IB Maths: Kinematics Essentials | IB 数学：运动学考点精讲

📚 IB Maths: Kinematics Essentials | IB 数学：运动学考点精讲

Kinematics is a core application of calculus in the IB Mathematics Analysis & Approaches and Applications & Interpretation syllabi. It models motion in a straight line, connecting displacement, velocity, and acceleration via differentiation and integration. This article delivers a thorough breakdown of key concepts, worked examples, and exam tactics to help you secure top marks on both Paper 1 and Paper 2.

运动学是 IB 数学分析与方法、应用与解释课程中微积分的核心应用。它模拟直线运动，通过微分和积分将位移、速度与加速度联系起来。本文深入解析关键概念、典型例题和应试技巧，助你在试卷一与试卷二中稳拿高分。

1. Fundamental Definitions: s, v, a | 基本定义：位移、速度、加速度

Displacement s(t) gives the position of a particle relative to an 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. All are functions of time in IB kinematics problems.

位移 s(t) 表示质点相对于原点在时刻 t 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。在 IB 运动学问题中，它们都是时间的函数。

Standard units: distance in metres (m), time in seconds (s). Velocity is in m s⁻¹ and acceleration in m s⁻². The relationships are rooted in calculus: v = ds/dt and a = dv/dt = d²s/dt².

标准单位：距离用米 (m)，时间用秒 (s)。速度单位 m s⁻¹，加速度单位 m s⁻²。这些关系植根于微积分： v = ds/dt 且 a = dv/dt = d²s/dt²。

2. Differentiation in Kinematics: Getting v and a from s | 微分法求速度与加速度

If displacement is given as a function of time, differentiate to find velocity. Use the power rule, chain rule, or product rule as needed. For example, s(t) = 3t³ − 2t² + 5 → v(t) = 9t² − 4t.

若位移以时间函数给出，微分可求速度。视需要应用幂法则、链式法则或乘法法则。例如 s(t) = 3t³ − 2t² + 5 → v(t) = 9t² − 4t。

Differentiate velocity to obtain acceleration: a(t) = dv/dt. For v(t) = 9t² − 4t, a(t) = 18t − 4. These steps are direct and typically tested in Paper 1 non‑calculator sections.

对速度再求导即得加速度： a(t) = dv/dt。对 v(t) = 9t² − 4t，有 a(t) = 18t − 4。这些步骤直接，常在试卷一非计算器部分考查。

3. Integration in Kinematics: Recovering s and v from a | 积分法恢复位移与速度

Given acceleration a(t), integrate to find velocity: v(t) = ∫ a(t) dt + C₁. A second integration yields displacement: s(t) = ∫ v(t) dt + C₂. The constants of integration are determined by initial conditions.

已知加速度 a(t)，积分得速度： v(t) = ∫ a(t) dt + C₁。再积分一次得到位移： s(t) = ∫ v(t) dt + C₂。积分常数由初始条件确定。

Common exam scenario: a = 6t − 2, at t = 0, v = 3, s = 0. Integrate a to get v = 3t² − 2t + C₁. Substitute t=0, v=3 → C₁ = 3, so v = 3t² − 2t + 3. Integrate v to get s = t³ − t² + 3t + C₂. With s(0)=0, C₂ = 0. Thus s = t³ − t² + 3t.

常见考题：a = 6t − 2，t=0 时 v=3，s=0。积分加速度：v = 3t² − 2t + C₁。代入 t=0, v=3 得 C₁=3，故 v = 3t² − 2t + 3。积分速度：s = t³ − t² + 3t + C₂。s(0)=0 推出 C₂=0，因此 s = t³ − t² + 3t。

4. Initial Conditions and Constants of Integration | 初始条件与积分常数

Initial conditions are vital when integrating. They link the general antiderivative to the specific physical situation. Always read “initially” or “at t = 0” to extract values for s, v, or a.

