📚 Mastering IGCSE Maths: Practice Book 194 | 精通IGCSE数学：练习册194
This article breaks down a classic IGCSE Mathematics problem from Practice Book 194, focusing on kinematics – the study of motion. We will explore displacement-time graphs, velocity-time graphs, and the key calculations that appear in exams.
本文深入解析练习册194中的一道经典IGCSE数学题，重点讨论运动学——即对运动的研究。我们将探索位移-时间图像、速度-时间图像，以及考试中常见的关键计算方法。
1. The Problem Statement | 问题陈述
In Practice Book 194, a train starts from rest at station A, accelerates uniformly for 10 seconds, then travels at a constant speed for 30 seconds, and finally decelerates uniformly to rest at station B in 20 seconds. The acceleration during the first stage is 2 m/s². Find the total distance between the two stations.
练习册194中，一列火车从A站静止出发，匀加速行驶10秒，然后以恒定速度行驶30秒，最后匀减速行驶20秒在B站停下。第一阶段的加速度为2 m/s²。求两站之间的总距离。
2. Breaking Down the Motion | 分解运动过程
The journey consists of three distinct stages. In stage 1, the train accelerates from rest. In stage 2, it moves with constant velocity. In stage 3, it decelerates back to rest. Each stage must be analysed separately before combining results.
整个旅程包含三个不同阶段。第一阶段，火车从静止开始加速；第二阶段，以恒定速度运动；第三阶段，减速至静止。必须先分别分析每个阶段，再综合结果。
3. Stage 1: Uniform Acceleration | 第一阶段：匀加速运动
We use the equation v = u + at. Here u = 0 m/s (starts from rest), a = 2 m/s², t = 10 s. So v = 0 + 2 × 10 = 20 m/s. The final velocity of this stage is 20 m/s.
我们使用公式 v = u + at。此处 u = 0 m/s（从静止开始），a = 2 m/s²，t = 10 s。因此 v = 0 + 2 × 10 = 20 m/s。该阶段的末速度为20 m/s。
v = u + at → v = 0 + 2 × 10 = 20 m/s
4. Distance during Acceleration | 加速阶段的距离
The distance s₁ travelled during uniform acceleration is given by s = ut + ½at². Alternatively, it equals the area under the velocity-time graph. For stage 1, s₁ = 0 × 10 + ½ × 2 × 10² = 100 m.
匀加速阶段行驶的距离 s₁ 由 s = ut + ½at² 给出。或者，它等于速度-时间图像下的面积。对于第一阶段，s₁ = 0 × 10 + ½ × 2 × 10² = 100 m。
s₁ = ½ × 2 × 10² = 100 m
5. Stage 2: Constant Velocity | 第二阶段：匀速运动
After the first 10 seconds, the train moves at a constant speed of 20 m/s for 30 seconds. The distance s₂ is simply speed × time: s₂ = 20 × 30 = 600 m. No acceleration occurs here.
在最初10秒后，火车以20 m/s的恒定速度运动30秒。距离 s₂ 就是速度 × 时间：s₂ = 20 × 30 = 600 m。此阶段没有加速度。
s₂ = 20 × 30 = 600 m
6. Stage 3: Uniform Deceleration | 第三阶段：匀减速运动
The train decelerates from 20 m/s to 0 m/s in 20 seconds. The deceleration a₃ = (v – u)/t = (0 – 20)/20 = -1 m/s². The distance s₃ can be found using s = vt – ½at², but it is easier to use the area formula for a triangle: s₃ = ½ × base × height = ½ × 20 × 20 = 200 m.
火车在20秒内从20 m/s减速到0 m/s。减速度 a₃ = (v – u)/t = (0 – 20)/20 = -1 m/s²。距离 s₃ 可以用 s = vt – ½at² 求出，但更简单使用三角形面积公式：s₃ = ½ × 底 × 高 = ½ × 20 × 20 = 200 m。
s₃ = ½ × 20 × 20 = 200 m
7. Total Distance | 总距离
Add the distances from the three stages: s_total = s₁ + s₂ + s₃ = 100 + 600 + 200 = 900 m. Therefore, the distance between stations A and B is 900 metres.
