Application to Mechanics | 力学应用

📚 Application to Mechanics | 力学应用

In A-Level Mathematics, mechanics is the branch that describes the motion of objects and the forces acting upon them. One of the most powerful applications of calculus is in linking displacement, velocity and acceleration, allowing you to model motion precisely whether acceleration is constant or varies with time. This article covers the core techniques of differentiation and integration in kinematics, the derivation of the SUVAT equations, handling variable acceleration, finding maxima and minima of velocity, and dealing with real-world problems such as free fall. Mastering these skills will give you the confidence to solve any mechanics problem involving motion along a straight line.

在A-Level数学中，力学是描述物体运动及所受作用力的分支。微积分最强大的应用之一就是将位移、速度和加速度联系起来，使你能够精确地建立运动模型，无论加速度是恒定的还是随时间变化的。本文涵盖了运动学中微分与积分的核心技巧、匀加速公式的推导、变加速问题的处理、速度最大值与最小值的求解，以及自由落体等实际问题的应对方法。掌握这些技能将使你能够自信地解决任何涉及直线运动的力学问题。

---

1. Kinematics and Calculus | 运动学与微积分

In one-dimensional motion, an object’s position is described by its displacement s from a fixed origin, usually taken as a function of time t. The rate at which displacement changes is velocity v, and the rate at which velocity changes is acceleration a. Calculus provides the essential links: velocity is the first derivative of displacement with respect to time, and acceleration is the derivative of velocity, or the second derivative of displacement.

在一维运动中，物体的位置用距离固定原点的位移 s 来描述，通常视为时间 t 的函数。位移变化的速率是速度 v，速度变化的速率是加速度 a。微积分提供了基本的联系：速度是位移对时间的一阶导数，加速度是速度的导数，也是位移的二阶导数。

• English: These relationships are expressed as v = ds/dt and a = dv/dt = d²s/dt². The reverse process, integration, gives displacement from velocity and velocity from acceleration.
• 中文：这些关系表示为 v = ds/dt 和 a = dv/dt = d²s/dt²。反过来，通过积分可以由速度得到位移，由加速度得到速度。

Because kinematics relies heavily on functions of time, being fluent in differentiation and integration of polynomials, trigonometric functions and simple exponentials is crucial for A-Level mechanics questions.

由于运动学高度依赖于时间函数，熟练地对多项式、三角函数和简单指数函数进行微积分运算对于A-Level力学题至关重要。

---

2. Displacement, Velocity, and Acceleration | 位移、速度和加速度

Displacement is a vector quantity measuring the change in position from a chosen origin, often denoted by s (or x) and measured in metres (m). Velocity v is the rate of change of displacement, measured in m s⁻¹. Acceleration a is the rate of change of velocity, measured in m s⁻². In simple rectilinear motion, positive and negative signs indicate direction. A common exam mistake is to confuse speed (the magnitude of velocity) with velocity, particularly when calculating total distance travelled.

位移是一个矢量，衡量从选定原点到当前位置的变化，通常用 s（或 x）表示，单位为米（m）。速度 v 是位移的变化率，单位为 m s⁻¹。加速度 a 是速度的变化率，单位为 m s⁻²。在简单的直线运动中，正负号表示方向。一个常见的考试错误是将速率（速度的大小）与速度混淆，尤其是在计算总路程时。

If an expression for displacement is given, say s = t³ – 6t² + 9t, you can immediately write down the velocity and acceleration by successive differentiation.

如果给出了位移的表达式，例如 s = t³ – 6t² + 9t，你可以通过逐次求导立即写出速度和加速度。

---

3. Differentiation in Mechanics | 力学中的微分

Differentiation allows you to move from displacement to velocity and from velocity to acceleration. Given s = f(t), differentiate term by term to find v = f ‘(t), then differentiate once more to find a = f ”(t). For the example above: v = ds/dt = 3t² – 12t + 9 and a = dv/dt = 6t – 12. This is a typical variable acceleration scenario; notice that a depends on t, not a constant.

微分使你能够从位移过渡到速度，以及从速度过渡到加速度。给定 s = f(t)，逐项求导得到 v = f ‘(t)，再求导一次得到 a = f ”(t)。以上例为例：v = ds/dt = 3t² – 12t + 9 和 a = dv/dt = 6t – 12。这是一个典型的变加速情景；注意加速度 a 依赖于时间 t，而不是常数。

You can also use differentiation to find when a particle is at rest (set v = 0), or to investigate turning points in velocity by solving a = 0. These techniques underpin many exam questions on particle motion.