积分时初始条件至关重要。它们将一般反导数与具体物理情境挂钩。务必留意“初始”或“在 t=0 时”来提取 s、v 或 a 的值。

If the wording gives “the particle passes through the origin with speed 4 m s⁻¹ at t = 2”, you have s(2)=0 and v(2)=±4. Pay attention to direction: speed is magnitude, velocity has sign.

若题目描述“质点在 t=2 时经过原点，速率为 4 m s⁻¹”，则有 s(2)=0 且 v(2)=±4。注意方向：速率是大小，速度带有正负号。

5. Direction of Motion and Turning Points | 运动方向与转向点

A particle changes direction when its velocity changes sign. This occurs at times when v(t)=0. Solve v(t)=0 to find the critical instants; these are potential turning points.

当速度改变符号时质点改变方向。这发生在 v(t)=0 的时刻。解 v(t)=0 求得关键瞬时；这些是潜在的转向点。

Confirm a direction change by checking the sign of v just before and after the root. If the sign flips from positive to negative or vice versa, a turn occurs. If the sign stays the same, it’s a stationary instant with no direction reversal.

通过检查该根前后 v 的符号确认方向改变。若符号由正变负或由负变正，则发生了转向。若符号不变，则为瞬时静止而无方向反转。

6. Displacement vs Total Distance Travelled | 位移与总路程的区别

Displacement is the net change in position; total distance is the sum of all movement regardless of direction. The two are equal only if the particle never changes direction.

位移是位置的净变化；总路程是所有运动距离之和，不考虑方向。只有当质点从未改变方向时两者才相等。

To find total distance over [t₁, t₂], integrate the absolute value of velocity: ∫ |v(t)| dt. Without absolute value, you need to split the integral at each turning point where v(t)=0 and use (v) or (−v) depending on the sign.

求 [t₁, t₂] 上的总路程，需对速度的绝对值积分：∫ |v(t)| dt。若无绝对值，则需在每个 v(t)=0 的转向点处分段，根据符号取 v 或 −v 积分。

Example: v = t² − 4t + 3 on [0,4]. Roots at t=1,3. Total distance = ∫₀¹ (v) dt − ∫₁³ (v) dt + ∫₃⁴ (v) dt (since v >0 on [0,1) and [3,4]; <0 on [1,3]).

例如 v = t² − 4t + 3 在 [0,4] 上。根为 t=1,3。总路程 = ∫₀¹ v dt − ∫₁³ v dt + ∫₃⁴ v dt（因为 v 在 [0,1) 和 [3,4] 为正，在 [1,3] 为负）。

7. Variable Acceleration and Non‑constant Forces | 变加速度问题

Many IB problems feature acceleration as a function of time, displacement, or velocity. For a(t) functions, integration is straightforward. For a(v) or a(s), use the chain rule: a = v dv/ds.

许多 IB 问题中加速度表示为时间、位移或速度的函数。对 a(t) 型函数积分直接。对于 a(v) 或 a(s)，应用链式法则： a = v dv/ds。

When a = f(v), separate variables: dv/dt = f(v) → dv/f(v) = dt. Then integrate. When a = f(s), use v dv = f(s) ds and integrate. These techniques are more common in AA HL.

当 a = f(v) 时，分离变量：dv/dt = f(v) → dv/f(v) = dt，然后积分。当 a = f(s) 时，利用 v dv = f(s) ds 并积分。这些技巧在 AA HL 中出现更多。

8. Finding Maximum and Minimum Velocity | 求最大与最小速度

To optimise velocity, find v(t) and set dv/dt = a(t) = 0. Solve for t and use the second derivative test or sign table to classify as max or min. The second derivative of v is da/dt, the jerk.

优化速度时，求出 v(t) 并令 dv/dt = a(t) = 0。解出 t，用二阶导数检验或符号表判断极大或极小。v 的二阶导数是 da/dt，即加加速度。

Remember that maximum speed may occur at the endpoints of a time interval as well as at critical points. Always check boundaries when the domain is restricted.

