Kinematics | 运动学

📚 Kinematics | 运动学

Kinematics is the branch of mechanics that describes the motion of objects using mathematical functions, without considering the forces that cause the motion. In IB Mathematics, kinematics brings calculus to life by connecting position, velocity, and acceleration through differentiation and integration. This topic appears in Analysis & Approaches as well as Applications & Interpretation, often in the context of polynomial or trigonometric functions, and extends naturally into vector motion and projectile problems.

运动学是力学的一个分支，它用数学函数描述物体的运动，而不考虑引起运动的原因。在 IB 数学中，运动学通过微分与积分将微积分赋予实际意义，将位置、速度与加速度联系起来。该主题同时出现在 “分析与方法” 和 “应用与解释” 课程中，常以多项式或三角函数为背景，并自然地延伸到矢量运动与抛体问题。

1. Position, Velocity and Acceleration as Functions | 用函数表示位置、速度和加速度

The position of a particle moving along a straight line is given by a function of time, typically denoted by s(t), x(t) or y(t). Here s represents the displacement from a fixed origin, measured in metres (m), and t is time in seconds (s). A positive value of s means the particle is to the right of the origin, while a negative value means it is to the left.

沿直线运动的粒子的位置由关于时间的函数给出，通常记为 s(t)、x(t) 或 y(t)。这里 s 表示相对于固定原点的位移，单位为米(m)，t 为时间，单位为秒(s)。s 取正值意味着粒子位于原点右侧，取负值则位于左侧。

Velocity is the instantaneous rate of change of position with respect to time. If the position function is s(t), then the velocity function is v(t) = ds/dt. Velocity has both magnitude and sign; a positive velocity indicates motion to the right, while a negative velocity indicates motion to the left. Speed is the absolute value of velocity, |v(t)|, and describes how fast the particle is moving regardless of direction.

速度是位置关于时间的瞬时变化率。若位置函数为 s(t)，则速度函数为 v(t) = ds/dt。速度既有大小也有符号；正速度表示向右运动，负速度表示向左运动。速率是速度的绝对值 |v(t)|，描述粒子运动的快慢而不考虑方向。

Acceleration measures how quickly the velocity changes. It is given by a(t) = dv/dt, the derivative of velocity with respect to time. Since velocity is itself the derivative of position, acceleration is also the second derivative of position: a(t) = d²s/dt². The SI unit of acceleration is m/s².

加速度衡量速度变化的快慢。它由 a(t) = dv/dt 给出，即速度对时间的导数。由于速度本身是位置的导数，加速度也是位置的二阶导数：a(t) = d²s/dt²。加速度的国际单位是 m/s²。

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2. Differentiation: From Position to Velocity and Acceleration | 微分：从位置到速度和加速度

Given a position function s(t), you obtain the velocity by differentiating once. For example, if s(t) = t³ − 6t² + 9t, then v(t) = 3t² − 12t + 9. To find the acceleration, differentiate again: a(t) = 6t − 12. These functions allow you to analyse the particle’s motion at any instant.

给定位置函数 s(t)，通过求一次导数得到速度。例如，若 s(t) = t³ − 6t² + 9t，则 v(t) = 3t² − 12t + 9。为求加速度，再次求导：a(t) = 6t − 12。这些函数使你可以分析粒子在任意瞬时的运动。

Key points on a motion graph are often found by setting v(t) = 0. These correspond to instants when the particle is momentarily at rest or changes direction. Solving v(t) = 0 gives the times when the velocity is zero; substituting these times into s(t) yields the turning points of the position function.

运动图像上的关键点通常通过令 v(t) = 0 来寻找。这些点对应粒子瞬时静止或改变方向的时刻。求解 v(t) = 0 得出速度为零的时刻；将这些时刻代入 s(t) 即得位置函数的转折点。

When the velocity is positive, the particle moves in the positive direction; when velocity is negative, it moves in the negative direction. A change in the sign of velocity indicates a reversal of motion. This often occurs at local maxima or minima of the position graph.

