Application of Time Functions in Kinematics | 时间函数在运动学中的应用

📚 Application of Time Functions in Kinematics | 时间函数在运动学中的应用

In A-Level Mathematics, kinematics describes the motion of objects without considering the forces that cause the motion. The key idea is to express position, velocity, and acceleration as functions of time, typically denoted by t. This approach turns physical motion into a problem of calculus, where differentiation and integration reveal the relationships between these fundamental quantities.

在 A-Level 数学中，运动学描述物体运动而不考虑引起运动的力。核心思想是将位置、速度和加速度表示为时间 t 的函数。这种方法将物理运动转化为微积分问题，通过微分和积分揭示这些基本量之间的关系。

1. Displacement, Velocity, and Acceleration | 位移、速度与加速度

Displacement, s(t), measures the position of an object relative to a fixed origin at time t. Unlike distance, displacement is a vector quantity and can be positive or negative. The instantaneous velocity, v(t), is the rate of change of displacement with respect to time, while the instantaneous acceleration, a(t), is the rate of change of velocity with respect to time.

位移 s(t) 表示物体在时刻 t 相对于固定原点的位置。与路程不同，位移是矢量，可以为正或负。瞬时速度 v(t) 是位移对时间的变化率，而瞬时加速度 a(t) 是速度对时间的变化率。

v(t) = ds/dt, a(t) = dv/dt = d²s/dt²

2. Differentiation: From Displacement to Acceleration | 微分：从位移到加速度

When the displacement function s(t) is known, we obtain velocity by differentiating once, and acceleration by differentiating twice. For example, if s(t) = t³ − 6t² + 9t, then v(t) = 3t² − 12t + 9 and a(t) = 6t − 12. These functions allow us to analyse the motion at any instant.

当位移函数 s(t) 已知时，对 s(t) 求一次导数得速度，求两次导数得加速度。例如，若 s(t) = t³ − 6t² + 9t，则 v(t) = 3t² − 12t + 9，a(t) = 6t − 12。这些函数使我们能够分析任意时刻的运动。

The velocity function is the gradient of the displacement-time graph.

速度函数是位移-时间图像上切线的斜率。
The acceleration function is the gradient of the velocity-time graph.

加速度函数是速度-时间图像上切线的斜率。

3. Integration: From Acceleration to Displacement | 积分：从加速度到位移

Integration is the reverse process. Given a(t), we integrate to find v(t), and integrate again to find s(t). Since indefinite integrals include an arbitrary constant, we need initial conditions to determine the specific solution. For example, if a(t) = 12t, and at t = 0, v = 5, then v(t) = 6t² + 5.

积分是微分的逆过程。已知 a(t)，对其积分可得 v(t)，再积分可得 s(t)。由于不定积分包含任意常数，我们需要初始条件来确定具体解。例如，若 a(t) = 12t，且当 t = 0，v = 5，则 v(t) = 6t² + 5。

v(t) = ∫ a(t) dt, s(t) = ∫ v(t) dt

4. Initial Conditions and Constants | 初始条件与积分常数

Each integration introduces an arbitrary constant, typically determined by the initial displacement and velocity. Without these conditions, the displacement function is only accurate up to a vertical shift. For instance, integrating v(t) = 4t + 3 gives s(t) = 2t² + 3t + C. If s(0) = 10, then C = 10.

每一次积分都会引入任意常数，通常由初始位移和初始速度确定。如果没有这些条件，位移函数只能精确到垂直平移。例如，对 v(t) = 4t + 3 积分得 s(t) = 2t² + 3t + C。若 s(0) = 10，则 C = 10。

In A-Level questions, the phrase “starts from rest” means v(0) = 0, while “is initially at the origin” means s(0) = 0. Always identify these clues before integrating.

在 A-Level 题目中，“从静止开始”表示 v(0) = 0，而“初始位置在原点”表示 s(0) = 0。积分前务必识别这些线索。

5. Worked Example 1: Given s(t) | 例题 1：已知 s(t)

A particle moves along a straight line such that s(t) = 2t³ − 15t² + 36t, where s is in metres and t is in seconds. Find the velocity and acceleration at t = 2.

质点沿直线运动，位移 s(t) = 2t³ − 15t² + 36t，其中 s 的单位为米，t 的单位为秒。求 t = 2 时的速度和加速度。

First differentiate: v(t) = 6t² − 30t + 36, a(t) = 12t − 30. Substituting t = 2 gives v(2) = 24 − 60 + 36 = 0, and a(2) = 24 − 30 = −6. So the particle is instantaneously at rest at t = 2, with acceleration −6 m/s².

首先求导：v(t) = 6t² − 30t + 36，a(t) = 12t − 30。代入 t = 2 得 v(2) = 24 − 60 + 36 = 0，a(2) = 24 − 30 = −6。因此在 t = 2 时质点瞬时静止，加速度为 −6 m/s²。

6. Worked Example 2: Given a(t) with Initial Conditions | 例题 2：已知 a(t) 且给出初始条件

The acceleration of a particle is given by a(t) = 6t − 4. At t = 0, the particle is at the origin with velocity 2 m/s. Find the displacement function s(t).

