📚 Variable Acceleration in One Dimension | 一维变加速度运动
In Edexcel A Level Mathematics, variable acceleration problems move beyond the constant acceleration equations (SUVAT) and require the use of calculus to relate displacement, velocity and acceleration. When acceleration is not constant, you must use differentiation and integration to model motion in one dimension accurately.
在 Edexcel A Level 数学中，变加速度问题超越了匀加速度方程（SUVAT），需要运用微积分来联系位移、速度和加速度。当加速度不是常量时，你必须使用微分和积分来准确建立一维运动模型。
1. From Constant to Variable Acceleration | 从匀加速度到变加速度
In mechanics, constant acceleration allows the use of the familiar SUVAT equations, but many real-life situations involve acceleration that changes with time, velocity or displacement. Variable acceleration therefore requires a calculus-based approach, where displacement x, velocity v and acceleration a are treated as functions rather than fixed quantities.
在力学中，匀加速度允许使用熟悉的 SUVAT 方程，但许多现实情况涉及随时间、速度或位移变化的加速度。因此，变加速度需要基于微积分的方法，将位移 x、速度 v 和加速度 a 视为函数而不是固定量。
The key idea is that velocity is the rate of change of displacement, and acceleration is the rate of change of velocity. This means calculus gives us the tools to move between these quantities even when acceleration is expressed as a function of time t or displacement x.
关键思想是速度是位移的变化率，加速度是速度的变化率。这意味着微积分为我们提供了在这些量之间转换的工具，即使加速度被表示为时间 t 或位移 x 的函数。
2. Key Definitions and Notation | 关键定义与符号
For one-dimensional motion along a straight line, we usually let x represent the displacement from a fixed origin, measured in metres (m). Velocity v is defined as the rate of change of displacement with respect to time, and acceleration a is the rate of change of velocity with respect to time.
对于沿直线的一维运动，我们通常用 x 表示从固定原点测量的位移，单位为米 (m)。速度 v 定义为位移随时间的变化率，加速度 a 定义为速度随时间的变化率。
These definitions are written in calculus notation as:
这些定义用微积分符号表示为：
v = dx/dt and a = dv/dt = d²x/dt²
Here dx/dt is the first derivative of displacement with respect to time, and d²x/dt² is the second derivative. In this notation, x, v and a may all be functions of time t, so differentiation and integration become essential skills.
这里 dx/dt 是位移对时间的一阶导数，d²x/dt² 是二阶导数。在这种符号中，x、v 和 a 都可能是时间 t 的函数，因此微分和积分成为基本技能。
3. Using Differentiation: From Displacement to Velocity to Acceleration | 使用微分：从位移到速度再到加速度
If you are given displacement x as a function of time t, then you can find the velocity v by differentiating x with respect to t once, and the acceleration a by differentiating v with respect to t, or by differentiating x twice.
如果给定位移 x 作为时间 t 的函数，那么你可以通过将 x 对 t 微分一次来求速度 v，再对 v 对 t 微分来求加速度 a，或者将 x 对 t 微分两次。
For example, suppose a particle moves along a straight line so that its displacement is given by x = 2t³ − 9t² + 12t. Then:
例如，假设一个粒子沿直线运动，其位移由 x = 2t³ − 9t² + 12t 给出。那么：
v = dx/dt = 6t² − 18t + 12
a = dv/dt = 12t − 18
This gives you the velocity and acceleration at any time t. To find the acceleration at a specific instant, substitute the value of t into the expression for a. For instance, at t = 2, a = 12(2) − 18 = 6 m s⁻².
这给出了任意时刻 t 的速度和加速度。要找到特定时刻的加速度，将 t 的值代入 a 的表达式。例如，在 t = 2 时，a = 12(2) − 18 = 6 m s⁻²。
4. Using Integration: From Acceleration to Velocity to Displacement | 使用积分：从加速度到速度再到位移
If acceleration is given as a function of time, a(t), then velocity is obtained by integrating acceleration with respect to time. A constant of integration appears, and its value is found using an initial condition such as the velocity at t = 0.
如果加速度作为时间的函数 a(t) 给出，那么速度通过对加速度关于时间积分得到。会出现一个积分常数，其值通过初始条件（如 t = 0 时的速度）来确定。
The same process applies when going from velocity to displacement. If v(t) is known, then displacement x is found by integrating v with respect to time, using an initial displacement to determine the constant.
