WJEC A2 Mathematics: Case Study in Action | WJEC A2 数学：案例分析实战演练

📚 WJEC A2 Mathematics: Case Study in Action | WJEC A2 数学：案例分析实战演练

In Year 13 WJEC Mathematics, the ability to apply pure, mechanics and statistics concepts to real-world scenarios is essential for top marks. This article presents a series of case study exercises that mirror the extended problem-solving questions found in A2 examinations. Each case study is broken down into modelling, mathematical manipulation and interpretation, helping you build confidence and fluency.

在WJEC A2数学中，将纯数、力学和统计概念应用于真实情境的能力是取得高分的关键。本文提供一系列案例分析练习，模拟A2考试中的长题。每个案例分为建模、数学运算和解释，帮助你建立信心与熟练度。

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1. Kinematics with Variable Acceleration | 运动学与可变加速度

the times when the particle is at rest, the acceleration at those times, and the total distance travelled in the first 5 seconds.

一质点沿直线运动，离固定原点位移 s 米由 s(t) = t³ − 6t² + 9t 给出 (t ≥ 0)。求质点静止的时刻、此时的加速度以及前5秒内的总路程。

Step 1: Differentiate s(t) to obtain velocity v and acceleration a. v = ds/dt = 3t² − 12t + 9 and a = dv/dt = 6t − 12.

步骤1：对 s(t) 求导得到速度 v 和加速度 a。v = ds/dt = 3t² − 12t + 9，a = dv/dt = 6t − 12。

Step 2: The particle is at rest when v = 0. Solving 3t² − 12t + 9 = 0 ⇒ t² − 4t + 3 = 0 ⇒ (t − 1)(t − 3) = 0, so t = 1 s and t = 3 s.

步骤2：质点静止时 v = 0。解 3t² − 12t + 9 = 0 ⇒ t² − 4t + 3 = 0 ⇒ (t − 1)(t − 3) = 0，得 t = 1 s 和 t = 3 s。

celeration at rest: at t = 1, a = 6(1) − 12 = −6 m s⁻²; at t = 3, a = 6(3) − 12 = 6 m s⁻².

步骤3：求静止时的加速度：t = 1 时 a = 6(1) − 12 = −6 m s⁻²；t = 3 时 a = 6(3) − 12 = 6 m s⁻²。

s(0)=0, s(1)=4, s(3)=0, s(5)=20. The particle changes direction at t=1 and t=3. Distance = |4−0| + |0−4| + |20−0| = 4 + 4 + 20 = 28 m.

步骤4：求路程，计算关键位移：s(0)=0, s(1)=4, s(3)=0, s(5)=20。质点在 t=1 和 t=3 改变方向。路程 = |4−0| + |0−4| + |20−0| = 4 + 4 + 20 = 28 m。

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2. Exponential Models: Population Growth & Radioactive Decay | 指数模型：人口增长与放射性衰变

A bacteria culture initially contains 500 bacteria. The rate of growth is proportional to the number present. After 3 hours, the count rises to 2000. Find the number after 5 hours and the doubling time.

某细菌培养液最初含有500个细菌，增长速度与当前数量成正比。3小时后计数为2000。求5小时后的数量以及倍增时间。

Let N(t) = N₀e^kt. With N₀ = 500, N(3) = 500 e^3k = 2000 ⇒ e^3k = 4 ⇒ 3k = ln 4 ⇒ k = (1/3) ln 4 ≈ 0.4621.

设 N(t) = N₀e^kt。N₀ = 500，N(3) = 500 e^3k = 2000 ⇒ e^3k = 4 ⇒ 3k = ln 4 ⇒ k = (1/3) ln 4 ≈ 0.4621。

After 5 hours: N(5) = 500 e^5k = 500 e^{(5/3) ln 4} = 500 × 4^5/3 ≈ 500 × 8.00 = 4000 bacteria.

