📚 Motion Analysis: A Year 11 CCEA Physics Case Study | Year 11 CCEA 物理运动案例实战演练
In CCEA GCSE Physics, analysing real-world motion is a key skill. This case study follows one complete journey of a car, from rest to stop, using velocity-time graphs, the equations of motion, Newton's laws and energy principles. By working through this example step by step, you will master the essential techniques needed for exam questions on motion and forces.
在 CCEA GCSE 物理中，分析实际运动是一项关键技能。本案例通过一辆汽车从静止到停止的完整过程，运用速度-时间图、运动方程、牛顿定律和能量原理逐步讲解。跟着这个例子一步步推演，你将掌握解决运动和力相关考题的核心方法。
1. The Scenario | 情景设定
celerates uniformly for 10 seconds until it reaches a speed of 20 m s⁻¹. It then travels at this constant speed for 30 seconds. Finally the driver applies the brakes and the car decelerates uniformly, coming to a stop after a further 5 seconds. The entire motion takes place in a straight line.
一辆质量为 1200 kg 的汽车在水平直路上从 t = 0 s 时静止出发。它匀加速 10 秒，达到 20 m s⁻¹ 的速度；然后以此速度匀速行驶 30 秒；最后司机刹车，汽车匀减速，经过 5 秒停下。整个运动沿直线进行。
The journey therefore splits into three stages. Stage 1 is uniform acceleration from 0 to 20 m s⁻¹. Stage 2 is constant velocity at 20 m s⁻¹. Stage 3 is uniform deceleration from 20 m s⁻¹ to rest. The word 'uniform' is important: it means the acceleration is constant, so the graphs are made of straight lines and we can use the standard equations of motion.
因此全程分为三个阶段：阶段 1 是速度从 0 匀加速到 20 m s⁻¹；阶段 2 是保持 20 m s⁻¹ 匀速；阶段 3 是从 20 m s⁻¹ 匀减速到静止。"匀"字很关键：它表示加速度恒定，所以图像由直线段构成，我们可以直接使用标准运动方程。
| Stage | Time interval / s | Duration / s | Motion | Acceleration / m s⁻² |
| 1 | 0 to 10 | 10 | uniform acceleration, 0 → 20 m s⁻¹ | +2 |
| 2 | 10 to 40 | 30 | constant speed, 20 m s⁻¹ | 0 |
| 3 | 40 to 45 | 5 | uniform deceleration, 20 → 0 m s⁻¹ | −4 |
The model assumes the road is level (so there is no gain or loss of gravitational potential energy), the direction of travel does not change, and resistance forces are either negligible or included inside the quoted values. These assumptions are typical of CCEA examination questions and should be stated if a question asks you to comment on the model.
该模型假设路面水平（因此没有重力势能的增减）、行驶方向不变，并且阻力要么忽略不计、要么已包含在所给数据中。这些假设是 CCEA 考题的常见设定，若题目要求评价模型，应把这些假设写出来。
2. Drawing the Velocity-Time Graph | 绘制速度-时间图
On a velocity-time graph the vertical axis represents velocity v in m s⁻¹ and the horizontal axis represents time t in s. The three stages appear as three line segments on the same axes.
在速度-时间图中，纵轴表示速度 v（单位 m s⁻¹），横轴表示时间 t（单位 s）。三个阶段在同一坐标系中表现为三条线段。
- Stage 1: a straight line from (0, 0) to (10, 20) - uniform acceleration.
- Stage 2: a horizontal line from (10, 20) to (40, 20) - constant speed.
- Stage 3: a straight line from (40, 20) to (45, 0) - uniform deceleration.
- 阶段 1：从 (0, 0) 到 (10, 20) 的直线——匀加速。
- 阶段 2：从 (10, 20) 到 (40, 20) 的水平线——匀速。
- 阶段 3：从 (40, 20) 到 (45, 0) 的直线——匀减速。
The gradient of a velocity-time graph gives the acceleration, and the area under the graph gives the displacement (the straight-line distance travelled). We will use both of these properties to solve the case. Note that a rising line means positive acceleration, a horizontal line means zero acceleration, and a falling line means negative acceleration (deceleration).
速度-时间图的斜率表示加速度，图线下方与时间轴围成的面积表示位移（即直线行驶的距离）。我们将同时利用这两条性质来解题。注意：上升的线表示正加速度，水平线表示加速度为零，下降的线表示负加速度（减速）。
3. Calculating Acceleration from the Gradient | 由斜率计算加速度
Acceleration is defined as the rate of change of velocity:
加速度定义为速度的变化率：
a = (v − u) / t
ting the values for the acceleration stage, u = 0 m s⁻¹, v = 20 m s⁻¹ and t = 10 s, gives:
其中 u 为初速度，v 为末速度，t 为所用时间。代入加速阶段的数值 u = 0 m s⁻¹、v = 20 m s⁻¹、t = 10 s，得：
a₁ = (20 − 0) / 10 = 2 m s⁻²
The same method applied to the braking stage uses u = 20 m s⁻¹, v = 0 m s⁻¹ and t = 5 s:
把同样的方法用于刹车阶段，取 u = 20 m s⁻¹、v = 0 m s⁻¹、t = 5 s：
a₃ = (0 − 20) / 5 = −4 m s⁻²
The negative sign tells us the car is slowing down, that is, the acceleration acts in the opposite direction to the motion. During the middle stage the velocity is constant, so the change in velocity is zero and the acceleration is 0 m s⁻².
