Kinematics: Displacement, Velocity, and Acceleration | 运动学：位移、速度与加速度

📚 Kinematics: Displacement, Velocity, and Acceleration | 运动学：位移、速度与加速度

Kinematics is the branch of mathematics that describes the motion of objects using functions of time. In IB Mathematics, the study of kinematics centres on the relationship between displacement, velocity, and acceleration, and it provides a powerful real-world application of differentiation and integration. By treating these quantities as linked through calculus, we can predict where a particle will be, how fast it is moving, and how its speed is changing at any instant.

运动学是描述物体运动随时间变化的数学分支。在 IB 数学中，运动学的核心是位移、速度和加速度之间的关系，它是微分和积分在现实世界中的一个有力应用。通过将这些量视为通过微积分相互联系的函数，我们可以预测粒子在任何时刻的位置、运动速度以及速度的变化率。

1. Fundamental Definitions of Motion | 运动的基本定义

In kinematics, the position of a particle is measured relative to a fixed origin along a straight line. Displacement, often denoted by s (or x), is the distance from the origin together with a direction, making it a vector quantity. Velocity v is the rate of change of displacement with respect to time, and acceleration a is the rate of change of velocity with respect to time. Speed is the magnitude of velocity and does not indicate direction.

在运动学中，粒子的位置是相对于固定原点沿直线测量的。位移通常用 s（或 x）表示，它是到原点的距离加上方向，因此是一个矢量。速度 v 是位移对时间的变化率，加速度 a 是速度对时间的变化率。速率是速度的大小，不表示方向。

2. Velocity and Acceleration as Derivatives | 速度和加速度作为导数

If displacement is given as a function of time, s = f(t), then the instantaneous velocity is the first derivative of displacement: v = ds/dt. Acceleration is the derivative of velocity with respect to time, or the second derivative of displacement: a = dv/dt = d²s/dt². This chain of derivatives allows us to move from position to velocity to acceleration. For example, if s(t) = t³ − 6t² + 9t, then v(t) = 3t² − 12t + 9 and a(t) = 6t − 12.

如果位移是时间函数 s = f(t)，那么瞬时速度是位移的一阶导数：v = ds/dt。加速度是速度对时间的导数，或者是位移的二阶导数：a = dv/dt = d²s/dt²。这一串导数让我们能够从位置过渡到速度，再到加速度。例如，若 s(t) = t³ − 6t² + 9t，则 v(t) = 3t² − 12t + 9，a(t) = 6t − 12。

3. Recovering Displacement and Velocity by Integration | 通过积分恢复位移和速度

When acceleration is given as a function of time, we can integrate to find velocity, and integrate velocity to find displacement. Since integration introduces an arbitrary constant, we need initial conditions—such as the velocity or displacement at t = 0—to determine the specific equation of motion.

当加速度作为时间的函数给出时，我们可以通过积分求出速度，再积分速度求出位移。由于积分会引入任意常数，我们需要初始条件——例如 t = 0 时的速度或位移——来确定具体的运动方程。

v(t) = ∫ a(t) dt + C₁

速度方程：v(t) = ∫ a(t) dt + C₁

s(t) = ∫ v(t) dt + C₂

位移方程：s(t) = ∫ v(t) dt + C₂

For instance, if a particle moves with acceleration a(t) = 4t and initially has velocity 2 m/s and displacement 5 m, then v(t) = ∫ 4t dt = 2t² + 2, and s(t) = ∫ (2t² + 2) dt = (2/3)t³ + 2t + 5.

例如，若一个粒子以加速度 a(t) = 4t 运动，初始速度为 2 m/s，位移为 5 m，则 v(t) = ∫ 4t dt = 2t² + 2，s(t) = ∫ (2t² + 2) dt = (2/3)t³ + 2t + 5。

4. Equations of Motion for Constant Acceleration | 匀加速运动方程

When acceleration is constant, we can derive a set of well-known SUVAT equations that relate displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t). These equations are particularly useful when acceleration does not change and are often applied in problems involving free fall or uniform acceleration along a line.