当速度为正值时，粒子沿正方向运动；当速度为负值时，粒子沿负方向运动。速度符号的改变表明运动方向反转，这通常发生在位置图的局部极大值或极小值处。

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3. Integration: From Acceleration to Velocity and Position | 积分：从加速度到速度和位置

If you know the acceleration function a(t) and the initial velocity, you can recover the velocity by integrating. The velocity at time t is given by v(t) = ∫ a(t) dt + C, where the constant of integration C is determined by the initial condition v(0) = v₀. Similarly, the position is found by integrating velocity: s(t) = ∫ v(t) dt + D, using the initial position s(0) = s₀ to find D.

若已知加速度函数 a(t) 和初速度，可以通过积分还原速度。t 时刻的速度由 v(t) = ∫ a(t) dt + C 给出，其中积分常数 C 由初始条件 v(0) = v₀ 确定。类似地，通过积分速度求得位置：s(t) = ∫ v(t) dt + D，并利用初始位置 s(0) = s₀ 求出 D。

For a constant acceleration a, the integration process yields the well‑known linear velocity function v(t) = u + at, where u is the initial velocity. Integrating again gives the quadratic position function s(t) = ut + ½at² + s₀, where s₀ is the initial displacement. This elegantly links calculus with the standard SUVAT equations.

对于恒定加速度 a，积分过程得到著名的线性速度函数 v(t) = u + at，其中 u 为初速度。再次积分给出二次位置函数 s(t) = ut + ½at² + s₀，其中 s₀ 为初始位移。这优雅地将微积分与标准的等加速度方程联系起来。

When acceleration is not constant, you must integrate the given function term by term. For instance, if a(t) = 6t − 4 and at t = 0 the particle has v = 2 and s = 5, then v(t) = 3t² − 4t + 2 and s(t) = t³ − 2t² + 2t + 5. Always check that the integrated functions satisfy the initial conditions.

当加速度不是常数时，需对给定函数逐项积分。例如，若 a(t) = 6t − 4，且在 t = 0 时粒子有 v = 2，s = 5，则 v(t) = 3t² − 4t + 2，s(t) = t³ − 2t² + 2t + 5。务必检验积分后的函数满足初始条件。

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4. Total Distance Traveled vs. Displacement | 路程与位移

Displacement is simply the change in position: Δs = s(t₂) − s(t₁). However, total distance traveled accounts for all movement, regardless of direction. If a particle reverses direction between t₁ and t₂, the distance is found by integrating the speed |v(t)| over the time interval, or by splitting the interval into subintervals where velocity does not change sign.

位移仅仅是位置的变化：Δs = s(t₂) − s(t₁)。然而，总路程计入了全部运动，不考虑方向。若粒子在 t₁ 与 t₂ 之间改变了方向，则需对速度的绝对值 |v(t)| 在时间区间上积分，或将区间划分为速度符号不变的若干子区间。

To compute total distance: first solve v(t) = 0 to find turning times, then evaluate the absolute displacement on each segment and sum them. For example, if s(t) = t³ − 3t² + 2t, with v(t) = 3t² − 6t + 2, roots of v(t) = 0 are t ≈ 0.423 and 1.577. The total distance from t = 0 to t = 3 is |s(0.423)−s(0)| + |s(1.577)−s(0.423)| + |s(3)−s(1.577)|.

计算总路程的方法：先解 v(t) = 0 求出改变方向的时刻，然后计算每一段上的绝对位移并求和。例如，若 s(t) = t³ − 3t² + 2t，v(t) = 3t² − 6t + 2，v(t) = 0 的根为 t ≈ 0.423 和 1.577，则从 t = 0 到 t = 3 的总路程为 |s(0.423)−s(0)| + |s(1.577)−s(0.423)| + |s(3)−s(1.577)|。

IB exam questions frequently ask for the distance traveled in a given time interval, not just the displacement. Always check for changes of direction by examining the sign of the velocity, and clearly separate distance from displacement in your answers.

IB 考题经常要求计算给定时间区间内的路程，而不仅是位移。一定要通过检查速度的符号来确认是否发生了方向改变，并在答案中明确区分路程与位移。

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5. Constant Acceleration Equations (SUVAT) | 等加速度方程 (SUVAT)

When acceleration is constant, five key formulae link the variables displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t). These are commonly remembered by the acronym SUVAT. You must be able to select the appropriate equation based on which quantities are known and which are required.

