📚 Applications of Differentiation in Kinematics | 微分在运动学中的应用
In A-Level Mathematics, differentiation provides a powerful framework for analysing the motion of particles. Kinematics, the branch of mechanics that describes motion without reference to its causes, relies fundamentally on derivatives to connect displacement, velocity and acceleration. This article presents a structured guide to applying differentiation in kinematic problems, with definitions, worked examples and examination tips aligned to the A-Level syllabus.
在A-Level数学中，微分为分析质点运动提供了强大的工具。运动学是力学中描述运动而不涉及其原因的分支，它以导数为核心纽带将位移、速度和加速度联系起来。本文将系统讲解微分在运动学问题中的应用，涵盖定义、例题解析和与A-Level考纲紧密对应的考试技巧。
1. Kinematics Basics | 运动学基础
Kinematics describes motion using three fundamental quantities: displacement (s), velocity (v) and acceleration (a). Displacement is the position of a particle measured from a fixed origin; velocity is the rate of change of displacement; acceleration is the rate of change of velocity. In A-Level mechanics, each quantity is treated as a function of time t, with displacement typically expressed in metres (m), velocity in metres per second (m/s) and acceleration in metres per second squared (m/s²).
运动学通过三个基本量描述运动：位移（s）、速度（v）和加速度（a）。位移是质点相对于固定原点的位置；速度是位移的变化率；加速度是速度的变化率。在A-Level力学中，每个量都是时间t的函数，位移通常以米（m）为单位，速度以米每秒（m/s）为单位，加速度以米每二次方秒（m/s²）为单位。
| Quantity | 物理量 | Symbol | 符号 | SI Unit | 国际单位 |
|---|---|---|
| Displacement | 位移 | s | metre (m) | 米 |
| Velocity | 速度 | v | m/s | 米/秒 |
| Acceleration | 加速度 | a | m/s² | 米/秒² |
When motion is one-dimensional and along a straight line, positive and negative signs indicate direction. A positive velocity means motion in the positive direction of the coordinate axis, while a negative velocity means motion in the opposite direction.
当运动是一维且沿直线进行时，正负号表示方向。速度为正意味着质点沿坐标轴正方向运动，速度为负则意味着沿相反方向运动。
2. Velocity as the Derivative of Displacement | 速度：位移的导数
The instantaneous velocity of a particle is defined as the derivative of displacement with respect to time. If the displacement function is s(t), then the velocity function is given by:
质点的瞬时速度定义为位移关于时间的导数。若位移函数为s(t)，则速度函数为：
v(t) = ds/dt
Geometrically, the derivative ds/dt represents the gradient of the tangent to the displacement–time graph at any instant. A steep positive gradient indicates rapid motion in the positive direction; a zero gradient indicates that the particle is momentarily at rest.
从几何角度看，导数ds/dt表示位移-时间图像上任意时刻切线的斜率。斜率很大且为正表示质点沿正方向快速运动；斜率为零则表示质点瞬时静止。
For example, if s(t) = 3t² + 2t, then v(t) = 6t + 2. At t = 1 s, the velocity is 8 m/s. Notice that differentiating the displacement function gives an expression valid for every instant of time, rather than merely an average over an interval.
例如，若s(t) = 3t² + 2t，则v(t) = 6t + 2。当t = 1 s时，速度为8 m/s。注意，对位移函数求导得到的是对任意时刻都成立的表达式，而不只是某个区间上的平均速度。
3. Acceleration as the Derivative of Velocity | 加速度：速度的导数
Acceleration measures how quickly velocity changes with time. It is defined as the derivative of velocity with respect to time:
加速度衡量速度随时间变化的快慢，定义为速度关于时间的导数：
a(t) = dv/dt = d²s/dt²
Thus, acceleration is also the second derivative of displacement with respect to time. This chain of relationships is fundamental: differentiating s gives v, and differentiating v gives a.
因此，加速度也是位移关于时间的二阶导数。这一关系链至关重要：对s求导得到v，对v求导得到a。
For instance, if s(t) = t³ − 6t² + 9t, then v(t) = 3t² − 12t + 9 and a(t) = 6t − 12. At t = 2 s, the acceleration is 0 m/s², which means the velocity is momentarily neither increasing nor decreasing.
例如，若s(t) = t³ − 6t² + 9t，则v(t) = 3t² − 12t + 9，a(t) = 6t − 12。当t = 2 s时，加速度为0 m/s²，意味着速度在该时刻既不增大也不减小。
4. The Meaning of the Second Derivative | 二阶导数的物理意义
The second derivative d²s/dt² provides valuable information about the shape of the displacement–time graph. When d²s/dt² > 0, the graph is concave upward, and the velocity is increasing; when d²s/dt² < 0, the graph is concave downward, and the velocity is decreasing.
二阶导数d²s/dt²提供了关于位移-时间图像形状的重要信息。当d²s/dt² > 0时，图像向下凹（凹向上），速度在增大；当d²s/dt² < 0时，图像向上凸（凹向下），速度在减小。
In kinematic terms, a positive acceleration indicates that the particle is speeding up in the positive direction or slowing down in the negative direction. A negative acceleration (often called deceleration) indicates the opposite. However, it is essential to compare the signs of velocity and acceleration to determine whether a particle is truly speeding up or slowing down. If v and a have the same sign, speed increases; if they have opposite signs, speed decreases.
