📚 Essential Formula Derivations for Oxford AQA Physics | 牛津AQA物理核心公式推导
Strong conceptual understanding in A-level Physics hinges on being able to derive key formulas from first principles. For the Oxford AQA specification, this skill not only helps you answer ‘show that’ questions but deepens your insight into the links between mechanics, electricity, and energy. In this guide, we walk through a selection of essential derivations—step by logical step—using only fundamental definitions, graphs, and algebra. Each section presents the English reasoning followed immediately by its Chinese translation, so you can master both the language and the logic.
A-level 物理的扎实概念理解依赖于能从基本原理推导出关键公式。针对牛津AQA考纲，这项技能不仅能帮助你应对“证明”类题目，还能加深你对力学、电学和能量之间联系的认识。在本指南中，我们精选了一系列核心推导，通过定义、图像和代数的逻辑步骤一一展开。每个部分先给出英文推导，紧接着提供对应的中文翻译，让你同时掌握语言与逻辑。
1. Derivation of v = u + at | 推导 v = u + at
Start with the definition of constant acceleration: acceleration is the rate of change of velocity. Symbolically, a = (v – u) / t, where v is the final velocity, u is the initial velocity, and t is the time taken for the change. Multiplying both sides by t gives a t = v – u. Finally, add u to both sides to obtain the first SUVAT equation: v = u + a t.
从匀加速度的定义入手：加速度是速度的变化率。用符号表示为 a = (v – u) / t，其中 v 是末速度，u 是初速度，t 是速度变化所用时间。两边同乘以 t 得到 a t = v – u。最后，两边加上 u 便得到第一个 SUVAT 方程：v = u + a t。
v = u + at
This linear relationship tells us that the final velocity is simply the initial velocity plus the increase from acceleration over time. Keep in mind that it only holds when acceleration is uniform.
这个线性关系表明，末速度等于初速度加上加速度在时间内的增量。务必注意，它仅在加速度恒定时成立。
2. Derivation of s = ut + ½ at² | 推导 s = ut + ½ at²
Displacement s can be found as the area under a velocity-time graph. For constant acceleration, the graph is a straight line starting at u and reaching v after time t. The area splits into a rectangular part of height u (area = u t) and a triangular part of height (v – u) and base t. The triangle’s area is ½ (v – u) t. Substituting v – u = a t from equation 1 gives the triangle area as ½ a t². Thus, total displacement s = u t + ½ a t².
位移 s 可以由速度-时间图像下的面积求得。对于匀加速度，图像是一条从 u 出发、经过时间 t 到达 v 的直线。面积分为两部分：高度为 u 的矩形（面积 = u t）和高度为 (v – u)、底边为 t 的三角形。三角形的面积是 ½ (v – u) t。将第一个公式 v – u = a t 代入，三角形面积变为 ½ a t²。因此，总位移 s = u t + ½ a t²。
s = ut + ½ a t²
This derivation beautifully connects graphical analysis with algebra. If initial velocity is zero, the relationship simplifies to s = ½ a t², which describes motion from rest under constant force.
这个推导巧妙地将图像分析与代数连接起来。若初速度为零，关系式简化为 s = ½ a t²，它描述了静止物体在恒力作用下的运动。
3. Derivation of v² = u² + 2as | 推导 v² = u² + 2as
We can eliminate time from the first two equations. From v = u + a t, rewrite as t = (v – u) / a. Substitute this expression for t into the displacement equation: s = u [(v – u)/a] + ½ a [(v – u)/a]². Multiply through by 2a to clear fractions: 2a s = 2u(v – u) + (v – u)². Expand: 2a s = 2uv – 2u² + v² – 2uv + u² = v² – u². Rearranging gives the time-independent link: v² = u² + 2 a s.
我们可以从前两个公式中消去时间。由 v = u + a t 改写为 t = (v – u) / a。将该 t 的表达式代入位移公式：s = u [(v – u)/a] + ½ a [(v – u)/a]²。两边乘以 2a 消去分母：2a s = 2u(v – u) + (v – u)²。展开得：2a s = 2uv – 2u² + v² – 2uv + u² = v² – u²。重新整理后便得到不含时间的关系式：v² = u² + 2 a s。
v² = u² + 2as
This equation is particularly useful when time is not given or required. For example, it helps calculate the braking distance of a car given its initial speed and deceleration.
