📚 IB & OCR Physics: Kinematics – Key Points for Exam | IB与OCR物理：运动学考点精讲
Kinematics is the branch of mechanics that describes motion without reference to its causes. For both IB and OCR Physics students, a solid grasp of kinematic concepts, equations, and graph interpretation is vital for success in exams. This guide walks you through every major topic, highlighting common pitfalls and providing clear examples.
运动学是力学的一个分支，描述物体的运动而不涉及运动的原因。对于 IB 和 OCR 物理考生而言，透彻掌握运动学概念、方程和图像解读是考试拿分的关键。本文将带你逐一梳理核心考点，点明常见误区，并提供清晰的示例。
1. Scalars and Vectors in Kinematics | 运动学中的标量与矢量
In kinematics, distance is a scalar quantity that measures the total path length travelled, whereas displacement is a vector that measures the shortest straight-line distance from the start to the end point, together with its direction.
在运动学中，路程是标量，测量物体运动轨迹的总长度；位移是矢量，测量从起点到终点的最短直线距离，并包含方向。
Speed is the scalar rate of change of distance, while velocity is the vector rate of change of displacement. Acceleration, being the rate of change of velocity, is also a vector. Always assign a consistent sign convention for direction (e.g., upwards positive, or rightwards positive).
速率是路程对时间的变化率，为标量；速度是位移对时间的变化率，为矢量。加速度是速度的变化率，同样为矢量。务必为方向规定一致的正负号（例如向上为正，或向右为正）。
2. Displacement, Velocity and Acceleration | 位移、速度与加速度
Average velocity is defined as vavg = Δs/Δt, where Δs is the change in displacement. Instantaneous velocity is the limit of average velocity as the time interval approaches zero, represented by the gradient of a displacement–time graph at a point.
平均速度定义为 vavg = Δs/Δt，其中 Δs 是位移的变化量。瞬时速度是时间间隔趋近于零时平均速度的极限，在位移–时间图像上表现为某点的切线斜率。
Acceleration is given by a = Δv/Δt. Constant acceleration produces a straight line on a velocity–time graph; non‑constant acceleration produces a curve. A negative acceleration does not always mean deceleration — it depends on the direction of velocity.
加速度为 a = Δv/Δt。匀加速运动在速度–时间图像上为一条直线；变加速则为曲线。负加速度并不总代表减速，需要结合速度方向判断。
3. Equations of Uniformly Accelerated Motion (SUVAT) | 匀加速运动方程（SUVAT）
The four SUVAT equations apply when acceleration is constant. They link displacement s, initial velocity u, final velocity v, acceleration a, and time t. You must know these by heart and be able to select the right one for the data given.
四个 SUVAT 方程适用于加速度恒定的情况，关联位移 s、初速度 u、末速度 v、加速度 a 和时间 t。必须熟记并能根据已知量选择正确的方程。
- v = u + at — relates velocities, acceleration and time.
中文：v = u + at，关联末速度、初速度、加速度和时间。 - s = ut + ½at² — calculates displacement when initial velocity and acceleration are known.
中文：s = ut + ½at²，已知初速度与加速度时计算位移。 - v² = u² + 2as — links velocities with displacement and acceleration, without explicit time.
中文：v² = u² + 2as，关联速度、位移与加速度，不含时间。 - s = ½(u + v)t — uses average velocity to find displacement.
中文：s = ½(u + v)t，利用平均速度求位移。
Always ensure the units are consistent (usually metres, seconds, m s⁻¹, m s⁻²). Choose the equation that includes the unknown you need and the quantities you already have.
务必保持单位一致（一般用米、秒、m s⁻¹、m s⁻²）。选择包含所求未知量且已知其他几个物理量的方程。
4. Free Fall and Vertical Motion | 自由落体与竖直上抛运动
In free fall near the Earth’s surface, the acceleration due to gravity g = 9.81 m s⁻² acts downwards. The same SUVAT equations apply, with acceleration a set to ±g depending on the sign convention. For objects thrown upwards, take care with the sign of u and g.
在地表附近的自由落体中，重力加速度 g = 9.81 m s⁻² 方向向下。同样适用 SUVAT 方程，加速度 a 根据正方向取 ±g。竖直上抛时需注意初速度 u 和 g 的符号。
At the highest point of a vertically launched object, the instantaneous velocity is zero, but the acceleration is still g downwards. The time to reach maximum height is t = u/g (if launched upward with speed u).
在竖直上抛的最高点，瞬时速度为零，但加速度仍为向下的 g。达到最高点的时间为 t = u/g （以速度 u 向上抛出时）。
5. Projectile Motion | 抛体运动
Projectile motion can be analysed by splitting it into independent horizontal and vertical components. Horizontally, there is no acceleration, so ux = u cosθ is constant. Vertically, acceleration is –g, giving SUVAT-governed motion with initial component uy = u sinθ.
抛体运动可通过分解为独立的水平与竖直分量来分析。水平方向无加速度，因此 ux = u cosθ 恒定。竖直方向有加速度 –g，初始分量为 uy = u sinθ，遵循 SUVAT 方程。
The time of flight for a projectile launched from and returning to the same vertical level is t = 2uy/g. The maximum height reached is H = uy²/(2g). The horizontal range is R = ux × t = (u² sin 2θ)/g.
若抛体从同一水平面起落，飞行时间为 t = 2uy/g。最大高度为 H = uy²/(2g)。水平射程为 R = ux × t = (u² sin 2θ)/g。
For a given launch speed, the maximum range is achieved at θ = 45° in the absence of air resistance. Air resistance, when considered qualitatively, reduces both the range and maximum height.
