📚 Variable Acceleration in One Dimension | 一维变加速运动
In A-Level Mechanics, we frequently extend the constant acceleration (suvat) equations to more realistic models where acceleration changes with time, velocity or displacement. This topic, variable acceleration in one dimension, is a fundamental part of the Edexcel specification and relies heavily on differential and integral calculus. Understanding how displacement, velocity and acceleration relate through rates of change and accumulation allows you to tackle a wide variety of motion problems, from simple variable forces to oscillations. Here, we will systematically explore the key concepts, techniques and common pitfalls, equipping you with the tools to handle any one-dimensional variable acceleration question confidently.
在A-Level力学中，我们常常将匀加速(suvat)方程扩展到加速度随时间、速度或位移变化的更真实模型。一维变加速运动是Edexcel考纲的基础部分，高度依赖微分和积分运算。理解位移、速度和加速度如何通过变化率与累积量相互关联，使你能够解决从简单变力到振动的各种运动问题。这里我们将系统梳理核心概念、求解技巧与常见误区，让你拥有应对任何一维变加速问题的能力。
1. The Calculus of Motion | 运动的微积分联系
In one-dimensional kinematics, we treat displacement s, velocity v and acceleration a as functions of time t. The fundamental relationships are differential: velocity is the rate of change of displacement, v = ds/dt, and acceleration is the rate of change of velocity, a = dv/dt = d2s/dt2. These derivatives are instantaneous measures, capturing how quickly the motion evolves at any given moment. When we need to reverse the process, integration becomes the key tool: given acceleration as a function of time, we integrate to recover velocity; given velocity, we integrate to recover displacement. Every integration introduces an arbitrary constant that must be pinned down by initial conditions, such as the velocity or position at t asp of this calculus backbone is essential before tackling variable acceleration problems.
在一维运动学中，我们将位移 s、速度 v 和加速度 a 视为时间 t 的函数。最基本的关系是微分：速度是位移的变化率，v = ds/dt；加速度是速度的变化率，a = dv/dt = d2s/dt2。这些导数捕捉了任意瞬时运动的快慢变化。当需要逆向求解时，积分成为核心工具：若加速度为时间函数，积分可求得速度；再由速度积分求得位移。每次积分引入一个任意常数，必须通过初始条件（如 t=0 时的速度或位置）来确定。在解决变加速问题之前，牢固掌握这一微积分基础至关重要。
2. Acceleration as a Function of Time | 加速度表示为时间的函数
The most straightforward variable acceleration scenario provides a explicitly as a function of t, for instance a = 6t + 2 or a = 4 – 3t2. Since the acceleration is only dependent on time, we can directly integrate to find the velocity function. Starting from a = dv/dt, we write dv = a(t) dt and integrate both sides: v = ∫ a(t) dt + C. This yields an expression for velocity with one unknown constant. A second integration, s = ∫ v(t) dt + K, gives the displacement. The algebra is usually simple, but the key is to carry the constants carefully and use the given boundary values to determine them.
最直接的变加速情形给出加速度 a 明确表示为 t 的函数，例如 a = 6t + 2 或 a = 4 – 3t2。因为加速度只依赖于时间，我们可以直接积分求得速度函数。从 a = dv/dt 出发，写出 dv = a(t) dt 并两边积分：v = ∫ a(t) dt + C。这样得到含有一个未知常数的速度表达式。再进行一次积分 s = ∫ v(t) dt + K 即得位移。代数过程通常不复杂，关键在于小心处理常数，并利用给定的边界值确定它们。
3. Finding Velocity from a(t) – Indefinite Integration | 从 a(t) 求速度——不定积分
Let's work through an example: a particle moves along a straight line with acceleration a = 12t – 4 m s-2. Integrating with respect to time gives the velocity: v = ∫ (12t – 4) dt = 6t2 – 4t + C. The constant C represents the initial velocity v(0). If we are not given an initial condition, the answer remains in terms of C. When the initial velocity is known, say v(0) = 3, we substitute t=0, v=3 into the expression: 3 = 0 – 0 + C ⇒ C=3, hence v = 6t2 – 4t + 3. Always remember that indefinite integration yields a family of functions, and the specific motion is selected by the initial condition.
我们来做一个例题：一个质点沿直线运动，加速度为 a = 12t – 4 m s-2。对时间积分得速度：v = ∫ (12t – 4) dt = 6t2 – 4t + C。常数 C 代表初始速度 v(0)。若未给出初始条件，答案保留含 C 的形式。当已知初速度，比如 v(0)=3，将 t=0, v=3 代入表达式：3 = 0 – 0 + C ⇒ C=3，于是 v = 6t2 – 4t + 3。务必记住，不定积分得到的是一族函数，具体运动由初始条件确定。
4. Using Initial Conditions to Find Constants | 利用初始条件确定常数
Initial conditions are the bridge between abstract integrals and the particular physical situation. Typically, you are given the velocity v (or displacement s) at a specific time, often t = 0. For a(t) problems, after the first integration we obtain v(t) containing C; plugging in t = 0 and v = v0 gives C = v0. After the second integration, we get s(t) with another constant K. Using a known displacement at t = 0, say s(0) = s0, determines K. Be careful: if the initial condition is given at a time other than zero, you must substitute that value for t. Always label your constants clearly – it is easy to confuse them when solving more complex problems.