当加速度恒定时，五个关键公式将位移(s)、初速度(u)、末速度(v)、加速度(a)和时间(t)联系起来。这些通常用首字母缩写 SUVAT 来记忆。你必须能够根据已知量和所求量选用合适的方程。

Equation	Situation
v = u + at	without s
s = ut + ½at²	without v
s = ½(u + v)t	without a
v² = u² + 2as	without t
s = vt − ½at²	without u (alternative form)

The equations apply only when the acceleration a is constant throughout the motion. If acceleration varies, you must use the calculus methods described earlier. Always define a positive direction and ensure that signs of u, v, a and s are consistent with that choice.

这些方程仅在加速度 a 始终恒定时适用。若加速度变化，则需使用前述微积分方法。务必定义正方向，并确保 u、v、a 和 s 的符号与所选方向一致。

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6. Deriving SUVAT Equations Using Calculus | 用微积分推导等加速度方程

The SUVAT equations can be derived directly from the definitions of velocity and acceleration as derivatives. Starting from a(t) = a (constant), integrate to get v(t) = ∫ a dt = at + C. With the initial condition v(0) = u, we obtain v = u + at. This is the first SUVAT formula.

等加速度方程可以直接由速度和加速度的导数定义推导出来。从 a(t) = a (常数) 出发，积分得 v(t) = ∫ a dt = at + C。利用初条件 v(0) = u，即得 v = u + at。这是第一个 SUVAT 公式。

Integrating velocity with respect to time gives position: s(t) = ∫ (u + at) dt = ut + ½at² + D. If s(0) = s₀, then D = s₀. When the origin is chosen so that s₀ = 0, we have s = ut + ½at². To obtain the time-independent equation, rearrange v = u + at as t = (v − u)/a and substitute into s = ut + ½at²; simplifying yields v² = u² + 2as.

对速度积分得到位置：s(t) = ∫ (u + at) dt = ut + ½at² + D。若 s(0) = s₀，则 D = s₀。若选取原点使 s₀ = 0，即得 s = ut + ½at²。为得到不含时间的方程，将 v = u + at 变形为 t = (v − u)/a 并代入 s = ut + ½at²，化简即得 v² = u² + 2as。

Additionally, the average velocity when acceleration is constant is ½(u + v), so displacement can be expressed as s = average velocity × time, giving s = ½(u + v)t. This clear derivation shows the deep connection between calculus and the traditional constant-acceleration formulas.

此外，当加速度恒定时，平均速度为 ½(u + v)，因此位移可表示为 s = 平均速度 × 时间，即 s = ½(u + v)t。这一清晰的推导展示了微积分与传统等加速度公式之间的深层联系。

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7. Vertical Motion Under Gravity | 重力作用下的竖直运动

Objects moving freely near the Earth’s surface experience a constant downward acceleration due to gravity, with magnitude g = 9.8 m/s² (or sometimes 9.81 or 10 in exam settings). In one‑dimensional vertical motion problems, it is conventional to take upwards as the positive direction. In that case the acceleration is a = −g = −9.8 m/s².

在地球表面附近自由运动的物体受到向下的恒定重力加速度，大小为 g = 9.8 m/s²（考试中有时取 9.81 或 10）。在一维竖直运动问题中，通常将向上取为正方向，此时加速度为 a = −g = −9.8 m/s²。

A typical problem states: “A ball is thrown vertically upwards from ground level with speed 20 m/s. Find the maximum height and the total time of flight.” Using v = u + at, at the maximum height v = 0, so 0 = 20 − 9.8t, giving t ≈ 2.04 s. The maximum height is found by s = ut + ½at² = 20(2.04) − 4.9(2.04)² ≈ 20.4 m. The total flight time to return to the ground is twice the time to the peak, about 4.08 s.

一个典型问题为：”一球以 20 m/s 的速度从地面竖直上抛。求最大高度和总飞行时间。” 使用 v = u + at，在最高点 v = 0，故 0 = 20 − 9.8t，得 t ≈ 2.04 s。由 s = ut + ½at² = 20(2.04) − 4.9(2.04)² ≈ 20.4 m 即得最大高度。返回地面的总飞行时间是到达最高点时间的两倍，约 4.08 s。

When the object is projected from a height, the initial position is not zero. For example, throwing a stone downwards from a cliff requires s = ut + ½at² combined with the initial height in a coordinate system that sets the origin conveniently. Sign consistency is essential; a negative s often indicates a position below the origin.