你还可以用微分来求质点何时静止（令 v = 0），或者通过解 a = 0 来研究速度的转折点。这些技巧是许多关于质点运动考题的基础。

---

4. Integration in Mechanics | 力学中的积分

While differentiation goes from displacement to acceleration, integration goes the opposite way. If acceleration is given as a function of time, integrating yields the velocity function: v = ∫ a dt. Integrating velocity gives displacement: s = ∫ v dt. Every integration introduces an arbitrary constant that must be determined using initial conditions, such as the velocity or displacement at t = 0.

微分是从位移到加速度进行下去，而积分则方向相反。如果加速度以时间的函数形式给出，积分可得到速度函数：v = ∫ a dt。对速度积分可得到位移：s = ∫ v dt。每次积分都会引入一个任意常数，该常数必须利用初始条件来确定，比如在 t = 0 时的速度或位移。

For example, if a = 6t – 2 and the initial velocity is v(0) = 3, then integrating gives v = ∫(6t – 2) dt = 3t² – 2t + C. Substituting t = 0 gives C = 3, so v = 3t² – 2t + 3. Neglecting the constant of integration is one of the most frequent errors in mechanics.

例如，若 a = 6t – 2 且初速度为 v(0) = 3，积分得 v = ∫(6t – 2) dt = 3t² – 2t + C。代入 t = 0 得到 C = 3，因此 v = 3t² – 2t + 3。遗漏积分常数是力学中最常见的错误之一。

---

5. Finding Displacement from Velocity | 由速度求位移

When velocity is known as a function of time, displacement is the area under the velocity–time graph, calculated by integration. For instance, given v = 4t + 1 and the particle is at the origin when t = 0, we find s = ∫ (4t + 1) dt = 2t² + t + C. Using s(0) = 0 gives C = 0, so s = 2t² + t. This is a pure displacement function; if you need total distance travelled when velocity changes sign, you must integrate the absolute value of velocity or split the time interval.

当速度作为时间的函数已知时，位移是速度-时间图下方的面积，可通过积分计算。例如，给定 v = 4t + 1 且质点在 t = 0 时位于原点，得到 s = ∫ (4t + 1) dt = 2t² + t + C。利用 s(0) = 0 得 C = 0，因此 s = 2t² + t。这是一个纯位移函数；若需要计算速度变号后的总路程，则必须对速度的绝对值积分或将时间区间分段。

In exam settings, always check whether the question asks for displacement or distance. Integrating velocity gives displacement; to obtain distance, you need to integrate the speed |v|.

在考试中，始终要看清题目问的是位移还是路程。对速度积分得到位移；要求路程，你必须对速率 |v| 进行积分。

---

6. Finding Velocity from Acceleration | 由加速度求速度

Similarly, when acceleration is given, velocity is found by integration plus one initial condition. A typical problem provides a = f(t) and the initial velocity u at t = 0. Integrate: v = ∫ a dt = F(t) + C, then use v(0) = u to find C. This is a fundamental routine for variable acceleration problems and is essentially the First Equation of Motion generalised.

类似地，当给定加速度时，速度通过积分加一个初始条件求得。一道典型的题目会给出 a = f(t) 和 t = 0 时的初速度 u。积分得：v = ∫ a dt = F(t) + C，然后用 v(0) = u 求出 C。这是变加速问题的基本流程，本质上是第一运动方程的推广。

Worked illustration: a particle starts from rest with a = 12t – 4. Then v = ∫(12t – 4)dt = 6t² – 4t + C. Since at t=0, v=0, we have C=0, hence v = 6t² – 4t. From there you can find the velocity at any future time, and then integrate again for displacement.

示例：一质点从静止出发，加速度为 a = 12t – 4。则 v = ∫(12t – 4)dt = 6t² – 4t + C。由于 t=0 时 v=0，得 C=0，故 v = 6t² – 4t。由此可求出任意时刻的速度，然后再积分求位移。

---

7. Constant Acceleration Formulae (SUVAT) | 匀加速运动公式 (SUVAT)

When acceleration is constant, the integrals simplify dramatically, giving the well-known SUVAT equations. These link displacement s, initial velocity u, final velocity v, acceleration a and time t. Starting from a = constant, integrate once to get v = u + at. Integrate velocity to get s = ut + ½at². You can also eliminate t to obtain v² = u² + 2as, and find the average velocity form s = ½(u + v)t.

当加速度恒定时，积分大大简化，得出著名的SUVAT方程。这些方程将位移 s、初速度 u、末速度 v、加速度 a 和时间 t 联系起来。从 a = 常数 出发，积分一次得到 v = u + at。对速度积分得到 s = ut + ½at²。也可以消去 t 得到 v² = u² + 2as，并求出平均速度形式 s = ½(u + v)t。

v = u + at    s = ut + ½at²    v² = u² + 2as    s = ½(u + v)t

Though these are often remembered as four or five formulas, they are direct consequences of calculus. Understanding their derivation prevents confusion about which formula applies when acceleration is constant. They cannot be used directly for variable acceleration; for that, calculus must be applied explicitly.

