Functions of Time in Kinematics and Parametric Equations | 运动学与参数方程中的时间函数

📚 Functions of Time in Kinematics and Parametric Equations | 运动学与参数方程中的时间函数

In Edexcel A-Level Mathematics, functions of time appear throughout both pure and applied modules. In mechanics, displacement x(t), velocity v(t), and acceleration a(t) are examples of time-dependent functions, and calculus links them through differentiation and integration. In pure mathematics, parametric equations often use t as the parameter defining a curve. Understanding how to manipulate these functions is essential for questions on variable acceleration, motion in a straight line, and parametric differentiation.

在 Edexcel A-Level 数学中，时间函数贯穿纯数学和应用模块。在力学中，位移 x(t)、速度 v(t) 和加速度 a(t) 都是随时间变化的函数，微积分通过求导和积分将它们联系起来。在纯数学中，参数方程常用 t 作为参数来定义曲线。掌握这些函数的处理方法是解决变加速运动、直线运动以及参数微分问题的关键。

1. What is a Function of Time? | 什么是时间的函数？

A function of time is any rule whose output depends on the variable t, usually measured in seconds. In mechanics, we often use x(t) or s(t) for displacement from a fixed origin, with t as the independent variable. Writing x = t³ − 6t² + 9t + 2 means the position of a particle changes continuously with time. This is different from constant acceleration problems where displacement is given by a quadratic from suvat equations.

时间的函数指的是输出依赖于变量 t 的规则，t 通常以秒为单位。在力学中，我们常用 x(t) 或 s(t) 表示相对于固定原点的位移，t 是自变量。写出 x = t³ − 6t² + 9t + 2 就意味着粒子的位置随时间连续变化。这与匀加速问题不同，匀加速问题中位移由 suvat 方程给出的二次函数表示。

x = x(t) or s = s(t)

The table below summarises the three main time functions in kinematics and their meanings.

下表总结了运动学中三个主要的时间函数及其含义。

2. Position, Velocity and Acceleration as Functions of Time | 位移、速度与加速度作为时间函数

In variable acceleration, position, velocity and acceleration are linked by two differentiation steps. Starting from displacement x(t), the first derivative gives velocity v(t), and the second derivative gives acceleration a(t). Conversely, starting from acceleration, integration recovers velocity and displacement.

在变加速运动中，位移、速度和加速度通过两步求导建立联系。从位移 x(t) 出发，一阶导数给出速度 v(t)，二阶导数给出加速度 a(t)。反过来，从加速度出发，积分可以还原速度和位移。

v(t) = dx/dt

a(t) = dv/dt = d²x/dt²

For the model x(t) = t³ − 6t² + 9t + 2, differentiating gives v(t) = 3t² − 12t + 9 and a(t) = 6t − 12. These are also functions of time, so we can evaluate them at any instant and analyse how motion changes.

对于模型 x(t) = t³ − 6t² + 9t + 2，求导可得 v(t) = 3t² − 12t + 9 和 a(t) = 6t − 12。它们同样是时间的函数，因此我们可以在任意时刻求值，并分析运动的变化情况。

3. Differentiating Position to Find Velocity and Acceleration | 对位移求导求速度与加速度

When you are given displacement as a function of time, use term-by-term differentiation to find velocity and acceleration. For x = t³ − 6t² + 9t + 2, the power rule gives the following results.

当题目给出位移随时间变化的函数时，使用逐项求导法则求速度和加速度。对于 x = t³ − 6t² + 9t + 2，使用幂法则可得到以下结果。

v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3)

a(t) = 6t − 12

Factorising the velocity is especially useful because it shows when the particle is at rest. Here v(t) = 0 when t = 1 or t = 3, so those are the two resting times. The acceleration can also be evaluated at these times to test whether the particle is speeding up or slowing down.

对速度进行因式分解尤其有用，因为它能显示粒子何时静止。这里当 t = 1 或 t = 3 时 v(t) = 0，所以这两个时刻是静止时刻。还可以在这些时刻计算加速度，以判断粒子正在加速还是减速。

4. Integrating Acceleration to Find Velocity and Position | 对加速度积分求速度与位移

If acceleration is given as a function of time, use integration to find velocity and displacement. Each integration introduces an arbitrary constant, so you will need initial conditions to determine those constants.

如果给出加速度随时间变化的函数，通过积分可以求出速度和位移。每次积分都会引入一个任意常数，因此你需要初始条件来确定这些常数。

v(t) = ∫ a(t) dt

x(t) = ∫ v(t) dt

For a(t) = 6t − 12, integrating gives v(t) = 3t² − 12t + C. If v(0) = 9, then C = 9, so v(t) = 3t² − 12t + 9. Integrating again gives x(t) = t³ − 6t² + 9t + D. If x(0) = 2, then D = 2, recovering the original displacement model.

对于 a(t) = 6t − 12，积分得到 v(t) = 3t² − 12t + C。如果 v(0) = 9，则 C = 9，所以 v(t) = 3t² − 12t + 9。再次积分得到 x(t) = t³ − 6t² + 9t + D。如果 x(0) = 2，则 D = 2，正好还原原来的位移模型。

5. Initial Conditions and Constants of Integration | 初始条件与积分常数

Initial conditions are normally given at t = 0, such as the initial velocity v(0) and initial position x(0). They allow you to find the constants of integration after each step. For example, if a(t) = 12t − 6, v(0) = 5 and x(0) = 0

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）