📚 Formula Derivation in Edexcel IGCSE Physics Student Book 2 | IGCSE Edexcel 物理：Edexcel IGCSE Physics Student Book 2 公式推导
Understanding how key formulas are derived is an essential part of mastering IGCSE Physics. It builds a deeper conceptual foundation and helps you apply equations confidently, even when problems become unfamiliar. In Edexcel IGCSE Physics Student Book 2, many important relationships in mechanics, electricity, waves, and fluids are introduced. This article walks through the step‑by‑step derivations of the most crucial formulas, clearly explaining the physics behind each one.
理解关键公式的推导过程是掌握 IGCSE 物理的重要环节。它能帮助你建立更深刻的概念基础，在面对不熟悉的题目时也能自信地运用公式。在 Edexcel IGCSE 物理学生用书第 2 册中，力学、电学、波和流体等部分引入了许多重要的关系式。本文逐步讲解这些核心公式的推导过程，清楚阐释每个公式背后的物理原理。
1. Derivation of Kinetic Energy | 动能公式推导
Kinetic energy is the energy an object possesses due to its motion. To derive its formula, imagine a constant net force F acting on an object of mass m. The object accelerates from rest to a velocity v while moving through a displacement s.
动能是物体由于运动而具有的能量。为了推导公式，设想一个恒定的净力 F 作用在质量为 m 的物体上，该物体从静止加速到速度 v，期间发生的位移为 s。
The work done by the net force is given by W = F s. Using Newton’s second law, we know that F = m a. From kinematics, the equation v² = u² + 2 a s can be applied. Since the object starts from rest, u = 0, and we obtain s = v² / (2a).
净力所做的功为 W = F s。根据牛顿第二定律，有 F = m a。由运动学公式 v² = u² + 2 a s，因为物体从静止出发，u = 0，可得到 s = v² / (2a)。
Substituting F and s into the work expression gives W = (m a) × (v² / (2a)) = ½ m v². This work transfers energy and becomes the kinetic energy of the object. Hence, the kinetic energy Eₖ is expressed as follows.
将 F 和 s 代入功的表达式，得 W = (m a) × (v² / (2a)) = ½ m v²。这部分功转移为能量，成为物体的动能。因此动能 Eₖ 的公式如下。
Eₖ = ½ m v²
2. Derivation of Gravitational Potential Energy | 重力势能公式推导
When an object is lifted at a constant speed, the work done against gravity is stored as gravitational potential energy. Consider an object of mass m raised through a vertical height Δh near the Earth’s surface, where the gravitational field strength g is approximately constant.
当物体以恒定速度被提升时，克服重力所做的功以重力势能的形式储存起来。考虑一个质量为 m 的物体，在地球表面附近，重力场强度 g 近似恒定的条件下，被竖直提升高度 Δh。
The force needed to lift the object without acceleration is equal to its weight, F = m g. The work done by this force over the distance Δh is W = F Δh = m g Δh. Since no kinetic energy is gained, all of this work becomes gravitational potential energy Eₚ.
在无加速度的情况下提升物体所需的力等于它的重量，F = m g。这个力在距离 Δh 上所做的功为 W = F Δh = m g Δh。由于动能没有增加，这些功全部转化为重力势能 Eₚ。
Eₚ = m g Δh
3. Derivation of Electrical Power Formulas | 电功率公式推导
Power P is defined as the rate of energy transfer, P = W / t. In an electric circuit, the work done when a charge Q moves through a potential difference V is W = V Q. Because current I is the rate of flow of charge, I = Q / t, we can substitute: P = V (I t) / t = V I. This gives the fundamental power equation.
功率 P 被定义为能量转换的速率，P = W / t。在电路中，当电荷 Q 通过电势差 V 时所做的功为 W = V Q。由于电流 I 是电荷流动的速率，I = Q / t，代入可得：P = V (I t) / t = V I。这就得到了基本的功率公式。
Using Ohm’s law, V = I R, we can express power in alternative forms that highlight energy dissipation in a resistor. Replacing V gives P = I² R. Alternatively, substituting I = V / R yields P = V² / R. All three expressions are equivalent under Ohmic conduction.