在运动学中，加速度为正表示质点沿正方向加速或沿负方向减速；加速度为负（常称为减速度）则相反。然而，必须比较速度与加速度的符号才能判断质点究竟是在加速还是减速。若v与a同号，速率增大；若异号，速率减小。
Consider a particle with v(t) = t² − 4t + 3. Its acceleration is a(t) = 2t − 4. At t = 0, v = 3 m/s and a = −4 m/s²: the particle is moving in the positive direction but slowing down. At t = 5, v = 8 m/s and a = 6 m/s²: both are positive, so the particle is speeding up.
考虑一个质点，其速度v(t) = t² − 4t + 3。其加速度a(t) = 2t − 4。在t = 0时，v = 3 m/s且a = −4 m/s²：质点沿正方向运动但在减速。在t = 5时，v = 8 m/s且a = 6 m/s²：两者均为正，因而质点在加速。
5. Finding Maximum and Minimum Displacement | 求位移的最大值与最小值
Differentiation is the standard tool for locating maximum and minimum values of displacement. At an extreme value of displacement, the velocity must be zero. Therefore, to find the time at which maximum displacement occurs, solve the equation:
微分是求位移最大值与最小值的标准工具。在位移的极值点处，速度必为零。因此，要求位移取得最大值的时刻，需解方程：
v(t) = ds/dt = 0
Once the candidate times are found, classify them using the second derivative. If d²s/dt² < 0 at a stationary point, the displacement is a local maximum; if d²s/dt² > 0, it is a local minimum. Alternatively, test the sign of v just before and after the stationary point.
找到候选时刻后，用二阶导数进行分类。若在驻点处d²s/dt² < 0，则位移为局部极大值；若d²s/dt² > 0，则为局部极小值。也可以检验驻点前后v的符号变化来确定极值类型。
For example, consider s(t) = 20t − 5t². Setting v(t) = 20 − 10t = 0 gives t = 2 s. Since d²s/dt² = −10 < 0, this is a maximum. The maximum displacement is s(2) = 20(2) − 5(2)² = 40 − 20 = 20 m.
例如，考虑s(t) = 20t − 5t²。令v(t) = 20 − 10t = 0，得t = 2 s。由于d²s/dt² = −10 < 0，这是极大值。最大位移为s(2) = 20(2) − 5(2)² = 40 − 20 = 20 m。
6. Finding Maximum Velocity and Minimum Speed | 求最大速度与最小速率
To find the time at which velocity reaches a maximum or minimum, differentiate the velocity function and set its derivative equal to zero:
要求速度达到最大值或最小值的时刻，对速度函数求导并令其导数为零：
dv/dt = d²s/dt² = 0
Solving dv/dt = 0 gives the times of stationary velocity. The second derivative of velocity, d³s/dt³, can be used to classify these points, though in practice examining the sign change of dv/dt around the stationary point is often simpler.
解方程dv/dt = 0可得速度驻点对应的时刻。速度的二阶导数d³s/dt³可用于对这些点进行分类，但在实际操作中，观察dv/dt在驻点附近的符号变化往往更简便。
It is also important to distinguish between velocity and speed. Speed is the magnitude of velocity, |v|. A particle may have zero velocity at a turning point, but its speed could be large elsewhere. When asked for minimum speed, always consider whether the velocity can equal zero, and also check endpoints of the given time interval.
还需要注意区分速度与速率。速率是速度的大小，即|v|。质点在转向点处速度可能为零，但在其他位置速率可能很大。当题目要求最小速率时，始终考虑速度是否可能为零，并同时检查给定时间区间的端点。
For instance, if v(t) = t² − 4t + 3 for 0 ≤ t ≤ 5, setting dv/dt = 2t − 4 = 0 gives t = 2, with v(2) = −1 m/s. The speed is |−1| = 1 m/s. At the endpoint t = 0, v = 3 m/s, and at t = 5, v = 8 m/s. Thus the minimum speed over the interval is 0 m/s, occurring at t = 1 and t = 3 where v(t) = 0.
例如，若v(t) = t² − 4t + 3且0 ≤ t ≤ 5，令dv/dt = 2t − 4 = 0得t = 2，此时v(2) = −1 m/s，速率为|−1| = 1 m/s。在端点t = 0处，v = 3 m/s；在t = 5处，v = 8 m/s。因此该区间上的最小速率为0 m/s，出现在t = 1和t = 3处，因为这两处v(t) = 0。
7. Worked Example 1: Cubic Displacement Function | 例题1：三次位移函数
A particle moves along a straight line such that its displacement is given by s(t) = t³ − 6t² + 9t, where s is measured in metres and t in seconds. Find: (a) the velocity when t = 2 s; (b) the times when the particle is at rest; (c) the acceleration when t = 1 s; (d) the total distance travelled in the first 4 seconds.