当加速度恒定时，我们可以推导出一组著名的 SUVAT 方程，它们联系了位移 (s)、初速度 (u)、末速度 (v)、加速度 (a) 和时间 (t)。这些方程在加速度不变时特别有用，常用于自由落体或沿直线匀加速运动的问题。

Equation (方程)	Notes (说明)
v = u + at	Connects velocity, time and acceleration
s = ut + ½at²	Gives displacement directly from initial velocity and acceleration
v² = u² + 2as	Eliminates time; useful for finding final speed
s = (u + v)t / 2	Uses the average velocity

Each equation assumes that acceleration is constant and that motion occurs along a straight line. In IB problems, you will often be asked to choose the appropriate equation based on the known and unknown quantities.

每个方程都假设加速度恒定且运动沿直线进行。在 IB 问题中，你经常会被要求根据已知和未知量选择合适的方程。

5. Motion with Variable Acceleration | 变加速运动

When acceleration is not constant, we must rely fully on calculus. Given a = g(t), integration yields velocity, and a second integration yields displacement. The process often leads to rational, trigonometric, or exponential velocity functions. Students must become comfortable with applying the initial conditions and interpreting the resulting functions to determine distances, times when the particle is at rest, and maximum speeds.

当加速度不是常数时，我们必须完全依赖微积分。给定 a = g(t)，积分得出速度，再积分得出位移。这一过程通常会导出有理函数、三角函数或指数函数形式的速度函数。学生需要熟练运用初始条件，并解读得到的函数来确定距离、粒子静止的时刻以及最大速率。

For example, a = 2 − 6t, with v(0) = 4, s(0) = 0, gives v(t) = ∫(2 − 6t)dt = 2t − 3t² + 4, and s(t) = ∫(2t − 3t² + 4)dt = t² − t³ + 4t. We can then find when the particle is at rest by solving v(t)=0: −3t² + 2t + 4 = 0.

例如，a = 2 − 6t，且 v(0) = 4，s(0) = 0，可得 v(t) = ∫(2 − 6t)dt = 2t − 3t² + 4，以及 s(t) = ∫(2t − 3t² + 4)dt = t² − t³ + 4t。然后通过解 v(t)=0，即 −3t² + 2t + 4 = 0，来求出粒子静止的时刻。

6. Turning Points, Rest and Extreme Velocities | 转向点、静止与速度极值

A particle is at rest when its velocity is zero. Solving v(t) = 0 gives the times when motion changes direction. At these turning points, the displacement may reach a local maximum or minimum. To determine whether the displacement is a maximum or minimum, we can examine the sign of the acceleration or use the second derivative of displacement. The maximum speed occurs either at endpoints of the given time interval or at times when acceleration is zero.

粒子在速度为零时处于静止状态。解 v(t) = 0 可以得到运动方向改变的时刻。在这些转向点处，位移可能达到局部最大值或局部最小值。为了判断位移是最大值还是最小值，我们可以检查加速度的符号或利用位移的二阶导数。最大速率要么出现在给定时间区间的端点，要么出现在加速度为零的时刻。

For s(t) = t³ − 6t² + 9t, v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3). The particle turns at t = 1 and t = 3. Substituting into s(t) gives the extreme positions; a(t) = 6t − 12 supports classifying them as a local maximum at t = 1 (a < 0) and a local minimum at t = 3 (a > 0).

对于 s(t) = t³ − 6t² + 9t，v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3)。粒子在 t = 1 和 t = 3 处转向。代入 s(t) 得到极值位置；a(t) = 6t − 12 可以通过 a < 0 在 t = 1 说明这是局部最大值，而 a > 0 在 t = 3 说明是局部最小值。

7. Distance Travelled versus Displacement | 行驶距离与位移的区别

Displacement is the net change in position from the starting point, whereas distance travelled is the total length of the path covered by the particle, ignoring direction. In calculus terms, distance travelled over a time interval [t₁, t₂] is found by integrating the absolute value of velocity: Total distance = ∫ |v(t)| dt. This is because the particle may move forwards and backwards, and the area under the speed-time curve gives the total distance.

