📚 IB AQA Maths: Kinematics Key Points | IB AQA 数学：运动学 考点精讲
Kinematics in IB Mathematics applies calculus to the motion of particles moving in a straight line. You will need to understand how displacement, velocity and acceleration relate through differentiation and integration, interpret motion graphs, and solve problems involving constant or variable acceleration. This article breaks down every essential concept into clear, paired explanations to help you master the topic and perform confidently in exams.
在 IB 数学中，运动学将微积分应用于沿直线运动的质点。你需要理解位移、速度和加速度如何通过微分和积分相互关联，解读运动图像，并解决涉及恒定或可变加速度的问题。本文将每一个核心概念分解为清晰的中英双语配对讲解，帮助你彻底掌握该主题并在考试中自信发挥。
1. Displacement, Velocity and Acceleration | 位移、速度和加速度
Displacement s(t) measures the position of a particle relative to a fixed origin at time t. Velocity v(t) is the rate of change of displacement, and acceleration a(t) is the rate of change of velocity. These three functions are connected by differentiation: v = ds/dt and a = dv/dt = d²s/dt². In kinematics problems you often start with one of them and derive the others.
位移 s(t) 描述质点在时间 t 相对于固定原点的位置。速度 v(t) 是位移的变化率，加速度 a(t) 是速度的变化率。这三个函数通过微分联系在一起：v = ds/dt，a = dv/dt = d²s/dt²。在运动学问题中，你通常从其中一个量出发，推导出其他量。
2. Using Differentiation to Find Velocity and Acceleration | 使用微分求速度和加速度
When you are given a displacement function s(t), differentiate once with respect to t to obtain velocity: v(t) = s'(t). Differentiate again to obtain acceleration: a(t) = v'(t) = s”(t). For example, if s(t) = t³ − 6t² + 9t metres, then v(t) = 3t² − 12t + 9 m s⁻¹ and a(t) = 6t − 12 m s⁻². This allows you to examine how the particle’s motion changes over time.
当给定位移函数 s(t) 时，对 t 求一次导数得到速度：v(t) = s'(t)。再求一次导数得到加速度：a(t) = v'(t) = s”(t)。例如，若 s(t) = t³ − 6t² + 9t 米，那么 v(t) = 3t² − 12t + 9 m s⁻¹，a(t) = 6t − 12 m s⁻²。这样你就能分析质点的运动如何随时间变化。
3. Using Integration to Recover Displacement and Velocity | 使用积分恢复位移和速度
If acceleration is given as a function of time, integrate to find velocity: v(t) = ∫ a(t) dt + C. Integrate velocity to find displacement: s(t) = ∫ v(t) dt + D. The constants of integration C and D are determined by initial conditions, such as the velocity or displacement at t = 0. This approach is essential when acceleration varies with time.
如果给定加速度关于时间的函数，先积分求得速度：v(t) = ∫ a(t) dt + C。再对速度积分得到位移：s(t) = ∫ v(t) dt + D。积分常数 C 和 D 由初始条件决定，例如 t = 0 时刻的速度或位移。当加速度随时间变化时，这一方法至关重要。
4. Constant Acceleration and the SUVAT Equations | 匀加速度与 SUVAT 公式
When acceleration is constant, five key quantities are linked: initial velocity u, final velocity v, acceleration a, displacement s and time t. The four SUVAT equations are:
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t
These only apply when acceleration is uniform. Choose the equation that contains the three known quantities and the one unknown you need to find.
当加速度恒定时，五个关键量联系在一起：初速度 u，末速度 v，加速度 a，位移 s 和时间 t。四个 SUVAT 方程如下：
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t
这些公式仅适用于匀加速度。选择包含三个已知量和待求未知量的方程即可。
5. Finding Maximum and Minimum Displacement | 求最大和最小位移
A particle reaches its extreme displacement when the velocity is momentarily zero and changes sign. Set v(t) = 0 and solve for t. Then use the second derivative test or analyse the sign change of v to confirm whether it is a maximum or minimum. Substitute the t value back into s(t) to find the extreme displacement.
当速度瞬时为零并改变符号时，质点达到位移的极值。令 v(t) = 0 并解出 t。接着使用二阶导数检验或分析 v 的符号变化，以确认是最大值还是最小值。再将 t 值代回 s(t) 即可求出极值位移。
6. Velocity‑Time Graphs and Their Interpretation | 速度‑时间图像及其解读
A velocity‑time graph plots v against t. The gradient of the curve gives the acceleration, while the area between the curve and the t‑axis gives the change in displacement (taking areas below the axis as negative). Straight horizontal lines represent constant velocity, sloped lines represent constant acceleration, and curves indicate variable acceleration.
