📚 IB Physics: Kinematics Essentials | IB 物理：运动学 考点精讲
Kinematics forms the foundation of mechanics in IB Physics, describing motion without reference to its causes. Mastering displacement, velocity, acceleration, and their graphical representations is essential for tackling more complex topics. This article breaks down the core concepts and problem-solving techniques you need to know.
运动学是 IB 物理力学的基础，它描述运动而不涉及引起运动的原因。掌握位移、速度、加速度以及它们的图像表示，是攻克更复杂专题的必要前提。本文为你拆解核心概念与解题技巧。
1. Scalar and Vector Quantities | 标量与矢量
Scalars have magnitude only, e.g. distance, speed, mass, time, energy. Vectors have both magnitude and direction, e.g. displacement, velocity, acceleration, force. Always pay attention to whether a quantity is scalar or vector when adding or resolving.
标量仅有大小，如距离、速率、质量、时间、能量。矢量具有大小和方向，如位移、速度、加速度、力。在相加或分解时，务必注意该量是标量还是矢量。
In one dimension, direction is signified by positive and negative signs. In two dimensions, vectors are added tip-to-tail or resolved into perpendicular components. The magnitude of a resultant vector R from two perpendicular components Aₓ and Aᵧ is given by R = √(Aₓ² + Aᵧ²), and direction θ = tan⁻¹(Aᵧ/Aₓ).
在一维中，方向用正负号表示。在二维中，矢量用首尾相接法相加，或分解为垂直分量。两个垂直分量 Aₓ 和 Aᵧ 的合矢量大小 R = √(Aₓ² + Aᵧ²)，方向 θ = tan⁻¹(Aᵧ/Aₓ)。
2. Displacement, Distance, Speed, Velocity | 位移、路程、速率、速度
Distance is the total length of the path travelled (scalar). Displacement is the straight‑line change in position from start to finish, with direction (vector). Average speed = total distance / total time. Average velocity = total displacement / total time.
路程是运动轨迹的总长度（标量）。位移是从起点到终点的直线位置变化，带有方向（矢量）。平均速率 = 总路程 / 总时间。平均速度 = 总位移 / 总时间。
Instantaneous speed is the magnitude of instantaneous velocity. In uniform motion, instantaneous velocity equals average velocity. In IB problems, distinguish carefully: a runner going around a 400 m track returns to start, displacement = 0, distance = 400 m, average velocity = 0, average speed ≠ 0.
瞬时速率是瞬时速度的大小。在匀速运动中，瞬时速度等于平均速度。在 IB 题中要仔细区分：绕 400 米跑道一圈的运动员回到起点，位移 = 0，路程 = 400 m，平均速度 = 0，平均速率 ≠ 0。
3. Acceleration | 加速度
Acceleration is the rate of change of velocity: a = Δv / Δt = (v – u) / t. It is a vector. Negative acceleration (opposite to direction of motion) reduces speed in one dimension. In uniform circular motion, speed is constant but velocity changes because direction changes, so there is centripetal acceleration.
加速度是速度的变化率：a = Δv / Δt = (v – u) / t。它是矢量。与运动方向相反的负加速度在一维中会减小速率。在匀速圆周运动中，速率恒定但因方向改变而速度变化，因此存在向心加速度。
Deceleration is simply acceleration in the opposite direction to velocity. Instantaneous acceleration is the gradient of the velocity‑time graph. Constant acceleration implies a linear v‑t relationship.
减速就是加速度与速度方向相反。瞬时加速度是速度‑时间图像的斜率。匀加速度意味着 v‑t 图呈线性关系。
4. Kinematic Equations (SUVAT) | 运动学方程（SUVAT）
For uniform acceleration in a straight line, five quantities are linked: s – displacement, u – initial velocity, v – final velocity, a – acceleration, t – time. The four equations are:
v = u + a t
s = ½ (u + v) t
s = u t + ½ a t²
v² = u² + 2 a s
对于直线上的匀加速度运动，有五个量：s – 位移，u – 初速度，v – 末速度，a – 加速度，t – 时间。四个方程为：
v = u + a t
s = ½ (u + v) t
s = u t + ½ a t²
v² = u² + 2 a s
To solve problems, identify the given quantities and choose the equation that contains only one unknown. Remember to use consistent signs for direction (e.g. upwards positive).
解题时，先找出已知量，选择只含一个未知量的方程。切记方向符号要一致（例如设向上为正）。
5. Free Fall | 自由落体
Free fall is motion under gravity alone, ignoring air resistance. The acceleration is g = 9.81 m s⁻² downwards. Use SUVAT equations with a = ± g depending on your sign convention. When an object is thrown upwards, at the highest point v = 0 momentarily, but a = g always.
自由落体是仅受重力作用、忽略空气阻力的运动。加速度 g = 9.81 m s⁻² 向下。使用 SUVAT 方程时根据正方向约定取 a = ± g。物体上抛时，最高点瞬间 v = 0，但加速度始终为 g。
If an object is dropped from rest, u = 0, displacement s = ½ g t². If thrown vertically upwards with speed u, max height h = u² / (2g), total time of flight T = 2u / g.
若物体从静止下落，u = 0，位移 s = ½ g t²。若以速度 u 竖直上抛，最大高度 h = u² / (2g)，全程时间 T = 2u / g。
6. Projectile Motion | 抛体运动
Projectile motion is analysed by resolving into horizontal and vertical components. Horizontal: constant velocity (aₓ = 0), thus sₓ = uₓ t. Vertical: constant acceleration aᵧ = –g (if up positive), use SUVAT: vᵧ = uᵧ – g t, sᵧ = uᵧ t – ½ g t².