同样的过程适用于从速度到位移。如果 v(t) 已知，则位移 x 通过对 v 关于时间积分得到，并使用初始位移来确定常数。
For example, suppose a particle has acceleration a = 6t − 4 m s⁻², and initially its velocity is 3 m s⁻¹ and displacement is 5 m. Then integrating the acceleration gives:
例如，假设一个粒子的加速度为 a = 6t − 4 m s⁻²，初始速度为 3 m s⁻¹，初始位移为 5 m。对加速度积分得到：
v = ∫(6t − 4) dt = 3t² − 4t + C
Using v(0) = 3 gives C = 3, so v = 3t² − 4t + 3. Integrating again gives x = ∫(3t² − 4t + 3) dt = t³ − 2t² + 3t + D, and using x(0) = 5 gives D = 5.
利用 v(0) = 3 得到 C = 3，所以 v = 3t² − 4t + 3。再次积分得到 x = ∫(3t² − 4t + 3) dt = t³ − 2t² + 3t + D，利用 x(0) = 5 得到 D = 5。
5. Finding Maximum and Minimum Velocity | 求最大速度和最小速度
When velocity is expressed as a function of time, the maximum or minimum velocity occurs when the acceleration is zero, provided the acceleration changes sign. To locate such points, set a = dv/dt = 0 and solve for t, then check the sign of a on either side or use the second derivative test.
当速度表示为时间的函数时，最大或最小速度出现在加速度为零的时刻，前提是加速度改变符号。要找到这些点，令 a = dv/dt = 0 并解出 t，然后检查 a 两侧的符号或使用二阶导数检验。
For example, if v = 6t² − 18t + 12, then a = 12t − 18. Setting a = 0 gives t = 1.5. Since a changes from negative to positive at t = 1.5, the velocity has a minimum at this time. Substituting t = 1.5 into v gives the minimum velocity.
例如，如果 v = 6t² − 18t + 12，则 a = 12t − 18。令 a = 0 得到 t = 1.5。由于 a 在 t = 1.5 处由负变正，速度在此刻有最小值。将 t = 1.5 代入 v 可得到最小速度。
The maximum or minimum velocity may also occur at the endpoints of the time interval considered, so always check the boundaries when a restricted time domain is given in the exam question.
最大或最小速度也可能出现在所考虑时间区间的端点，因此当考题给出受限时间域时，务必检查边界值。
6. Finding Maximum Displacement and Turning Points | 求最大位移与折返点
A particle changes direction when its velocity is zero. At that instant, the displacement may reach a local maximum or minimum value, depending on whether the particle is turning back towards the origin or moving past it. To find the maximum displacement from the origin, set v = 0 and solve for t, then substitute these times into the displacement function x(t).
当粒子速度为零时，它会改变方向。在那一刻，位移可能达到局部最大值或最小值，这取决于粒子是返回原点还是通过原点。要找到离原点的最大位移，令 v = 0 并解出 t，然后将这些时刻代入位移函数 x(t)。
For instance, if x = t³ − 6t² + 9t, then v = 3t² − 12t + 9 = 3(t − 1)(t − 3). Setting v = 0 gives t = 1 and t = 3. The values x(1) = 4 and x(3) = 0 indicate that the particle reaches a maximum displacement of 4 m at t = 1 before turning back.
例如，如果 x = t³ − 6t² + 9t，则 v = 3t² − 12t + 9 = 3(t − 1)(t − 3)。令 v = 0 得到 t = 1 和 t = 3。值 x(1) = 4 和 x(3) = 0 表明粒子在 t = 1 时达到 4 m 的最大位移，然后折返。
To determine whether a turning point gives a maximum or minimum displacement, examine the sign of velocity before and after the point, or use the acceleration at that instant. If v changes from positive to negative, the displacement is a local maximum.
要确定折返点给出的是最大位移还是最小位移，检查该点前后速度的符号，或使用该时刻的加速度。如果 v 从正变为负，则位移为局部最大值。
7. Distance Travelled Versus Displacement | 路程与位移的区别
Displacement is the vector quantity giving the position relative to the origin, while distance travelled is the scalar quantity measuring the total ground covered regardless of direction. When a particle changes direction, the total distance travelled is not the same as the final displacement.
位移是表示相对于原点位置的矢量，而路程是测量无论方向如何的总路径长度的标量。当粒子改变方向时，总路程与最终位移不同。
To find the total distance travelled over a time interval, first identify the times when velocity is zero, as these indicate possible direction changes. Then integrate the absolute value of velocity over each subinterval, or compute the displacement over each subinterval and add their absolute values.
