📚 AQA Maths: Kinematics Key Points | AQA 数学：运动学 考点精讲
Kinematics is a fundamental branch of mechanics that describes the motion of objects without considering the forces causing it. In AQA Maths, mastering key concepts such as displacement, velocity, acceleration, and the equations of motion for constant acceleration is essential for tackling mechanics problems. This article covers the core topics, common pitfalls, and exam tips to help you succeed.
运动学是力学的基础分支，描述物体的运动而不考虑引起运动的力。在 AQA 数学中，掌握位移、速度、加速度以及匀加速运动方程等核心概念，对于处理力学问题至关重要。本文将涵盖核心考点、常见错误和应试技巧，帮助你取得成功。
1. Displacement, Velocity and Acceleration | 位移、速度和加速度
Displacement (s) is a vector quantity: it has both magnitude and direction, whereas distance is a scalar. It is measured in metres (m). Velocity (v) is the rate of change of displacement with respect to time, also a vector. Speed is its scalar counterpart.
位移 (s) 是矢量：既有大小又有方向，而路程是标量。它的单位是米 (m)。速度 (v) 是位移随时间的变化率，也是矢量。速率是其对应的标量。
Acceleration (a) is the rate of change of velocity. In AQA mechanics, if acceleration is constant, we can use the SUVAT equations. Average velocity is total displacement divided by total time.
加速度 (a) 是速度的变化率。在 AQA 力学中，若加速度恒定，我们可以使用 SUVAT 方程。平均速度等于总位移除以总时间。
a = Δv / Δt
加速度定义为速度变化量除以时间间隔：a = Δv / Δt。
2. The SUVAT Equations | 匀加速直线运动方程
When acceleration is constant, five 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.
当加速度恒定时，五个方程将位移 (s)、初速度 (u)、末速度 (v)、加速度 (a) 和时间 (t) 联系起来。它们由速度和加速度的定义推导而来。
v = u + at
s = ½(u + v)t
s = ut + ½at²
s = vt – ½at²
v² = u² + 2as
这些方程中不含有最终位置以外的未知力，是解决匀加速直线运动问题的核心工具。注意每个量的符号必须与选定的正方向一致。
These equations contain no force unknowns apart from the final position and are the core toolkit for constant-acceleration problems. Note that the sign of each quantity must be consistent with the chosen positive direction.
3. Problem-Solving with SUVAT | 应用匀加速方程解题策略
Start by listing the known values: s, u, v, a, t. Identify which one is unknown and which equation does not involve that unknown. Always check that all units are consistent (e.g., convert km h⁻¹ to m s⁻¹).
首先列出已知量：s, u, v, a, t。确定哪个是未知量，并选择不包含该未知量的方程。务必确保所有单位一致（例如，将 km h⁻¹ 转换为 m s⁻¹）。
Draw a diagram to define the positive direction. If an object decelerates, use a negative value for acceleration. For multi-stage motion, split the journey into sections where acceleration is constant and apply SUVAT to each stage.
画出示意图确定正方向。如果物体减速，加速度取负值。对于多阶段运动，将过程分成加速度恒定的若干阶段，分别应用 SUVAT 方程。
4. Vertical Motion under Gravity | 重力作用下的垂直运动
For objects moving freely under gravity, the acceleration is g = 9.8 m s⁻² vertically downwards. If upwards is taken as positive, a = –9.8 m s⁻². The same SUVAT equations apply, with a replaced by g or –g.
对于仅在重力作用下的物体，加速度为 g = 9.8 m s⁻²，方向竖直向下。若取向上为正方向，则 a = –9.8 m s⁻²。相同的 SUVAT 方程适用，只需将 a 替换为 g 或 –g。
At the highest point of an upward throw, the final velocity is momentarily zero, but acceleration is still g. Time to reach maximum height can be found from v = u + at with v = 0.
在竖直上抛的最高点，末速度瞬间为零，但加速度仍为 g。到达最高点的时间可通过 v = u + at 并令 v = 0 求得。
Symmetry: The time to go up equals the time to come down, and the speed at the same height during ascent and descent is equal, provided air resistance is negligible.
对称性：忽略空气阻力时，上升时间等于下落时间，且在相同高度处上升与下降的速率相等。
5. Displacement-Time Graphs | 位移-时间图
The gradient of a displacement-time graph gives the velocity. A straight line means constant velocity; a curve means changing velocity (acceleration). A horizontal line indicates the object is stationary.
位移-时间图的斜率表示速度。直线表示速度恒定；曲线表示速度在变化（存在加速度）。水平线段表示物体静止。
If the graph crosses the time axis, the object is passing through the origin. The steepness of the tangent at any point gives the instantaneous velocity. Always label axes with units.
若图线与时间轴相交，表示物体经过原点。任意点处切线的倾斜程度给出瞬时速度。始终在坐标轴上标注单位。
6. Velocity-Time Graphs | 速度-时间图
The gradient of a velocity-time graph gives the acceleration. The area between the graph and the time axis represents displacement. Areas above the axis are positive displacement; areas below are negative.
速度-时间图的斜率表示加速度。图线与时间轴之间的面积表示位移。时间轴上方面积对应正位移，下方面积对应负位移。
To find the total distance travelled, add the absolute values of all area segments. For a straight line graph, the area can often be split into triangles and rectangles.
要计算通过的总路程，将所有面积区域的绝对值相加。对于直线图，面积通常可分解为三角形和矩形计算。
7. Acceleration-Time Graphs | 加速度-时间图
The area under an acceleration-time graph gives the change in velocity (Δv). A horizontal line indicates constant acceleration. If the graph is beneath the axis, the velocity is decreasing.
