📚 Variable Acceleration | 变加速度
In Edexcel A Level Mechanics, many motion problems involve acceleration that is not constant. The standard SUVAT equations fail in these cases. Instead, students must use calculus to link displacement, velocity and acceleration. This article covers differentiation, integration, initial conditions, turning points and the difference between displacement and distance travelled.
在 Edexcel A Level 力学中，许多运动问题的加速度并不是恒定的。此时标准的 SUVAT 公式不再适用，学生必须使用微积分来联系位移、速度和加速度。本文涵盖微分、积分、初始条件、静止时刻、转向点以及位移与路程的区别。
1. What Is Variable Acceleration? | 什么是变加速度？
en as a function such as a = 6t − 2, a = 4 − t², or a = −ω²s. Because a is not constant, equations such as v = u + at and s = ut + ½at² are no longer valid.
变加速度是指加速度随时间 t、位移 s 或速度 v 变化。在很多考题中，加速度会以函数形式给出，例如 a = 6t − 2、a = 4 − t² 或 a = −ω²s。由于 a 不是常数，v = u + at 和 s = ut + ½at² 等公式不再成立。
Calculus provides the exact relationship between motion quantities: velocity is the derivative of displacement, and acceleration is the derivative of velocity. This gives a direct chain from s to v to a, and also allows us to reverse the process by integration.
微积分给出了运动量之间的精确关系：速度是位移的导数，加速度是速度的导数。这样就能从位移 s 求速度 v，再求加速度 a，也可以通过积分反过来求解。
A key exam warning is that SUVAT only works when acceleration is constant. If a depends on t or s, you must use differentiation or integration. This single decision often determines whether a mechanics question is solved correctly.
考试中的一个重要提醒是：只有当加速度恒定时才能使用 SUVAT。如果 a 依赖于 t 或 s，就必须使用微分或积分。这个判断往往决定了一道力学题能否做对。
2. Core Definitions and Notation | 核心定义与记号
Displacement s is the position of a particle relative to a fixed origin, so it is a signed quantity. Velocity v is the rate of change of displacement. Acceleration a is the rate of change of velocity. Time t is usually the independent variable. Distance travelled is the total length of the path and is always positive.
位移 s 是质点相对于固定原点的位置，因此是有正负方向的量。速度 v 是位移的变化率。加速度 a 是速度的变化率。时间 t 通常是自变量。路程是运动路径的总长度，始终为正。
| Quantity | Symbol and Unit | Calculus Relation |
|---|---|---|
| Displacement 位移 | s, m | s = s(t) |
| Velocity 速度 | v, m s⁻¹ | v = ds/dt |
| Acceleration 加速度 | a, m s⁻² | a = dv/dt = d²s/dt² |
| Time 时间 | t, s | independent variable |
terms of t, use integration. If it gives a in terms of s, use the chain rule a = v dv/ds.
书写解答时要注意记号。如果题目给出 s 关于 t 的表达式，就用微分。如果给出 a 关于 t 的表达式，就用积分。如果给出 a 关于 s 的表达式，就用链式法则 a = v dv/ds。
3. Differentiation: From Displacement to Velocity and Acceleration | 微分：由位移求速度与加速度
tiating again: a = dv/dt = d²s/dt². This is the most direct type of variable acceleration question.
如果位移以时间的函数给出，即 s = f(t)，那么速度通过一次微分得到：v = ds/dt。加速度通过再次微分得到：a = dv/dt = d²s/dt²。这是最常见的变加速度题型。
egative velocity tells us the particle is moving in the negative direction at that instant.
例如，若 s = t³ − 4t² + 3t，则 v = 3t² − 8t + 3，a = 6t − 8。当 t = 2 时，v = 3(4) − 16 + 3 = −1 m s⁻¹，a = 12 − 8 = 4 m s⁻²。负速度说明该时刻质点沿负方向运动。
e to zero, and sums or differences are differentiated term by term. For trigonometric or exponential displacement functions, use the standard calculus results for sin t, cos t and eᵗ, and remember that calculus with trigonometric functions uses radians.
主要工具是幂函数法则：d/dt (tⁿ) = n tⁿ⁻¹。常数微分为零，和与差可以逐项微分。如果位移函数涉及三角函数或指数函数，就要使用 sin t、cos t 和 eᵗ 的标准微积分结果，并记住三角函数的微积分使用弧度。
4. Integration: From Acceleration to Velocity and Displacement | 积分：由加速度求速度与位移
city: s = ∫ v dt + D. The constants C and D are not optional; they must be found from initial conditions.
