Applications of Integration in Kinematics | 积分在运动学中的应用

📚 Applications of Integration in Kinematics | 积分在运动学中的应用

In A-Level Mathematics, kinematics describes the motion of a particle in terms of displacement, velocity and acceleration. Given an expression for acceleration or velocity, integration allows us to recover the other quantities, and graph interpretations help visualise the motion. This article covers the essential integration techniques used in kinematics, supported by worked examples and exam-style advice.

在A-Level数学中，运动学用位移、速度和加速度描述质点的运动。已知加速度或速度的表达式时，积分可以帮助我们求出其余的运动学量，图像解读则有助于直观理解运动过程。本文将介绍运动学中涉及积分的基本方法，并配合完整例题和考试技巧。

1. Kinematics and the Role of Integration | 运动学与积分的作用

In one-dimensional motion, the basic definitions are: velocity is the rate of change of displacement, v = ds/dt, and acceleration is the rate of change of velocity, a = dv/dt. Differentiation moves from displacement to velocity to acceleration; integration reverses this direction.

在一维运动中，基本定义如下：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。微分过程从位移得到速度、再得到加速度；而积分则沿相反方向进行。

Integration therefore enables us to find displacement from velocity, and velocity from acceleration. When acceleration is given as a function of time, two successive integrations are needed to obtain displacement.

因此，积分可以帮助我们从速度求出位移，并从加速度求出速度。当加速度以时间函数给出时，需要连续两次积分才能得到位移。

2. From Acceleration to Velocity | 从加速度到速度

If acceleration a(t) is known, velocity is obtained by integrating with respect to time:

若已知加速度 a(t)，对时间积分即可得到速度：

v(t) = ∫ a(t) dt

Because integration introduces an arbitrary constant, the initial velocity must be used to fix it. For a constant acceleration a, this gives the familiar formula v = u + at, where u is the velocity at t = 0.

由于积分会产生任意常数，必须利用初速度来确定它。对于匀加速度 a，可得熟悉的速度公式 v = u + at，其中 u 是 t = 0 时的速度。

Example: if a(t) = 6t – 4 and v(0) = 2, then v(t) = 3t² – 4t + C. Substituting v(0) = 2 gives C = 2, so v(t) = 3t² – 4t + 2.

例如：若 a(t) = 6t – 4，且 v(0) = 2，则 v(t) = 3t² – 4t + C。代入 v(0)=2 得 C = 2，因此 v(t) = 3t² – 4t + 2。

3. From Velocity to Displacement | 从速度到位移

Once velocity is known, displacement is found by integrating velocity with respect to time:

求得速度之后，对速度关于时间积分即可得到位移：

s(t) = ∫ v(t) dt

The constant of integration is determined from the initial displacement s(0). For example, continuing from v(t) = 3t² – 4t + 2, integration gives s(t) = t³ – 2t² + 2t + C. If s(0) = 1, then C = 1, so s(t) = t³ – 2t² + 2t + 1.

积分常数由初始位移 s(0) 决定。例如，继续使用 v(t) = 3t² – 4t + 2，积分得到 s(t) = t³ – 2t² + 2t + C。若 s(0) = 1，则 C = 1，所以 s(t) = t³ – 2t² + 2t + 1。

4. Using Initial Conditions | 使用初始条件

In indefinite integration, the constant C is unknown, but physical problems always provide initial conditions such as v(0) or s(0). These conditions allow us to find a unique solution.

在不定积分中，常数 C 是未知的，但实际问题通常会给出初始条件，如 v(0) 或 s(0)。利用这些条件可以确定唯一解。

When limits are known, definite integrals are often more convenient. The displacement between t = 0 and t = T is:

当积分限已知时，使用定积分往往更方便。从 t = 0 到 t = T 的位移为：

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