当题目未给出或不要求时间时，该方程尤为实用。例如，它可用于计算给定初速度和减速度时汽车的刹车距离。
4. Derivation of Kinetic Energy (Eₖ = ½ mv²) | 推导动能 (Eₖ = ½ mv²)
Consider an object of mass m accelerating from rest under a constant net force F over a displacement s. The work done on the object is W = F s. By Newton’s second law, F = m a. From the kinematic equation v² = u² + 2a s, and taking u = 0, we get v² = 2a s, so a s = v² / 2. Substituting into the work expression: W = F s = (m a) s = m (v² / 2) = ½ m v². Since the work done is transferred to kinetic energy, Eₖ = ½ m v².
考虑一个质量为 m 的物体在恒定合力 F 作用下从静止开始加速，移动位移 s。对物体做的功为 W = F s。根据牛顿第二定律，F = m a。由运动学方程 v² = u² + 2a s，令 u = 0 得 v² = 2a s，因此 a s = v² / 2。代入功的表达式：W = F s = (m a) s = m (v² / 2) = ½ m v²。由于所做的功转化为动能，所以 Eₖ = ½ m v²。
Eₖ = ½ m v²
This shows that kinetic energy depends on the square of velocity. Doubling the speed quadruples the kinetic energy, a fact visible in vehicle stopping distances.
这显示出动能取决于速度的平方。速度加倍将使动能变为原来的四倍，这一点在车辆刹车距离中显而易见。
5. Derivation of Gravitational Potential Energy (Eₚ = mgh) | 推导重力势能 (Eₚ = mgh)
To lift an object of mass m vertically at a constant speed, an upward force equal to its weight (mg) must be applied. The work done against gravity over a height h is W = force × distance = mg × h. This work is stored as gravitational potential energy, provided the height change is small enough that g remains constant. Therefore, the change in gravitational potential energy ΔEₚ = m g h.
要匀速竖直提升一个质量为 m 的物体，必须施加一个与其重力 (mg) 大小相等的向上的力。克服重力在高度 h 上所做的功为 W = 力 × 距离 = mg × h。只要高度变化足够小、g 保持恒定，这些功便以重力势能的形式储存。因此，重力势能的变化量 ΔEₚ = m g h。
Eₚ = mgh
In Oxford AQA problems, you often take the reference point where h = 0 as ground level. This simple linear relationship underpins hydroelectric power calculations and energy conservation problems.
在牛津AQA的题目中，常取 h = 0 处为地面参考点。这一简单线性关系是水力发电计算和能量守恒问题的基础。
6. Derivation of Elastic Potential Energy (E = ½ kx²) | 推导弹性势能 (E = ½ kx²)
When a spring is stretched or compressed by a displacement x, the restoring force follows Hooke’s law F = k x, where k is the spring constant. The force varies linearly from 0 to k x, so the average force exerted during the stretch is ½ k x. The work done to deform the spring is average force × displacement = (½ k x) × x = ½ k x². This work is stored as elastic potential energy: E = ½ k x².
当弹簧被拉伸或压缩一位移 x 时，恢复力遵循胡克定律 F = k x，其中 k 为劲度系数。力从 0 线性增加到 k x，因此拉伸过程中的平均力为 ½ k x。使弹簧形变所做的功等于平均力 × 位移 = (½ k x) × x = ½ k x²。这些功以弹性势能的形式储存：E = ½ k x²。
E = ½ k x²
Note the quadratic dependence on extension. The area under the force-extension graph (a triangle) also gives the same result: ½ (base x height) = ½ x (k x) = ½ k x².
注意能量与伸长量的二次关系。力-伸长量图像下的面积（一个三角形）也可得到相同结果：½ × (底 × 高) = ½ x (k x) = ½ k x²。
7. Derivation of Resistivity Formula (R = ρ L / A) | 推导电阻率公式 (R = ρ L / A)
Resistance R of a conductor is directly proportional to its length L and inversely proportional to its cross-sectional area A. Introducing the constant of proportionality, resistivity ρ, gives R ∝ L / A → R = ρ L / A. To derive this, consider a wire under a potential difference V. The electric field E = V / L, and current density J = I / A. Using the microscopic form of Ohm’s law, E = ρ J, we substitute: V / L = ρ (I / A). Rearranging for V / I = R yields R = ρ L / A.