当物体从某一高度抛出时，初始位置不为零。例如，从悬崖向下掷石块，需在适当坐标系中结合初始高度使用 s = ut + ½at²。符号一致性至关重要；负的 s 通常表示位于原点下方。

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8. Projectile Motion in Two Dimensions | 二维抛体运动

Projectile motion is analysed by treating the horizontal and vertical components independently. The horizontal acceleration is zero, so the horizontal velocity remains constant at u cos θ, where θ is the angle of projection above the horizontal. The vertical motion is subject to constant downward acceleration −g, with initial vertical velocity u sin θ.

抛体运动通过独立分析水平和竖直分量来研究。水平加速度为零，因此水平速度保持恒定，为 u cos θ，其中 θ 是初速度与水平面的夹角。竖直运动受向下的恒定加速度 −g 影响，初始竖直速度为 u sin θ。

The position of the projectile at time t can be written parametrically:

x(t) = (u cos θ) t

y(t) = (u sin θ) t − ½gt²

where the origin is taken as the launch point. Eliminating the parameter t gives the trajectory equation, which is a quadratic in x and therefore a parabola.

抛体在 t 时刻的位置可写为参数形式：

x(t) = (u cos θ) t

y(t) = (u sin θ) t − ½gt²

其中原点取在发射点。消去参数 t 得到轨迹方程，它是关于 x 的二次式，因此曲线为抛物线。

The time of flight is found by setting y = 0 (assuming launch and landing on the same horizontal level): T = (2u sin θ)/g. The horizontal range R is the x‑coordinate at that time: R = (u² sin 2θ)/g. The maximum height H occurs when the vertical velocity is zero: H = (u² sin² θ)/(2g). These expressions assume air resistance is negligible.

飞行时间由令 y = 0 求得（假设发射与着陆在同一水平面）：T = (2u sin θ)/g。水平射程 R 是此时对应的 x 坐标：R = (u² sin 2θ)/g。最大高度 H 出现在竖直速度为零时：H = (u² sin² θ)/(2g)。以上表达式均忽略空气阻力。

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9. Vector Kinematics | 矢量运动学

When motion is not confined to a straight line, position, velocity and acceleration are described by vectors. In IB Mathematics, a common representation is the position vector r(t) = x(t)i + y(t)j (and possibly a z‑component in 3D). The velocity vector is v(t) = r‘(t) = x'(t)i + y'(t)j, and the acceleration vector a(t) = r”(t).

当运动不限于直线时，位置、速度和加速度用矢量描述。在 IB 数学中，一种常见表示是位置矢量 r(t) = x(t)i + y(t)j（三维时还可能包含 z 分量）。速度矢量为 v(t) = r‘(t) = x'(t)i + y'(t)j，加速度矢量 a(t) = r”(t)。

The magnitude of the velocity vector is the speed: |v(t)| = √(x'(t)² + y'(t)²). The direction of motion is given by the unit vector v/|v|. Vector kinematics allows us to handle problems where the particle moves along a curved path, such as a parametric curve, and to investigate whether the particle is speeding up or slowing down using the dot product v • a.

速度矢量的大小是速率：|v(t)| = √(x'(t)² + y'(t)²)。运动方向由单位矢量 v/|v| 给出。矢量运动学使我们能够处理粒子沿曲线（如参数曲线）运动的问题，并可通过点积 v • a 考察粒子是在加速还是在减速。

For example, if r(t) = (t² + 1)i + (t³ − 2t)j, then v(t) = 2ti + (3t² − 2)j and a(t) = 2i + 6tj. At t = 1, the speed is √(2² + 1²) = √5 ≈ 2.24 m/s. The dot product v • a = 2·2 + 1·6 = 10 > 0, indicating the speed is increasing at that instant.

