📚 Essential AS Physics Formula Derivations | 国际AS物理核心公式推导
In International AS Physics, understanding how key formulas are derived is just as important as using them. This article walks through the step‑by‑step derivations of the most fundamental equations in mechanics, energy, momentum and gravitation. Each derivation is presented clearly, using only symbols and explanations that appear on standard formula sheets, so you can master both the reasoning and the result.
在国际AS物理课程中，理解核心公式的推导过程与运用公式同样重要。本文将逐步推演力学、能量、动量和引力中最基础的方程。每个推导都只用标准公式表上的符号和清晰的说明，帮助你同时掌握推理过程和最终结果。
1. Derivation of v = u + at | 推导速度‑时间关系式
Start from the definition of acceleration as the rate of change of velocity: a = (v − u) / t, where a is constant acceleration, u is initial velocity, v is final velocity after time t. Rearranging this relation directly gives v = u + a t. This is the first of the SUVAT equations for uniformly accelerated motion along a straight line.
从加速度的定义——速度的变化率出发：a = (v − u) / t，其中a是恒定加速度，u是初速度，v是时间t后的末速度。直接移项就得到v = u + a t。这是匀加速直线运动 SUVAT 方程组中的第一式。
2. Derivation of s = ut + ½ at² | 推导位移‑时间关系式
For constant acceleration, average velocity vₐᵥ = (u + v) / 2. Substituting the expression for v from the first equation, vₐᵥ = (u + u + at) / 2 = u + ½ a t. Displacement s equals average velocity multiplied by time: s = vₐᵥ × t = (u + ½ a t) t, hence s = u t + ½ a t². This relation gives displacement without needing the final velocity.
在恒定加速度下，平均速度 vₐᵥ = (u + v) / 2。将第一式中的 v 代入，得 vₐᵥ = (u + u + at) / 2 = u + ½ a t。位移 s 等于平均速度乘以时间：s = vₐᵥ × t = (u + ½ a t) t，因此 s = u t + ½ a t²。该关系式可以在不知道末速度时求出位移。
3. Derivation of v² = u² + 2as | 推导速度‑位移关系式
Eliminate time t from the first two equations. From v = u + at, we have t = (v − u) / a. Substitute this into s = u t + ½ a t²:
s = u[(v − u)/a] + ½ a[(v − u)/a]²
= (u(v − u))/a + (v − u)²/(2a).
Multiply throughout by 2a:
2a s = 2u(v − u) + (v − u)²
= 2u v − 2u² + v² − 2u v + u²
= v² − u².
Rearrange to obtain v² = u² + 2 a s. This formula is particularly useful when time is not known.
从前两式中消去时间t。由 v = u + at 得 t = (v − u) / a，代入 s = u t + ½ a t²：
s = u[(v − u)/a] + ½ a[(v − u)/a]²
= (u(v − u))/a + (v − u)²/(2a)。
两边同乘以 2a：
2a s = 2u(v − u) + (v − u)²
= 2u v − 2u² + v² − 2u v + u²
= v² − u²。
整理得 v² = u² + 2 a s。在不涉及时间的问题中，这个公式非常实用。
4. Deriving Kinetic Energy Formula Eₖ = ½mv² | 推导动能公式
Consider a constant net force F acting on a mass m that accelerates from rest to speed v over a displacement s. The work done by the force is W = F s. From Newton’s second law, F = m a, and from v² = u² + 2a s with u = 0, we have v² = 2 a s, so s = v² / (2a). Substituting both into the work expression: W = (m a) × (v² / (2a)) = ½ m v². This work is stored as kinetic energy, so Eₖ = ½ m v².
若一个恒定的合外力 F 作用在质量 m 上，使其从静止加速到位移 s 后达到速度 v。力做的功 W = F s。由牛顿第二定律 F = m a，并由 v² = u² + 2a s 取 u = 0 得 v² = 2 a s，即 s = v² / (2a)。代入功的表达式：W = (m a) × (v² / (2a)) = ½ m v²。这些功储存为动能，因此 Eₖ = ½ m v²。
5. Derivation of Momentum and Impulse Relationship | 推导动量与冲量关系
Newton’s second law in its general form states that net force equals the rate of change of momentum: F = Δp / Δt. For a constant force acting over a time interval Δt, impulse J = F Δt. Since Δp = m v − m u, the impulse‑momentum theorem follows directly: J = F Δt = Δp = m(v − u). This shows that the impulse applied to an object equals its change in momentum.
