📚 Year 13 CIE Mathematics: Case Study Practice Drills | Year 13 CIE 数学：案例分析实战演练
Case study and modelling questions are an integral part of the CIE A-Level Mathematics syllabus, requiring you to bring together calculus, algebra, and interpretation skills in a single real-world scenario. This article walks through a detailed case study of a particle moving along a straight line, demonstrating how to approach such problems step by step. You will learn how to derive motion quantities, handle distances and optimisation, and avoid common errors—all essential for achieving high marks in your Year 13 exams.
案例分析及建模题目是 CIE A-Level 数学大纲的重要组成部分，要求你在一道真实情境题目中综合运用微积分、代数和解释能力。本文以一个沿直线运动的质点为例，详细展示如何一步步攻克这类问题。你将学会怎样推导运动量、处理路程与优化，并避开常见错误——这些对于在 Year 13 考试中取得高分至关重要。
1. Understanding the Case Study Scenario | 理解案例背景
The case study begins with a clearly defined physical situation. A particle P moves along a horizontal line. Its displacement s metres from a fixed origin O at time t seconds (t ≥ 0) is given by s = t³ – 6t² + 9t + 1. The problem asks for velocity and acceleration expressions, times when the particle is at rest, the total distance travelled in the first 4 seconds, the extreme values of velocity on [0,4], and a sketch of the velocity-time graph. Later, we extend this to minimise a cost function depending on velocity—a typical modelling twist.
本案例从一个明确定义的物理情境开始。质点 P 沿水平直线运动，其离开固定原点 O 的位移 s（米）由时间 t 秒（t ≥ 0）给定：s = t³ – 6t² + 9t + 1。问题要求求出速度和加速度表达式、质点瞬时静止的时刻、前 4 秒内通过的总路程、区间 [0,4] 上速度的极值，以及速度-时间草图。之后我们延伸出一个依赖于速度的成本函数最小化问题——典型的建模环节。
Before diving into calculations, always identify the independent variable (t), the dependent variable (s), and the domain of interest. Highlight key words like ‘displacement’, ‘distance’, ‘instantaneously at rest’, and ‘maximum velocity’ so you can select the correct mathematical tool for each part.
在开始计算之前，务必明确自变量（t）、因变量（s）以及考虑的时域。圈出 ‘displacement’、’distance’、’instantaneously at rest’、’maximum velocity’ 等关键词，以便为每一小问选择正确的数学工具。
2. Deriving Velocity and Acceleration | 推导速度与加速度
Velocity is the first derivative of displacement with respect to time, and acceleration is the second derivative. Differentiating s = t³ – 6t² + 9t + 1 gives the velocity function v(t) = 3t² – 12t + 9. Factorising this reveals v(t) = 3(t² – 4t + 3) = 3(t – 1)(t – 3). For acceleration, differentiate again: a(t) = 6t – 12 = 6(t – 2). Always simplify these expressions as fully as possible, because the factorised forms will help you find zeros and analyse sign changes later.
速度是位移关于时间的一阶导数，加速度则是二阶导数。对 s = t³ – 6t² + 9t + 1 求导，得到速度函数 v(t) = 3t² – 12t + 9。因式分解后为 v(t) = 3(t² – 4t + 3) = 3(t – 1)(t – 3)。继续求导得加速度 a(t) = 6t – 12 = 6(t – 2)。尽可能化简这些表达式，因为因式分解后的形式有助于后续寻找零点和分析符号变化。
v = ds/dt = 3t² – 12t + 9 = 3(t – 1)(t – 3)
a = dv/dt = 6t – 12 = 6(t – 2)
Write the final formulas clearly in your solution, and double-check your differentiation. A simple slip here can propagate errors through the entire case study. Remember that velocity can be positive or negative, indicating direction, while speed is the absolute value.
在解答中清晰地写出最终公式，并仔细核对导数。这里的一个小错误会蔓延到整个案例。请记住速度可以为正或负，代表运动方向，而速率是速度的绝对值。
3. Times When the Particle Is Instantaneously at Rest | 质点瞬时静止的时刻
A particle is instantaneously at rest when its velocity is zero. Set v(t) = 0: 3(t – 1)(t – 3) = 0, giving t = 1 s and t = 3 s. At both instants, the particle changes direction. Checking the sign of v on intervals (0,1), (1,3), and (3,4] confirms the direction of motion: for t < 1, v > 0 (moving away from origin); between 1 and 3, v < 0 (moving back towards origin); for t > 3, v > 0 again (moving away). This sign analysis is crucial for calculating total distance later.
质点瞬时静止时速度为零。令 v(t) = 0：3(t – 1)(t – 3) = 0，解得 t = 1 秒和 t = 3 秒。在这两个时刻质点均改变运动方向。检查 v 在区间 (0,1)、(1,3) 和 (3,4] 上的符号可以确定运动方向：t < 1 时 v > 0（背离原点移动）；1 到 3 之间 v < 0（向原点返回）；t > 3 时 v > 0（再次背离原点）。这种符号分析对后续计算总路程至关重要。
| Interval | t < 1 | 1 < t < 3 | t > 3 |
|---|---|---|---|
| Sign of v(t) | Positive | Negative | Positive |
| Direction | Away from O | Towards O | Away from O |
Many students mistakenly think a zero velocity implies the particle stops permanently—it is only an instantaneous change of direction. Always label these as ‘instantaneously at rest’ and describe the subsequent motion.
