Top Speed | 最大速度（终端速度）

📚 Top Speed | 最大速度（终端速度）

In CIE A-Level Physics, “top speed” usually refers to terminal velocity—the constant maximum speed reached by an object moving through a fluid when the resultant force becomes zero. It appears in topics such as motion, forces, energy and power, and it is a common exam focus because it tests both conceptual understanding and mathematical treatment of drag forces.

在 CIE A-Level 物理中，“最大速度”通常指终端速度——物体在流体中运动时，当合力为零时所达到的恒定最大速度。它出现在运动、力、能量和功率等主题中，也是常见的考试重点，因为它同时考查对阻力概念的定性理解和定量处理。

1. Defining Top Speed and Terminal Velocity | 最大速度与终端速度的定义

Top speed is the greatest velocity an object reaches in a given situation. For a falling object moving through air or liquid, this steady maximum speed is called terminal velocity.

最大速度是物体在特定情况下所能达到的最大速度。对于在空气或液体中下落的物体，这个稳定的最大速度称为终端速度。

Terminal velocity is not an extra force; it is the result of balanced forces. When the downward weight equals the total upward resistive force, the net force becomes zero and acceleration becomes zero.

终端速度不是一种额外的力，而是力平衡的结果。当向下的重力等于向上的总阻力时，合力为零，加速度为零。

In an exam, you should describe terminal velocity as the constant velocity reached when the driving force is equal and opposite to the total resistive force.

在考试中，你应该将终端速度描述为：当驱动力与总阻力大小相等、方向相反时所达到的恒定速度。

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2. Forces on a Falling Object | 下落物体的受力分析

Consider an object released from rest in air. At the instant of release its velocity is zero, so there is no drag force. The only force is weight, W = mg, so the initial acceleration is the free-fall acceleration g.

考虑一个物体从静止开始在空气中释放。释放瞬间速度为零，因此没有阻力。唯一的力是重力 W = mg，所以初始加速度为自由落体加速度 g。

As the object speeds up, the fluid resistance or drag force increases. The resultant downward force is weight minus drag, so the resultant force becomes smaller.

随着物体加速，流体阻力或阻力增大。向下的合力等于重力减阻力，因此合力变小。

Newton’s second law gives the acceleration at any instant: a = (mg – F_drag) / m. Since F_drag increases with speed, the acceleration decreases.

根据牛顿第二定律，任意时刻的加速度为 a = (mg – F_drag) / m。由于 F_drag 随速度增大，加速度不断减小。

When the drag force grows until it equals the weight, the net force is zero. The object can no longer accelerate, so it continues at constant speed—the terminal velocity.

当阻力增大到等于重力时，合力为零。物体不能再加速，因此保持恒定速度运动——这就是终端速度。

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3. Drag Force and Its Velocity Dependence | 阻力及其与速度的关系

For a body moving through a fluid at moderate to high speed, the drag force is approximately proportional to the square of the speed. A useful model is:

对于在流体中以中高速运动的物体，阻力近似与速度的平方成正比。一个常用的模型是：

F_drag = ½ ρ v² A C

where ρ is the fluid density, v is the speed, A is the cross-sectional area perpendicular to motion, and C is the drag coefficient, which depends on the shape of the object.

其中 ρ 是流体密度，v 是速度，A 是垂直于运动方向的横截面积，C 是阻力系数，取决于物体的形状。

At very low speeds, drag may be proportional to v rather than v². However, most CIE A-Level questions on terminal velocity use the quadratic model, so you should be familiar with the v² relationship.

在极低速度下，阻力可能与 v 成正比而不是 v²。但大多数 CIE A-Level 关于终端速度的题目使用二次方模型，因此你应熟悉 v² 关系。

Because drag depends on v², doubling the speed makes the drag force four times larger. This nonlinear relationship explains why acceleration falls off quickly as speed builds up.

由于阻力与 v² 有关，速度加倍会使阻力增至四倍。这种非线性关系解释了为什么随着速度增大，加速度会迅速减小。

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4. Why Acceleration Decreases | 为什么加速度会减小

Acceleration is proportional to the resultant force. As an object falls, its speed increases, so the upward drag force increases while the weight remains constant.

加速度与合力成正比。物体下落时速度增加，因此向上的阻力增大，而重力保持不变。

The resultant downward force is W – F_drag. Since F_drag grows with v, the resultant force decreases, and therefore the acceleration decreases.

向下的合力为 W – F_drag。由于 F_drag 随 v 增大，合力减小，因此加速度减小。

This does not mean the object is slowing down. The velocity is still increasing, but the rate at which it increases becomes smaller and smaller.

这并不意味着物体在减速。速度仍在增加，但速度增加的速率越来越小。

Eventually the acceleration becomes zero. At this point the object has reached its maximum speed, because there is no longer any net force to increase the velocity.

