📚 Case Study Practice for Year 10 Edexcel Physics | Year 10 Edexcel 物理：案例分析实战演练
This case study approach shows how to combine concepts of motion, forces, energy, and graph interpretation – just as you will in Edexcel Year 10 Physics exams. We walk through a realistic braking scenario, solving it step by step to build your confidence and problem-solving routine.
本案例分析展示了如何将运动、力、能量和图像解读这些概念综合运用——就像你在 Edexcel Year 10 物理考试中要做的那样。我们通过一个真实的制动情景，逐步求解，帮你树立信心并形成解题习惯。
1. Case Scenario | 案例场景
A driver in a 1200 kg car travels at a constant 20 m/s on a straight road. A hazard appears 45 m ahead. The driver's reaction time is 0.7 s, after which the brakes produce a constant deceleration of 6 m/s². The key question: will the car stop before hitting the hazard?
一位驾驶员驾驶一辆 1200 kg 的汽车，在笔直的道路上以 20 m/s 匀速行驶。前方 45 m 处突然出现障碍物。驾驶员的反应时间为 0.7 s，随后刹车产生恒定的 6 m/s² 减速度。关键问题：汽车能在撞上障碍物前停下来吗？
2. Identifying Key Variables | 识别关键变量
tities: initial speed u = 20 m/s, final speed v = 0 m/s, mass m = 1200 kg, reaction time tr e to hazard = 45 m. Using standard notation keeps your working clear and reduces mistakes.
先列出已知量：初速度 u = 20 m/s，末速度 v = 0 m/s，质量 m = 1200 kg，反应时间 tr = 0.7 s，加速度 a = −6 m/s²，障碍物距离 = 45 m。使用标准符号能让你的解题过程清晰，减少错误。
3. Speed and Kinetic Energy | 速度与动能换算
Before braking, the car possesses kinetic energy Ek = ½ m u². Substituting gives Ek his entire amount of energy must be removed by the braking force to bring the car to rest.
制动前，汽车具有动能 Ek = ½ m u²。代入数值：Ek = ½ × 1200 × (20)² = 600 × 400 = 240 000 J。要想让汽车停下来，制动力必须消除全部这些能量。
4. Calculating Thinking Distance | 反应距离计算
During the reaction time the car continues at constant speed. Thinking distance = u × tr = 20 m/s × 0.7 s = 14 m. So the car covers 14 metres before braking even begins – a crucial part of the stopping process.
在反应时间内，汽车保持原速前进。反应距离 = u × tr = 20 m/s × 0.7 s = 14 m。因此，在真正开始刹车之前，汽车已经前进了 14 米——这是停车过程中关键的一环。
5. Braking Distance and Deceleration | 制动距离与减速
= 400 → s = 400/12 ≈ 33.3 m. This is the distance travelled while the brakes are applied.
使用方程 v² = u² + 2 a s，其中 s 是制动距离。代入：0 = (20)² + 2 × (−6) × s → 0 = 400 − 12 s → 12 s = 400 → s = 400/12 ≈ 33.3 m。这就是施加刹车后汽车移动的距离。
6. Total Stopping Distance | 总停车距离
3 m = 47.3 m. Because the hazard is only 45 m away, the car will collide with the obstacle – it needed at least 47.3 m to stop safely.
总停车距离 = 反应距离 + 制动距离 = 14 m + 33.3 m = 47.3 m。由于障碍物仅在 45 m 外，汽车将会撞上——它至少需要 47.3 m 才能安全停下。
7. Analysis of Forces and Friction | 力与摩擦的分析
The decelerating force is friction between tyres and road. Magnitude: F = m |a| g = 1200 × 9.8 = 11 760 N, so the coefficient of friction μ = F / N = 7200 / 11 760 ≈ 0.61 – typical for dry asphalt.
减速力来自轮胎与路面之间的摩擦力。大小：F = m |a| = 1200 × 6 = 7200 N。法向反作用力 N = m g = 1200 × 9.8 = 11 760 N，因此摩擦系数 μ = F / N = 7200 / 11 760 ≈ 0.61——这是干燥沥青路面的典型值。
8. Energy Transfers and Work Done | 能量转化与功
J. This matches the initial kinetic energy exactly, confirming that kinetic energy is converted into thermal energy (and some sound) in the brakes and tyres.
制动力所做的功 = F × 制动距离 = 7200 N × 33.3 m ≈ 240 000 J。这正好等于初始动能，证实动能被转化为刹车和轮胎中的热能（以及部分声能）。
9. Velocity-Time Graph | 速度-时间图
Draw a horizontal line at 20 m/s for 0.7 s, then a straight line descending to zero. Braking time tb is found from v = u + a t: 0 = 20 + (−6) tb → tb e area under the graph (rectangle 20×0.7 plus triangle ½×20×3.33) gives 14 + 33.3 = 47.3 m – confirming the total distance.
先画出 0.7 s 内 20 m/s 的水平线，然后是一条倾斜下降至零的直线。制动时间 tb 可由 v = u + a t 求得：0 = 20 + (−6) tb → tb = 20/6 ≈ 3.33 s。图像在 0.7 + 3.33 = 4.03 s 结束。图线下的面积（矩形 20×0.7 加三角形 ½×20×3.33）给出 14 + 33.3 = 47.3 m——再次验证了总距离。
10. Extension: Influence of Road Conditions | 扩展思考：路面状况的影响
Wet roads can reduce deceleration to 4 m/s². Then braking distance = u² / (2|a| 0 / 3 ≈ 133 m. This shows why drivers must increase following distance in poor conditions.
湿滑路面可能会使减速度降至 4 m/s²。此时制动距离 = u² / (2|a|) = 400 / 8 = 50 m，总距离 = 14 + 50 = 64 m。在结冰路面（减速度约 1.5 m/s²），制动距离飙升至 400 / 3 ≈ 133 m。这说明了为什么在恶劣条件下驾驶员必须增大跟车距离。
11. Summary and Exam Tips | 总结与备考要点
This case study blends equations of motion, energy concepts, force analysis, and graph skills. In the exam, always underline given values, pick the correct SUVAT equation, keep units consistent, and relate energy transfers to work done. Practice with different initial speeds, masses, and road conditions to master the method.
本案例融合了运动方程、能量概念、力的分析和图像技能。考试时，一定要在已知数值下划线，选择正确的 SUVAT 方程，保持单位一致，并将能量转化与做功联系起来。用不同的初速度、质量和路面条件反复练习，就能熟练驾驭这种方法。
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