不少同学误以为速度为零意味着质点永久停止——那只是瞬时改变方向。解题时务必标注 ‘instantaneously at rest’ 并描述随后的运动状态。
4. Displacement vs Distance Travelled | 位移与路程
Displacement measures the net change in position from start to finish, while total distance accounts for the entire path length. For t from 0 to 4 seconds, the displacement is simply s(4) – s(0) = (64 – 96 + 36 + 1) – (1) = 4 m. Total distance, however, requires integrating the absolute value of velocity. Since the velocity changes sign at t=1 and t=3, split the integral into three parts: from 0 to 1, v is positive; from 1 to 3, v is negative (take the negative of the integral to get distance); from 3 to 4, v is positive again.
位移衡量从起点到终点位置的净变化，而总路程则考虑完整的运动路径长度。在 0 到 4 秒内，位移仅由 s(4) – s(0) = (64 – 96 + 36 + 1) – (1) = 4 米得出。然而总路程需要积分速度的绝对值。因为速度在 t=1 和 t=3 时变号，因此将积分分成三段：从 0 到 1，v 为正；1 到 3，v 为负（取积分的相反数以获得路程）；3 到 4，v 再次为正。
Distance = ∫₀¹ v(t) dt + ∫₁³ (-v(t)) dt + ∫₃⁴ v(t) dt
Calculate each definite integral: ∫ v dt = [t³ – 6t² + 9t] (since antiderivative of 3t²-12t+9 is t³-6t²+9t). The distance from 0 to 1 is [1 – 6 + 9] – 0 = 4 m. From 1 to 3, using the same antiderivative, (27 – 54 + 27) – (1 – 6 + 9) = 0 – 4 = -4; the absolute contribution is 4 m. From 3 to 4: (64 – 96 + 36) – 0 = 4 m. Summing gives a total distance of 4 + 4 + 4 = 12 m. Note that the net displacement of 4 m is much smaller than the total path length—a perfect illustration of why distance and displacement must not be confused.
计算每个定积分：∫ v dt = [t³ – 6t² + 9t]（因为 3t²-12t+9 的原函数是 t³-6t²+9t）。0 到 1 段的路程为 [1 – 6 + 9] – 0 = 4 米。1 到 3 段： (27 – 54 + 27) – (1 – 6 + 9) = 0 – 4 = -4，取绝对值得 4 米。3 到 4 段： (64 – 96 + 36) – 0 = 4 米。总路程为 4 + 4 + 4 = 12 米。注意到净位移只有 4 米，远小于总路程——充分说明为何不能将位移与路程混为一谈。
5. Maximum and Minimum Velocity on [0,4] | 在区间 [0,4] 上的最大与最小速度
To find the extreme values of a continuous function on a closed interval, check stationary points inside the interval and the endpoints. Here the velocity function is v(t) = 3t² – 12t + 9. Set its derivative, a(t) = 6t – 12, equal to zero: t = 2. Evaluate v at t = 0, 2, and 4: v(0) = 9, v(2) = 3(4) – 24 + 9 = -3, v(4) = 3(16) – 48 + 9 = 9. Therefore, on [0,4], the maximum velocity is 9 m/s (occurring at t=0 and t=4), and the minimum velocity is -3 m/s (at t=2 s).
要寻找闭区间上连续函数的极值，需检查区间内的驻点与端点。这里速度函数为 v(t) = 3t² – 12t + 9。令其导数 a(t) = 6t – 12 为零，得 t = 2。分别计算 t = 0, 2, 4 处的 v 值：v(0) = 9, v(2) = 3(4) – 24 + 9 = -3, v(4) = 3(16) – 48 + 9 = 9。因此在 [0,4] 上，最大速度为 9 米/秒（出现在 t=0 和 t=4），最小速度为 -3 米/秒（在 t=2 秒）。
Notice that the ‘maximum’ velocity refers to the greatest value, which is positive, while the ‘minimum’ velocity is the most negative. If the question asks for maximum speed, you would take absolute values, which would give 9 m/s as well, but the time of maximum speed would be at endpoints. Always read the wording carefully.