速度‑时间图像绘制 v 关于 t 的变化。曲线的斜率表示加速度，曲线与 t 轴之间的面积表示位移的变化量（轴下方的面积取负值）。水平直线代表匀速，倾斜直线代表匀加速度，曲线则表示变加速度。
7. Displacement‑Time Graphs | 位移‑时间图像
On a displacement‑time graph, the gradient at any point equals the instantaneous velocity. A straight line indicates constant velocity, while a curve shows changing velocity. Points where the gradient is zero correspond to stationary points, which may be maxima or minima of displacement. The graph never returns information about acceleration directly; you must differentiate twice.
在位移‑时间图像上，任意点的斜率等于瞬时速度。直线表示匀速，曲线则表示变速。斜率为零的点对应质点静止的位置，可能是位移的极大值或极小值。该图像不直接给出加速度信息，需要经过两次微分才能获得。
8. Acceleration‑Time Graphs | 加速度‑时间图像
An acceleration‑time graph directly shows how acceleration varies with time. The area under the curve between two times equals the change in velocity over that interval. A horizontal line at zero indicates constant velocity, while a positive/negative line shows increasing/decreasing velocity. For variable acceleration, the graph helps visualise when the particle experiences the greatest change in velocity.
加速度‑时间图像直接展示加速度如何随时间变化。曲线在两个时刻之间的面积等于该时间段内速度的变化量。零水平线表示匀速，正/负水平线表示速度在增加/减少。对于变加速度，图像有助于可视化质点何时经历最大的速度变化。
9. Using Initial Conditions with Integration | 利用积分初始条件求解
When integrating acceleration or velocity, always include a constant of integration. Use given values such as v(0) = v₀ or s(0) = s₀ to find the constant. Without initial conditions, the displacement or velocity function can only be determined up to an arbitrary constant. In an exam, carefully read the wording: ‘starts from rest’ means v = 0 at t = 0; ‘passes through the origin’ means s = 0 at that instant.
在对加速度或速度积分时，务必加上积分常数。利用已知条件，如 v(0) = v₀ 或 s(0) = s₀，确定常数值。若没有初始条件，位移函数或速度函数只能确定到差一个任意常数。考试时需仔细审题：“从静止开始”意味着 t = 0 时 v = 0；“经过原点”意味着在那一时刻 s = 0。
10. Describing the Direction of Motion | 描述运动方向
A positive velocity means the particle is moving in the positive direction (away from the origin in the defined sense), while a negative velocity indicates motion in the opposite direction. If velocity changes sign, the particle changes direction. To find when the particle changes direction, solve v(t) = 0 and check for a sign change. The total distance travelled differs from displacement when direction changes; you must sum the absolute values of displacement in each interval.
正速度表示质点沿正方向运动（按定义背离原点），负速度则表示沿反方向运动。若速度的符号改变，质点即改变方向。要找出改变方向的时刻，需解方程 v(t) = 0 并检查符号变化。当方向改变时，总路程与位移不同；你需要将各时间段内的位移绝对值相加。
11. Variable Acceleration Problems | 变加速度问题
When acceleration is not constant, differentiation and integration techniques become essential. You might be given a(t) and asked to find v(t) and s(t), or be given s(t) and asked to find the times when the particle is instantaneously at rest. Typical IB problems combine calculus with algebraic manipulation, sometimes requiring you to solve quadratic or cubic equations after setting v = 0. Always check that your solutions lie within the given time interval.
当加速度不恒定时，微分和积分技巧变得必不可少。你可能得到 a(t) 并被要求求 v(t) 和 s(t)，或者得到 s(t) 并被要求找出质点瞬时静止的时刻。典型的 IB 题目将微积分与代数计算相结合，有时需要在令 v = 0 后求解二次或三次方程。务必检查解是否在给定的时间区间内。
12. Exam Tips for Kinematics | 运动学考试技巧
Always distinguish between displacement and distance. Show clear steps when differentiating or integrating, and do not forget the constants. Label axes on sketch graphs and indicate key features such as intercepts and turning points. When using SUVAT, write down the values of u, v, a, s, t you know and choose the equation with a single missing variable. Finally, check your units: displacement in metres, time in seconds gives velocity in m s⁻¹ and acceleration in m s⁻².
始终区分位移和路程。微分或积分时写出清晰步骤，并且不要遗漏常数。在草图上标注坐标轴，并标出关键特征，如截距和驻点。使用 SUVAT 时，先写出已知的 u, v, a, s, t 的值，再选择只含一个未知量的方程。最后，检查单位：位移用米，时间用秒，则速度为 m s⁻¹，加速度为 m s⁻²。
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