📚 AP Physics C Mechanics: Kinematics Summary | AP物理C力学：运动学知识点总结
Kinematics is the foundation of mechanics, describing motion without regard to its causes. In AP Physics C, you must analyze motion using calculus and vectors. This summary covers all essential concepts, formulas, and problem-solving strategies for the exam. Mastering these will prepare you for free-response and multiple-choice questions that demand precise mathematical reasoning.
运动学是力学的基础，它描述运动而不追究其原因。在 AP 物理 C 中，你必须运用微积分和矢量分析运动。本文总结了考试涉及的所有核心概念、公式和解题策略。掌握这些内容，你将能应对需要严密数学推理的自由回答和选择题。
1. Position, Displacement, and Average Velocity | 位置、位移与平均速度
Position x(t) gives the location of a particle along a chosen coordinate axis as a function of time t. Displacement Δx = x(t₂) – x(t₁) is the net change in position, a vector quantity that can be positive, negative, or zero.
位置 x(t) 表示粒子在选定坐标轴上随时间 t 的位置。位移 Δx = x(t₂) – x(t₁) 是位置的净变化，它是一个矢量，可以是正值、负值或零。
Average velocity is defined as vₐᵥ = Δx / Δt. It indicates the overall rate of change of position and has the same sign as displacement. Do not confuse it with average speed, which is total distance divided by Δt and is always non-negative.
平均速度定义为 vₐᵥ = Δx / Δt。它反映位置变化的整体快慢，其符号与位移相同。不要将其与平均速率混淆，平均速率是总路程除以 Δt，且始终非负。
In one dimension, the path of motion may involve turning, so total distance and displacement magnitude can differ significantly. Average speed > |average velocity| unless motion is unidirectional.
在一维运动中，运动路径可能包含折返，因此总路程与位移大小可能相差很大。除非单向运动，否则平均速率 > |平均速度|。
2. Instantaneous Velocity and Acceleration | 瞬时速度与加速度
Instantaneous velocity is the limit of average velocity as Δt approaches zero: v(t) = lim_{Δt→0} Δx/Δt = dx/dt. It is the derivative of position with respect to time and can be read as the slope of an x-t graph.
瞬时速度是平均速度在 Δt 趋于零时的极限：v(t) = lim_{Δt→0} Δx/Δt = dx/dt。它是位置对时间的导数，也可视作 x-t 图线的斜率。
Instantaneous acceleration is defined similarly: a(t) = dv/dt = d²x/dt². It measures how quickly velocity changes and is the slope of a v-t graph. A particle can have zero velocity but non-zero acceleration, e.g., at the peak of a vertical throw.
瞬时加速度类似地定义为：a(t) = dv/dt = d²x/dt²。它衡量速度变化的快慢，是 v-t 图线的斜率。粒子可以速度为零但加速度不为零，例如竖直上抛的最高点。
Instantaneous speed is the magnitude of instantaneous velocity, |v(t)|. It tells how fast the particle is moving regardless of direction.
瞬时速率是瞬时速度的大小 |v(t)|。它只反映粒子运动的快慢，不考虑方向。
3. Kinematic Equations for Constant Acceleration | 匀加速直线运动方程
When acceleration a is constant, we can integrate the definitions to obtain four fundamental equations. These equations relate time t, initial velocity v₀, final velocity v, displacement Δx (or position x with x₀ as initial position), and acceleration a.
当加速度 a 恒定时，我们可以通过积分定义得到四个基本方程。这些方程将时间 t、初速度 v₀、末速度 v、位移 Δx（或位置 x，其中 x₀ 为初始位置）以及加速度 a 联系起来。
The following table summarizes the equations and their typical usage. All equations assume a is constant and motion is along a straight line.
下表总结了这些方程及其典型用法。所有方程均假设 a 恒定且沿直线运动。
| Equation | Common Use / 常见用途 |
|---|---|
| v = v₀ + at | Find velocity after time t / 已知时间求速度 |
| x = x₀ + v₀t + ½at² | Find position when time is known / 已知时间求位置 |
| v² = v₀² + 2aΔx | Relates velocity and displacement without time / 不含时间的速度-位移关系 |
| Δx = ½(v₀ + v)t | Uses average velocity when a is constant / 用平均速度求位移 |
Always choose the equation that contains the unknown you need and all the given quantities. Remember to define a coordinate system and assign proper signs to v₀, v, a, and Δx.
始终选择包含所求未知量和所有已知量的方程。必须建立坐标系，并给 v₀、v、a 和 Δx 赋予正确的正负号。
4. Using Calculus to Relate Position, Velocity, and Acceleration | 用微积分联系位置、速度与加速度
In AP Physics C, you must be comfortable deriving or integrating kinematics quantities. Given x(t), differentiate to find v(t) and a(t). Given a(t), integrate to find v(t) and then x(t), using initial conditions to determine constants.