负号说明汽车在减速，即加速度方向与运动方向相反。在中间阶段速度不变，速度变化量为零，因此加速度为 0 m s⁻²。
4. Distance Travelled: The Area Under the Graph | 位移：图线下方的面积
The area under a velocity-time graph equals the displacement. Each stage forms a simple shape, so we can find the three distances separately and then add them.
速度-时间图下方的面积等于位移。每个阶段都构成简单图形，因此可以先分别求出三段距离再相加。
- Stage 1 is a triangle: area = ½ × base × height = ½ × 10 × 20 = 100 m.
- Stage 2 is a rectangle: area = base × height = 30 × 20 = 600 m.
- Stage 3 is a triangle: area = ½ × 5 × 20 = 50 m.
- 阶段 1 是三角形：面积 = ½ × 底 × 高 = ½ × 10 × 20 = 100 m。
- 阶段 2 是矩形：面积 = 底 × 高 = 30 × 20 = 600 m。
- 阶段 3 是三角形：面积 = ½ × 5 × 20 = 50 m。
Adding the three areas gives the total distance for the whole journey:
把三个面积相加即得全程总距离：
s(total) = 100 + 600 + 50 = 750 m
7 m s⁻¹. Notice that this average is much lower than the cruising speed of 20 m s⁻¹, because the car spent 15 s accelerating or braking.
全程平均速度等于总距离除以总时间 45 s，即 750 / 45 ≈ 16.7 m s⁻¹。注意这个平均值明显低于 20 m s⁻¹ 的巡航速度，因为汽车有 15 s 处于加速或刹车状态。
5. Checking the Answers with the Equations of Motion | 用运动方程检验答案
acceleration is uniform. For the first stage, with u = 0 m s⁻¹, a = 2 m s⁻² and t = 10 s:
只要加速度恒定，就可以使用标准运动方程（常称为 SUVAT 方程）。对第一阶段，取 u = 0 m s⁻¹、a = 2 m s⁻²、t = 10 s：
s = ut + ½at² = 0 + ½ × 2 × 10² = 100 m
y with the area of the triangle. We can also check the final velocity using v² = u² + 2as:
这与三角形的面积完全一致。也可以用 v² = u² + 2as 检验末速度：
v² = 0² + 2 × 2 × 100 = 400, so v = 20 m s⁻¹
For the braking stage, v = 0 m s⁻¹, u = 20 m s⁻¹ and a = −4 m s⁻², so:
对刹车阶段，v = 0 m s⁻¹、u = 20 m s⁻¹、a = −4 m s⁻²，因此：
0² = 20² + 2 × (−4) × s, so s = 400 / 8 = 50 m
Both methods agree, which is exactly what a well-written exam answer should show. Using the graph and the equations together is a powerful way to catch arithmetic slips.
两种方法结果一致，这正是规范答卷应体现的。图像法与方程法结合使用，是发现计算错误的有力手段。
6. Newton's Laws and the Forces on the Car | 牛顿定律与汽车受力
Newton' leration: F = ma. During the acceleration stage the resultant force on the 1200 kg car is:
牛顿第二定律指出合力等于质量乘以加速度：F = ma。在加速阶段，这辆 1200 kg 汽车受到的合力为：
F = 1200 × 2 = 2400 N
During the constant-speed stage the car is not accelerating, so by Newton's first law the resultant force is zero. The forward driving force from the engine is exactly balanced by the resistive forces of air resistance and friction, even though the car is still moving.
在匀速阶段，汽车没有加速度，根据牛顿第一定律合力为零。发动机提供的向前的驱动力恰好被空气阻力和摩擦阻力平衡，尽管汽车仍在运动。
iving force. Examiners often reward candidates who distinguish between the resultant force and the individual forces acting on a body.
实际上，2400 N 是合力。若阻力合计为 600 N，那么发动机必须提供 2400 + 600 = 3000 N 的驱动力。区分合力与作用在物体上的各个分力，往往是考官给分的要点。
- Vertical forces: weight (about 11 800 N) and the normal reaction are equal and opposite, so they cancel.
- Horizontal forces: driving force forwards, resistive forces and braking force backwards.
- 竖直方向：重力（约 11 800 N）与地面支持力大小相等、方向相反，互相抵消。
- 水平方向：向前的驱动力，向后的阻力和制动力。
7. The Braking Phase and the Stopping Force | 刹车阶段与制动力
During stage 3 the car decelerates at 4 m s⁻², so the resultant force acting on it is:
在第三阶段，汽车以 4 m s⁻² 减速，因此作用在它上面的合力为：
F = ma = 1200 × (−4) = −4800 N
The minus sign simply means the force acts backwards, opposite to the direction of travel. The braking force, of magnitude 4800 N, is produced mainly by friction between the brake pads and the brake discs, and by friction between the tyres and the road surface.
负号只表示力的方向与运动方向相反。制动力的大小为 4800 N，主要来自刹车片与刹车盘之间的摩擦，以及轮胎与路面之间的摩擦。
Notice that the braking force is twice the accelerating force, even though both stages involve the same change in speed of 20 m s⁻¹. This is because the braking time (5 s) is only half the acceleration time (10 s), and a shorter time must produce a larger force for the same change in velocity.
注意制动力是加速力的两倍，尽管两个阶段速度变化量都是 20 m s⁻¹。原因是刹车时间（5 s）只有加速时间（10 s）的一半，在速度变化量相同时，时间越短所需的力越大。
8. Energy Analysis: Kinetic Energy and Work Done | 能量分析：动能与功
culated from E(k) = ½mv². During the constant-speed stage the car travels at 20 m s⁻¹, so:
动能由 E(k) = ½mv² 计算。在匀速阶段，汽车以 20 m s⁻¹ 行驶，因此：
