；v 是末速度（m s⁻¹）；a 是恒定加速度（m s⁻²）；t 是时间间隔（s）。位移是矢量，依据所选正方向可为正或负。

It is common to express acceleration as the rate of change of velocity: a = (v – u) / t, provided the acceleration is constant. This expression will be our starting point for the first equation. The other equations can then be derived by combining this relationship with definitions of average velocity and displacement.

通常将加速度表示为速度的变化率：a = (v – u) / t，前提是加速度恒定。此表达式将作为我们推导第一个方程的出发点。其他方程则可通过将此关系与平均速度和位移的定义结合而得出。

4. Equation 1: v = u + at | 方程1：v = u + at

From the definition of constant acceleration, we know that acceleration equals the change in velocity divided by the time taken. That is, a = (v – u) / t. Rearranging this to make v the subject gives v = u + at. This is the first SUVAT equation, linking final velocity, initial velocity, acceleration and time.

由恒定加速度的定义可知，加速度等于速度的变化量除以所用时间，即 a = (v – u) / t。将此式整理得 v = u + at。这就是第一个 SUVAT 方程，它将末速度、初速度、加速度和时间联系起来。

v = u + a t

5. Equation 2: s = ½ (u + v) t | 方程2：s = ½ (u + v) t

For an object moving with constant acceleration, the velocity changes linearly with time. Therefore the average velocity during the time interval t is simply the arithmetic mean of the initial and final velocities: v_avg = (u + v) / 2. Displacement s is the product of average velocity and time: s = v_avg t. Substituting the expression for average velocity yields s = ½ (u + v) t.

对于匀加速运动的物体，速度随时间线性变化。因此，时间间隔 t 内的平均速度就是初速度与末速度的算术平均值：v_avg = (u + v) / 2。位移 s 等于平均速度乘以时间：s = v_avg t。将平均速度的表达式代入，即得 s = ½ (u + v) t。

s = ½ (u + v) t

6. Deriving s = ut + ½ at² | 推导 s = ut + ½ at²

This equation can be obtained by substituting v = u + at from Equation 1 into Equation 2. Start with s = ½ (u + v) t, replace v with u + at, giving s = ½ (u + u + at) t = ½ (2u + at) t. Simplify to get s = ut + ½ at². This is the third SUVAT equation, which gives displacement directly in terms of initial velocity, acceleration and time, without needing the final velocity.

此方程可通过将方程1中的 v = u + at 代入方程2而得到。从 s = ½ (u + v) t 出发，用 u + at 替换 v，得 s = ½ (u + u + at) t = ½ (2u + at) t。化简后即得 s = ut + ½ at²。这就是第三个 SUVAT 方程，它用初速度、加速度和时间直接表示位移，无需末速度。

s = u t + ½ a t²

7. Deriving v² = u² + 2as (Algebraic Method) | 推导 v² = u² + 2as（代数法）

To derive the equation that does not involve time, we can eliminate t from Equation 1 and Equation 2. From v = u + at, we express time as t = (v – u) / a. Substituting this into s = ½ (u + v) t gives s = ½ (u + v) × (v – u) / a. Since (u + v)(v – u) = v² – u², the expression becomes s = (v² – u²) / (2a). Multiplying both sides by 2a and rearranging gives v² = u² + 2as. This equation is particularly useful when time is not known or not required.

为了推导出不包含时间的方程，我们可以从方程1和方程2中消去 t。由 v = u + at 得 t = (v – u) / a。代入 s = ½ (u + v) t，得到 s = ½ (u + v) × (v – u) / a。由于 (u + v)(v – u) = v² – u²，表达式变为 s = (v² – u²) / (2a)。两边同乘 2a 并整理，即得 v² = u² + 2as。当时间未知或无需计算时，这个方程特别有用。

v² = u² + 2 a s

8. Graphical Approach: the v–t Graph | 图像方法：速度–时间图

All these equations can also be derived from a velocity–time graph. For uniform acceleration, the v–t graph is a straight line with slope equal to acceleration a, starting at (0, u) and ending at (t, v). The gradient of the line is (v – u)/t = a, which immediately gives v = u + at. The area under the graph represents the displacement s. Understanding the graphical interpretation provides a powerful visual check and can be used in exams to derive the equations if you forget the algebraic forms.

