AQA Year 13 Mathematics: Case Study in Action | AQA 高三年级数学：案例分析实战演练

📚 AQA Year 13 Mathematics: Case Study in Action | AQA 高三年级数学：案例分析实战演练

In this in-depth case study, we will model the acceleration of a car subject to engine thrust and air resistance. The problem integrates pure mathematics (differential equations, separable ODEs, analytical solution) with numerical methods (Euler's method), reflecting the AQA Year 13 specification. You will see how a single realistic scenario can span multiple topic areas and develop the practical problem‑solving skills essential for the exam.

在这个深度案例分析中，我们将对一辆受到引擎推力和空气阻力作用的汽车加速过程进行建模。该问题融合了纯数学（微分方程、可分离常微分方程、解析解）与数值方法（欧拉法），紧密贴合 AQA Year 13 的考试要求。你将看到一个真实的场景如何横跨多个知识板块，并培养对考试至关重要的实际问题解决能力。

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1. Problem Scenario: Car Acceleration | 问题场景：汽车加速

iences air resistance (drag) that is directly proportional to its velocity. The drag coefficient is k = 200 N s m⁻¹. At time t = 0 s, the car starts from rest. We want to find how the velocity v (in m s⁻¹) varies with time t (in seconds), both analytically and using a step‑by‑step numerical approximation.

一辆质量为 1000 kg 的汽车在平直水平路面上加速。引擎提供恒定的前向驱动力 4000 N。随着车速增加，汽车受到与速度成正比的空气阻力，阻力系数为 k = 200 N s m⁻¹。在 t = 0 s 时，汽车从静止开始运动。我们需要求出速度 v（m s⁻¹）随时间 t（秒）的变化规律，既要通过解析方法，也要用逐步数值逼近求解。

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2. Formulating the Differential Equation | 建立微分方程

Applying Newton' nce, the net force is 4000 − 200v. With m = 1000 kg and acceleration a = dv/dt, we obtain:

沿运动方向应用牛顿第二定律：合外力 = 质量 × 加速度。大小为 200v 的阻力与运动方向相反。因此合外力为 4000 − 200v。已知 m = 1000 kg，加速度 a = dv/dt，可列出方程：

1000 dv/dt = 4000 − 200v

Dividing through by 1000 yields the core governing differential equation:

两边同除以 1000，得到核心的控制微分方程：

dv/dt = 4 − 0.2v

The initial condition is v(0) = 0. This is a first‑order linear ordinary differential equation that can be solved by separation of variables.

初始条件为 v(0) = 0。这是一个一阶线性常微分方程，可用分离变量法求解。

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3. Analytical Solution by Separation of Variables | 分离变量法求解析解

Rewrite the equation to isolate v and t:

重写方程，将 v 和 t 分离：

1/(4 − 0.2v) dv = dt

Integrate both sides. The left‑hand integral leads to a natural logarithm:

对两边积分。左边的积分会产生自然对数：

∫ 1/(4 − 0.2v) dv = ∫ dt

−5 ln|4 − 0.2v| = t + C

ine the constant C. Since 4 − 0 = 4, we have −5 ln 4 = C. Substitute back and solve for v:

代入初始条件 t = 0 时 v = 0 确定常数 C。因为 4 − 0 = 4，有 −5 ln 4 = C。代回并解出 v：

−5 ln|4 − 0.2v| = t − 5 ln 4

ln|4 − 0.2v| = ln 4 − 0.2t

4 − 0.2v = 4 e⁻⁰·²ᵗ

Finally, rearrange to obtain the explicit velocity function:

最后整理得到显式的速度函数：

v(t) = 20(1 − e⁻⁰·²ᵗ)

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4. Understanding the Terminal Velocity | 理解终端速度

nd drag exactly balance: 4000 N = 200 N s m⁻¹ × 20 m s⁻¹. For any initial speed, the car would asymptotically approach this limit.

当 t → ∞ 时，指数项 e⁻⁰·²ᵗ 衰减为零，因此 v(t) 趋近于 20 m s⁻¹。这就是终端（极限）速度，此时驱动力与阻力恰好平衡：4000 N = 200 N s m⁻¹ × 20 m s⁻¹。无论初速如何，汽车都会渐近地趋于这个极限。

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5. Introduction to Euler's Numerical Method | 欧拉数值方法介绍

When an analytical solution is difficult or impossible, Euler' uation dv/dt = f(t, v) and a starting point (t₀, v₀), we choose a step size h and iterate:

当解析解难以求得或不存在时，欧拉法提供了一种简单的数值逼近。对于微分方程 dv/dt = f(t, v) 和初始点 (t₀, v₀)，我们选取步长 h，并重复迭代：

vₙ₊₁ = vₙ + h f(tₙ, vₙ)

tₙ₊₁ = tₙ + h

nces the solution by using the slope at the current point to estimate the next value. The smaller the step size h, the closer the approximation to the true solution (but at the cost of more steps).

此处 f(t, v) = 4 − 0.2v。该方法利用当前点的斜率来估计下一个值，从而推进解。步长 h 越小，逼近值越接近真实解（但需要更多步数）。

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6. Implementing Euler's Method Step‑by‑Step | 逐步实施欧拉法

Let us apply Euler' te the first five steps and compare them with the exact values from v(t) = 20(1 − e⁻⁰·²ᵗ).

我们用步长 h = 1 s 来应用欧拉法，从 t₀ = 0, v₀ = 0 开始。递推公式为 vₙ₊₁ = vₙ + 1×(4 − 0.2vₙ)。我们将计算前五步，并与精确值 v(t) = 20(1 − e⁻⁰·²ᵗ) 进行对比。

The table below summarises the results (all velocities in m s⁻¹, rounded to four decimal places).

下表汇总了计算结果（所有速度单位 m s⁻¹，保留四位小数）。

t (s)	vexact	vEuler (h=1)	Absolute error
0	0	0	0
1	3.6254	4.0000	0.3746
2	6.5936	7.2000	0.6064
3	9.0238	9.7600	0.7362
4	11.0134	11.8080	0.7946
5	12.6424	13.4464	0.8040

Even with this relatively large step size, Euler's method captures the trend but overestimates the velocity. The error grows with time because the method uses the slope from the beginning of each interval, which over‑predicts the acceleration when the drag is increasing.

即便在这个较大的步长下，欧拉法也能捕捉到趋势，但会高估速度。误差随时间增大，因为该方法用的是每个区间起点的斜率，而当阻力增加时，该斜率会高估

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