在 AP 物理 C 中，你必须能熟练推导或积分运动学量。已知 x(t)，求导可得 v(t) 和 a(t)。已知 a(t)，积分可得 v(t) 再得 x(t)，并利用初始条件确定积分常数。
For example, if a(t) = 3t + 2 (m/s²) and v(0) = 5 m/s, then v(t) = ∫(3t+2)dt = (3/2)t² + 2t + C. Using v(0)=5 gives C=5, so v(t) = 1.5t² + 2t + 5. Position is found by integrating v(t).
例如，若 a(t) = 3t + 2 (m/s²)，且 v(0) = 5 m/s，则 v(t) = ∫(3t+2)dt = (3/2)t² + 2t + C。代入 v(0)=5 得 C=5，因此 v(t) = 1.5t² + 2t + 5。再积分 v(t) 可得位置。
Displacement during a time interval can also be computed directly as the definite integral of velocity: Δx = ∫_{t₁}^{t₂} v(t) dt. This corresponds to the area under a v-t curve, which may be evaluated geometrically or analytically.
某段时间内的位移也可直接通过速度的定积分计算：Δx = ∫_{t₁}^{t₂} v(t) dt。这对应 v-t 曲线下的面积，可用几何法或解析法求出。
Similarly, change in velocity is the area under an a-t curve: Δv = ∫ a(t) dt. Understanding these integral relations is crucial for problems where acceleration is not constant.
类似地，速度的变化量是 a-t 曲线下的面积：Δv = ∫ a(t) dt。理解这些积分关系对于加速度不恒定的问题至关重要。
5. Graphical Analysis of Motion | 运动图像分析
Position-time graph: The slope of the tangent line gives instantaneous velocity v. A straight line indicates constant velocity; a curve indicates acceleration. The concavity (second derivative) reveals the sign of acceleration.
位置-时间图：切线斜率给出瞬时速度 v。直线表示匀速；曲线表示有加速度。曲线的凹凸方向（二阶导数）揭示加速度的正负。
Velocity-time graph: The slope equals acceleration a. The area between the curve and the time axis (with signs) gives displacement Δx. A horizontal line means constant velocity, i.e., zero acceleration.
速度-时间图：斜率等于加速度 a。曲线与时间轴之间的面积（带符号）给出位移 Δx。水平线表示速度恒定，即加速度为零。
Acceleration-time graph: The area gives change in velocity Δv. These graphs help visualize relationships when functional forms are given or when interpreting experimental data.
加速度-时间图：面积给出速度变化量 Δv。当给出函数形式或解读实验数据时，这些图有助于直观理解相互关系。
For AP exam questions, you may be asked to sketch graphs, identify corresponding motion, or extract information such as turning points (v=0) and maximum velocity (a=0, changing sign).
在 AP 考试中，你可能需要绘制草图、识别对应的运动状态，或提取信息如折返点（v=0）及最大速度点（a=0 且变号）。
6. Vectors and Component Method | 矢量与分量法
In two or three dimensions, position is described by a position vector r = x i + y j + z k. Displacement is Δr = r₂ – r₁. Velocity and acceleration are the first and second derivatives of r: v = dr/dt, a = dv/dt = d²r/dt².
在二维或三维空间中，位置由位置矢量 r = x i + y j + z k 描述。位移为 Δr = r₂ – r₁。速度和加速度是 r 的一阶和二阶导数：v = dr/dt，a = dv/dt = d²r/dt²。
Motion along each coordinate axis is independent. Resolve all vectors into components using trigonometry: Aₓ = A cos θ, Aᵧ = A sin θ. Apply kinematic equations separately to x- and y-directions, then recombine results if needed.
各坐标轴方向的运动是独立的。用三角学将矢量分解为分量：Aₓ = A cos θ，Aᵧ = A sin θ。对 x 和 y 方向分别应用运动学方程，必要时再合成结果。
Unit vectors i, j, k have magnitude 1 and indicate direction. Vector addition and subtraction are performed component-wise. The magnitude of a vector is |A| = √(Aₓ² + Aᵧ²) and its direction is given by tan θ = Aᵧ/Aₓ.
单位矢量 i, j, k 大小为 1 并指示方向。矢量的加减按分量进行。矢量的大小为 |A| = √(Aₓ² + Aᵧ²)，方向由 tan θ = Aᵧ/Aₓ 给出。
7. Projectile Motion | 抛体运动
A projectile launched near Earth’s surface experiences constant downward acceleration g = 9.8 m/s² (ignoring air resistance). Horizontal acceleration is zero, so horizontal velocity vₓ remains constant.
在地表附近发射的抛体，受恒定向下的加速度 g = 9.8 m/s² 作用（忽略空气阻力）。水平加速度为零，因此水平速度 vₓ 保持不变。
The standard equations, assuming launch from origin with initial speed v₀ at angle θ₀:
标准方程（假设从原点以初速率 v₀、仰角 θ₀ 发射）：
x = v₀ cosθ₀ t
y = v₀ sinθ₀ t – ½gt²
vᵧ = v₀ sinθ₀ – gt
The trajectory is parabolic. Time of flight for level ground is T = (2 v₀ sinθ₀)/g. Maximum height H = (v₀² sin²θ₀)/(2g) occurs at t = T/2. Range R = (v₀² sin 2θ₀)/g, which is maximized at θ₀ = 45° for level ground.
