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IB Math: Using Integration to Solve Kinematics Problems | IB数学：用积分解决运动学问题

2026-08-30 15:09 · 本文部分内容由AI辅助创作

📚 IB Math: Using Integration to Solve Kinematics Problems | IB数学：用积分解决运动学问题

Kinematics describes the motion of objects without considering the forces that cause that motion. In IB Mathematics, integration is a powerful tool for working backwards from acceleration to velocity and displacement, and for calculating exact values such as total distance traveled. This article explains the key rules, common traps, and exam-style techniques you need to master.

运动学研究物体的运动而不考虑引起运动的力。在IB数学中，积分是一个强大的工具，可以从加速度反推速度和位移，并计算总路程等精确数值。本文将讲解关键法则、常见陷阱以及你需要在考试中掌握的技巧。

1. Displacement vs Distance | 位移与路程

Displacement is a vector quantity that measures the straight-line change in position from the starting point. It can be positive, negative, or zero.

位移是矢量，衡量从起点到终点的直线位置变化，可以是正、负或零。

Distance is a scalar quantity that measures the total length of the path traveled. It is always non-negative.

路程是标量，衡量经过的路径总长度，永远非负。

For example, a runner who completes one lap around a 400 m track has a displacement of 0 m but has traveled a distance of 400 m.

例如，一名跑者沿400米跑道跑完一圈，位移为0米，但路程为400米。

This difference is essential when interpreting velocity and speed: velocity is the derivative of displacement, while speed is the derivative of distance.

这一区别在理解速度与速率时至关重要：速度是位移的导数，而速率是路程的导数。

2. The Fundamental Relations | 基本关系

If s(t) is displacement, v(t) is velocity, and a(t) is acceleration at time t, then calculus gives the following relations:

设s(t)为位移，v(t)为速度，a(t)为t时刻的加速度，则微积分给出以下关系：

v(t) = ds/dt, a(t) = dv/dt = d²s/dt²

Velocity is the gradient of the displacement-time graph, and acceleration is the gradient of the velocity-time graph.

速度是位移-时间图像的斜率，加速度是速度-时间图像的斜率。

By the Fundamental Theorem of Calculus, integration reverses differentiation, so we can recover velocity from acceleration and displacement from velocity.

由微积分基本定理，积分是微分的逆运算，因此我们可以由加速度恢复速度，由速度恢复位移。

3. Integration of Acceleration to Velocity | 由加速度积分求速度

Given acceleration a(t), the velocity satisfies v(t) = ∫a(t)dt.

已知加速度a(t)，速度满足v(t) = ∫a(t)dt。

Since an indefinite integral produces a constant of integration C, you need an initial condition such as v(0) to determine C.

由于不定积分会产生积分常数C，你需要一个初始条件（如v(0)）来确定C。

Example: If a(t) = 6t − 4 and v(0) = 5, then:

例：若a(t) = 6t − 4且v(0) = 5，则：

v(t) = ∫(6t − 4)dt = 3t² − 4t + C; v(0)=5 ⇒ C=5

Therefore v(t) = 3t² − 4t + 5.

因此v(t) = 3t² − 4t + 5。

the initial velocity is u, then C = u, so v = u + at, one of the familiar SUVAT equations.

对于匀加速度a，积分得到v = at + C。若初速度为u，则C = u，于是v = u + at，这是常见的SUVAT方程之一。

4. Integration of Velocity to Displacement | 由速度积分求位移

Similarly, displacement is given by s(t) = ∫v(t)dt, with an initial position s(0).

类似地，位移由s(t) = ∫v(t)dt给出，并需初始位置s(0)。

mple, if v(t) = 10t − t² + 3, then the general displacement is s(t) = 5t² − t³/3 + 3t + C.

例如，若v(t) = 10t − t² + 3，则一般位移为s(t) = 5t² − t³/3 + 3t + C。

This gives the net position of the particle relative to its starting point, not the total path length.

这给出质点相对于起点的净位置，而不是路径总长度。

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5. Definite Integrals: Change in Displacement | 定积分与位移变化

The definite integral of velocity over a time interval gives the net change in displacement:

速度在时间区间上的定积分给出位移的净变化：

Δs = ∫ₐᵇ v(t)dt = s(b) − s(a)

Note that this value can be negative if the particle moves back toward the starting point.

注意，如果质点向起点方向运动，这个值可能为负。

Example: If v(t) = 3t² − 2, then the change in displacement from t = 1 to t = 3 is:

例：若v(t) = 3t² − 2，则从t = 1到t = 3的位移变化为：

∫₁³ (3t² − 2)dt = [t³ − 2t]₁³ = (27 − 6) − (1 − 2) = 22

This value is the final displacement minus initial displacement, which can be less than the total distance traveled if the particle changes direction.

该值是末位移减去初位移，如果质点改变方向，它可能小于总路程。

6. Initial Conditions and Constant of Integration | 初始条件与积分常数

IB examiners often expect you to write the general integrated expression first, then substitute the initial condition.

IB考官通常希望你先写出积分后的通式，再代入初始条件。

For example, from a(t) = 12t² + 2, we get v(t) = 4t³ + 2t + C. If v(1) = 10, then C = 4.

例如，由a(t) = 12t² + 2，得到v(t) = 4t³ + 2t + C。若v(1) = 10，则C = 4。

Forgetting C is one of the most common mistakes on IB paper questions.

忘记常数C是IB笔试中最常见的错误之一。

For displacement, the same idea applies: after integrating v(t), use s(0) to find the second constant.

对于位移也是如此：对v(t)积分后，利用s(0)求出第二个常数。

7. Distance Traveled: Integrating Absolute Value | 路程：对绝对值积分

To find the total distance traveled, integrate the speed (magnitude of velocity):

要求总路程，应对速率（速度的大小）积分：

Distance = ∫ₐᵇ |v(t)| dt

locating the times when v(t) = 0, splitting the interval, and summing the absolute areas.

这需要找出v(t) = 0的时刻，将区间分段，并将各段面积的绝对值相加。

Follow these steps:

按以下步骤操作：