📚 Derivation of Kinematic Equations | 运动学公式推导
Understanding how the standard equations of motion are derived is fundamental for mastering A‑Level mechanics. This article walks through each step of the derivations, starting from the definition of constant acceleration, and connects them to real‑world examples.
理解标准运动学方程的推导过程，是掌握 A‑Level 力学的基础。本文将从匀加速度的定义出发，逐步推导各个公式，并结合实际例子说明其应用。
1. Constant Acceleration & Basic Definitions | 匀加速度与基本定义
We assume an object moves in a straight line with constant acceleration a. Let initial velocity be u, final velocity after time t be v, and displacement in that time be s.
我们假设物体沿直线运动，加速度 a 恒定。设初速度为 u，经时间 t 后的末速度为 v，这段时间内的位移为 s。
Acceleration is defined as the rate of change of velocity: a = (v − u)/t. This relation is the starting point for all kinematic derivations.
加速度定义为速度的变化率：a = (v − u)/t。这一关系是所有运动学推导的起点。
2. Deriving v = u + at | 推导 v = u + at
From the definition of acceleration a = (v − u)/t, we can directly rearrange to express final velocity. Multiply both sides by t: at = v − u.
根据加速度的定义 a = (v − u)/t，可以直接变形得到末速度表达式。两边同时乘以 t：at = v − u。
Then add u to both sides to obtain the familiar first equation:
然后两边加上 u，得到熟悉的第一个方程：
v = u + at
This equation tells us how velocity changes linearly with time when acceleration is constant. It does not involve displacement.
该方程表明，加速度恒定时，速度随时间线性变化。该式不涉及位移。
3. Displacement from Average Velocity | 从平均速度求位移
When acceleration is constant, velocity changes at a uniform rate, so the average velocity vₐᵥ over the interval is simply the arithmetic mean of initial and final velocities:
加速度恒定意味着速度均匀变化，因此这段时间内的平均速度 vₐᵥ 就是初速度和末速度的算术平均值：
vₐᵥ = (u + v)/2
Displacement is average velocity multiplied by time: s = vₐᵥ × t. Substituting the average velocity gives a direct relationship between s, u, v and t:
位移等于平均速度乘以时间：s = vₐᵥ × t。代入平均速度表达式，得到 s、u、v 和 t 之间的直接关系：
s = (u + v)t / 2
This equation is useful when the final velocity is known but acceleration itself is not required.
当已知末速度而不需要加速度时，该方程非常有用。
4. Deriving s = ut + ½at² | 推导 s = ut + ½at²
To eliminate v from the displacement formula, substitute v = u + at into s = (u + v)t/2.
为了从位移公式中消去 v，将 v = u + at 代入 s = (u + v)t/2。
Start with s = (u + v)t/2. Replace v:
从 s = (u + v)t/2 出发，替换 v：
s = (u + [u + at])t / 2 = (2u + at)t / 2
Separate the fraction: (2u t)/2 + (at²)/2, which simplifies to:
拆分分数：(2u t)/2 + (at²)/2，化简得到：
s = ut + ½at²
This is one of the most widely used equations. It gives displacement directly in terms of initial velocity, time, and acceleration.
这是最常用的方程之一，它直接用初速度、时间和加速度表示位移。
5. Deriving v² = u² + 2as | 推导 v² = u² + 2as
Sometimes we need to relate velocity and displacement without involving time. We can eliminate t by rearranging v = u + at to express t: t = (v − u)/a.
有时我们需要在未知时间的条件下关联速度与位移。我们可以通过变换 v = u + at 来表示 t：t = (v − u)/a。
Substitute this expression for t into s = (u + v)t/2:
将该 t 表达式代入 s = (u + v)t/2：
s = (u + v) × ((v − u)/a) / 2
Simplify the numerator: (u + v)(v − u) = v² − u² (difference of squares). So:
化简分子：(u + v)(v − u) = v² − u²（平方差公式）。于是：
s = (v² − u²) / (2a)
Multiply both sides by 2a and rearrange to get the final form:
两边乘 2a 并整理，得到最终形式：
v² = u² + 2as
This equation is particularly handy when time is not mentioned, such as in stopping‑distance calculations or free‑fall velocity at a given height.
该方程在题目未提及时间时特别方便，例如计算刹车距离或给定高度处的自由落体速度。
6. Alternative Derivation Using Calculus | 用微积分推导（拓展）
For students taking Further Maths or curious about a more rigorous approach, the same equations follow directly from integrating constant acceleration.
对于学习进阶数学或希望了解更严谨方法的同学，上述公式可以通过对匀加速度积分直接得出。
Acceleration is the derivative of velocity: dv/dt = a. Integrating from 0 to t with initial velocity u gives v = u + at.
加速度是速度的导数：dv/dt = a。在 0 到 t 间积分并代入初速 u，得到 v = u + at。
Velocity is the derivative of displacement: ds/dt = v = u + at. Integrating again yields s = ut + ½at² (assuming initial displacement s₀ = 0).
速度是位移的导数：ds/dt = v = u + at。再次积分得到 s = ut + ½at²（设初始位移 s₀ = 0）。
Eliminating t by using the chain rule a = v dv/ds and integrating gives v² = u² + 2as. This calculus‑based derivation unifies the entire set.