某质点的加速度为 a(t) = 6t − 4。在 t = 0 时，质点位于原点，速度为 2 m/s。求位移函数 s(t)。

Integrate to get v(t) = 3t² − 4t + C. Using v(0) = 2 gives C = 2, so v(t) = 3t² − 4t + 2. Integrate again: s(t) = t³ − 2t² + 2t + D. Since s(0) = 0, D = 0. Hence s(t) = t³ − 2t² + 2t.

积分得 v(t) = 3t² − 4t + C。利用 v(0) = 2 得 C = 2，所以 v(t) = 3t² − 4t + 2。再次积分：s(t) = t³ − 2t² + 2t + D。因为 s(0) = 0，所以 D = 0。因此 s(t) = t³ − 2t² + 2t。

7. Graphical Interpretation | 图像解释

Displacement-time graphs have gradients equal to velocity. A horizontal tangent means zero velocity, i.e., the particle is momentarily at rest. Velocity-time graphs have gradients equal to acceleration, while the area under the velocity-time curve gives the change in displacement over that interval.

位移-时间图像的斜率等于速度。水平切线表示速度为零，即质点瞬时静止。速度-时间图像的斜率等于加速度，而曲线下方的面积表示该时间段内位移的变化量。

Graph	Gradient	Area under curve
s-t graph	velocity	not used
v-t graph	acceleration	change in displacement
a-t graph	jerk (not required)	change in velocity

8. Maxima and Minima in Kinematics | 运动学中的最大值与最小值

To find when a particle reaches its maximum displacement, set v(t) = 0 and solve for t. Check the second derivative or sign change to confirm it is a maximum. Similarly, maximum speed occurs when a(t) = 0, provided the velocity function is differentiable and the critical point lies within the domain.

要求质点何时达到最大位移，令 v(t) = 0 并解出 t。通过二阶导数或符号变化确认是否为最大值。类似地，最大速度出现在 a(t) = 0 时，前提是速度函数可导且临界点在定义域内。

Stationary points of s(t) correspond to v(t) = 0.

s(t) 的驻点对应 v(t) = 0。
Stationary points of v(t) correspond to a(t) = 0.

v(t) 的驻点对应 a(t) = 0。

9. Constant Acceleration and the suvat Equations | 匀加速度与 suvat 方程

When acceleration is constant, the time functions take a particularly simple form. Integrating a constant acceleration a gives v = u + at, and s = ut + ½at². These are the suvat equations for motion in a straight line, where u is initial velocity, v is final velocity, s is displacement, and t is time.

当加速度恒定时，时间函数呈现特别简单的形式。对恒定加速度 a 积分得 v = u + at，s = ut + ½at²。这就是直线运动中的 suvat 方程，其中 u 是初速度，v 是末速度，s 是位移，t 是时间。

v = u + at, s = ut + ½at², v² = u² + 2as

These formulas should only be used when the acceleration is constant. In A-Level questions, the phrase “uniform acceleration” is the signal for applying them directly.

这些公式仅适用于匀加速度情况。在 A-Level 题目中，“匀加速度”或“uniform acceleration”是直接使用的信号。

10. Motion Under Gravity | 重力作用下的运动

A common application is vertical motion under gravity, where acceleration is approximately g = 9.8 m/s² downward. For an object projected upward, s(t) = ut − ½gt², with displacement measured upward from the launch point. The maximum height is reached when v = 0, i.e., t = u/g.

一个常见应用是重力作用下的竖直运动，加速度约等于 g = 9.8 m/s² 向下。对于向上抛出的物体，s(t) = ut − ½gt²，位移向上为正。最大高度出现在 v = 0 时，即 t = u/g。

At the highest point, the velocity is zero, but the acceleration remains g downward. This is because acceleration is the rate of change of velocity, not the velocity itself.

在最高点，速度为零，但加速度仍为 g 向下。因为加速度是速度的变化率，而不是速度本身。

11. Common Pitfalls and Examination Tips | 常见错误与考试要点

Students often confuse speed with velocity. Speed is the magnitude of velocity and is always non-negative. When asked for ‘distance travelled’, you must integrate |v(t)| over the time interval, accounting for direction changes.

学生经常混淆速率与速度。速率是速度的大小，永远非负。当问及“总路程”时，必须对时间区间内的 |v(t)| 积分，并考虑方向变化。

Always write units in final answers.

最终答案务必写上单位。
Check that t values are sensible: negative time is often invalid.

检查 t 值是否合理：负时间通常无效。
When a particle changes direction, v(t) changes sign.

当质点改变方向时，v(t) 会改变符号。

12. Summary | 总结

Time functions lie at the heart of A-Level kinematics. By using differentiation and integration, we can move seamlessly between displacement, velocity, and acceleration. Correctly applying initial conditions, interpreting graphs, and recognising the special case of constant acceleration are essential skills for solving exam problems reliably.

时间函数是 A-Level 运动学的核心。通过微分和积分，我们可以在位移、速度和加速度之间顺畅转换。正确应用初始条件、解读图像，以及识别匀加速这一特殊情况，是可靠解决考试题目的关键技能。

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Application of Time Functions in Kinematics | 时间函数在运动学中的应用