一质点沿直线运动，其位移为s(t) = t³ − 6t² + 9t，其中s以米为单位，t以秒为单位。求：(a) t = 2 s时的速度；(b) 质点静止的时刻；(c) t = 1 s时的加速度；(d) 前4秒内经过的总路程。
(a) Velocity at t = 2 | t = 2时的速度
Differentiating: v(t) = 3t² − 12t + 9. At t = 2: v(2) = 3(2)² − 12(2) + 9 = 12 − 24 + 9 = −3 m/s. The negative sign indicates that the particle is moving in the negative direction at this instant.
求导得：v(t) = 3t² − 12t + 9。在t = 2处：v(2) = 3(2)² − 12(2) + 9 = 12 − 24 + 9 = −3 m/s。负号表示此刻质点沿负方向运动。
(b) Times at rest | 静止时刻
At rest means v(t) = 0: 3t² − 12t + 9 = 0. Dividing by 3: t² − 4t + 3 = 0, which factors as (t − 1)(t − 3) = 0. Hence t = 1 s or t = 3 s.
静止即v(t) = 0：3t² − 12t + 9 = 0。除以3得：t² − 4t + 3 = 0，因式分解为(t − 1)(t − 3) = 0。因此t = 1 s或t = 3 s。
(c) Acceleration at t = 1 | t = 1时的加速度
Differentiate v(t): a(t) = 6t − 12. At t = 1: a(1) = 6(1) − 12 = −6 m/s². The acceleration is 6 m/s² in the negative direction.
对v(t)求导：a(t) = 6t − 12。在t = 1处：a(1) = 6(1) − 12 = −6 m/s²。加速度沿负方向，大小为6 m/s²。
(d) Total distance in the first 4 seconds | 前4秒的总路程
Compute displacements at key times: s(0) = 0, s(1) = 1 − 6 + 9 = 4 m, s(3) = 27 − 54 + 27 = 0 m, s(4) = 64 − 96 + 36 = 4 m. The particle moves from 0 to 4 m between t = 0 and t = 1, back to 0 m between t = 1 and t = 3, then to 4 m between t = 3 and t = 4. Total distance = 4 + 4 + 4 = 12 m.
计算关键时刻的位移：s(0) = 0，s(1) = 1 − 6 + 9 = 4 m，s(3) = 27 − 54 + 27 = 0 m，s(4) = 64 − 96 + 36 = 4 m。质点在t = 0到t = 1之间从0运动到4 m，在t = 1到t = 3之间返回0 m，再在t = 3到t = 4之间运动到4 m。总路程 = 4 + 4 + 4 = 12 m。
8. Worked Example 2: Maximum Height of a Projectile | 例题2：抛体的最大高度
A ball is thrown vertically upwards from ground level. Its height above the ground is given by h(t) = 25t − 5t², where h is in metres and t in seconds. Find the maximum height reached and the time taken, and determine the velocity with which the ball hits the ground.
一个小球从地面竖直上抛。其离地高度为h(t) = 25t − 5t²，其中h以米为单位，t以秒为单位。求达到的最大高度及所需时间，并求小球落地时的速度。
Maximum height | 最大高度
The maximum height occurs when the vertical velocity is zero. The velocity is v(t) = dh/dt = 25 − 10t. Setting v(t) = 0 gives 25 − 10t = 0, so t = 2.5 s. The second derivative d²h/dt² = −10 < 0 confirms a maximum. Hence h_max = 25(2.5) − 5(2.5)² = 62.5 − 31.25 = 31.25 m.
最大高度出现在竖直速度为零时。速度为v(t) = dh/dt = 25 − 10t。令v(t) = 0得25 − 10t = 0，故t = 2.5 s。二阶导数d²h/dt² = −10 < 0确认这是极大值。因此h_max = 25(2.5) − 5(2.5)² = 62.5 − 31.25 = 31.25 m。
Time of flight and impact velocity | 飞行时间与落地速度
The ball returns to the ground when h(t) = 0: t(25 − 5t) = 0, so t = 0 or t = 5 s. The impact velocity is v(5) = 25 − 10(5) = −25 m/s. The negative sign indicates downward motion. Note that the impact speed equals the initial speed of 25 m/s, demonstrating the symmetry of projectile motion under uniform gravity.
小球落地时h(t) = 0：t(25 − 5t) = 0，故t = 0或t = 5 s。落地速度为v(5) = 25 − 10(5) = −25 m/s。负号表示向下运动。注意落地速率与初速率25 m/s相等，体现了匀重力作用下抛体运动的对称性。
9. Common Exam Pitfalls | 常见考试易错点
The following errors are frequently observed in A-Level kinematics questions involving differentiation.
以下错误在涉及微分的A-Level运动学题目中频繁出现，务必警惕。
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Confusing displacement and distance. Displacement is a vector quantity with direction; distance is the total length of the path travelled. To find distance, you must identify intervals where velocity changes sign and sum the absolute displacements over each interval.
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混淆位移与路程。位移是带方向的矢量；路程是实际经过的路径总长度。求
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