Functions of Time: Motion Described by Time-Dependent Functions | 时间函数：用时间变量描述运动

📚 Functions of Time: Motion Described by Time-Dependent Functions | 时间函数：用时间变量描述运动

In A-Level Mechanics, the motion of a particle along a straight line is often described by expressing displacement, velocity and acceleration as functions of time t. Mastering these time-dependent functions and the calculus linking them is essential for solving kinematics problems with variable acceleration. This article will guide you through the key concepts, worked examples and common pitfalls.

在A-Level力学中，一个质点沿直线运动时，位移、速度和加速度常常被表示为时间t的函数。掌握这些以时间为自变量的函数以及它们之间的微积分关系，是解决变加速度运动学问题的关键。本文将带你梳理核心概念、典型示例和常见误区。

---

1. Understanding Functions of Time | 理解时间函数

In mechanics, we treat kinematic quantities as functions of t. Instead of using constant-acceleration formulae, we work with expressions like s(t), v(t) and a(t) that can vary with time. This allows us to model more realistic motion, such as a car accelerating non-uniformly or a particle moving under a variable force.

在力学中，我们将运动学量视为t的函数。我们不用匀加速公式，而是使用会随时间变化的s(t)、v(t)和a(t)表达式。这使得我们可以模拟更真实的运动，例如汽车非均匀加速或质点在变力作用下运动。

The key relationships are built on differentiation and integration. Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity. Therefore, if one function is known, the others can be derived.

核心关系建立在求导和积分之上。速度是位移的变化率，加速度是速度的变化率。因此，只要已知其中一个函数，就能推导出其他函数。

---

2. Displacement s(t): The Starting Point | 位移s(t)：起点

Displacement s(t) gives the position of a particle relative to a fixed origin at time t. It is usually provided as a polynomial, such as s = 2t³ – 9t² + 12t + 3. The variable t is typically measured in seconds and s in metres.

位移s(t)给出了质点在t时刻相对于固定原点的位置。它通常以多项式的形式给出，例如 s = 2t³ – 9t² + 12t + 3。t通常以秒为单位，s以米为单位。

Displacement can be positive, negative or zero. A positive value means the particle is on one side of the origin, while a negative value means it is on the opposite side. The total distance travelled is not the same as displacement when the particle changes direction.

位移可以为正、为负或为零。正值表示质点在原点的一侧，负值表示在另一侧。当质点改变运动方向时，总路程与位移并不相同。

---

3. Velocity v(t) = ds/dt | 速度v(t) = ds/dt

Velocity is obtained by differentiating displacement with respect to time. If s(t) = 2t³ – 9t² + 12t + 3, then

v(t) = ds/dt = 6t² – 18t + 12.

速度通过位移对时间求导得到。如果 s(t) = 2t³ – 9t² + 12t + 3，那么

v(t) = ds/dt = 6t² – 18t + 12.

The sign of velocity indicates the direction of motion: positive v means moving in the positive direction; negative v means moving in the negative direction. The particle is instantaneously at rest when v = 0.

速度的正负表示运动方向：v为正表示向正方向运动；v为负表示向负方向运动。当v=0时，质点瞬时静止。

---

4. Acceleration a(t) = dv/dt = d²s/dt² | 加速度a(t) = dv/dt = d²s/dt²

Acceleration is the derivative of velocity, or the second derivative of displacement. Continuing the example,

a(t) = dv/dt = 12t – 18.

加速度是速度的导数，或位移的二阶导数。继续上面的例子，

a(t) = dv/dt = 12t – 18.

A positive value of acceleration means the velocity is increasing (in the positive sense), while a negative acceleration means the velocity is decreasing. However, whether the particle is speeding up or slowing down depends on whether velocity and acceleration have the same sign.

加速度为正表示速度在（正向上）增加，加速度为负表示速度在减小。但质点是加速还是减速，要看速度与加速度是否同号。

---

5. Recovering Displacement from Velocity | 由速度恢复位移

If velocity v(t) is given, displacement s(t) is found by integration. For instance, if v = 6t² – 18t + 12, then

s(t) = ∫ v dt = 2t³ – 9t² + 12t + C.

如果给出了速度v(t)，通过积分可求得位移s(t)。例如，若 v = 6t² – 18t + 12，则

s(t) = ∫ v dt = 2t³ – 9t² + 12t + C.

The constant of integration C represents the initial displacement when t = 0. To find C, substitute a known position, typically given in the problem.

积分常数C代表t=0时的初始位移。求C时需代入一个已知位置，通常题目会给出。

---

6. Recovering Velocity from Acceleration | 由加速度恢复速度

Similarly, if acceleration a(t) is provided, velocity is the integral of acceleration. For a = 12t – 18,

v(t) = ∫ a dt = 6t² – 18t + D.

类似地，若给出加速度a(t)，速度就是加速度的积分。对于 a = 12t – 18，

v(t) = ∫ a dt = 6t² – 18t + D.

The constant D is determined using the initial velocity, often given as ‘at t = 0, v = 12’ or similar. Always check the initial conditions before finalising your expression.

常数D用初始速度来确定，题目常给出‘t = 0时，v = 12’之类的条件。在敲定表达式之前，务必检查初始条件。

---

7. The Role of Initial Conditions | 初始条件的作用

Initial conditions make the solution unique. Without them, an expression for s(t) or v(t) contains an unknown constant and cannot be used for specific predictions. For example, if you know that at t = 0, s = 3, then C = 3, giving s(t) = 2t³ – 9t² + 12t + 3.