在给定初速度下，无空气阻力时射程最大对应的抛射角为 45°。定性地考虑空气阻力时，射程和最大高度都会减小。
6. Motion Graphs (s–t, v–t, a–t) | 运动图像 (s–t, v–t, a–t 图)
A displacement–time graph (s–t) reveals the position of an object over time. The gradient gives the instantaneous velocity. A straight line represents constant velocity; a curved line indicates acceleration.
位移–时间图像 (s–t) 展示物体位置随时间的变化。图像斜率代表瞬时速度。直线表示匀速运动，曲线表示有加速度。
A velocity–time graph (v–t) has a gradient equal to acceleration and the area under the graph equal to the displacement. A horizontal line means constant velocity, while a sloping line means constant acceleration.
速度–时间图像 (v–t) 的斜率等于加速度，图像与时间轴所围的面积等于位移。水平线表示匀速，斜线表示匀加速。
An acceleration–time graph (a–t) shows how acceleration varies with time. The area under an a–t graph gives the change in velocity (Δv). For constant acceleration, the graph is a horizontal line.
加速度–时间图像 (a–t) 显示加速度随时间的变化。a–t 图像下的面积对应速度变化量 Δv。匀加速时，a–t 图为一条水平线。
7. Graph Skills: Slopes and Areas | 图像解读：斜率与面积
Reading motion graphs accurately is essential. For an s–t graph, draw a tangent at a point to find instantaneous velocity. For a v–t graph, the displacement obtained from the area should be calculated with attention to shapes (rectangles, triangles, or trapezoids).
准确解读运动图像至关重要。在 s–t 图上，可通过作切线求某点的瞬时速度。在 v–t 图上，根据图形形状（矩形、三角形或梯形）计算面积得到位移。
Be careful with negative areas on v–t or a–t graphs when the line lies below the time axis: they represent motion in the negative direction. Total displacement is the algebraic sum of areas, while total distance travelled is the sum of their absolute values.
注意 v–t 或 a–t 图中位于时间轴下方的部分，面积代表负方向的量。计算总位移时需取面积的代数和，而总路程则要取各面积的绝对值之和。
8. Relative Velocity | 相对速度
Relative velocity describes the velocity of one object as observed from another moving frame. For two objects A and B moving along a straight line, the velocity of A relative to B is: vA/B = vA – vB.
相对速度描述从另一个运动参考系观察某一物体的速度。对沿同一直线运动的 A、B 两物体，A 相对于 B 的速度为：vA/B = vA – vB。
In two dimensions, relative velocity is found by vector subtraction. Draw the vectors tip‑to‑tail or resolve into components, then calculate the magnitude and direction. IB students are often asked to solve problems involving boats crossing rivers or aeroplanes in wind, where the resultant velocity matters.
在二维运动中，相对速度需通过矢量减法求解。可作矢量三角形或将速度分解为分量后计算大小和方向。IB 考生常会遇到小船过河或飞机在风中航行的问题，需要求合速度。
9. Common Misconceptions and Exam Tips | 常见误区与考试技巧
Confusing distance with displacement and speed with velocity is a frequent mistake — always check whether a vector or scalar is required. Negative acceleration does not automatically mean slowing down; deceleration only occurs when velocity and acceleration have opposite signs.
混淆路程与位移、速率与速度是常见错误——务必看清题目要求的是矢量还是标量。负加速度不一定代表减速；只有速度与加速度异号时物体才减速。
Another pitfall is forgetting to convert all values to consistent SI units. Also, in projectile problems, do not use the total time of flight in horizontal equations unless you are certain the vertical motion starts and ends at the same height.
另一个陷阱是忘记将所有数值统一成国际单位制。在抛体问题中，除非确认竖直方向起落高度相同，否则不要将总飞行时间直接代入水平运动方程。
Examiners often test the ability to derive SUVAT equations from a velocity–time graph. Practise sketching graphs for different motion scenarios and calculating areas and gradients quickly but accurately.
考官常会考察学生从 v–t 图推导 SUVAT 方程的能力。建议多练习根据运动情景画草图，并迅速且准确地计算面积与斜率。
10. Summary of Key Formulas | 关键公式总结
The list below brings together the most important kinematic equations. Use these as a quick refresher before exams.
以下汇总了最重要的运动学方程，可作为考前快速回顾资料。
- v = u + at — velocity after time t.
中文：v = u + at，t 时刻的末速度。 - s = ut + ½at² — displacement with constant acceleration.
中文：s = ut + ½at²，匀加速运动下的位移。 - v² = u² + 2as — velocity–displacement relation independent of time.
中文：v² = u² + 2as，不含时间的速度–位移关系。 - s = ½(u + v)t — displacement via average velocity.
中文：s = ½(u + v)t，用平均速度求位移。 - tflight = 2uy/g — time of flight for symmetric projectile.
中文：tflight = 2uy/g，对称抛体的飞行时间。 - R = (u² sin 2θ)/g — range of a projectile.
中文：R = (u² sin 2θ)/g，抛体的水平射程。 - H = uy²/(2g) — maximum height.
中文：H = uy²/(2g)，最大高度。 - vA/B = vA – vB — relative velocity (1D).
中文：vA/B = vA – vB，一维相对速度。
Solid understanding of these equations, combined with meticulous graph analysis and vector manipulation, will give you a strong foundation for tackling kinematics questions in IB or OCR Physics exams.
扎实理解这些方程，配合细致的图像分析与矢量运算，能为你在 IB 或 OCR 物理考试中攻克运动学题目打下坚实基础。
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