如果加速度以时间的函数给出，即 a = f(t)，那么速度通过积分得到：v = ∫ a dt + C。位移通过积分速度得到：s = ∫ v dt + D。常数 C 和 D 不是可有可无的，必须由初始条件确定。
t³ − t² + Ct + D. Without initial values for v and s, the motion is not fully determined.
考虑 a = 6t − 2。则 v = ∫ (6t − 2) dt = 3t² − 2t + C。再积分得到 s = ∫ (3t² − 2t + C) dt = t³ − t² + Ct + D。如果没有 v 和 s 的初始值，运动就无法完全确定。
or example, if v = 3t² − 2t, then dv/dt = 6t − 2, which matches the original acceleration.
积分是微分的逆过程。这意味着你总可以把积分结果微分回去检查。例如，若 v = 3t² − 2t，则 dv/dt = 6t − 2，正好等于原来的加速度。
5. Initial Conditions and Constants of Integration | 初始条件与积分常数
ases include initially at rest, which means v = 0 when t = 0, and initially at the origin, which means s = 0 when t = 0.
每个不定积分都会引入一个常数。在力学中，这些常数由初始条件确定，通常是在 t = 0 时给出的条件。常见表述包括 initially at rest，表示 t = 0 时 v = 0；initially at the origin，表示 t = 0 时 s = 0。
3t² − 2t + 8. Then s = t³ − t² + 8t + D. Using s(0) = 0 gives D = 0, so s = t³ − t² + 8t.
例如，设 a = 6t − 2，t = 0 时 v = 8，且 t = 0 时 s = 0。由 v = 3t² − 2t + C，代入 v(0) = 8 得 C = 8，所以 v = 3t² − 2t + 8。再积分得 s = t³ − t² + 8t + D。代入 s(0) = 0 得 D = 0，所以 s = t³ − t² + 8t。
n time t may not appear directly. In that case, use the chain rule to change the variable:
如果加速度以位移的函数给出，即 a = f(s)，那么时间 t 可能不会直接出现。此时要用链式法则换变量：
a = dv/dt = v dv/ds
This converts the problem into an integration with respect to s, which is often the only practical method for questions such as simple harmonic motion.
这样就把问题转化为关于 s 的积分，这通常是解决简谐运动等问题的唯一实用方法。
6. Worked Example: a = 6t − 2 | 例题：a = 6t − 2
s from rest at the origin. Find expressions for v and s, and then find v and s when t = 3.
一个质点沿直线运动，加速度为 a = 6t − 2 m s⁻²。它从原点由静止出发。求 v 和 s 的表达式，并求 t = 3 时的 v 和 s。
² − 2t + C. Since it starts from rest, v = 0 when t = 0, so C = 0. Therefore v = 3t² − 2t.
积分加速度：v = ∫ (6t − 2) dt = 3t² − 2t + C。由于由静止出发，t = 0 时 v = 0，所以 C = 0。因此 v = 3t² − 2t。
t² + D. Since it starts at the origin, s = 0 when t = 0, so D = 0. Therefore s = t³ − t².
积分速度：s = ∫ (3t² − 2t) dt = t³ − t² + D。由于从原点出发，t = 0 时 s = 0，所以 D = 0。因此 s = t³ − t²。
− 9 = 18 m. The particle is moving in the positive direction and is 18 m from the origin.
当 t = 3 时，v = 3(9) − 6 = 21 m s⁻¹，s = 27 − 9 = 18 m。质点沿正方向运动，距原点 18 m。
2/3 s. At t = 2/3, s = (2/3)³ − (2/3)² = 8/27 − 12/27 = −4/27 m. This is a turning point.
求质点瞬时静止的时刻，令 v = 0：3t² − 2t = 0，即 t(3t − 2) = 0。因此 t = 0 或 t = 2/3 s。当 t = 2/3 时，s = (2/3)³ − (2/3)² = 8/27 − 12/27 = −4/27 m。这就是一个转向点。
7. Stationary Points and Turning Points | 静止点与转向点
ariable acceleration problem, this usually requires solving a polynomial or trigonometric equation. These times are called stationary points of the displacement function.
当速度为 0 时，质点瞬时静止：v = 0。在变加速度问题中，这通常需要解多项式方程或三角方程。这些时刻称为位移函数的静止点。
To decide whether a stationary point is a turning point, check whether the velocity changes sign on either side of the point. If v changes from positive to negative, the particle reverses direction and moves back. If v changes from negative to positive, it also reverses direction. If v does not change sign, the particle momentarily stops but continues in the same direction.
要判断静止点是否为转向点，需要检查该点两侧速度是否变号。如果 v 由正变负，质点反向运动；如果 v 由负变正，也反向运动。如果 v 不变号，质点只是瞬时停下，然后继续沿原方向运动。
Acceleration can also help. If v = 0 and a ≠ 0, the
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