📚 IB Math Exercise 14K.1: Kinematics and Integration | IB数学练习14K.1：运动学与积分
Integration is a cornerstone of calculus in the IB Mathematics Analysis and Approaches course. One of its most tangible applications appears in kinematics—the study of motion. In Exercise 14K.1, students explore how indefinite and definite integrals can be used to move from acceleration to velocity, and from velocity to displacement. Understanding these relationships in both directions sharpens your ability to model real-world motion and to solve exam problems confidently.
积分是IB数学分析与解释课程中微积分的基石，其最具体的应用之一出现在运动学中。在练习14K.1中，学生将探索如何利用不定积分和定积分，从加速度推导向速度，再从速度推导出位移。双向掌握这些关系能提升你模拟真实运动的能力，并让你在考试中充满信心地解题。
1. The Two-Way Street: Differentiation and Integration in Motion | 双向通道：运动中的微分与积分
In kinematics, the primary quantities are displacement s(t), velocity v(t), and acceleration a(t). Differentiation gives us v(t) = ds/dt and a(t) = dv/dt. Conversely, integration allows us to recover the original functions when we know a derivative and an initial condition. This inverse relationship is the foundation of problems in Exercise 14K.1.
在运动学中，基本物理量是位移 s(t)、速度 v(t) 和加速度 a(t)。求导可以给出 v(t) = ds/dt 与 a(t) = dv/dt。反之，当我们已知导数和一个初始条件时，积分能帮助我们复原原函数。这一互逆关系是练习14K.1所有问题的基础。
2. Core Kinematic Equations through Integration | 核心运动学方程的积分形式
Given the acceleration function a(t), we find velocity by integrating:
v(t) = ∫ a(t) dt
Once the velocity function is known, displacement is obtained by another integration:
s(t) = ∫ v(t) dt
These indefinite integrals produce constants of integration, which are determined by initial values such as the velocity at t = 0 or the position at t = 0. Definite integrals can directly give the change in velocity or displacement over a time interval without first finding the whole function.
若已知加速度函数 a(t)，可以通过积分求出速度：
v(t) = ∫ a(t) dt
一旦得到速度函数，再积分一次即可得到位移：
s(t) = ∫ v(t) dt
这些不定积分会产生积分常数，其值由初始条件确定，例如 t = 0 时的速度或位置。而使用定积分，则可以直接求出某一时间段内速度或位移的变化量，无需先完整写出原函数。
3. Handling Initial Conditions in Exercise 14K.1 | 练习14K.1中的初始条件处理
Initial conditions are the key to locking down the constant of integration. A typical problem states: “At t = 0, the particle is at the origin with velocity 5 m s⁻¹.” After integrating a(t) to obtain v(t) = F(t) + C, you substitute t = 0 and set v(0) = 5 to solve for C. The same logic applies when finding displacement: integrate v(t) and use the initial position s(0) to determine the integration constant.
初始条件是锁定积分常数的关键。一个典型的题目会这样给出：“在 t = 0 时，质点位于原点，速度为 5 m s⁻¹。” 当你积分 a(t) 得到 v(t) = F(t) + C 后，代入 t = 0 并令 v(0) = 5，即可解出 C。求位移时使用同样的逻辑：积分 v(t)，并利用初始位置 s(0) 来确定积分常数。
4. Definite Integrals: Change over an Interval | 定积分：某一区间内的变化量
Instead of finding complete functions, you can directly compute the change in velocity or displacement. For velocity:
v(t₂) − v(t₁) = ∫t₁t₂ a(t) dt
For displacement:
s(t₂) − s(t₁) = ∫t₁t₂ v(t) dt
This approach is particularly useful when the question asks for the total distance travelled or the change in position between two specific times, and it appears frequently throughout Exercise 14K.1.
