📚 Kinematics | 动力学 考点精讲
Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion. In the CCEA A-Level Physics specification, you need to master the language of displacement, velocity, and acceleration, understand graphical representations of motion, and confidently apply the equations of uniformly accelerated motion to both one-dimensional and two-dimensional scenarios. This revision guide breaks down every critical concept, provides step-by-step examples, and highlights common pitfalls to help you achieve top marks.
动力学是力学的分支，描述物体的运动而不考虑引起运动的力。在 CCEA A-Level 物理考试大纲中，你需要掌握位移、速度和加速度的语言，理解运动的图形表示，并能自信地将匀加速运动方程应用于一维和二维情景。本复习指南将分解每个关键概念，提供分步示例，并突出常见误区，帮助你取得高分。
1. Scalar and Vector Quantities | 标量与矢量
In kinematics, it is essential to distinguish between scalar quantities, which have magnitude only, and vector quantities, which have both magnitude and direction. Distance is a scalar, while displacement is a vector. Speed is a scalar; velocity is a vector. The distinction affects how we combine measurements: vectors add using tip-to-tail geometry or component resolution, while scalars add arithmetically. Always pay attention to signs in one-dimensional motion – a negative velocity simply means motion in the opposite direction to the chosen positive axis.
在动力学中，区分只有大小的标量和既有大小又有方向的矢量至关重要。路程是标量，位移是矢量。速率是标量，速度是矢量。这一区别影响我们如何合成测量值：矢量使用三角形法则或分量分解相加，而标量则直接算术相加。在一维运动中要始终注意正负号——负速度仅表示运动方向与选定的正方向相反。
2. Displacement, Velocity and Acceleration Definitions | 位移、速度与加速度的定义
Average velocity vₐᵥ = Δs/Δt, and instantaneous velocity is the gradient of a displacement– itutes acceleration. In uniform circular motion, for instance, speed is constant but the direction changes continuously, so there is a centripetal acceleration. Always use the SI units: metre (m) for displacement, metre per second (m s⁻¹) for velocity, and metre per second squared (m s⁻²) for acceleration.
位移 s 是从参考点出发在给定方向上的直线距离。平均速度 vₐᵥ = Δs/Δt，瞬时速度是位移-时间图线的斜率。加速度 a = Δv/Δt；它是速度的变化率。由于速度是矢量，其大小或方向的任何变化都构成加速度。例如，在匀速圆周运动中，速率不变但方向不断变化，因此存在向心加速度。始终使用国际单位制：位移用米（m），速度用米每秒（m s⁻¹），加速度用米每二次方秒（m s⁻²）。
3. Graphs of Motion | 运动图线
Interpreting displacement–time (s–t), velocity–time (v–t) and acceleration–time (a–t) graphs is a fundamental skill. The slope of an s–t graph gives velocity; a curved s–t graph indicates changing velocity, i.e. acceleration. The slope of a v–t graph gives acceleration, and the area under a v–t graph between two times represents the change in displacement. The area under an a–t graph gives the change in velocity. For uniformly accelerated motion, a v–t graph is a straight line with constant gradient; the area under it can be split into a rectangle and a triangle to derive the suvat equations.
解读位移-时间 (s–t)、速度-时间 (v–t) 和加速度-时间 (a–t) 图线是一项基本技能。s–t 图线的斜率给出速度；弯曲的 s–t 图线表示速度在变化，即存在加速度。v–t 图线的斜率给出加速度，而 v–t 图线下两时刻之间的面积表示位移的变化量。a–t 图线下的面积给出速度的变化量。对于匀加速运动，v–t 图线是一条斜率恒定的直线；其下的面积可分为矩形和三角形，从而推导出 suvat 方程。
4. The 'suvat' Equations of Uniformly Accelerated Motion | 匀加速运动的 ‘suvat’ 方程
For motion in a straight line with constant acceleration a, the five standard equations link displacement s, initial velocity u, final velocity v, acceleration a, and time t. They are derived from the definitions of velocity and acceleration and the area under a v–t graph. The equations are:
对于加速度 a 恒定的直线运动，五个标准方程将位移 s、初速度 u、末速度 v、加速度 a 和时间 t 联系起来。它们源自速度和加速度的定义以及 v–t 图线下的面积。方程如下：
v = u + at
s = ut + ½at²
s = ½(u + v)t
v² = u² + 2as
s = vt − ½at²
When using these equations, choose the one that contains only the unknowns you need. Always define a positive direction and assign signs to vector quantities accordingly – for example, upward positive for falling objects makes acceleration due to gravity negative. A common mistake is to mix sign conventions within one calculation.