最终加速度变为零。此时物体已达到最大速度，因为没有合力再使速度增加。

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5. The Terminal Velocity Equation | 终端速度公式

At terminal velocity, the forces are balanced:

在终端速度下，力达到平衡：

mg = ½ ρ vₜ² A C

Rearranging for terminal velocity gives:

整理后得到终端速度表达式：

vₜ = √(2mg / (ρ A C))

This equation shows that terminal velocity increases with mass and decreases with fluid density, cross-sectional area and drag coefficient.

该公式表明，终端速度随质量增大而增大，随流体密度、横截面积和阻力系数增大而减小。

If the weight is written as W, the same equation can be written as vₜ = √(2W / (ρ A C)). Both forms are acceptable in calculations.

如果重力写作 W，同一公式也可写成 vₜ = √(2W / (ρ A C))。在计算中两种形式都可以。

You do not need to memorise the derivation, but you must be able to set weight equal to drag force and rearrange confidently.

你不需要背诵推导过程，但必须能够令重力等于阻力，并熟练地进行变形。

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6. From Release to Terminal Velocity: A Skydiver | 从释放到终端速度：跳伞者

A skydiver is the classic example of terminal velocity. When the skydiver jumps from a stationary aircraft, the initial velocity is zero and the only significant force is weight, so the initial acceleration is about 9.81 m s⁻².

跳伞者是终端速度的典型例子。当跳伞者从静止的飞机上跳下时，初始速度为零，唯一重要的力是重力，因此初始加速度约为 9.81 m s⁻²。

As speed builds up, air resistance increases. The skydiver’s acceleration decreases, but the speed still grows until the first terminal velocity is reached—typically around 50 m s⁻¹ in a spread-eagle position.

随着速度增大，空气阻力增大。跳伞者的加速度减小，但速度仍会增大，直到达到第一个终端速度——在四肢伸展姿势下通常约为 50 m s⁻¹。

When the parachute opens, the cross-sectional area and drag coefficient increase sharply. The upward drag force suddenly becomes much larger than the weight, causing a large upward acceleration and rapid deceleration.

当降落伞打开时，横截面积和阻力系数急剧增大。向上的阻力突然远大于重力，产生很大的向上加速度，使跳伞者迅速减速。

As the speed falls, the drag force decreases again. A new, much lower terminal velocity is reached, typically around 5 m s⁻¹, allowing a safe landing.

随着速度下降，阻力再次减小。最终达到一个新的、低得多的终端速度，通常约为 5 m s⁻¹，从而实现安全着陆。

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7. Velocity–Time and Acceleration–Time Graphs | 速度–时间图像与加速度–时间图像

For an object falling from rest through a fluid, the velocity–time graph starts at the origin with a steep gradient equal to g. The gradient gradually decreases and the curve levels off at the terminal velocity.

对于从静止开始通过流体下落的物体，速度–时间图像从原点开始，初始斜率等于 g。斜率逐渐减小，曲线在终端速度处趋于水平。

The acceleration–time graph starts at g, then decreases smoothly and approaches zero. It is not a straight line, because the drag force depends on v², so the acceleration changes nonlinearly.

加速度–时间图像从 g 开始，然后平滑下降并趋近于零。它并不是直线，因为阻力与 v² 有关，所以加速度呈非线性变化。

A common error is to think the velocity–time graph is a straight line that suddenly becomes horizontal. The correct shape is a curve that asymptotically approaches the terminal velocity.

一个常见错误是认为速度–时间图像是一条直线，然后突然变为水平。正确的形状是一条逐渐趋近于终端速度的曲线。

You may also be asked to sketch displacement–time or acceleration–time graphs. Displacement–time should start curved and become nearly straight with constant gradient at terminal velocity.

你可能还需要绘制位移–时间或加速度–时间图像。位移–时间图像开始时是曲线，到达终端速度后趋近为斜率恒定的直线。

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8. Factors That Increase or Decrease Top Speed | 影响最大速度的因素

From the terminal velocity equation, vₜ = √(2mg / (ρ A C)), you can predict how changing physical quantities affects the top speed.

根据终端速度公式 vₜ = √(2mg / (ρ A C))，你可以预测改变物理量会如何影响最大速度。

Factor change | 因素变化	Effect on vₜ | 对 vₜ 的影响	Reason | 原因
Larger mass | 质量增大	Increases | 增大	Greater weight needs greater drag to balance it | 更大的重力需要更大的阻力来平衡
Larger area | 面积增大	Decreases | 减小	Drag force is larger at any given speed | 相同速度下阻力更大
Denser fluid | 流体密度增大	Decreases | 减小	Greater fluid resistance | 流体阻力更大
Streamlined shape | 流线型形状	Increases | 增大	Smaller drag coefficient C | 阻力系数 C 更小

In many questions, only one factor is changed while the others are kept constant. Be careful: if mass increases but shape and size also change, the result may be different.

在许多题目中，只有某一个因素改变，其他因素保持不变。注意：如果质量增大但形状和尺寸也改变，结果可能不同。

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9. Worked Example: Falling Sphere | 例题：下落小球

A metal sphere of mass 0.20 kg and radius 0.10 m falls through air of density 1.2 kg m⁻³. The drag coefficient is 0.47. Calculate its terminal velocity.