5小时后：N(5) = 500 e^5k = 500 e^{(5/3) ln 4} = 500 × 4^5/3 ≈ 500 × 8.00 = 4000 个细菌。

Doubling time T: e^kT = 2 ⇒ T = (ln 2)/k = (ln 2) / ( (1/3) ln 4 ) = (3 ln 2) / (2 ln 2) = 1.5 hours.

倍增时间 T：e^kT = 2 ⇒ T = (ln 2)/k = (ln 2) / ((1/3) ln 4) = (3 ln 2)/(2 ln 2) = 1.5 小时。

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3. Statistical Testing: Binomial Hypothesis Test | 统计检验：二项分布假设检验

A manufacturer claims that only 10% of its light bulbs are defective. In a sample of 20 bulbs, 4 are defective. Test at the 5% significance level whether the true proportion of defectives is greater than 10%.

某制造商声称灯泡次品率仅10%。在20个灯泡样本中有4个次品。在5%显著性水平下检验真实次品率是否大于10%。

Let X ~ 0.1). H₀: p = 0.1, H₁: p > 0.1. Test statistic: observed defective count = 4. Find P(X ≥ 4 | p=0.1).

设 X ~ B(20, 0.1)。H₀: p = 0.1，H₁: p > 0.1。检验统计量：观测次品数 = 4。计算 P(X ≥ 4 | p=0.1)。

)=190×0.1²×0.9¹⁸≈0.2852, P(X=3)=1140×0.1³×0.9¹⁷≈0.1901. Sum ≈0.8670, so P(X ≥ 4) ≈ 0.1330.

P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − [P(X=0)+P(X=1)+P(X=2)+P(X=3)]。公式：P(X=0)=0.9²⁰≈0.1216，P(X=1)=20×0.1×0.9¹⁹≈0.2702，P(X=2)=190×0.1²×0.9¹⁸≈0.2852，P(X=3)=1140×0.1³×0.9¹⁷≈0.1901。总和≈0.8670，所以 P(X ≥ 4) ≈ 0.1330。

Since 0.133 > 0.05, we do not reject H₀. There is insufficient evidence that the defective rate exceeds 10%.

由于 0.133 > 0.05，不拒绝 H₀。没有充分证据表明次品率超过10%。

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4. Newton's Second Law on an Inclined Plane | 牛顿第二定律与斜面

ension T. If the block accelerates at 1.2 m s⁻² up the plane, find T. (Take g = 9.8 m s⁻²)

质量5 kg的物块置于光滑斜面（与水平成30°）上，用平行于斜面的绳子以拉力 T 沿斜面向上拉。若物块以1.2 m s⁻²向上加速，求 T。(g = 9.8 m s⁻²)

ine: component of weight down slope = mg sin 30° = 5 × 9.8 × 0.5 = 24.5 N. Applying Newton's second law: T − 24.5 = 5 × 1.2 ⇒ T = 6 + 24.5 = 30.5 N.

分解平行于斜面的力：重力沿斜面的分量 = mg sin 30° = 5 × 9.8 × 0.5 = 24.5 N。应用牛顿第二定律：T − 24.5 = 5 × 1.2 ⇒ T = 6 + 24.5 = 30.5 N。

iction with coefficient μ = 0.2 were present, normal reaction R = mg cos 30° = 5 × 9.8 × (√3/2) ≈ 42.44 N. Friction F = μR ≈ 8.49 N. Then T − 24.5 − 8.49 = 6 ⇒ T = 39.0 N (approx).

若存在摩擦系数 μ = 0.2，法向反力 R = mg cos 30° = 5 × 9.8 × (√3/2) ≈ 42.44 N。摩擦力 F = μR ≈ 8.49 N。则 T − 24.5 − 8.49 = 6 ⇒ T ≈ 39.0 N。

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5. Pure Mathematics: Proof by Contradiction | 纯数：反证法证明

Prove by contradiction that √2 is irrational.