使用这些方程时，选择只包含所需未知量的那一个。始终定义正方向，并相应地为矢量赋值正负号——例如，对于下落的物体取向上为正，则重力加速度为负。一个常见错误是在同一次计算中混用不同的符号约定。
5. Free Fall Under Gravity | 重力作用下的自由落体
ownward throughout the flight. The time to rise equals the time to fall back to the original level. Use the suvat equations with a = −g when up is positive, or a = +g when down is positive. Displacement, velocity and time can be found for any point in the trajectory.
当物体在地表附近被释放或抛出时，它会受到恒定的向下加速度 g ≈ 9.81 m s⁻²。这是匀加速运动的一种特殊情况。如果一个物体垂直上抛，到达最高点时速度为零，但在整个飞行过程中加速度始终为向下的 g。上升时间等于落回原水平面的下降时间。取向上为正时使用 a = −g，或取向下为正时使用 a = +g 应用 suvat 方程。可以求出轨迹中任意点的位移、速度和时间。
6. Projectile Motion – Horizontal and Vertical Components | 抛体运动——水平与竖直分量
. Vertically, the motion is uniformly accelerated with a = g downward. Use the suvat equations for the vertical component: initial vertical velocity uᵧ = u sin θ, and vertical displacement y = uᵧ t − ½gt² (with upward positive). The time of flight is determined by the vertical motion; the horizontal range is then uₓ × total time of flight. The trajectory is parabolic.
抛体运动通过将初速度分解为独立的水平分量和竖直分量来分析。水平方向上不受加速度（忽略空气阻力），因此水平分速度 uₓ = u cos θ 保持恒定，水平位移 x = uₓ t。竖直方向是加速度向下的匀加速运动，a = g。对竖直分量使用 suvat 方程：初始竖直速度 uᵧ = u sin θ，竖直位移 y = uᵧ t − ½gt²（取向上为正）。飞行时间由竖直运动决定；水平射程则为 uₓ × 总飞行时间。轨迹为抛物线。
7. Calculating Time of Flight, Maximum Height and Range | 计算飞行时间、最大高度与射程
T = (u² sin 2θ)/g. Note that the maximum range for a given initial speed is achieved at θ = 45°. If the landing level is different from the launch level, the time of flight must be solved from the quadratic in t using the full vertical displacement equation.
对于从同一水平面发射并落回该水平面的抛体，飞行时间 T 通过设定竖直位移 y = 0 求得：0 = uᵧ T − ½gT²，得 T = 2uᵧ / g = (2u sin θ)/g。最大高度 H 发生在竖直速度为零时：0 = uᵧ² − 2gH，所以 H = uᵧ²/(2g)。水平射程 R = uₓ T = (u² sin 2θ)/g。注意，给定初速率，最大射程在 θ = 45° 时达到。如果落地点与发射点水平高度不同，则必须使用完整的竖直位移方程求解关于 t 的二次方程来得到飞行时间。
8. Relative Velocity in One and Two Dimensions | 一维和二维中的相对速度
The velocity of object A relative to object B is given by vA/B = vA − vB. In one dimension, treat the velocities as signed scalars. In two dimensions, vector subtraction is needed, often by resolving into components. This concept appears in problems with moving walkways, boats crossing rivers, or wind affecting aircraft. For a boat crossing a river, the resultant velocity is the vector sum of the boat's velocity relative to water and the water's velocity relative to the ground. Drift and time to cross are calculated from the components perpendicular and parallel to the bank.