虽然这些通常被记作四或五个公式，但它们都是微积分的直接结果。理解其推导过程可以避免混淆何时使用哪个公式。对于变加速运动不能直接使用它们；必须显式地应用微积分。

---

8. Variable Acceleration Problems | 变加速问题

In Edexcel A-Level Mechanics, variable acceleration problems are common. You are typically given acceleration as a function of time, or occasionally as a function of displacement. The standard approach is to integrate a(t) to find v(t), using the initial velocity to find the constant, then integrate v(t) to find s(t), using the initial displacement. Many questions ask for the velocity or displacement at a specific time, or for the maximum speed.

在爱德思A-Level力学中，变加速问题很常见。你通常会得到加速度作为时间的函数，有时也作为位移的函数。标准方法是积分 a(t) 求 v(t)，用初始速度求积分常数，然后积分 v(t) 求 s(t)，利用初始位移。许多题目要求求特定时刻的速度或位移，或求最大速率。

Consider: a particle moves along a straight line with acceleration a = 8 – 2t m s⁻², initial velocity 5 m s⁻¹ and initial displacement +2 m. First, integrate acceleration: v = ∫(8 – 2t) dt = 8t – t² + C. Using v(0)=5 gives C=5, so v = 8t – t² + 5. Then integrate velocity: s = ∫(8t – t² + 5) dt = 4t² – (1/3)t³ + 5t + D. With s(0)=2, D=2. Thus the motion is fully described.

考虑：一个质点沿直线运动，加速度为 a = 8 – 2t m s⁻²，初速度 5 m s⁻¹，初始位移 +2 m。首先积分加速度：v = ∫(8 – 2t) dt = 8t – t² + C。利用 v(0)=5 得 C=5，所以 v = 8t – t² + 5。然后积分速度：s = ∫(8t – t² + 5) dt = 4t² – (1/3)t³ + 5t + D。代入 s(0)=2 得 D=2。于是运动被完全描述。

---

9. Maximum and Minimum Velocity | 速度的最大值与最小值

In variable acceleration scenarios, velocity can pass through maximum or minimum values. To find these, treat velocity as a function of time and differentiate to find its stationary points: set dv/dt = a = 0. Then verify whether each point is a maximum or minimum, either by checking the sign of acceleration on either side or by finding the second derivative d²v/dt² = da/dt. If a = 0 and the acceleration changes from positive to negative, velocity is a local maximum; if from negative to positive, a local minimum.

在变加速情景中，速度可以经过最大值或最小值。要求出这些，将速度视为时间的函数并求导找出驻点：令 dv/dt = a = 0。然后通过检查加速度在两侧的符号变化，或求二阶导数 d²v/dt² = da/dt，来验证该点是最大值还是最小值。若 a = 0 且加速度由正变负，速度取局部最大值；若由负变正，则为局部最小值。

Example: given v = t³ – 9t² + 24t – 10, find the maximum velocity. Differentiate: a = 3t² – 18t + 24. Solve 3t² – 18t + 24 = 0 => t² – 6t + 8 = 0 => t = 2 or t = 4. The second derivative da/dt = 6t – 18: at t=2 it is negative (max), at t=4 positive (min). So maximum velocity occurs at t=2, v(2) = 8 – 36 + 48 – 10 = 10 m s⁻¹. This technique is regularly tested.

示例：给定 v = t³ – 9t² + 24t – 10，求最大速度。求导得：a = 3t² – 18t + 24。解 3t² – 18t + 24 = 0，得 t² – 6t + 8 = 0，即 t = 2 或 t = 4。二阶导数 da/dt = 6t – 18：在 t=2 处为负（极大值），在 t=4 处为正（极小值）。所以最大速度出现在 t=2 时，v(2) = 8 – 36 + 48 – 10 = 10 m s⁻¹。这种技巧常被考查。

---

10. Using Boundary Conditions | 边界条件的运用

The constants of integration obtained when moving from acceleration to velocity and from velocity to displacement are uniquely determined by boundary conditions. Typically these are the initial velocity and initial displacement. In some problems, however, the boundary conditions may be given at different times. For instance, ‘the particle is at rest at t = T‘ or ‘the particle passes through the origin when t = 5‘. Substituting these values into the integrated expression allows you to solve for the constants.