抛体运动通过分解为水平与竖直分量来分析。水平方向：匀速 (aₓ = 0)，因此 sₓ = uₓ t。竖直方向：匀加速度 aᵧ = –g（设向上为正），使用 SUVAT：vᵧ = uᵧ – g t，sᵧ = uᵧ t – ½ g t²。
For a projectile launched at angle θ with initial speed u: uₓ = u cos θ, uᵧ = u sin θ. Time of flight = 2 u sin θ / g, range = (u² sin 2θ) / g, max height = (u² sin²θ) / (2g). Maximum range occurs at 45°.
对于以初速度 u、角度 θ 抛出的物体：uₓ = u cos θ，uᵧ = u sin θ。飞行时间 = 2 u sin θ / g，射程 = (u² sin 2θ) / g，最大高度 = (u² sin²θ) / (2g)。最大射程出现在 45°。
7. Projectile Example – Horizontal Launch | 抛体实例 – 平抛
When a projectile is launched horizontally from height H: uᵧ = 0. Time to fall: t = √(2H / g). Horizontal displacement: sₓ = u×√(2H / g). The vertical component of velocity just before hitting the ground: vᵧ = g t = √(2 g H).
当物体从高度 H 水平抛出时：uᵧ = 0。下落时间：t = √(2H / g)。水平位移：sₓ = u×√(2H / g)。落地前瞬间的竖直分速度：vᵧ = g t = √(2 g H)。
This is a common IB problem. The path is a parabola. The speed at impact is √(u² + 2gH). Use energy considerations as a quick check.
这是 IB 常见题型。轨迹为抛物线。落地速度大小为 √(u² + 2gH)。可用能量观点快速验证。
8. Motion Graphs (Displacement–Time, Velocity–Time) | 运动图像（s‑t 图、v‑t 图）
Displacement–time graph: slope = velocity. Straight line = constant velocity; curved line = changing velocity (acceleration). Area has no physical meaning for s‑t.
位移‑时间图：斜率 = 速度。直线表示匀速；曲线表示速度变化（加速）。s‑t 图的面积无物理意义。
Velocity–time graph: slope = acceleration; area between graph and time axis = displacement. A horizontal line means constant velocity. A sloping line shows uniform acceleration. If the graph crosses the time axis, velocity changes direction.
速度‑时间图：斜率 = 加速度；图线与时间轴围成的面积 = 位移。水平线表示匀速。斜线表示匀加速。若图线穿过时间轴，速度方向改变。
9. Acceleration–Time Graphs | 加速度–时间图（a‑t 图）
The area under an a‑t graph gives the change in velocity. A horizontal a‑t line indicates constant acceleration. The slope of a‑t graph is jerk (rate of change of acceleration), rarely required in IB but conceptually useful.
a‑t 图下的面积表示速度变化量。水平的 a‑t 线表示匀加速度。a‑t 图的斜率是加加速度（加速度的变化率），IB 很少直接要求，但有助于概念理解。
When converting between graphs, remember: the slope of s‑t gives v‑t; the slope of v‑t gives a‑t. The area of a‑t gives Δv; the area of v‑t gives Δs.
在图像之间转换时记住：s‑t 的斜率得 v‑t；v‑t 的斜率得 a‑t。a‑t 的面积得 Δv；v‑t 的面积得 Δs。
10. Relative Velocity in One Dimension | 一维相对速度
The velocity of A relative to B is vA/B = vA – vB. Use vector signs. If two cars move in the same direction, relative speed is the difference of their speeds. In opposite directions, relative speed is the sum.
A 相对于 B 的速度为 vA/B = vA – vB。使用矢量符号。若两车同向行驶，相对速率是速率之差；反向行驶，相对速率是速率之和。
For IB, relative velocity problems are usually solved in one dimension, but the concept extends to two dimensions using vector subtraction.
在 IB 中，相对速度问题多在一维中求解，但概念可扩展至二维，使用矢量减法。
11. Motion with Variable Acceleration (Calculus) | 变加速度运动（微积分）
When acceleration is not constant, kinematic functions are related by calculus: v = ds/dt, a = dv/dt = d²s/dt². Conversely: s = ∫ v dt, v = ∫ a dt. Given a function for displacement, velocity, or acceleration, you can differentiate or integrate to find others.
当加速度不恒定时，运动学函数通过微积分联系：v = ds/dt，a = dv/dt = d²s/dt²。反之：s = ∫ v dt，v = ∫ a dt。若已知位移、速度或加速度的函数，可微分或积分求其他量。
These problems appear in the HL syllabus. For example, given a(t) = 6t – 2, find v(t) and s(t) with initial conditions. Use definite integrals for changes over intervals.
此类问题出现在 HL 大纲中。例如，给出 a(t) = 6t – 2，结合初始条件求 v(t) 和 s(t)。使用定积分求某段时间内的变化量。
12. Common Pitfalls and IB Exam Tips | 常见易错点与 IB 考试技巧
Always define a positive direction and stick to it. Mixing up displacement and distance is a frequent error. In projectile motion, vertical and horizontal motions are independent; time links them. Double-check signs for acceleration due to gravity. When using v² = u² + 2as, ensure s is the displacement, not distance.
务必事先定义正方向并始终遵守。混淆位移和路程是常见错误。在抛体运动中，竖直与水平运动独立，由时间联系。务必检查重力加速度的符号。使用 v² = u² + 2as 时，确保 s 是位移而非路程。
For graph interpretation, read axes carefully. A common trick: a v‑t graph of a ball thrown upwards and caught at the same height shows a straight line crossing the t‑axis. The area from going up and down may cancel if displacement is asked; for total distance, add absolute areas.
解读图像时要仔细阅读坐标轴。常见陷阱：上抛后落回原高度的 v‑t 图是一条穿过 t 轴的直线。若问位移，上下面积会抵消；若问总路程，则需将面积的绝对值相加。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导