轨迹为抛物线。同高度起落时，飞行时间 T = (2 v₀ sinθ₀)/g。最大高度 H = (v₀² sin²θ₀)/(2g) 发生在 t = T/2 处。水平射程 R = (v₀² sin 2θ₀)/g，在 θ₀ = 45° 时达到最大。
For non-level ground, you must solve the quadratic for y(t) = final height. Always treat x and y motions independently and connect them through time t.
若起落点高度不同，需解二次方程 y(t) = 末高度。始终独立处理 x 和 y 方向的运动，并用时间 t 关联它们。
8. Relative Motion | 相对运动
The velocity of object A relative to object B is defined as v_AB = v_A – v_B, where v_A and v_B are velocities measured in a common frame (usually the ground). For position, r_AB = r_A – r_B.
物体 A 相对于物体 B 的速度定义为 v_AB = v_A – v_B，其中 v_A 和 v_B 是同一参考系（通常为地面）中测得的速度。对于位置，r_AB = r_A – r_B。
In one dimension, relative velocity is simply the difference with appropriate signs. For example, if car A moves at +20 m/s and car B at +15 m/s, v_AB = 20 – 15 = +5 m/s (A pulls away from B).
一维情况下，相对速度就是带正负号的速度差。例如，若 A 车以 +20 m/s 运动，B 车以 +15 m/s 运动，则 v_AB = 20 – 15 = +5 m/s（A 远离 B）。
In two dimensions, use vector subtraction. This concept is essential for problems involving moving walkways, boats in currents, and airplanes in wind. The observer’s frame determines which velocities are added or subtracted.
二维情况使用矢量减法。该概念对于解决涉及传送带、流水行船和风中飞行等问题至关重要。观测者所在参考系决定速度的加减关系。
The Galilean transformation for low speeds: r’ = r – Vt, v’ = v – V, where V is the constant relative velocity between two inertial frames. Acceleration is invariant: a’ = a.
低速下的伽利略变换：r’ = r – Vt，v’ = v – V，其中 V 是两个惯性系之间的恒定相对速度。加速度不变：a’ = a。
9. Uniform Circular Motion | 匀速圆周运动
A particle moving in a circle of radius r at constant speed v undergoes centripetal (center-seeking) acceleration directed toward the center: a_c = v²/r. The speed v is related to angular speed ω (rad/s) by v = rω, so a_c = ω²r.
以恒定速率 v 沿半径 r 的圆周运动的粒子，具有指向圆心的向心加速度：a_c = v²/r。速率 v 与角速度 ω（rad/s）的关系为 v = rω，故 a_c = ω²r。
The period T is the time for one full revolution: T = 2πr/v = 2π/ω. Frequency f = 1/T (Hz). Although speed is constant, velocity is not, because direction changes continuously.
周期 T 是完成一整圈所需的时间：T = 2πr/v = 2π/ω。频率 f = 1/T（Hz）。虽然速率恒定，但由于方向不断变化，速度并不恒定。
Centripetal acceleration is perpendicular to velocity and does not change speed; it only changes direction. In vector form: a = -(v²/r) r̂ (radially inward). This acceleration is caused by a net centripetal force, but in kinematics we describe the motion itself.
向心加速度垂直于速度，不改变速率，只改变方向。矢量形式为：a = -(v²/r) r̂（径向向内）。该加速度由净向心力引起，但运动学中我们只描述运动本身。
10. Non-Uniform Circular Motion | 非匀速圆周运动
If the speed along a circular path changes, the particle has both centripetal (radial) acceleration a_c = v²/r and tangential acceleration a_t = dv/dt. The total acceleration vector is a = a_c (radial inward) + a_t (tangential). Its magnitude is a = √(a_c² + a_t²).
若圆周运动中的速率变化，则粒子同时具有向心（径向）加速度 a_c = v²/r 和切向加速度 a_t = dv/dt。总加速度矢量为 a = a_c（径向向内） + a_t（切向）。其大小为 a = √(a_c² + a_t²)。
Angular acceleration α = dω/dt = a_t / r. Kinematic equations for constant angular acceleration are analogous to those for linear motion. Replace x with θ, v with ω, and a with α.
角加速度 α = dω/dt = a_t / r。恒定角加速度下的运动学方程与直线运动方程类似，只需将 x 替换为 θ，v 替换为 ω，a 替换为 α。
| Linear Equation | Angular Analog |
|---|---|
| v = v₀ + at | ω = ω₀ + αt |
| Δx = v₀t + ½at² | Δθ = ω₀t + ½αt² |
| v² = v₀² + 2aΔx | ω² = ω₀² + 2αΔθ |
These relationships are critical for problems with pulleys, rolling objects, and any system where rotation is coupled to translation. Always associate the tangential components with the linear motion along the circle.
这些关系对于滑轮、滚动体以及任何转动与平动耦合的系统至关重要。始终将切向分量与沿圆周的线运动联系起来。
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