初始条件让解变得唯一。没有初始条件，s(t)或v(t)的表达式中含有未知常数，无法用于具体预测。举例来说，若已知t=0时s=3，则C=3，从而s(t) = 2t³ – 9t² + 12t + 3。

Common given conditions include ‘the particle passes through the origin at t = 0’, ‘initially at rest’ (v = 0 at t = 0), or ‘initial velocity is 5 m/s’. Translate these into mathematical equations to find the constant.

常见的给定条件有‘质点在t=0时经过原点’、‘初始静止’（t=0时v=0）或‘初速度为5 m/s’。将这些条件转化为数学方程即可求出常数。

---

8. Analysing Motion: When is Velocity Zero? | 分析运动：何时速度为零？

Solving v(t) = 0 gives the times when the particle is instantaneously at rest. These moments often correspond to changes in direction. For v = 6t² – 18t + 12, factorising gives

6(t² – 3t + 2) = 6(t – 1)(t – 2) = 0,

so v = 0 at t = 1 s and t = 2 s.

解v(t)=0可以得到质点瞬时静止的时刻。这些时刻通常对应着运动方向的改变。对于 v = 6t² – 18t + 12，因式分解得

6(t² – 3t + 2) = 6(t – 1)(t – 2) = 0，

因此 t = 1 s 和 t = 2 s 时 v = 0。

To determine whether the particle changes direction, check the sign of velocity just before and after the root. If the sign changes, the direction reverses.

要判断质点是否改变方向，可以检查根两侧速度的符号。如果符号变化，则方向反转。

---

9. Maximum and Minimum Velocity | 最大和最小速度

To find extreme values of velocity, we set a(t) = 0 because acceleration is the derivative of velocity. From a = 12t – 18 = 0, we get t = 1.5 s. Then we evaluate v at t = 1.5, and at the boundaries of the interval considered.

要寻找速度的极值，可令a(t)=0，因为加速度是速度的导数。由 a = 12t – 18 = 0 得 t = 1.5 s。然后计算 t = 1.5 以及所考虑区间端点处的v值。

We can also use the second derivative test: differentiate a to get da/dt = 12 (positive), so the velocity has a minimum at t = 1.5. Indeed, v(1.5) = 6(2.25) – 27 + 12 = –1.5 m/s, a minimum.

我们也可以用二阶导数检验：对a求导得 da/dt = 12（正），因此速度在 t = 1.5 处取得极小值。事实上，v(1.5) = 6(2.25) – 27 + 12 = –1.5 m/s，是极小值。

---

10. Real-World Context and Interpretation | 现实背景与解释

When interpreting functions of time, remember that s(t) gives position, not distance. To find the total distance travelled between t = 0 and t = 3, you must check for changes of direction (v = 0) and sum the absolute displacements for each segment. This is a classic exam requirement.

在解释时间函数时，记住s(t)给出的是位置，而不是路程。要求从t=0到t=3的总路程，必须检查运动方向的变化（v=0的时刻），并对每一段位移取绝对值求和。这是经典的考试要求。

Acceleration sign can be misleading. When a particle is moving in the positive direction and a is negative, it is decelerating. But if the particle is moving in the negative direction and a is negative, the particle is speeding up in the negative direction. Understanding the product of v and a is essential.

加速度的符号可能具有误导性。当质点向正方向运动且a为负时，它在减速。但如果质点向负方向运动且a为负，那么它是在负方向上加速。理解v与a的乘积至关重要。

---

11. Common Mistakes with Functions of Time | 时间函数常见错误

• Forgetting the constant of integration when finding s or v from a derivative. Always include +C and use initial conditions.

  在由导数求s或v时忘记积分常数。务必加上+C并利用初始条件。

• Confusing displacement with distance. If a question asks ‘find the distance travelled’, do not simply substitute t into s(t) without checking for direction changes.

  混淆位移与路程。如果题目要求‘求运动的路程’，不能未经方向变化检查就直接将t代入s(t)。

• Misapplying the sign of acceleration. A negative a does not always mean ‘slowing down’. Always compare the direction of motion with the direction of acceleration.

  错误理解加速度的符号。a为负并不总是表示‘减速’。要始终比较运动方向与加速度方向。

• Algebraic errors when differentiating or integrating powers of t. Remember: d/dt (tⁿ) = n tⁿ⁻¹ and ∫ tⁿ dt = tⁿ⁺¹/(n+1) + C.

  在求导或积分t的幂次时犯代数错误。记住：d/dt (tⁿ) = n tⁿ⁻¹， ∫ tⁿ dt = tⁿ⁺¹/(n+1) + C。

---

12. Summary and Key Takeaways | 总结与要点

Functions of time provide a precise description of motion with variable acceleration. The core toolkit includes:

Relationship	Formula
Displacement → Velocity	v = ds/dt
Velocity → Acceleration	a = dv/dt = d²s/dt²
Velocity → Displacement	s = ∫ v dt + C
Acceleration → Velocity	v = ∫ a dt + D

时间函数为描述变加速度运动提供了精确的方式。核心工具包括：

关系	公式
位移 → 速度	v = ds/dt
速度 → 加速度	a = dv/dt = d²s/dt²
速度 → 位移	s = ∫ v dt + C
加速度 → 速度	v = ∫ a dt + D

In exam questions for Edexcel A-Level Mathematics, you will routinely be asked to form these functions, find constants, determine when a particle is at rest, and calculate distance travelled. Practise with polynomials, then extend to simple trigonometric or exponential functions of time for a deeper understanding.

在Edexcel A-Level数学的考题中，你会被要求建立这些函数、求出常数、判断质点何时静止并计算路程。先用多项式多加练习，再延伸到简单的三角函数或指数函数形式，以获得更深入的理解。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）