要找到一段时间内通过的总路程，首先确定速度为零的时刻，因为这些时刻表示可能的方向改变。然后对每个子区间积分速度的绝对值，或计算每个子区间的位移并加上它们的绝对值。
For example, if a particle moves so that v = 3t² − 12t + 9, the direction changes at t = 1 and t = 3. Between t = 0 and t = 1 the velocity is positive, between t = 1 and t = 3 it is negative, and after t = 3 it is positive again. The total distance is |x(1) − x(0)| + |x(3) − x(1)| + |x(4) − x(3)| if the interval is from 0 to 4.
例如，如果粒子运动使得 v = 3t² − 12t + 9，方向在 t = 1 和 t = 3 时改变。在 t = 0 到 t = 1 之间速度为正，在 t = 1 到 t = 3 之间速度为负，在 t = 3 之后速度再次为正。如果区间为 0 到 4，则总路程为 |x(1) − x(0)| + |x(3) − x(1)| + |x(4) − x(3)|。
8. Acceleration as a Function of Displacement: Using a = v dv/dx | 加速度作为位移的函数：使用 a = v dv/dx
Sometimes acceleration is given as a function of displacement x rather than time t. In such cases, the chain rule provides a very useful relationship: a = dv/dt = (dv/dx)(dx/dt) = v dv/dx. This allows you to relate v, x and a without involving time directly.
有时加速度作为位移 x 的函数而不是时间 t 的函数给出。在这种情况下，链式法则提供了一个非常有用的关系：a = dv/dt = (dv/dx)(dx/dt) = v dv/dx。这使你能够直接联系 v、x 和 a，而不直接涉及时间。
This relation is often written in the form v dv/dx = a. To solve problems, you separate the variables v and x and integrate both sides. For example, if a = 2x and the particle starts from rest at x = 0, then v dv/dx = 2x, giving ∫v dv = ∫2x dx, so ½v² = x² + C. Using v = 0 when x = 0 gives C = 0, hence v² = 2x².
这个关系通常写成 v dv/dx = a。要解这类问题，需要分离变量 v 和 x 并对两边积分。例如，如果 a = 2x 且粒子在 x = 0 处从静止出发，则 v dv/dx = 2x，得到 ∫v dv = ∫2x dx，所以 ½v² = x² + C。利用当 x = 0 时 v = 0 得到 C = 0，因此 v² = 2x²。
This method is especially useful for questions where acceleration depends on position, such as motion under a spring force or in a variable gravitational field. It also emphasises the importance of correct integration constants.
这种方法对于加速度依赖于位置的问题特别有用，例如在弹簧力或变重力场下的运动。它还强调了正确确定积分常数的重要性。
9. Worked Example 1: Polynomial Displacement Function | 例题 1：多项式位移函数
A particle moves in a straight line with displacement x = t³ − 5t² + 8t metres, where t is the time in seconds. Find the velocity, the acceleration, and the time at which the particle first comes to rest.
一个粒子沿直线运动，位移为 x = t³ − 5t² + 8t 米，其中 t 为时间（秒）。求速度、加速度以及粒子第一次静止的时间。
Differentiating the displacement with respect to t gives the velocity:
将位移对 t 微分得到速度：
v = dx/dt = 3t² − 10t + 8
Differentiating again gives the acceleration:
再次微分得到加速度：
a = dv/dt = 6t − 10
To find when the particle is at rest, set v = 0: 3t² − 10t + 8 = 0. This factorises to (3t − 4)(t − 2) = 0, giving t = 4/3 s and t = 2 s. The particle first comes to rest at t = 4/3 seconds.
要找到粒子静止的时刻，令 v = 0：3t² − 10t + 8 = 0。此式可分解为 (3t − 4)(t − 2) = 0，得到 t = 4/3 秒和 t = 2 秒。粒子第一次静止在 t = 4/3 秒。
At t = 4/3, the acceleration is a = 6(4/3) − 10 = 8 − 10 = −2 m s⁻², confirming that the particle is decelerating at that instant.
在 t = 4/3 时，加速度为 a = 6(4/3) − 10 = 8 − 10 = −2 m s⁻²，这证实了粒子在该时刻正在减速。
10. Worked Example 2: Distance Travelled with Change of Direction | 例题 2：含折返的路程计算
A particle moves along a straight line with velocity v = t² − 4t + 3 m s⁻¹ for t ≥ 0. Find the total distance travelled in the first 5 seconds.