从加速度到速度以及从速度到位移所得到的积分常数，是由边界条件唯一确定的。这些通常是初始速度和初始位移。然而，在某些问题中，边界条件可能在不同时刻给出。例如，“质点在 t = T 时静止”或“质点在 t = 5 时经过原点”。将这些值代入积分表达式即可求解常数。

A particularly tricky case involves total distance travelled. Because the displacement function s(t) obtained by integration gives net change in position, you must first find when velocity is zero to identify any changes in direction, then sum the absolute displacements over each subinterval. The total distance is ∫ |v| dt over the time interval, which often requires breaking the integral at roots of v(t)=0. Always distinguish clearly between ‘displacement’ and ‘distance’ in your working.

一种特别棘手的情形涉及总路程。由于积分得到的位移函数 s(t) 给出的是位置的净变化，你必须首先求出速度为零的时刻以确定方向是否改变，然后对每个子区间上的位移绝对值求和。总路程是在时间区间上对 ∫ |v| dt 积分，这通常需要在 v(t)=0 的根处分段积分。在作答时，始终要清晰区分“位移”和“路程”。

---

11. Practical Examples: Free Fall and Projectiles | 实例：自由落体与抛体运动

One of the most common applications is vertical motion under gravity. Taking upwards as positive, the acceleration due to gravity is a = -g, where g = 9.8 m s⁻² (or exactly 9.8 unless stated otherwise). For an object thrown vertically upwards with initial speed u, the velocity function is obtained directly from constant acceleration: v = u – gt, and displacement s = ut – ½gt² (if starting from s=0). The maximum height occurs when v=0, giving t = u/g and height H = u²/(2g). These results are frequently required in mechanics problems.

最常见的应用之一是重力作用下的竖直运动。取向上为正方向，重力加速度为 a = -g，其中 g = 9.8 m s⁻²（除非另作说明，一般取9.8）。对于以初速度 u 竖直上抛的物体，可直接由匀加速得到速度函数：v = u – gt，以及位移 s = ut – ½gt²（若从 s=0 开始）。最大高度出现在 v=0 时，得 t = u/g，高度 H = u²/(2g)。这些结果在力学题中经常用到。

In projectile problems where horizontal and vertical motions are independent, calculus is used for vertical motion while horizontal motion has constant velocity. For a particle projected at an angle, the vertical component is handled with a = -g, and integration yields the classic parabolic path. Though these are often solved with SUVAT, the calculus route reinforces the physics.

在水平和竖直运动相互独立的抛体问题中，竖直运动使用微积分处理，而水平运动具有恒定速度。对于以一定角度抛出的质点，竖直分量用 a = -g 处理，积分得出经典的抛物线轨迹。尽管这类题目常通过SUVAT求解，但使用微积分可以加深对物理本质的理解。

---

12. Summary and Exam Tips | 总结与应试技巧

To excel in mechanics questions involving calculus, always follow a systematic approach: identify what is given (displacement, velocity or acceleration as a function of time) and what you need to find. Write down the appropriate differential or integral relations. For variable acceleration, integrate stepwise and use initial conditions to determine constants. When tackling a problem, clearly note positive direction and be meticulous with signs.

要在涉及微积分的力学题中脱颖而出，务必遵循系统的方法：明确已知量（位移、速度或加速度作为时间的函数）和求解目标。写出相应的微分或积分关系式。对于变加速问题，逐步积分并用初始条件确定常数。处理问题时，要清楚地标明正方向，并严谨对待正负号。

• English: If the acceleration is constant, you may use SUVAT equations; otherwise, you must use calculus.
• 中文：如果加速度恒定，可以使用SUVAT方程；否则必须使用微积分。
• English: Remember that integrating velocity yields displacement, not distance. For distance, integrate the absolute velocity or split the interval at turning points.
• 中文：记住，对速度积分得到的是位移而非路程。求路程需对速度的绝对值积分或在转向点处分段。
• English: When finding maximum velocity, set a = 0 and check the nature of the stationary point.
• 中文：求最大速度时，令 a = 0 并检查驻点性质。
• English: Show clear steps and include units; Edexcel rewards method marks generously.
• 中文：展示清晰的步骤并写明单位；爱德思考试局对解题方法分给予得很慷慨。

Mastering the application of calculus to mechanics not only secures high marks in this topic but also builds a solid foundation for further study in physics and engineering. Practise with past papers, focusing on variable acceleration, sign interpretation and distance versus displacement distinctions, and you will approach the exam with great confidence.

掌握微积分在力学中的应用不仅能确保在该主题上获得高分，还能为物理和工程的后续学习打下坚实基础。通过练习历年真题，重点关注变加速、符号分析以及路程与位移的区分，你将充满信心地迎接考试。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）