除了寻找完整函数，你还可以直接计算速度或位移的变化量。对于速度：
v(t₂) − v(t₁) = ∫t₁t₂ a(t) dt
对于位移：
s(t₂) − s(t₁) = ∫t₁t₂ v(t) dt
当题目问及两个特定时刻之间的总路程或位置变化时，这种方法特别有用，且在整个练习14K.1中出现频率很高。
5. Displacement vs. Distance Travelled: A Critical Distinction | 位移与路程：一个关键区别
Displacement is the net change in position, while distance travelled is the total length of the path covered, regardless of direction. If velocity changes sign, the particle reverses direction. To find distance travelled over a time interval, you must integrate the absolute value of velocity or split the integral at the times when v(t) = 0. Exercise 14K.1 often includes a part asking for distance, not just displacement, testing your conceptual understanding.
位移是位置的净变化量，而路程是质点实际经过的路径总长度，不考虑方向。如果速度改变符号，质点会反向运动。要计算某一时间段内的路程，你必须对速度的绝对值进行积分，或在 v(t) = 0 的时刻将积分分段。练习14K.1中常会出现要求路程而非位移的题目，以此检验你对概念的透彻理解。
6. Reading the Signs: Positive and Negative Motion | 解读符号：正向与负向运动
In IB kinematics problems, a positive velocity usually indicates motion to the right or upward, and a negative velocity indicates motion to the left or downward. Acceleration that has the same sign as velocity means the particle is speeding up; opposite signs mean it is slowing down. This sign analysis is essential when you need to describe the particle’s behaviour at a specific instant—another skill reinforced in 14K.1.
在IB运动学问题中，正速度通常表示向右或向上运动，负速度表示向左或向下运动。加速度与速度符号相同时，质点在加速；符号相反时，质点在减速。当你需要描述质点在某一瞬间的运动状态时，这种符号分析至关重要——这也是练习14K.1强化的另一项技能。
7. Common Functional Forms in Exercise 14K.1 | 练习14K.1中常见的函数形式
The acceleration function is typically provided as a polynomial, trigonometric function, or exponential expression. Typical examples include:
- a(t) = 6t − 2 (linear acceleration)
- a(t) = 12t² − 4 (quadratic in time)
- a(t) = 3 sin t or a(t) = cos(2t) (trigonometric acceleration)
- a(t) = 2e−t (exponential decay)
加速度函数通常以多项式、三角函数或指数表达式的形式给出。常见例子包括：
- a(t) = 6t − 2（线性加速度）
- a(t) = 12t² − 4（时间的二次函数）
- a(t) = 3 sin t 或 a(t) = cos(2t)（三角函数型加速度）
- a(t) = 2e−t（指数衰减）
8. Worked Example 1: Polynomial Acceleration | 例题1：多项式加速度
A particle moves along a straight line with acceleration a(t) = 6t − 2 (in m s⁻²). At t = 0, its velocity is v = 5 m s⁻¹ and its displacement is s = 0 m.
Step 1: Find v(t).
Integrate acceleration: v(t) = ∫ (6t − 2) dt = 3t² − 2t + C.
Apply initial condition: v(0) = 3(0)² − 2(0) + C = 5 → C = 5.
Therefore, v(t) = 3t² − 2t + 5.
Step 2: Find s(t).
Integrate velocity: s(t) = ∫ (3t² − 2t + 5) dt = t³ − t² + 5t + D.
Apply initial condition: s(0) = 0³ − 0² + 5(0) + D = 0 → D = 0.
Therefore, s(t) = t³ − t² + 5t.
Step 3: Find displacement at t = 3.
s(3) = 27 − 9 + 15 = 33 m.
一质点沿直线运动，加速度为 a(t) = 6t − 2（单位 m s⁻²）。在 t = 0 时，速度为 v = 5 m s⁻¹，位移为 s = 0 m。
步骤1：求 v(t)。
对加速度积分：v(t) = ∫ (6t − 2) dt = 3t² − 2t + C。
代入初始条件：v(0) = 3(0)² − 2(0) + C = 5 → C = 5。
因此，v(t) = 3t² − 2t + 5。
步骤2：求 s(t)。
对速度积分：s(t) = ∫ (3t² − 2t + 5) dt = t³ − t² + 5t + D。
代入初始条件：s(0) = 0³ − 0² + 5(0) + D = 0 → D = 0。
因此，s(t) = t³ − t² + 5t。
步骤3：求 t = 3 时的位移。
s(3) = 27 − 9 + 15 = 33 m。
9. Worked Example 2: Trigonometric Acceleration | 例题2：三角函数型加速度
A particle’s acceleration is given by a(t) = 4 cos(2t) m s⁻². Initially the particle is at rest at the origin. Find v(t) and s(t).