例如，若 r(t) = (t² + 1)i + (t³ − 2t)j，则 v(t) = 2ti + (3t² − 2)j，a(t) = 2i + 6tj。在 t = 1 时，速率为 √(2² + 1²) = √5 ≈ 2.24 m/s。点积 v • a = 2·2 + 1·6 = 10 > 0，表明该瞬时速率在增加。

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10. Interpreting Graphs in Kinematics | 运动学图像解读

Graphical analysis is a powerful tool in kinematics. A position–time graph shows the displacement s(t); its gradient at any point gives the instantaneous velocity. A straight line indicates constant velocity, while a curve indicates acceleration. Turning points on the graph correspond to v = 0.

图像分析是运动学中的有力工具。位置–时间图显示位移 s(t)；其上任意一点的斜率给出瞬时速度。直线表示匀速运动，曲线则表示存在加速度。图像上的转折点对应于 v = 0。

On a velocity–time graph, the gradient represents acceleration, and the area between the graph and the t‑axis gives the displacement. If a velocity–time graph crosses the t‑axis, the areas above and below must be considered separately when calculating total distance, because the integral of velocity gives net displacement, while the integral of the absolute value gives distance.

在速度–时间图中，斜率代表加速度，图线与 t 轴之间的面积表示位移。若速度–时间图穿过 t 轴，计算总路程时必须分别考虑上方面积和下方面积，因为速度的积分给出净位移，而绝对值的积分给出路程。

An acceleration–time graph directly shows how acceleration changes; the area under it equals the change in velocity. IB students should be prepared to move between these graphs, using differentiation to go from s to v to a, and integration to go in reverse, and to interpret features such as points of inflection in the s–t graph where acceleration changes sign.

加速度–时间图直接显示加速度如何变化；图线下的面积等于速度的变化量。IB 学生应能够在这些图像间进行转换，即利用微分从 s 到 v 再到 a，利用积分反向操作，并解读 s–t 图中的拐点等特征，这些点对应加速度变号。

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11. Solving Kinematics Problems with Technology | 利用技术求解运动学问题

IB examinations encourage the use of a graphical calculator for kinematic analysis. You can graph position, velocity or acceleration functions, use the built‑in derivative and integral commands, and find roots or intersection points numerically. This is especially helpful when functions are not simple polynomials, for instance when s(t) = e⁻⁰·²ᵗ sin(πt).

IB 考试鼓励使用图形计算器进行运动学分析。你可以绘制位置、速度或加速度函数的图像，使用内置的求导和积分命令，并通过数值方法求根或交点。当函数不是简单多项式时，如 s(t) = e⁻⁰·²ᵗ sin(πt)，这一功能尤为有用。

When using technology, always record the commands used (for example, nSolve for finding when v = 0, or fnInt for integrating speed). Present your working clearly, even if some steps are performed on the calculator, so that the examiner can follow your reasoning. Sketching the graphs with key features labelled is good practice.

使用技术时，务必记录所用的命令（例如用 nSolve 求 v = 0 的时刻，或用 fnInt 积分速率）。即使某些步骤在计算器上完成，也应清晰呈现解答过程，使考官能够理解你的推理。绘制并标注关键特征的草图是一种良好习惯。

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12. Common Pitfalls and Tips | 常见误区与提示

One frequent mistake is confusing displacement and total distance. Always read the question carefully: if it says “find the distance traveled”, you must account for any change in direction. Another pitfall is forgetting to include constants of integration when moving from acceleration to position, leading to incorrect functions.

一个常见误区是混淆位移与总路程。务必仔细审题：若问题要求 “求经过的路程”，必须考虑所有方向改变。另一个陷阱是在从加速度积分至位置时忘记加上积分常数，从而得到错误函数。

Sign errors in SUVAT problems often arise from an inconsistent choice of positive direction. Decide on a positive direction at the start, and apply it uniformly to displacement, velocity and acceleration. Finally, when dealing with vector kinematics, remember that the speed is the magnitude of the velocity vector; to find when the particle is at rest, set the velocity vector equal to the zero vector, not just one component.

在 SUVAT 问题中，符号错误常源自正方向选择不一致。从一开始就确定正方向，并对位移、速度和加速度统一应用。最后，在处理矢量运动学时，记住速率是速度矢量的大小；求粒子静止的时刻应令速度矢量等于零矢量，而不仅仅是一个分量为零。

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