利用链式法则 a = v dv/ds 消去 t 并积分，得到 v² = u² + 2as。这一基于微积分的推导将整套公式统一起来。
7. Vector Nature of the Equations | 方程的矢量性
All four equations are vector equations. In one‑dimensional problems, we choose a positive direction and treat u, v, a, and s as signed scalars.
这四个方程均为矢量方程。在一维问题中，我们选定正方向后将 u、v、a 和 s 视为带正负号的标量。
If acceleration opposes the direction of motion, a takes a negative value. This naturally produces deceleration in the equations, e.g. a negative v or a decreasing v over time.
如果加速度与运动方向相反，a 取负值，方程会自然地体现减速效果，例如速度 v 随时间减小。
Throw an object upwards: choose up as positive, then a = −g (g = 9.81 m s⁻²).
向上抛物体：选向上为正，则 a = −g (g = 9.81 m s⁻²)。
The equations then describe the rise, momentary rest, and fall seamlessly.
然后方程就能无缝地描述上升、瞬间静止和下落的全过程。
8. Worked Example – Free Fall from Rest | 实例分析 – 静止释放的自由落体
cliff. After 3.0 seconds, find its velocity and the distance fallen. Take g = 9.81 m s⁻².
一块石头从悬崖静止释放。3.0 秒后，求它的速度和下落距离。取 g = 9.81 m s⁻²。
Known values: u = 0, a = 9.81 m s⁻², t = 3.0 s.
已知量：u = 0，a = 9.81 m s⁻²，t = 3.0 s。
Using v = u + at: v = 0 + 9.81 × 3.0 = 29.4 m s⁻¹ (downwards).
用 v = u + at：v = 0 + 9.81 × 3.0 = 29.4 m s⁻¹（向下）。
Using s = ut + ½at²: s = 0 + 0.5 × 9.81 × (3.0)² = 44.1 m.
用 s = ut + ½at²：s = 0 + 0.5 × 9.81 × (3.0)² = 44.1 m。
These steps show how the derived equations directly solve typical exam problems.
这些步骤展示了推导出的方程如何直接求解典型的考试题目。
9. Worked Example – Stopping Distance | 实例分析 – 刹车距离
A car travelling at 20 m s⁻¹ brakes with constant deceleration of 4.0 m s⁻². Find the stopping distance.
一辆汽车以 20 m s⁻¹ 的速度行驶，以恒定减速度 4.0 m s⁻² 刹车。求刹车距离。
Here u = 20 m s⁻¹, v = 0, a = −4.0 m s⁻² (negative because opposite to motion). We need s without time, so use v² = u² + 2as.
这里 u = 20 m s⁻¹，v = 0，a = −4.0 m s⁻²（因为是减速，方向与运动相反）。我们需要求 s 而无需时间，故使用 v² = u² + 2as。
Rearrange to s = (v² − u²)/(2a). Substitute: (0 − 20²) / (2 × −4.0) = (−400)/(−8.0) = 50 m.
变形得 s = (v² − u²)/(2a)。代入数值：(0 − 20²) / (2 × −4.0) = (−400)/(−8.0) = 50 m。
The negative signs cancel, giving a positive displacement as expected. The stopping distance is 50 metres.
负号相互抵消，位移为正值，符合预期。刹车距离为 50 米。
10. Common Mistakes to Avoid | 常见错误规避
Sign errors: always define a positive direction before inserting values. If acceleration is downward and you have chosen upward as positive, a must be negative.
符号错误：代入数值前一定要规定正方向。若加速度向下而你选向上为正，则 a 必须为负。
Misusing average velocity: the formula s = (u+v)t/2 only holds for constant acceleration, not for varying acceleration.
误用平均速度：公式 s = (u+v)t/2 仅在加速度恒定时成立，加速度变化时不适用。
Forgetting squared terms: in s = ut + ½at² the ½ factor and the square on t are crucial; careless omission leads to large errors.
遗漏平方项：在 s = ut + ½at² 中，½ 因子和 t 的平方至关重要，疏忽遗漏会导致巨大误差。
11. Summary of Equations | 公式汇总
The four kinematic equations for constant acceleration in a straight line are:
匀加速直线运动的四个运动学方程为：
| Equation | Quantities involved | Useful when... |
| v = u + at | u, v, a, t | No displacement s |
| s = (u+v)t/2 | u, v, t, s | No acceleration a |
| s = ut + ½at² | u, a, t, s | No final velocity v |
| v² = u² + 2as | u, v, a, s | No time t |
Memorising these equations is only the first step; understanding their derivations allows you to adapt them confidently in unfamiliar situations.
记住这些公式只是第一步，理解它们的推导过程才能使你在陌生情境中自信地灵活运用。
12. Practice & Extension | 练习与拓展
Try deriving the equations yourself starting from a = (v − u)/t and the definition of average velocity. Then test your understanding with problems involving projectiles, where two independent directions must be considered separately.
试着自己从 a = (v − u)/t 和平均速度的定义出发推导这些方程。然后通过涉及抛体运动的问题检验理解，抛体问题需分别考虑两个独立的方向。
For further practice, modify the worked examples: what if the stone is thrown downwards with an initial speed? What if the car brakes on a slope? The same principles apply, but vector signs become even more critical.
进一步练习时可以修改上述实例：若石头以一定初速度向下抛出会怎样？若汽车在斜坡上刹车呢？基本原理相同，但矢量的正负号将更为关键。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图