初始条件是连接抽象积分与具体物理情境的桥梁。通常题目给出特定时刻（多为 t=0）的速度 v 或位移 s。对于 a(t) 类问题，第一次积分后得到含 C 的 v(t)；代入 t=0 及 v=v0 即得 C=v0。第二次积分后得到含另一常数 K 的 s(t)，利用已知的初始位移 s(0)=s0 求出 K。注意：若初始条件给出的是非零时刻，就必须代入相应的 t 值。务必清晰标注常数——在更复杂的问题中很容易混淆它们。
5. Determining Displacement from Velocity | 由速度确定位移
Once the full velocity function v(t) is known, displacement is obtained by a further integration: s(t) = ∫ v(t) dt + K. For example, using the earlier velocity v = 6t2 – 4t + 3, we integrate to find s = 2t3 – 2t2 + 3t + K. If the particle starts at the origin, s(0)=0 ⇒ K=0. If it starts 5 m to the right, s(0)=5 ⇒ K hen tells us the position at any time. Remember, displacement is a vector in one dimension – ive the opposite. This is crucial when interpreting whether the particle passes through the origin or changes direction.
一旦求出了完整的速度函数 v(t)，再经一次积分即得位移：s(t) = ∫ v(t) dt + K。例如，用之前的速度 v = 6t2 – 4t + 3，积分得 s = 2t3 – 2t2 + 3t + K。若质点从原点出发，s(0)=0 ⇒ K=0；若从右侧5米处出发，s(0)=5 ⇒ K=5。由此得到的位移函数告诉我们任意时刻的位置。记住，位移是带有方向的一维矢量——正号表示某一方向，负号表示相反方向。这在判断质点是否经过原点或改变运动方向时至关重要。
6. Acceleration as a Function of Displacement | 加速度表示为位移的函数
Some of the most challenging Edexcel questions give acceleration in terms of displacement, e.g. a = –4x or a = 2x3. Here we cannot simply integrate with respect to time because a is not directly a function of t. Instead, we use the alternative form of acceleration derived from the chain rule: a = v dv/dx. This allows us to relate velocity and displacement without time. We rearrange as v dv = a(x) dx and integrate both sides. The resulting expression links v2 to x, much like the energy-conservation idea in physics. The constant of integration is again determined by a known pairing of velocity and displacement, typically an initial or boundary condition.
部分最具挑战的Edexcel问题给出加速度关于位移的函数，如 a = –4x 或 a = 2x3。此时不能简单对时间积分，因为加速度并不是时间的显函数。我们转而应用由链式法则导出的加速度另一形式：a = v dv/dx。这让我们无须时间就能建立速度与位移的关系。重排得到 v dv = a(x) dx，两边积分。得出的表达式将 v2 与 x 联系起来，与物理学中的能量守恒思想类似。积分常数同样由已知的速度-位移对应关系（通常为初始或边界条件）确定。
7. The Chain Rule: a = v dv/dx | 链式法则：a = v dv/dx
The equivalence a = v dv/dx is not merely a trick – it is a direct consequence of a = dv/dt and v = dx/dt. By the chain rule, dv/dt = (dv/dx)(dx/dt) = v dv/dx. Thus, when a is given as a function of x, we set v dv/dx = a(x) and integrate. A neat integrated form is ∫ v dv = ∫ a(x) dx, which yields ½v2 = ∫ a(x) dx + constant. This form is particularly useful for finding the speed at a given position without solving for time. In exams, presenting clear steps showing the separation of variables v and x will gain method marks even if the integration is challenging.
表达式 a = v dv/dx 并非只是技巧，它是 a = dv/dt 和 v = dx/dt 的直接推论。由链式法则，dv/dt = (dv/dx)(dx/dt) = v dv/dx。因此，当 a 已知为 x 的函数时，设 v dv/dx = a(x) 并积分。一个简洁的积分形式是 ∫ v dv = ∫ a(x) dx，得出 ½v2 = ∫ a(x) dx + 常数。这一形式对于求给定位置的速度而无需解出时间特别有用。考试时，清晰地展示分离变量 v 和 x 的步骤，即使积分较复杂也能获得方法分。
8. Solving Problems with a(x) | 解决 a(x) 相关问题
Consider a particle where a = –9x. Using v dv/dx = –9x, we separate: ∫ v dv = ∫ –9x dx → ½v2 = –(9/2)x2 + C. If the particle is released from rest at x = 2 m, then v=0 when x=2 gives 0 = –18 + C ⇒ C=18. Hence ½v2 = 18 – 4.5x2, so v = ±√36 – 9x2. The sign depends on the direction of motion. This describes simple harmonic-like behaviour, and you can then go further to find v at x=1, or the maximum displacement where v ns the door to a whole class of mechanics problems that move beyond constant acceleration.
考虑一质点，其加速度 a = –9x。利用 Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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