物体 A 相对于物体 B 的速度由 vA/B = vA − vB 给出。在一维情况下，将速度视为带符号的标量。在二维情况下，需要进行矢量减法，通常通过分量分解来实现。这一概念出现在移动步行道、船过河或风对飞机的影响等问题中。对于过河的船，合速度是船相对于水的速度与水相对于地面的速度的矢量和。漂移和过河时间可根据垂直于河岸和平行于河岸的分量计算。
9. Interpreting Kinematic Graphs Numerically | 数值化解读运动图线
CCEA exam questions often present a v–t graph and ask you to find acceleration (from gradient), distance travelled (from area under graph), or average speed (total distance ÷ total time). For non‑uniform acceleration, areas may be estimated by counting squares or by using geometric shapes like triangles and trapeziums. Instantaneous velocity is read directly from the graph. An s–t graph with decreasing gradient indicates deceleration; an increasing gradient indicates acceleration. Be precise with units and show all working when calculating areas – break complex regions into rectangles and triangles.
CCEA 考试题目常给出 v–t 图线，要求你求出加速度（通过斜率）、移动距离（通过图线下的面积）或平均速率（总路程 ÷ 总时间）。对于非匀加速的情况，可通过数方格或使用三角形和梯形等几何形状来估算面积。瞬时速度直接从图线上读取。斜率减小的 s–t 图线表示减速；斜率增大表示加速。在计算面积时要精确使用单位并展示所有计算步骤——将复杂区域分割为矩形和三角形。
10. Using Suvat with Galileo’s Odd‑Numbers Rule & Thinking Steps | 使用 suvat 与伽利略奇数律及思考步骤
Galileo' , 2nd, 3rd seconds are ½g, 3×½g, 5×½g. This can quickly verify answers. A reliable problem‑ (4) Solve algebraically before substituting numbers. (5) Check units and sign consistency.
伽利略的观察——从静止开始的匀加速物体在连续相等时间间隔内经过的距离之比为 1：3：5：7 ……——是一个有用的检查工具。例如，从静止自由落体，第 1 秒、第 2 秒、第 3 秒内下落的距离为 ½g, 3×½g, 5×½g。这可以快速验证答案。一个可靠的问题解决策略是：(1) 画出示意图并标注正方向。(2) 列出已知量 s、u、v、a、t，标出未知量。(3) 选择不含某个未知量的 suvat 方程。(4) 先代数求解，再代入数字。(5) 检查单位和符号一致性。
11. Tackling Multi‑Stage Kinematics Problems | 处理多阶段运动学问题
Many real CCEA questions involve two or more stages of motion – for example, a car accelerates uniformly, then travels at constant speed, then decelerates to rest. Break the motion into segments where acceleration is constant. Write a suvat list for each segment. The final velocity of one stage becomes the initial velocity of the next. Use the total time or total displacement constraints to form equations linking the stages. Drawing a v–t graph for the whole journey often clarifies the relationships and makes area calculations straightforward.
许多真实的 CCEA 题目涉及两个或多个运动阶段——例如，一辆汽车先匀加速，然后匀速行驶，再匀减速至静止。将运动分解为加速度恒定的若干段。为每一段列出 suvat 清单。前一段的末速度成为下一段的初速度。利用总时间或总位移的约束条件，建立联系各段的方程。画出整个运动过程的 v–t 图线通常可以理清关系，并使面积计算变得简单直接。
12. Avoiding Common Mistakes | 避免常见错误
is moving upward. Another is misinterpreting the sign of displacement in projectile motion – e velocity: average speed = total distance / total time, while average velocity = total displacement / total time. Finally, always convert units to SI before calculation.
最常见的错误之一是忘记重力加速度即使物体向上运动时也向下。另一个常见错误是误解抛体运动中位移的符号——位移是从发射点到最终位置的有向线段，而不是总路径长度。学生还经常在加速度不恒定时误用 s = ut + ½at²；在使用任何 suvat 方程之前，务必确认运动是匀加速的。此外，不要混淆平均速率与平均速度：平均速率 = 总路程 / 总时间，而平均速度 = 总位移 / 总时间。最后，计算前务必把单位转换为国际单位制。
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