将三阶段距离相加：s_total = s₁ + s₂ + s₃ = 100 + 600 + 200 = 900 m。因此，A站与B站之间的距离为900米。
s_total = 100 + 600 + 200 = 900 m
8. Velocity-Time Graph | 速度-时间图像
Plotting velocity on the y-axis and time on the x-axis gives a trapezoid. The first segment rises from (0,0) to (10,20). The second is horizontal from (10,20) to (40,20). The third falls from (40,20) to (60,0). The area under this graph equals the total distance (900 m), confirming our calculation.
以速度为纵轴、时间为横轴作图，得到梯形。第一段从(0,0)上升到(10,20)。第二段从(10,20)到(40,20)为水平直线。第三段从(40,20)下降到(60,0)。图像下的面积等于总距离（900 m），验证了我们的计算。
| Segment | Time (s) | Velocity (m/s) | Shape |
| 1 | 0 to 10 | 0 to 20 | Line sloping up |
| 2 | 10 to 40 | 20 | Horizontal line |
| 3 | 40 to 60 | 20 to 0 | Line sloping down |
9. Average Speed | 平均速度
The average speed is total distance divided by total time. Total time = 10 + 30 + 20 = 60 s. Average speed = 900 / 60 = 15 m/s. This is not simply the arithmetic mean of speeds because the train spends different times at different speeds.
平均速度等于总距离除以总时间。总时间 = 10 + 30 + 20 = 60 s。平均速度 = 900 / 60 = 15 m/s。这不是速度的简单算术平均值，因为火车在不同速度上花费的时间不同。
vavg = 900 ÷ 60 = 15 m/s
10. Displacement-Time Graph | 位移-时间图像
If we plot displacement against time, the graph is a curve for the first stage (quadratic), a straight line for the second stage (constant slope), and another curve for the third stage. The gradient at any point gives the instantaneous velocity.
如果以位移对时间作图，第一阶段是曲线（二次函数），第二阶段是直线（斜率恒定），第三阶段又是曲线。任意一点的切线斜率给出瞬时速度。
- Stage 1: gradient increases from 0 to 20 m/s – concave up.
- Stage 1中文：斜率从0增加到20 m/s，图像上凹。
- Stage 2: constant gradient = 20 m/s – straight line.
- Stage 2中文：斜率恒为20 m/s，是直线。
- Stage 3: gradient decreases back to 0 – concave down.
- Stage 3中文：斜率减小至0，图像下凹。
11. Common Pitfalls | 常见陷阱
Students often confuse velocity-time and displacement-time graphs. In a velocity-time graph, the gradient is acceleration, and the area is distance. In a displacement-time graph, the gradient is velocity, and there is no area interpretation.
学生经常混淆速度-时间图像和位移-时间图像。在速度-时间图像中，斜率是加速度，面积是距离；在位移-时间图像中，斜率是速度，没有面积的含义。
Another common error: forgetting that a deceleration is still an acceleration with a negative sign. Always define a positive direction and treat signs consistently.
另一个常见错误：忘记减速仍然是加速度，只是符号为负。始终定义正方向并保持符号一致。
12. Full Solution Summary | 完整解题总结
To solve any kinematics problem like Practice Book 194, follow these steps:
要解决像练习册194这样的运动学问题，请遵循以下步骤：
- Identify the stages and note known values (u, v, a, t, s).
- 确定各阶段并记录已知量（初速度、末速度、加速度、时间、距离）。
- Use v = u + at to find missing velocities or accelerations.
- 使用 v = u + at 求未知速度或加速度。
- Use s = ut + ½at² or area under the velocity-time graph to find distances.
- 使用 s = ut + ½at² 或速度-时间图像下的面积求距离。
- Add all distances to get the total.
- 将所有距离相加得到总量。
Final answer: Distance between stations = 900 m
最终答案：两站间距离 = 900 米
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