注意最大速率可能出现在时间区间的端点以及临界点。当定义域有限制时，务必检查边界。

9. Interpreting s–t, v–t, a–t Graphs | 解读位移、速度、加速度图像

Key graph facts: The slope of an s–t graph gives velocity; the slope of a v–t graph gives acceleration. The area under a v–t graph gives change in displacement; area under an a–t graph gives change in velocity.

关键图像事实：s–t 图的斜率给出速度；v–t 图的斜率给出加速度。v–t 图下的面积给出位移的变化量；a–t 图下的面积给出速度的变化量。

When a v–t graph crosses the t‑axis, the area must be split into signed regions to compute displacement or total distance. Negative area means motion in the negative direction.

当 v–t 图像穿过 t 轴时，必须将面积分成带符号的区域来计算位移或总路程。负面积表示朝负方向运动。

10. Word Problems and Structured Questions | 应用题与结构化问题

IB exam questions often embed kinematics in a real‑world context, such as a particle along a line, a car braking, or an object falling with air resistance (modelled by a(v)). Read the whole text, identify the given functions or conditions, and set out the calculus systematically.

IB 考题常将运动学嵌入实际语境，如沿直线运动的质点、汽车制动或受空气阻力的落体（用 a(v) 建模）。通读全文，识别给定函数或条件，并系统地列出微积分步骤。

Always define which direction is positive. If a question says “velocity 5 m s⁻¹ upwards”, but you set upwards as positive, then v = +5. If downwards is positive, v = −5 for the same motion. Explicitly state your sign convention.

始终明确哪个方向为正。若题述“速度 5 m s⁻¹ 向上”，而你设定向上为正，则 v = +5。若向下为正，则同一运动 v = −5。明确陈述你的正负约定。

11. Common Mistakes and Exam Tips | 常见错误与考试技巧

Mixing up speed and velocity: speed is |v|, velocity has sign. For distance, use |v| or split the integral. Many candidates lose marks by incorrectly evaluating total distance without considering direction changes.

混淆速率与速度：速率是 |v|，速度带符号。求路程需用 |v| 或分段积分。许多考生因未考虑方向变化而错误计算总路程，导致失分。

Missing constants: after integration, always write +C. Use initial conditions immediately. Double‑check that your integrated expression satisfies the given s(0), v(0) values. Rounding errors can accumulate; leave answers exact until the final step.

漏掉常数：积分后务必写上 +C。立即使用初始条件。仔细检查积分后的表达式是否满足给定的 s(0)、v(0) 值。舍入误差会累积；在最后一步之前保留精确答案。

When a function is piecewise or changes formula, treat each piece separately. For graph questions, label intercepts and turning points clearly to support your reasoning.

当函数为分段或公式变化时，分段处理。对于图像题，清晰标注截距和转向点以支持你的推理。

12. Quick Reference Summary | 速查总结

v = ds/dt, a = dv/dt = d²s/dt². Integrate a to get v, integrate v to get s. Constants determined by initial values. Turning points at v=0 (with sign change). Total distance = ∫ |v| dt. For a=f(v), separate variables. For a=f(s), use v dv = f(s) ds.

v = ds/dt，a = dv/dt = d²s/dt²。积分 a 得 v，积分 v 得 s。常数由初值确定。转向点在 v=0 且符号改变处。总路程 = ∫ |v| dt。a=f(v) 时分离变量，a=f(s) 时用 v dv = f(s) ds。

Draw or interpret graphs: slope of s–t is v; area under v–t is Δs. Remember that acceleration can be variable – do not assume constant acceleration unless stated.

绘制或解读图像：s–t 斜率即 v；v–t 下方面积为 Δs。记住加速度可能变化——除非明确说明，否则不要假定匀加速度。

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IB Maths: Kinematics Essentials | IB 数学：运动学 考点精讲