用反证法证明 √2 是无理数。

Assume √2 is rational, so √2 = a/b where a, b are integers with no common factors (b ≠ 0). Squaring gives 2 = a²/b² ⇒ a² = 2b².

假设 √2 是有理数，则 √2 = a/b，其中 a 和 b 为互质整数 (b ≠ 0)。平方得 2 = a²/b² ⇒ a² = 2b²。

a² is even, so a must be even. Write a = 2k for some integer k. Substituting: (2k)² = 2b² ⇒ 4k² = 2b² ⇒ b² = 2k². Hence b² is even, so b is even.

因此 a² 为偶数，则 a 必为偶数。令 a = 2k。代入：(2k)² = 2b² ⇒ 4k² = 2b² ⇒ b² = 2k²。所以 b² 为偶数，b 也为偶数。

Both a and b are even, contradicting the assumption that a/b is in its simplest form. Hence √2 cannot be rational; it is irrational.

a 和 b 均为偶数，与 a/b 为最简分数矛盾。因此 √2 不可能是有理数，是无理数。

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6. Differential Equations: Newton's Law of Cooling | 微分方程：冷却定律

A metal rod is heated to 90 °C and left in a room at 20 °C. After 5 minutes its temperature is 60 °C. Assuming Newton's law of cooling, find the temperature after 10 minutes and the time to reach 30 °C.

一金属棒加热至90 °C后置于20 °C室内。5分钟后温度为60 °C。根据牛顿冷却定律，求10分钟后的温度和达到30 °C的时间。

Let θ be temperature excess: θ(t) = T(t) − 20. Then dθ/dt = −kθ, so θ = θ₀e^−kt. Initially θ₀ = 70. After 5 min, θ = 40, so 40 = 70 e^−5k ⇒ e^−5k = 4/7 ⇒ −5k = ln(4/7) ⇒ k = −(1/5) ln(4/7) ≈ 0.1116.

设剩余温度 θ(t) = T(t) − 20，则 dθ/dt = −kθ，解得 θ = θ₀e^−kt。初始 θ₀ = 70。5分钟后 θ = 40，故 40 = 70 e^−5k ⇒ e^−5k = 4/7 ⇒ −5k = ln(4/7) ⇒ k ≈ 0.1116。

At t = 10 min: θ = 70 e^−10k = 70(e^−5k)² = 70 × (4/7)² ≈ 70 × 16/49 ≈ 22.86 °C. Temperature T = 20 + 22.86 = 42.9 °C.

t = 10 min：θ = 70 e^−10k = 70(e^−5k)² = 70 × (4/7)² ≈ 70 × 16/49 ≈ 22.86 °C。温度 T = 20 + 22.86 = 42.9 °C。

For T = 30 °C, θ = 10. 10 = 70 e^−kt ⇒ e^−kt = 1/7 ⇒ −kt = ln(1/7) = −ln 7 ⇒ t = (ln 7)/k ≈ 1.9459 / 0.1116 ≈ 17.4 minutes.

当 T = 30 °C，θ = 10。10 = 70 e^−kt ⇒ e^−kt = 1/7 ⇒ t = (ln 7)/k ≈ 1.9459 / 0.1116 ≈ 17.4 分钟。

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7. Parametric Equations and Projectiles | 参数方程与抛体运动

of motion are x = 20t, y = 15t − 4.9t². Find the maximum height, time of flight and range.

一抛体从地面以速率25 m s⁻¹、角度 α (tan α = 3/4)发射。运动参数方程为 x = 20t，y = 15t − 4.9t²。求最大高度、飞行时间和射程。

Maximum height when dy/dt = 0: 15 − 9.8t = 0 ⇒ t = 15/9.8 ≈ 1.5306 s. Then y = 15(1.5306) − 4.9(1.5306)² ≈ 22.96 − 11.48 = 11.48 m.

最大高度时 dy/dt = 0：15 − 9.8t = 0 ⇒ t ≈ 1.5306 s。此时 y = 15(1.5306) − 4.9(1.5306)² ≈

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