一个粒子沿直线运动，速度 v = t² − 4t + 3 m s⁻¹，t ≥ 0。求前 5 秒内通过的总路程。
First find when the particle changes direction by setting v = 0: t² − 4t + 3 = 0, so (t − 1)(t − 3) = 0, giving t = 1 and t = 3. The velocity is positive from t = 0 to 1, negative from t = 1 to 3, and positive after t = 3.
首先通过令 v = 0 找到粒子改变方向的时刻：t² − 4t + 3 = 0，所以 (t − 1)(t − 3) = 0，得到 t = 1 和 t = 3。速度从 t = 0 到 1 为正，从 t = 1 到 3 为负，t = 3 之后为正。
Now find the displacement function by integrating v:
现在通过积分 v 求位移函数：
x = ∫(t² − 4t + 3) dt = (1/3)t³ − 2t² + 3t + C
Assuming the initial displacement at t = 0 is zero, C = 0. The key positions are x(0) = 0, x(1) = 1/3 − 2 + 3 = 4/3, x(3) = 9 − 18 + 9 = 0, and x(5) = 125/3 − 50 + 15 = 20/3.
假设 t = 0 时初始位移为零，则 C = 0。关键位置为 x(0) = 0，x(1) = 1/3 − 2 + 3 = 4/3，x(3) = 9 − 18 + 9 = 0，x(5) = 125/3 − 50 + 15 = 20/3。
The total distance is the sum of the absolute changes in displacement over each subinterval: |4/3 − 0| + |0 − 4/3| + |20/3 − 0| = 4/3 + 4/3 + 20/3 = 28/3 metres, which is approximately 9.33 m.
总路程是每个子区间位移变化绝对值的和：|4/3 − 0| + |0 − 4/3| + |20/3 − 0| = 4/3 + 4/3 + 20/3 = 28/3 米，约等于 9.33 m。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
One common mistake is forgetting to add the constant of integration when finding velocity or displacement from acceleration. Always use the given initial conditions to determine these constants, and show your working clearly.
一个常见错误是从加速度求速度或位移时忘记加上积分常数。务必使用给定的初始条件来确定这些常数，并清晰展示解题过程。
Another frequent error is confusing displacement with distance travelled. Remember that distance travelled is the sum of the magnitudes of displacement over each time interval where the direction does not change, not simply the final displacement.
另一个常见错误是将位移与路程混淆。记住路程是在方向不变的每个时间区间内位移大小之和，而不是简单的最终位移。
When maximum or minimum values are required, do not just set acceleration to zero and assume the velocity is a maximum. Check that the acceleration actually changes sign, and also consider the endpoints of the time interval if one is specified.
当需要求最大值或最小值时，不要只令加速度为零并假设速度是最大值。检查加速度是否确实改变符号，并且如果指定了时间区间，还要考虑区间的端点。
Finally, when acceleration is given as a function of displacement, always use the chain rule relation a = v dv/dx and separate variables properly. Misapplying SUVAT equations to variable acceleration problems is a serious error.
最后，当加速度作为位移的函数给出时，始终使用链式法则关系 a = v dv/dx，并正确分离变量。将 SUVAT 方程误用于变加速度问题是一个严重错误。
12. Summary and Key Formulae | 总结与关键公式
Variable acceleration in one dimension is fully described by the calculus relationships between displacement, velocity and acceleration. The core formulae you must remember are:
一维变加速度运动完全由位移、速度和加速度之间的微积分关系描述。你必须记住的核心公式为：
v = dx/dt
a = dv/dt = d²x/dt²
v = ∫ a dt
x = ∫ v dt
a = v dv/dx
These five expressions form the backbone of all variable acceleration questions in the Edexcel mechanics component. Practice moving fluently between them using differentiation and integration, and always pay careful attention to initial conditions and direction changes.
这五个表达式构成了 Edexcel 力学部分所有变加速度问题的核心。通过微分和积分练习熟练地在它们之间转换，并始终仔细关注初始条件和方向变化。
Mastering variable acceleration not only prepares you for exam success but also builds the foundational understanding needed for further study in physics, engineering and applied mathematics.
掌握变加速度不仅为考试成功做好准备，还为进一步学习物理、工程和应用数学打下基础。
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