请注意，这里的 ‘maximum’ velocity 指代数值上的最大值，是正数，而 ‘minimum’ velocity 是数值上的最小值，是负数。如果题目问到最大速率（maximum speed），则需取绝对值，结果仍是 9 米/秒，但出现时刻仍在端点。务必仔细审题。
6. Sketching and Interpreting the Velocity-Time Graph | 速度-时间图的绘制与解读
A velocity-time graph helps visualise the motion. v(t) = 3t² – 12t + 9 is a concave-up parabola with intercepts at t=1 and t=3, and vertex at t=2, v=-3. Plot key points: (0,9), (1,0), (2,-3), (3,0), (4,9). The graph lies above the t-axis for t<1 and t>3, and below for 1 速度-时间图有助于直观感受运动过程。v(t) = 3t² – 12t + 9 是一条开口向上的抛物线，与 t 轴交于 t=1 和 t=3，顶点在 t=2 处 v=-3。标出关键点：(0,9), (1,0), (2,-3), (3,0), (4,9)。当 t<1 和 t>3 时图像位于 t 轴上方，1 Interpretation bridges mathematics and real motion. From t=0 to 1, the particle moves away with decreasing speed; it stops at t=1, then reverses direction from t=1 to 3, reaching maximum speed away at t=2 but moving back towards O; then it stops again at t=3 and moves away once more. Writing a concise description alongside your graph earns valuable communication marks in CIE exams. 解读环节将数学与现实运动衔接起来。从 t=0 到 1，质点以递减的速率背离原点；在 t=1 处静止，随后 1 到 3 秒反向运动，t=2 时反向速率最大，且返回靠近原点；接着在 t=3 处再次静止并重新背离原点。在图像旁附上一段简洁的描述能为你在 CIE 考试中赢得珍贵的表达分。 Real-world case studies often ask you to adapt the model. Imagine the engine power cost C (in suitable units) depends on velocity as C = v² – 6v + 10. Replacing v with the earlier expression gives C(t) = (3t² – 12t + 9)² – 6(3t² – 12t + 9) + 10. While this can be expanded and differentiated, a smarter approach is to recognise that C = (v – 3)² + 1. Since the squared term is non-negative, C is minimised when v(t) is as close to 3 as possible. Solve v(t) = 3: 3t² – 12t + 9 = 3 → 3t² – 12t + 6 = 0 → t² – 4t + 2 = 0, giving t = 2 ± √2. Both times lie within [0,4]: t ≈ 0.586 s and t ≈ 3.414 s. At these instants, the cost drops to its minimum value of 1. 真实世界的案例分析常常要求你调整模型。假设引擎功耗成本 C（适当单位）依赖于速度，关系为 C = v² – 6v + 10。将 v 代入之前的表达式得 C(t) = (3t² – 12t + 9)² – 6(3t² – 12t + 9) + 10。虽然可以展开再求导，但更聪明的做法是配方：C = (v – 3)² + 1。由于平方式非负，当 v(t) 尽可能接近 3 时 C 最小。解 v(t) = 3：3t² – 12t + 9 = 3 → 3t² – 12t + 6 = 0 → t² – 4t + 2 = 0，得 t = 2 ± √2。这两个时刻均在 [0,4] 内：约 t = 0.586 秒与 t = 3.414 秒。此时成本降至最小值 1。 This extension demonstrates how optimisation links to kinematics. Instead of mechanically differentiating a complicated composite function, we exploited the structure of the problem. Always look for such simplifications—they save time and reduce algebraic errors. The final answer should include both the times and the minimum cost, and a brief interpretation (e.g., ‘the engine operates most efficiently at these two instants when velocity equals 3 m/s’). 这一扩展展示了优化如何与运动学挂钩。我们没有机械地对复杂复合函数求导，而是利用了问题的结构。时常寻找此类简化方法——它们能节省时间并减少代数错误。最终答案应给出对应的时刻和最小成本，并附上简短说明（例如 ‘当速度等于 3 米/秒时发动机效率最高，出现于两个瞬时’）。 When tackling CIE case study questions, watch out for these frequent mistakes: (1) Forgetting to split integrals when velocity changes sign when computing distance; (2) Confusing displacement with distance; (3) Neglecting to check endpoints for maxima/minima; (4) Misinterpreting a negative velocity as speed; (5) Failing to simplify derivatives before solving equations; (6) Not providing final interpretation or units. To counter these, always read the question twice, underline action verbs (find, show, hence, evaluate), and present your working logically. 在应对 CIE 案例分析题时，请警惕这些常见失误：(1) 计算路程时忘记在速度变号处分解积分；(2) 混淆位移与路程；(3) 求极值时忽略检查端点；(4) 将负的速度误解为速率；(5) 解方程前没有先化简导数；(6) 未提供最终解释或单位。对策是：仔细审题两遍，划出行动词（find, show, hence, evaluate），并逻辑清晰地进行验算。 Time management is also key. Allocate roughly 1 minute per mark, and if a part stumps you, move on and return later. In the CIE exam, modelling questions often carry high marks but are accessible if you break them into small, manageable steps—exactly as demonstrated in this drill. Practising with past papers under timed conditions will boost both speed and confidence. 时间管理同样关键。大约按 1 分钟 1 分的节奏分配时间，若被某一小问卡住，先跳过再回来。在 CIE 考试中，建模题往往分值高，但只要你将其拆解为可控的小步骤——正如本次演练所示——就能从容应对。用真题在限时条件下勤加练习，将显著提升速度和信心。 Published by TutorHao | Mathematics Revision Series | aleveler.com 更多咨询请联系16621398022（同微信）
7. Extending the Model: Optimising a Cost Function | 模型扩展：优化成本函数
8. Common Pitfalls and Exam Strategies | 常见失误与应试策略
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导