📚 AS Physics Unit 2 June 2022: Formula Derivations | AS物理单元2 2022年6月：公式推导
Understanding how key physics equations are derived is essential for mastering AS Unit 2 topics such as mechanics, materials, waves, and electricity. Instead of simply memorising formulas, we can explore the logical steps that connect fundamental principles to the equations you encounter in exam papers. This article walks you through the derivations of the most important formulas, using clear reasoning and consistent mathematical steps. Each section presents a derivation with a concise explanation, followed by a Chinese version of the same content to support bilingual learning.
理解关键物理方程的推导过程对于掌握 AS 单元 2（力学、材料、波与电学）至关重要。与其单纯记忆公式，不如通过逻辑步骤将基本原理与考试题目中的方程式连接起来。本文将一步步推导最重要的公式，并使用清晰的推理和连贯的数学步骤。每一节先用英文给出简要解释和推导，再用中文复述相同内容，以支持双语学习。
1. Deriving v = u + at | 推导 v = u + at
u and the final velocity is v after a time interval t, the acceleration is given by a = (v – SUVAT equations and forms the foundation for describing motion with constant acceleration.
加速度定义为速度的变化率。对于匀加速度 a，若初速度为 u，经过时间 t 后末速度为 v，则由定义可得 a = (v – u)/t。重新整理此式即得到 v = u + at。这是运动学 SUVAT 方程组的第一个方程，也是描述匀加速运动的基础。
2. Deriving s = ut + ½at² | 推导 s = ut + ½at²
+ ½at². This equation links displacement to initial velocity, acceleration, and time without involving the final velocity.
为求位移 s，我们利用匀加速运动中平均速度等于 (u + v)/2 这一事实。位移等于平均速度乘以时间：s = ((u + v)/2) × t。将 v = u + at 代入，得 s = ((u + (u + at))/2) × t = ((2u + at)/2) × t = ut + ½at²。该方程将位移与初速度、加速度和时间联系起来，不涉及末速度。
3. Deriving v² = u² + 2as | 推导 v² = u² + 2as
ate time t from the two preceding equations. Start with v = u + at and solve for t: t = (v – u)/a. Substitute this into s = ut + ½at²: s = u×(v – u)/a + ½a × ((v – u)/a)². Multiply both sides by 2a to simplify: 2as = 2u(v – u) + (v – u)² = 2uv – 2u² + v² – 2uv + u² = v² – arranging yields v² = u² + 2as. This equation is particularly useful when time is unknown.
我们可以从前面两个方程中消去时间 t。由 v = u + at 解出 t = (v – u)/a。将此式代入 s = ut + ½at²：s = u×(v – u)/a + ½a × ((v – u)/a)²。两边同乘 2a 化简：2as = 2u(v – u) + (v – u)² = 2uv – 2u² + v² – 2uv + u² = v² – u²。整理得 v² = u² + 2as。当时间未知时，该方程十分有用。
4. Deriving F = ma from Momentum | 从动量推导 F = ma
t is equal to the rate of change of its momentum: F = Δp/Δt. For a constant mass, Δp = m(v – u), so F = m(v – u)/Δt. But (v – = ma is a special case of the more general momentum principle when mass remains constant.
动量 p 定义为质量与速度的乘积：p = mv。牛顿第二定律指出，物体所受的合外力等于其动量变化率：F = Δp/Δt。在质量恒定的情况下，Δp = m(v – u)，因此 F = m(v – u)/Δt。而 (v – u)/Δt 就是加速度 a，故 F = ma。这表明，当质量不变时，F = ma 是更普遍的动量原理的特例。
5. Deriving Impulse–Momentum Theorem | 推导冲量–动量定理
entum form, F = Δp/Δt, so multiplying both sides by Δt gives FΔt = Δp. This is the impulse– is especially valuable when forces vary over short time intervals, such as in collisions.
冲量定义为合外力与作用时间的乘积：Impulse = FΔt。由牛顿第二定律的动量形式 F = Δp/Δt，两边乘以 Δt 得 FΔt = Δp。这就是冲量–动量定理：作用在物体上的冲量等于其动量的变化。该定理在处理碰撞等短时间内力变化的问题时特别有用。
6. Deriving Kinetic Energy Formula | 推导动能公式
gy, so KE = ½mv². The derivation can be extended to an initial velocity u, giving the work–energy theorem: net work = ½mv² – ½mu².
