Applying Calculus to Mechanics: Kinematics, Forces and Variable Acceleration | 微积分在力学中的应用：运动学、力与变加速度

📚 Applying Calculus to Mechanics: Kinematics, Forces and Variable Acceleration | 微积分在力学中的应用：运动学、力与变加速度

In Edexcel A-Level Mathematics, Mechanics questions often require you to move between displacement, velocity and acceleration by using differentiation and integration. When acceleration is not constant, the familiar suvat equations are no longer valid, so calculus becomes the main tool. This article reviews the core ideas, worked examples and common exam pitfalls for applying calculus to straight-line motion, forces, work, impulse and vector kinematics.

在 Edexcel A-Level 数学中，力学题常要求你通过微分和积分在位移、速度与加速度之间转换。当加速度不是常量时，熟悉的 suvat 公式不再适用，微积分便成为主要工具。本文复习核心概念、例题及考试常见易错点，涵盖直线运动、力、功、冲量与向量运动学。

1. Differentiation: From Displacement to Acceleration | 微分：从位移到加速度

If a particle moves in a straight line and its displacement s metres from a fixed origin at time t seconds is given by s(t), then velocity v and acceleration a are the first and second derivatives of s with respect to t.

如果质点沿直线运动，其在时间 t 秒时相对固定原点的位移 s（米）由函数 s(t) 给出，则速度 v 是 s 对 t 的一阶导数，加速度 a 是二阶导数。

v = ds/dt

a = dv/dt = d²s/dt²

For example, if s = t³ – 6t² + 9t, then differentiating once gives v = 3t² – 12t + 9, and differentiating again gives a = 6t – 12.

例如，若 s = t³ – 6t² + 9t，则求导一次得 v = 3t² – 12t + 9，再求导得 a = 6t – 12。

Always quote the derivative rule when moving from s to v or from v to a in an exam, because Edexcel mark schemes often award method marks for the correct derivative relationship even if the simplification goes wrong.

考试中从 s 求 v 或从 v 求 a 时，一定要写出导数关系式，因为 Edexcel 评分标准通常会给方法分，即使后续化简有误。

2. Integration: From Acceleration Back to Displacement | 积分：从加速度回到位移

To reverse the process, use integration. If acceleration a(t) is known and the initial velocity at t = 0 is u, then the velocity at time t is given by a definite integral from 0 to t.

反过来则使用积分。如果已知加速度 a(t)，且 t = 0 时的初速度为 u，则任意时间 t 的速度可由 0 到 t 的定积分表示。

v(t) = u + ∫₀ᵗ a(τ) dτ

Similarly, if the initial displacement at t = 0 is s₀, then displacement is obtained by integrating velocity.

类似地，若 t = 0 时的初位移为 s₀，则位移由速度积分得到。

s(t) = s₀ + ∫₀ᵗ v(τ) dτ

Using a dummy variable τ inside the integral avoids confusion with the upper limit t. This notation is especially useful when acceleration is given as a function of t.

在积分内使用哑变量 τ 可以避免与上限 t 混淆。当加速度以 t 的函数给出时，这种写法尤为有用。

3. Using Initial Conditions to Fix Constants | 利用初始条件确定常数

When you integrate indefinitely, you must add an arbitrary constant and then use boundary information to find it. A typical Edexcel question gives acceleration a = 6t – 12, initial velocity v(0) = 9 and initial displacement s(0) = 0.

不定积分时必须加上任意常数，再利用边界条件求该常数。Edexcel 常见题会给出 a = 6t – 12，初速度 v(0) = 9，初位移 s(0) = 0。

First integrate acceleration to find velocity.

首先对加速度积分求速度。

v = ∫(6t – 12)dt = 3t² – 12t + C

Use v(0) = 9 to obtain C = 9, so v = 3t² – 12t + 9.

利用 v(0) = 9 得 C = 9，因此 v = 3t² – 12t + 9。

Then integrate velocity to find displacement.

再对速度积分求位移。

s = ∫(3t² – 12t + 9)dt = t³ – 6t² + 9t + D

Since s(0) = 0, the constant D = 0, giving s = t³ – 6t² + 9t.

由于 s(0) = 0，常数 D = 0，因此 s = t³ – 6t² + 9t。

Do not use suvat here: those equations only apply when acceleration is constant. In this example a = 6t – 12 depends on t, so calculus is required.