所有这些方程也可以从速度–时间图中推导出来。对于匀加速运动，v–t 图是一条斜率等于加速度 a 的直线，起点为 (0, u)，终点为 (t, v)。直线斜率为 (v – u)/t = a，这就直接给出了 v = u + at。图线下面积表示位移 s。理解图像解释可以提供强大的直观检验，若在考试中忘记代数形式，也可以用此方法推导方程。

9. Displacement from Area under the v–t Graph | 由 v–t 图下面积求位移

The area under the v–t graph between 0 and t is a trapezium. Its area can be calculated as the average of the parallel sides (u and v) multiplied by the horizontal width t: area = ½ (u + v) t. Since area equals displacement s, we immediately have s = ½ (u + v) t, which is Equation 2. Moreover, the trapezium can be split into a rectangle of area u t and a triangle of area ½ (v – u) t. Substituting (v – u) = a t gives s = u t + ½ a t², which is the third equation derived purely from geometry.

0 到 t 之间 v–t 图线下的面积是一个梯形。其面积可计算为平行边（u 和 v）的平均值乘以水平宽度 t：面积 = ½ (u + v) t。由于面积等于位移 s，我们立即得到 s = ½ (u + v) t，即方程2。此外，该梯形可分解为一个面积为 u t 的矩形和一个面积为 ½ (v – u) t 的三角形。将 (v – u) = a t 代入，即得 s = u t + ½ a t²，这便是纯几何推导出的第三个方程。

10. Obtaining v² = u² + 2as from v–t and Algebra | 结合 v–t 图和代数得出 v² = u² + 2as

Using the v–t graph we have the area expression s = ½ (u + v) t, and from the slope we have a = (v – u)/t, which gives t = (v – u)/a. Substituting this t into s = ½ (u + v) t yields s = ½ (u + v) × (v – u)/a. Simplifying the numerator using the difference of squares leads to s = (v² – u²)/(2a) and finally v² = u² + 2as. Thus the graph combined with simple algebra reproduces the same fourth equation, reinforcing the consistency of the kinematic model.

利用 v–t 图，我们有面积表达式 s = ½ (u + v) t，从斜率可得 a = (v – u)/t，从而 t = (v – u)/a。将此 t 代入 s = ½ (u + v) t，得 s = ½ (u + v) × (v – u)/a。利用平方差公式化简分子，得 s = (v² – u²)/(2a)，最终推出 v² = u² + 2as。这样，图像与简单代数相结合再现了同一个第四方程，巩固了运动学模型的一致性。

11. Summary of the Four SUVAT Equations | 四个 SUVAT 方程总结

The four equations are: (1) v = u + at, (2) s = ½ (u + v) t, (3) s = ut + ½ at², and (4) v² = u² + 2as. Each equation involves four of the five SUVAT variables, enabling you to solve for one unknown provided you know three others. When solving problems, always list the known quantities and the unknown, then select the equation that does not involve the missing variable.

这四个方程是：(1) v = u + at，(2) s = ½ (u + v) t，(3) s = ut + ½ at²，以及 (4) v² = u² + 2as。每个方程都包含五个 SUVAT 变量中的四个，因此只要已知其中三个量，便可求解另一个未知量。解题时，要列出已知量和未知量，然后选择不包含缺失变量的那个方程。

It is essential to remember the restrictions: constant acceleration, straight-line motion, and careful sign conventions for direction. The derivations show that these equations are not independent; they emerge from the definitions of acceleration and average velocity under the constant acceleration condition.

必须牢记其限制条件：加速度恒定、直线运动，并注意方向的符号约定。这些推导表明，方程并非相互独立，它们是在恒定加速度条件下从加速度和平均速度的定义中产生的。

12. Common Pitfalls and Exam Tips | 常见错误与考试建议

A typical mistake is using these equations when acceleration is not constant, such as in circular motion or when forces change. Another trap is neglecting the direction of vectors; if an object is slowing down, acceleration has the opposite sign to velocity. In exams, derivation questions might ask you to start from basic definitions, so it is wise to practise deriving all four equations from either a = (v–u)/t and average velocity, or from the v–t graph. Always double-check that you have used the correct sign for each quantity.

一个典型错误是在加速度不恒定的情况下使用这些方程，例如圆周运动或受力变化时。另一个陷阱是忽视矢量的方向性：如果物体在减速，加速度的符号与速度相反。在考试中，推导题可能要求你从基本定义出发，因此明智的做法是练习从 a = (v–u)/t 和平均速度，或从 v–t 图推导所有四个方程。务必反复检查每个量的符号是否正确。

When handling free-fall under gravity, the acceleration a is usually replaced by g (9.81 m s⁻²) and taken as positive or negative depending on the chosen coordinate system. Practising these derivations with numerical examples helps reinforce the concepts and reduces reliance on rote memorisation.

在处理重力作用下的自由落体时，加速度 a 通常用 g (9.81 m s⁻²) 替代，并根据所选坐标系取正或负。通过数值例子练习这些推导有助于巩固概念，减少对死记硬背的依赖。

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Deriving the Equations of Motion | 运动学方程的推导