考虑质量为 m 的物体在恒合外力 F 作用下从静止开始加速，位移为 s。合力做功 W = Fs。利用 F = ma 和运动学关系 v² = 2as（因 u = 0），代入 a = v²/(2s)，得 W = m × (v²/(2s)) × s = ½mv²。这部分功以动能形式储存起来，因此 KE = ½mv²。该推导可推广至初速度为 u 的情形，得到动能定理：合外力做功 = ½mv² – ½mu²。
7. Deriving Work Done by a Constant Force | 推导恒力做功
r, cosθ = 0 and no work is done. This formula connects mechanical work to energy transfer.
当恒力以与位移方向成 θ 角作用在物体上时，做功等于位移与力沿位移方向分量的乘积，即 W = Fs cosθ。若力与位移平行，cosθ = 1，W = Fs；若相互垂直，cosθ = 0，不做功。该公式将机械功与能量传递联系起来。
8. Deriving Resistance and Resistivity | 推导电阻与电阻率
sidering that doubling the length doubles the number of obstacles electrons encounter, while doubling the area halves the resistance by providing a wider path. The formula is essential for designing circuits and understanding material properties.
一段均匀导体的电阻 R 与其长度 L 成正比，与横截面积 A 成反比，比例常数即为材料的电阻率 ρ。因此 R = ρL/A。从微观角度可理解为：长度加倍使电子遇到的碰撞次数加倍，电阻翻倍；面积加倍则提供了更宽的导电通道，电阻减半。该公式对电路设计和理解材料性质至关重要。
9. Deriving Series Resistance Formula | 推导串联电阻公式
individual p.d.s: V = V₁ + V₂ + V₃ + ... Using Ohm’s law V = IR for each term, we have IRtotal = IR₁ + IR₂ + IR₃ + ... Dividing by I gives Rtotal ... This simple additive formula applies only when components share the same current path.
对于串联的电阻器，通过每个电阻器的电流 I 相同。整个组合两端的总电势差 V 等于各个电势差之和：V = V₁ + V₂ + V₃ + ... 对每一项应用欧姆定律 V = IR，得 IRtotal = IR₁ + IR₂ + IR₃ + ... 两边除以 I，得 Rtotal = R₁ + R₂ + R₃ + ... 该简单的相加公式仅适用于各元件处于同一电流通路的情形。
10. Deriving Parallel Resistance Formula | 推导并联电阻公式
plits into the branch currents: I = I₁ + I₂ + I₃ + ... Applying Ohm’s law, I = V/R, so V/Rtotal = V/R₁ + V/R₂ + V/R₃ + ... Cancelling V yields 1/Rtotal = 1/R₁ + 1/R₂ + 1/R₃ + ... For two resistors, this simplifies to Rtotal (R₁R₂)/(R₁ + R₂). The derivation highlights the reciprocal nature of parallel resistance.
在并联连接中，各电阻器两端的电势差 V 相同，但总电流 I 分流到各支路：I = I₁ + I₂ + I₃ + ... 应用欧姆定律 I = V/R，于是 V/Rtotal = V/R₁ + V/R₂ + V/R₃ + ... 消去 V 得 1/Rtotal = 1/R₁ + 1/R₂ + 1/R₃ + ... 对于两个电阻，可简化为 Rtotal = (R₁R₂)/(R₁ + R₂)。这一推导体现了并联电阻的倒数关系。
11. Deriving Electrical Power Formulas | 推导电功率公式
P = I²R, or using I = V/R to get P = V²/R. These three equivalent formulas allow flexibility when analysing circuits.
功率是能量转换的速率。在电器元件中，电势差 V 定义为单位电荷 Q 所转移的能量 W：V = W/Q。电流 I 是单位时间流过的电荷：I = Q/t。因此，功率 P = W/t = (VQ)/t = V × (Q/t) = VI。利用欧姆定律 V = IR 代入得 P = I²R，或代入 I = V/R 得 P = V²/R。这三个等价公式为电路分析提供了灵活性。
12. Deriving Young’s Modulus | 推导杨氏模量
/ (ΔL/L) = FL/(AΔL). This equation is valid within the linear elastic region of the stress–strain graph, where Hooke’s law applies and the modulus is constant.
杨氏模量 E 是材料刚度的量度，定义为拉伸应力与拉伸应变的比值。应力 σ 是单位横截面积 A 上所施加的力 F：σ = F/A。应变 ε 是单位原始长度 L 上的伸长量 ΔL：ε = ΔL/L。因此 E = σ/ε = (F/A) / (ΔL/L) = FL/(AΔL)。该方程在应力–应变图的线弹性区域内成立，此时胡克定律适用，模量为定值。
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