位移是从起点开始位置的净变化量，而行驶距离是粒子所经路径的总长度，与方向无关。用微积分的话说，在时间区间 [t₁, t₂] 内的行驶距离可以通过对速度的绝对值积分来求得：总距离 = ∫ |v(t)| dt。这是因为粒子可能前后移动，而速率-时间曲线下的面积给出总距离。

If v(t) changes sign, we split the integral into intervals where v(t) is positive and where it is negative, taking the absolute value of each integrated portion. This is a common IB question: “Find the total distance travelled in the first 5 seconds,” which requires careful sign analysis of v(t).

如果 v(t) 改变符号，我们需要把积分分段，在 v(t) 为正和为负的区间分别积分，并取各部分的绝对值。这是 IB 中常见的问题：“求在前 5 秒内行驶的总距离”，这就需要对 v(t) 的符号进行仔细分析。

8. Interpreting Graphs of Motion | 运动图的解读

Kinematics heavily relies on interpreting displacement–time (s-t), velocity–time (v-t), and acceleration–time (a-t) graphs. The gradient of an s-t graph gives the instantaneous velocity; the gradient of a v-t graph gives the acceleration. The area under a v-t graph gives the change in displacement (or distance if sign is considered). The area under an a-t graph gives the change in velocity.

运动学高度依赖于对位移–时间图 (s-t)、速度–时间图 (v-t) 和加速度–时间图 (a-t) 的解读。s-t 图的斜率给出瞬时速度；v-t 图的斜率给出加速度。v-t 图下的面积给出位移的变化（若考虑符号则为距离）。a-t 图下的面积给出速度的变化。

• s-t graph: horizontal line → at rest; straight sloping line → constant velocity; curve → changing velocity.
• s-t 图：水平线 → 静止；倾斜直线 → 匀速；曲线 → 变速。
• v-t graph: horizontal line → constant velocity; sloping line → constant acceleration; area → displacement.
• v-t 图：水平线 → 匀速；倾斜线 → 匀加速；面积 → 位移。
• a-t graph: zero line → constant velocity; area → change in velocity.
• a-t 图：零线 → 匀速；面积 → 速度的变化。

Students should be able to sketch one graph given another, and to extract times of rest, maximum speed, and total distance from these graphs without necessarily knowing the algebraic functions.

学生需要能够根据一个函数的图像画出另一图像，并能在不确切知道代数表达式的情况下，从这些图像中提取静止时间、最大速率和总距离等信息。

9. Initial Conditions and Boundary Constraints | 初始条件与边界约束

In integration problems, the role of initial conditions is crucial. The velocity at t = 0, usually denoted by u or v₀, fixes the constant of integration from the acceleration. The displacement at t = 0, often denoted by s₀ or x₀, fixes the constant after integrating velocity. Without these conditions, the motion is described only up to an arbitrary shift.

在积分问题中，初始条件的作用至关重要。t = 0 时的速度，通常记作 u 或 v₀，确定了从加速度积分后得到的积分常数。t = 0 时的位移，通常记作 s₀ 或 x₀，确定了从速度积分后得到的常数。如果没有这些条件，运动只能被描述为相差任意一个平移。

It is also possible for conditions to be given at times other than t = 0. For example, s(3) = 10 can be used to determine the constant. This requires careful substitution into the integrated expression. IB exam questions often combine both types of constraints.

条件也可能在 t = 0 以外的时刻给出。例如，s(3) = 10 可以用来确定常数。这需要仔细代入积分后的表达式。IB 考试题目经常结合这两种约束条件。

10. Real-World Application Problems | 实际应用问题

Kinematics problems in IB often model real situations: a car braking, a particle projected vertically upward, a falling object under gravity, or a ball rolling to rest. In these scenarios, students must interpret the mathematical solution in context, such as explaining why a negative time solution is disregarded or why the particle returns to the origin. The gravitational acceleration is often taken as 9.8 m/s² or 10 m/s² downwards.