Integrate to find velocity: v(t) = ∫ 4 cos(2t) dt = 2 sin(2t) + C.
Since v(0) = 0: 2 sin(0) + C = 0 → C = 0. So v(t) = 2 sin(2t).
Integrate to find displacement: s(t) = ∫ 2 sin(2t) dt = −cos(2t) + D.
Use s(0) = 0: −cos(0) + D = −1 + D = 0 → D = 1.
Therefore, s(t) = 1 − cos(2t). This function oscillates between 0 m and 2 m, showing periodic motion.
某质点的加速度为 a(t) = 4 cos(2t) m s⁻²。初始时质点静止于原点。求 v(t) 与 s(t)。
积分求速度：v(t) = ∫ 4 cos(2t) dt = 2 sin(2t) + C。
由 v(0) = 0 得：2 sin(0) + C = 0 → C = 0，所以 v(t) = 2 sin(2t)。
积分求位移：s(t) = ∫ 2 sin(2t) dt = −cos(2t) + D。
利用 s(0) = 0：−cos(0) + D = −1 + D = 0 → D = 1。
因此，s(t) = 1 − cos(2t)。该函数在0 m与2 m之间振荡，展现周期运动。
10. Checking Your Answers with Differentiation | 通过求导检验答案
A reliable way to verify your integration is to differentiate the resulting function and confirm you obtain the given function. If you found v(t) = 3t² − 2t + 5, then dv/dt = 6t − 2, which matches the original a(t). Similarly, if s(t) = t³ − t² + 5t, differentiating gives v(t) = 3t² − 2t + 5. This quick check can catch sign errors or missing constants before you move on.
检验积分结果的一种可靠方法是，对所得函数求导，并确认是否得到已知函数。若你得到 v(t) = 3t² − 2t + 5，则 dv/dt = 6t − 2，与原始的 a(t) 一致。类似地，若 s(t) = t³ − t² + 5t，求导后得到 v(t) = 3t² − 2t + 5。在继续答题前，这种快速检查可以帮你排查符号错误或遗漏的常数。
11. Addressing the Common Errors in 14K.1 | 练习14K.1中的典型错误
Students often forget to add the constant of integration on the first step, incorrectly setting it to zero. Another pitfall is confusing displacement with distance; always check whether the velocity changes sign before calculating total distance. Also, when using definite integrals, ensure the limits are substituted correctly—especially with trigonometric functions where the argument can be 2t instead of t. Reviewing these points after solving a 14K.1 question will solidify your technique.
学生常犯的第一个错误是忘记在第一步添加积分常数，错误地默认其为零。另一个陷阱是将位移与路程混淆；在计算总路程之前，务必检查速度是否改变符号。此外，使用定积分时，要确保正确地代入上下限——尤其是对于三角函数，其内部可能是 2t 而非单纯的 t。在解完一道14K.1题目后回顾这些要点，有助于巩固你的解题技巧。
12. Linking 14K.1 to the Broader IB Syllabus | 将14K.1联系到更广阔的IB大纲
Exercise 14K.1 is not an isolated drill; it prepares you for paper 2 questions that combine kinematics with equation solving, graphing, and optimisation. You may be asked to find when a particle returns to the origin or when the speed is minimal. These problems often require setting v(t) = 0 or s(t) = 0 and solving trigonometric or polynomial equations. A solid command of integration in motion contexts also supports your understanding of differential equations later in the course.
练习14K.1并非孤立的训练；它为你应对试卷二中结合运动学与方程求解、图像分析及优化的问题做好了准备。你可能需要求出质点何时返回原点，或何时速率最小。此类问题常需要令 v(t) = 0 或 s(t) = 0，并求解三角函数或多项式方程。在运动情境中扎实地掌握积分，也将支持你后续在课程中理解微分方程。
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