一个质量为 0.20 kg、半径为 0.10 m 的金属球在密度为 1.2 kg m⁻³ 的空气中下落。阻力系数为 0.47。计算其终端速度。

Step 1: Calculate the cross-sectional area. A = πr² = π × (0.10)² = 0.0314 m².

步骤 1：计算横截面积。A = πr² = π × (0.10)² = 0.0314 m²。

Step 2: Set weight equal to drag force at terminal velocity.

步骤 2：在终端速度下，令重力等于阻力。

mg = ½ ρ vₜ² A C

Step 3: Rearrange and substitute values.

步骤 3：变形并代入数值。

vₜ = √(2 × 0.20 × 9.81 / (1.2 × 0.0314 × 0.47))

vₜ ≈ 14.9 m s⁻¹

This means the sphere reaches a steady speed of about 14.9 m s⁻¹, at which point the air resistance exactly balances its weight.

这意味着小球达到约 14.9 m s⁻¹ 的稳定速度，此时空气阻力恰好与重力平衡。

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10. Top Speed of Vehicles and Power | 车辆的最大速度与功率

For a car or bicycle, top speed is reached when the engine or driving force equals the total resistive force. The resistive forces include air drag and rolling resistance.

对于汽车或自行车，当发动机驱动力等于总阻力时，达到最大速度。阻力包括空气阻力和滚动阻力。

At a constant speed, the net force is zero, so the useful power output from the engine is given by:

在恒定速度下，合力为零，因此发动机的有用功率输出为：

P = F v

where F is the driving force. At top speed, F equals the total resistive force, so the top speed can be estimated from P / F.

其中 F 是驱动力。在最大速度时，F 等于总阻力，因此最大速度可通过 P / F 估算。

However, air drag increases with v², so the required power to overcome drag increases with v³. This is why a car needs much more power to increase top speed from 40 m s⁻¹ to 60 m s⁻¹ than from 20 m s⁻¹ to 40 m s⁻¹.

然而，空气阻力随 v² 增大，因此克服阻力所需功率随 v³ 增大。这就是为什么汽车将最大速度从 40 m s⁻¹ 提高到 60 m s⁻¹ 所需功率远大于从 20 m s⁻¹ 提高到 40 m s⁻¹ 的原因。

In calculations, if the resistive force is given as a constant, you can simply use v = P / F. If the resistive force depends on speed, you may need to solve a more complex equation.

在计算中，如果阻力给定为常数，你可以直接使用 v = P / F。如果阻力与速度有关，则可能需要解更复杂的方程。

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11. Energy Considerations at Terminal Velocity | 终端速度下的能量分析

Once an object reaches terminal velocity, its kinetic energy remains constant because its speed is constant. The gravitational potential energy lost during fall is not being converted into kinetic energy.

一旦物体达到终端速度，由于速度恒定，其动能保持不变。下落过程中损失的重力势能不会转化为动能。

Instead, the lost gravitational potential energy is transferred to the surrounding fluid as thermal energy due to work done against drag. This is why falling through a fluid can cause heating.

相反，损失的重力势能通过对阻力做功而转化为流体的热能。这就是为什么物体穿过流体下落时会产生热效应。

The rate of energy dissipation at terminal velocity is P = F v = mg vₜ. This power is transferred to the fluid and ultimately dissipated as heat.

在终端速度下，能量耗散率 P = F v = mg vₜ。该功率传递给流体并最终以热量形式耗散。

This energy analysis is useful in explaining why meteors and spacecraft heat up when entering the atmosphere at high speed.

这种能量分析有助于解释为什么流星和航天器高速进入大气层时会发热。

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12. Exam Tips and Common Errors | 考试技巧与常见错误

Do not say that a falling object has zero acceleration from the moment it is dropped. At the instant of release, acceleration is g, not zero.

不要说下落物体从释放瞬间起加速度为零。在释放瞬间，加速度为 g，而不是零。

Do not state that “gravity stops” at terminal velocity. Weight still acts downward, but it is balanced by drag, so the resultant force is zero.

不要说在终端速度下“重力停止作用”。重力仍然向下作用，只是被阻力平衡，所以合力为零。

When sketching graphs, label the terminal velocity clearly and show the gradient decreasing smoothly rather than abruptly becoming zero.

画图时，要清楚标出终端速度，并展示斜率平滑减小，而不是突然变为零。

Always use the correct cross-sectional area in drag calculations. For a sphere it is πr²; for a parachute it is the area of the open canopy, not the surface area of the fabric.

在阻力计算中始终使用正确的横截面积。对于球体是 πr²；对于降落伞是张开伞面的面积，而不是伞布的表面积。

Finally, quote the correct unit for terminal velocity: m s⁻¹. Remember that terminal velocity is a speed, so it should be positive in most one-dimensional falling problems.

最后，要写出终端速度的正确单位：m s⁻¹。记住终端速度是速度大小，在大多数一维下落问题中应为正值。

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