IB 中的运动学问题通常会模拟真实情境：汽车刹车、粒子竖直上抛、物体在重力作用下下落，或者滚动直至停止的球。在这些情境中，学生必须结合实际情况来解释数学解答，例如说明为何舍去负的时间解，或粒子为何回到原点。重力加速度通常取为 9.8 m/s² 或 10 m/s² 向下。

For vertical motion, the equations are the same but the acceleration is constant: a = −g (taking upwards as positive). Velocities upward are positive, downward negative. The maximum height is reached when v = 0. Total time of flight and maximum height can be derived using the SUVAT equations or integration.

对于竖直运动，方程相同，但加速度为常数：a = −g（取向上为正）。向上的速度为正，向下为负。当 v = 0 时达到最大高度。飞行总时间和最大高度可以利用 SUVAT 方程或积分推导出来。

A typical problem: a stone is thrown vertically upward with speed 20 m s⁻¹ from a cliff 50 m above the sea. Find its height above sea level after t seconds, the maximum height, and the time it hits the sea. By setting s(t) = 0 and using s = 50 + 20t − ½gt², we solve for relevant t.

一个典型的问题是：一块石头以 20 m s⁻¹ 的速度从离海面 50 m 的悬崖上竖直向上抛出。求 t 秒后它距海面的高度、最大高度以及它入水的时间。设 s(t) = 0，并利用 s = 50 + 20t − ½gt² 解出相关的 t。

11. Common Pitfalls and Key Tips | 常见错误与关键提示

When tackling kinematics questions, students frequently confuse displacement and distance, forget to include the initial constants after integration, or misapply the sign of acceleration. Always check the direction implied by the sign of velocity. A velocity-time graph crossing the time axis indicates a change in direction; total distance requires splitting the integral at these roots. Ensure that the units are consistent throughout the calculation.

在处理运动学问题时，学生经常混淆位移与距离，积分后忘记加上初始常数，或者误用加速度的符号。务必检查由速度正负号所表示的方向。速度–时间图穿过时间轴表明方向发生了改变；求总距离时需要在这些零点将积分分段。确保整个计算过程中单位一致。

For variable acceleration problems, do not assume that average velocity equals (u+v)/2 unless acceleration is constant. Similarly, the formula s = vt is only valid when velocity is constant. When in doubt, rely on the fundamental definitions: v = ds/dt and a = dv/dt, and their integral forms.

对于变加速问题，除非加速度恒定，否则不要假设平均速度等于 (u+v)/2。同样，公式 s = vt 只在速度恒定时成立。若有疑问，请依靠基本定义：v = ds/dt 和 a = dv/dt，以及它们的积分形式。

12. Summary and Connection to the IB Syllabus | 总结与 IB 大纲的联系

Kinematics beautifully ties together the concepts of differentiation and integration in a tangible way. In the IB Mathematics: Analysis and Approaches course, you will encounter these techniques in the context of straight-line motion, often within the calculus section. In Mathematics: Applications and Interpretation, kinematics problems may also appear, sometimes paired with technology for solving polynomial equations. Mastering this topic requires fluency in differentiating and integrating polynomials, rational functions, and basic trigonometric functions, as well as a solid understanding of graphical interpretation.

运动学以直观的方式将微分和积分的概念紧密结合在一起。在 IB 数学：分析与方法的课程中，你会在直线运动的背景下遇到这些技巧，通常是在微积分章节。在数学：应用与解释中，运动学问题也可能出现，有时会结合技术工具求解多项式方程。掌握这一主题需要熟练对多项式、有理函数和基本三角函数进行微分和积分，同时扎实理解图像解读。

Regular practice with past paper questions, particularly those that ask for both the total distance travelled and the times of rest, will build confidence. Remember that calculus is not just an abstract tool—it is the language of motion.

经常练习历年真题，尤其是那些既要求总行驶距离又要求静止时间的题目，将帮助你建立信心。请记住，微积分不仅仅是一个抽象的工具——它是描述运动的语言。

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