牛顿第二定律的普遍形式为合外力等于动量的变化率：F = Δp / Δt。对于在时间间隔 Δt 内作用的恒力，冲量 J = F Δt。因为 Δp = m v − m u，直接得到冲量‑动量定理：J = F Δt = Δp = m(v − u)。这表明物体受到的冲量等于其动量的变化。
6. Derivation of Power in Terms of Force and Velocity | 推导功率的力与速度表达式
Power P is the rate of doing work: P = W / t. For a constant force F acting in the direction of motion, the work done over a small displacement Δs is W = F Δs. Therefore P = F Δs / t = F v, where v = Δs / t is the instantaneous velocity. When the force and velocity are not parallel, the dot product form P = F v cos θ is used.
功率 P 是做功的快慢：P = W / t。对于沿运动方向的恒力 F，在微小位移 Δs 上做的功为 W = F Δs。因此 P = F Δs / t = F v，其中 v = Δs / t 是瞬时速度。当力与速度不平行时，采用点积形式 P = F v cos θ。
7. Derivation of Centripetal Acceleration a = v²/r | 推导向心加速度公式
For an object moving at constant speed v in a circle of radius r, the position vector rotates through an angle Δθ in time Δt. The velocity vector also rotates by the same Δθ, changing direction but not magnitude. The change in velocity Δv has magnitude v Δθ (for small angles). Since Δθ = v Δt / r, we have Δv = v (v Δt / r) = v² Δt / r. Acceleration magnitude a = Δv / Δt = v² / r, directed toward the centre.
一物体以恒定速率 v 沿半径为 r 的圆周运动，其位置矢量在时间 Δt 内转过角度 Δθ。速度矢量也转过相同的 Δθ，方向改变而大小不变。速度变化 Δv 的大小为 v Δθ（小角度下）。由于 Δθ = v Δt / r，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。
8. Derivation of Gravitational Potential Energy Near Earth’s Surface | 推导近地表重力势能
Near the Earth’s surface, the gravitational force on a mass m is approximately constant: F = m g downward. To lift the mass a height h at constant speed, an equal and opposite force must be applied. The work done by the lifting force is W = F d = m g h. This work is stored as gravitational potential energy, so ΔEₚ = m g h. The choice of zero level is arbitrary; only changes in Eₚ are physically meaningful.
在地球表面附近，质量为 m 的物体所受重力近似恒定：F = m g 竖直向下。若要匀速将其举升高度 h，需施加大小相等方向相反的力。举升力做的功 W = F d = m g h。这些功储存为重力势能，因此 ΔEₚ = m g h。零势能面的选取是任意的，只有势能的变化才有物理意义。
9. Derivation of Escape Velocity | 推导逃逸速度
Escaping a planet’s gravity means reaching an infinite distance with zero final kinetic energy. Using conservation of energy: initial K.E. + U = final K.E. + U∞. At the surface, K.E. = ½ m vₑ² and gravitational potential energy U = − G M m / R, where M is the planet’s mass and R its radius. At infinity, both K.E. and U are zero (taking U∞ = 0). Thus ½ m vₑ² − G M m / R = 0. Solve for vₑ: vₑ = √(2 G M / R). Since g = G M / R², this can also be written as vₑ = √(2 g R).
挣脱行星引力意味着到达无穷远处时末动能为零。运用能量守恒：初始动能 + 势能 = 末动能 + 无穷远处势能。在行星表面，动能 K.E. = ½ m vₑ²，引力势能 U = − G M m / R，其中 M 为行星质量，R 为其半径。无穷远处动能和势能均为零（取 U∞ = 0）。因此 ½ m vₑ² − G M m / R = 0。解出 vₑ：vₑ = √(2 G M / R)。由于 g = G M / R²，也可写成 vₑ = √(2 g R)。
10. Derivation of Kepler’s Third Law for Circular Orbits | 推导圆轨道开普勒第三定律
For a planet (or satellite) of mass m in a circular orbit of radius r around a central mass M, the gravitational force provides the centripetal force: G M m / r² = m v² / r. Cancel m and multiply by r: v² = G M / r. The orbital period T = 2π r / v, so v = 2π r / T. Substituting gives (2π r / T)² = G M / r, which simplifies to 4π² r² / T² = G M / r. Rearranging yields T² = (4π² / G M) r³. This shows that the square of the period is proportional to the cube of the orbital radius.
对于在中心质量 M 的引力场中以圆轨道半径 r 运行的行星（或卫星）质量 m，引力提供向心力：G M m / r² = m v² / r。约去 m 并两边乘 r 得 v² = G M / r。轨道周期 T = 2π r / v，所以 v = 2π r / T。代入得 (2π r / T)² = G M / r，简化后为 4π² r² / T² = G M / r。整理即得 T² = (4π² / G M) r³。这表明周期的平方与轨道半径的立方成正比。
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