此处不能使用 suvat：这些公式仅在加速度为常量时成立。本例中 a = 6t – 12 依赖于 t，因此必须使用微积分。

4. Rest, Direction Changes and Maximum Speed | 静止、转向与最大速度

A particle is at rest when its velocity is zero. Solving v = 0 gives the times when the particle changes direction, because velocity changes sign at these points.

质点静止时速度为零。解 v = 0 可得到质点改变方向的时刻，因为速度在这些点会变号。

For the velocity v = 3t² – 12t + 9, factorise to get v = 3(t – 1)(t – 3), so the particle is at rest at t = 1 s and t = 3 s.

对于速度 v = 3t² – 12t + 9，因式分解得 v = 3(t – 1)(t – 3)，因此质点在 t = 1 s 和 t = 3 s 时静止。

To find the maximum or minimum velocity, set acceleration equal to zero. Here a = 6t – 12, so a = 0 at t = 2. Substituting gives v(2) = 3(4) – 12(2) + 9 = -3 m s⁻¹, which is the minimum velocity.

要求速度的最大值或最小值，令加速度为零。此处 a = 6t – 12，因此 t = 2 时 a = 0。代入得 v(2) = 3(4) – 12(2) + 9 = -3 m s⁻¹，这是最小速度。

In a velocity-time graph, the particle moves in one direction when v > 0 and the opposite direction when v < 0. Marking the intervals between rest times helps you see the journey clearly.

在速度-时间图中，v > 0 时质点沿正方向运动，v < 0 时沿反方向运动。在静止时刻之间标出区间可以帮助你清晰地理解运动过程。

5. Distance Versus Displacement | 路程与位移的区别

Displacement is the net change in position, so it is found directly from s(t₂) – s(t₁). Distance travelled is the total length of the path, so you must integrate the absolute value of velocity.

位移是位置的净变化，因此直接由 s(t₂) – s(t₁) 求出。路程是运动路径的总长度，因此必须对速度的绝对值积分。

distance = ∫ |v(t)| dt

Consider s = t³ – 6t² + 9t from t = 0 to t = 4. The velocity v = 3(t – 1)(t – 3) has sign changes at t = 1 and t = 3.

考虑 s = t³ – 6t² + 9t 从 t = 0 到 t = 4。速度 v = 3(t – 1)(t – 3) 在 t = 1 和 t = 3 处变号。

Interval	Sign of v	Direction
0 < t < 1	positive	forward
1 < t < 3	negative	backward
3 < t < 4	positive	forward

The displacement is s(4) – s(0) = 4 – 0 = 4 m. The total distance is found by splitting the integral at the rest times.

位移为 s(4) – s(0) = 4 – 0 = 4 m。总路程需要在静止时刻处将积分分段计算。

distance = ∫₀¹ v dt + |∫₁³ v dt| + ∫₃⁴ v dt

= 4 + |-4| + 4 = 12 m

Many candidates lose marks by giving 4 m as the distance. Whenever velocity changes sign, displacement and distance are different.

许多考生因把 4 m 当作路程而失分。只要速度变号，位移和路程就不同。

6. Newton’s Second Law with a Variable Force | 变力下的牛顿第二定律

Newton’s second law states that force F, mass m and acceleration a are linked by F = ma. If the force is given as a function of time, then acceleration is also a function of time and can be integrated.

牛顿第二定律指出力 F、质量 m 和加速度 a 满足 F = ma。若力以时间的函数给出，则加速度也是时间的函数，可以进行积分。

A particle of mass 2 kg moves in a straight line under a force F = 6t N. The initial velocity is 3 m s⁻¹, and the particle starts at the origin.

一个质量为 2 kg 的质点沿直线在力 F = 6t N 作用下运动。初速度为 3 m s⁻¹，质点从原点出发。

First find acceleration: a = F/m = 6t/2 = 3t m s⁻².

先求加速度：a = F/m = 6t/2 = 3t m s⁻²。

Then integrate: v = 3 + ∫₀ᵗ 3τ dτ = 3 + 1.5t².

然后积分：v = 3 + ∫₀ᵗ 3τ dτ = 3 + 1.5t²。

Integrate again for displacement: s = ∫₀ᵗ (3 + 1.5τ²) dτ = 3t + 0.5t³.

再次积分求位移：s = ∫₀

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Applying Calculus to Mechanics: Kinematics, Forces and Variable Acceleration | 微积分在力学中的应用：运动学、力与变加速度