加速度-时间图下的面积表示速度的变化量 (Δv)。水平线段表示加速度恒定。若图线位于轴下方，则速度在减小。
Knowing the initial velocity, you can determine the velocity at any time by adding the area up to that time. This is especially useful when acceleration is not constant but given as a function of time.
知道初速度后，通过累加某时刻之前的面积即可确定该时刻的速度。这在加速度非恒定但作为时间函数给出时尤其有用。
8. Kinematics in Two Dimensions (Vectors) | 二维运动学（矢量）
In two dimensions, position, velocity and acceleration are expressed as vectors using unit vectors i and j. The horizontal and vertical components are independent.
在二维空间中，位置、速度和加速度使用单位向量 i 和 j 的矢量表示。水平与竖直分量相互独立。
r = x i + y j
v = vₓ i + v_y j
a = aₓ i + a_y j
Differentiation of the position vector gives velocity, and differentiating velocity gives acceleration. Integration reverses the process, providing the constant of integration can be found from initial conditions.
对位置向量求导得到速度，对速度求导得到加速度。积分则是逆过程，积分常数可由初始条件确定。
To find the speed, calculate the magnitude of the velocity vector: |v| = √(vₓ² + v_y²). Direction is given by the angle from the positive i direction.
速率是速度矢量的大小：|v| = √(vₓ² + v_y²)。方向由与 i 正方向的夹角给出。
9. Projectile Motion | 抛体运动
A projectile moves under constant vertical acceleration (g) and zero horizontal acceleration (neglecting air resistance). Its path is a parabola. Resolve the initial velocity into horizontal and vertical components.
抛体在恒定的竖直加速度 (g) 和零水平加速度（忽略空气阻力）下运动，轨迹为抛物线。需将初速度分解为水平和竖直分量。
uₓ = u cos θ, u_y = u sin θ
Horizontal motion: x = uₓ t. Vertical motion: y = u_y t – ½ g t², v_y = u_y – g t. Time of flight is found when y returns to the initial vertical position.
水平方向：x = uₓ t。竖直方向：y = u_y t – ½ g t², v_y = u_y – g t。飞行时间通过令 y 等于初始竖直位置求得。
Maximum height occurs when v_y = 0. The range is the horizontal distance travelled during the flight, and is maximised when the launch angle is 45° on level ground.
最大高度发生在 v_y = 0 时。射程是飞行期间的水平位移，在水平地面上发射角为 45° 时射程最大。
10. Relative Motion | 相对运动
Relative position of B with respect to A is given by rB/A = rB – rA. Similarly, relative velocity vB/A = vB – vA. This concept is useful for interception and overtaking problems.
B 相对于 A 的相对位置为 r_B/A = r_B – r_A。类似地，相对速度 v_B/A = v_B – v_A。此概念在追击与相遇问题中很有用。
If two objects are moving towards each other, the relative speed is the sum of their speeds. When they move in the same direction, the relative speed is the difference.
若两物体相向运动，相对速度大小为两者速率之和。若同向运动，相对速度大小为两者速率之差。
To find when and where one object catches another, set their position vectors equal, or use the relative velocity and initial separation.
求追及的时刻和位置时，令两者的位置向量相等，或利用相对速度和初始距离求解。
11. Using Calculus in Kinematics | 运动学中的微积分应用
If displacement s is given as a function of time, velocity v = ds/dt and acceleration a = dv/dt = d²s/dt². Conversely, velocity is the integral of acceleration, and displacement is the integral of velocity.
若位移 s 作为时间函数给出，速度 v = ds/dt，加速度 a = dv/dt = d²s/dt²。相反地，速度是加速度的积分，位移是速度的积分。
v = ∫ a dt, s = ∫ v dt
Evaluating the constant of integration requires initial conditions, e.g. at t = 0, v = u. This method allows you to find displacement even when acceleration is not constant.
确定积分常数需要初始条件，例如当 t = 0 时 v = u。这种方法即使加速度不恒定，也能求出位移。
You may be asked to find the maximum displacement by setting v = 0, or to find the distance travelled by integrating speed (|v|) over time.
你可能需要令 v = 0 以求最大位移，或通过对速率 (|v|) 积分求通过的路程。
12. Common Mistakes and Exam Tips | 常见错误与应试技巧
One of the most frequent errors is confusing distance with displacement. Remember that displacement can be negative, but distance is always positive. Always check the direction convention.
最常见的错误之一是混淆路程与位移。记住位移可为负，而路程恒为正。务必检查正方向约定。
Forgetting to convert units (e.g., time in minutes, distance in km) costs valuable marks. Practise reading the question carefully to extract s, u, v, a, t correctly.
忘记换算单位（如时间用分钟、距离用千米）会白白丢分。仔细读题，正确提取 s, u, v, a, t。
Using the wrong SUVAT equation is another pitfall. When in doubt, write down all five variables and cross out the one you don’t need; choose the equation that avoids it.
用错 SUVAT 方程也是一大陷阱。有疑问时，列出全部五个变量，划掉不需要的那个，再选择不含它的方程。
In vector kinematics, treat i and j components separately. Do not mix horizontal and vertical calculations in the same step unless combining final results.
在矢量运动学中，将 i 和 j 分量分开处理。除非合成最终结果，否则不要在同一步骤中混合水平与竖直运算。
Finally, show all working clearly and double-check that your answer is reasonable in the context of the problem.
最后，清晰展示所有步骤，并检查答案是否在题目背景下合理。
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