利用欧姆定律 V = I R，我们可以将功率表示为其他形式，以突出电阻上的能量耗散。用 I R 替换 V 得到 P = I² R。或者，代入 I = V / R 得到 P = V² / R。在欧姆导电条件下，这三个表达式是等价的。
P = V I = I² R = V² / R
4. Resistors in Series | 串联电阻公式推导
When resistors are placed in series, the same current I flows through each one because there is only one path for the charge. The total potential difference V_total supplied by the source is equal to the sum of the individual potential differences: V_total = V₁ + V₂ for two resistors.
当电阻串联时，由于电荷只有一条路径，相同的电流 I 流过每一个电阻。电源提供的总电势差 V_total 等于各单个电阻两端电势差之和：对于两个电阻，V_total = V₁ + V₂。
Applying Ohm’s law to each resistor and to the whole combination, V_total = I R_total, V₁ = I R₁, and V₂ = I R₂. Substituting these into the voltage sum equation gives I R_total = I R₁ + I R₂.
对每个电阻以及整个组合应用欧姆定律，有 V_total = I R_total、V₁ = I R₁ 和 V₂ = I R₂。将这些代入电压求和方程可得 I R_total = I R₁ + I R₂。
Since the current I is common, it cancels out, leaving R_total = R₁ + R₂. This rule extends to any number of series resistors: R_total = R₁ + R₂ + R₃ + …
由于电流 I 相同，可以消去，得到 R_total = R₁ + R₂。这一规则可推广到任意数量的串联电阻：R_total = R₁ + R₂ + R₃ + …
Rtotal = R₁ + R₂ + R₃ + …
5. Resistors in Parallel | 并联电阻公式推导
In a parallel circuit, each resistor is connected directly across the source, so the potential difference V across each branch is the same. By contrast, the total current I_total from the source splits at the junctions: I_total = I₁ + I₂ for two parallel resistors.
在并联电路中，每个电阻直接并接在电源两端，因此各支路两端的电势差 V 相同。而来自电源的总电流 I_total 在节点分流：对于两个并联电阻，I_total = I₁ + I₂。
Applying Ohm’s law to the whole circuit and to each branch gives I_total = V / R_total, I₁ = V / R₁, and I₂ = V / R₂. Substituting into the current equation: V / R_total = V / R₁ + V / R₂.
对整个电路及各支路应用欧姆定律，有 I_total = V / R_total、I₁ = V / R₁ 和 I₂ = V / R₂。代入电流方程：V / R_total = V / R₁ + V / R₂。
Voltage V is the same throughout, so it cancels, resulting in 1 / R_total = 1 / R₁ + 1 / R₂. For more resistors in parallel, the reciprocals add: 1 / R_total = 1 / R₁ + 1 / R₂ + 1 / R₃ + …
由于各处电压 V 相同，消去后得到 1 / R_total = 1 / R₁ + 1 / R₂。更多电阻并联时，倒数相加：1 / R_total = 1 / R₁ + 1 / R₂ + 1 / R₃ + …
1 / Rtotal = 1 / R₁ + 1 / R₂ + 1 / R₃ + …
6. Wave Speed Equation | 波速公式推导
The speed of a wave v relates how fast the disturbance travels through a medium. Consider a continuous wave with a steady frequency f and a well‑defined wavelength λ. Frequency is the number of complete waves passing a point per second, and wavelength is the distance between two consecutive identical points on the wave.
波速 v 描述波在介质中传播的快慢。考虑一个具有稳定频率 f 和确定波长 λ 的连续波。频率是每秒钟通过某一点的完整波的个数，波长则是波上两个相邻相同点之间的距离。
One complete wavelength passes a fixed point in a time interval equal to the period T. By definition, T = 1 / f. Therefore, in one second, f complete wavelengths move past a point. The total distance covered by the wave in one second is f × λ.
一个完整的波长在一个周期 T 的时间间隔内通过某个固定点。根据定义，T = 1 / f。因此，在一秒钟内有 f 个完整的波长经过一点，波在一秒内传播的总距离就是 f × λ。
Since speed
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