导体的电阻 R 与其长度 L 成正比、与其横截面积 A 成反比。引入比例常数电阻率 ρ，得到 R ∝ L / A → R = ρ L / A。为推导该式，考虑一根存在电势差 V 的导线。电场 E = V / L，电流密度 J = I / A。利用欧姆定律的微观形式 E = ρ J，代入得：V / L = ρ (I / A)。整理 V / I = R 即得 R = ρ L / A。
R = ρ L / A
This derivation links macroscopic resistance to the material’s intrinsic property. Longer wires have greater resistance, while thicker wires (larger A) have lower resistance.
这个推导将宏观电阻与材料的固有性质联系起来。导线越长，电阻越大；而较粗的导线（A 更大）电阻更小。
8. Derivation of Electrical Power (P = I²R) | 推导电功率 (P = I²R)
Power is defined as the rate of energy transfer: P = W / t. The work done moving a charge Q through a potential difference V is W = Q V. Hence, P = (Q V) / t. Since current I = Q / t, we have P = I V. If the component obeys Ohm’s law, V = I R, then substituting gives P = I (I R) = I² R. Power can also be written as P = V² / R using Ohm’s law to eliminate I.
功率定义为能量转移的速率：P = W / t。将电荷 Q 移动通过电势差 V 所做的功为 W = Q V。因此，P = (Q V) / t。由于电流 I = Q / t，可得 P = I V。若元件遵循欧姆定律 V = I R，代入后得 P = I (I R) = I² R。利用欧姆定律消去 I，功率也可写作 P = V² / R。
P = I² R = V² / R
These alternative forms are handy for analysing power loss in transmission lines (P ∝ I²) or for comparing appliances operating at the same voltage (P ∝ 1/R).
这些替代形式便于分析输电线中的功率损耗 (P ∝ I²) 或比较在相同电压下工作的电器的功率 (P ∝ 1/R)。
9. Derivation of Impulse-Momentum Theorem (F Δt = Δp) | 推导冲量-动量定理 (F Δt = Δp)
Newton’s second law states that the resultant force on an object equals the rate of change of its momentum: F = Δp / Δt. Multiply both sides by the time interval Δt to get F Δt = Δp. The quantity F Δt is called impulse, and it equals the change in momentum. For constant mass m, Δp = m(v – u), so F Δt = m v – m u.
牛顿第二定律指出，物体所受合外力等于其动量的变化率：F = Δp / Δt。两边同乘以时间间隔 Δt 得到 F Δt = Δp。物理量 F Δt 称为冲量，它等于动量的变化量。对于质量不变的物体，Δp = m(v – u)，因此 F Δt = m v – m u。
F Δt = Δp = m v – m u
This theorem is crucial for understanding collisions and safety features like airbags, which increase Δt to reduce the average force for a given momentum change.
该定理对于理解碰撞行为和像安全气囊这样的安全装置至关重要——安全气囊通过延长作用时间 Δt 来减少某一给定动量变化下的平均作用力。
10. Derivation of Centripetal Acceleration (a = v² / r) | 推导向心加速度 (a = v² / r)
An object moving at constant speed v in a circle of radius r experiences a continuous change in direction. In a short time Δt, the object moves through an angle Δθ, tracing an arc of length v Δt. By geometry, arc length = r Δθ, so Δθ = v Δt / r. The velocity vector rotates by the same angle Δθ. The magnitude of the velocity change is |Δv| ≈ v Δθ (using small-angle approximation). Substituting for Δθ gives |Δv| = v (v Δt / r) = v² Δt / r. The acceleration magnitude a = |Δv| / Δt = v² / r, directed toward the centre.
以恒定速率 v 在半径为 r 的圆周上运动的物体，其方向在不断改变。在短暂的时间 Δt 内，物体转过角度 Δθ，描出一条弧长 v Δt。由几何关系，弧长 = r Δθ，故 Δθ = v Δt / r。速度矢量转过相同的角度 Δθ。速度变化量的大小为 |Δv| ≈ v Δθ（利用小角度近似）。代入 Δθ 得 |Δv| = v (v Δt / r) = v² Δt / r。加速度大小 a = |Δv| / Δt = v² / r，方向指向圆心。
a = v² / r
This result reveals that centripetal acceleration depends on the square of the speed and inversely on the radius. It is essential for studying satellite orbits, banked tracks, and the loop-the-loop problem.
这一结果表明向心加速度取决于速度的平方，并与半径成反比。它是研究卫星轨道、倾斜弯道和竖直圆环问题不可或缺的基础。
Published by TutorHao | Physics Revision Series | aleveler.com
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