Motion in 2 Dimensions | 二维运动

📚 Motion in 2 Dimensions | 二维运动

In A-Level Mechanics, motion in two dimensions extends the concepts of kinematics from one dimension to a plane, using vectors to describe displacement, velocity and acceleration. This topic is central to understanding projectile motion, relative motion, and any scenario where an object moves under gravity while following a curved path. Mastering vector notation and component analysis is essential for success in the Edexcel mechanics modules.

在A-Level力学中，二维运动将运动学的概念从一维扩展到平面，使用向量来描述位移、速度和加速度。此主题是理解抛体运动、相对运动以及任何物体在重力作用下沿曲线轨迹运动的关键。掌握向量记法和分量分析对于在Edexcel力学模块中取得成功至关重要。

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1. Introduction to 2D Motion | 二维运动简介

When motion occurs in a single straight line, we use scalar equations with positive and negative signs to indicate direction. In two dimensions, position, velocity and acceleration must be treated as vector quantities. A vector has both magnitude and direction, and is often expressed in terms of unit vectors i (horizontal) and j (vertical), or as column vectors.

当运动发生在单一直线时，我们使用带有正负号的标量方程来表示方向。在二维中，位置、速度和加速度必须作为向量来处理。向量既有大小又有方向，通常用单位向量 i（水平）和 j（垂直）表示，或者表示为列向量。

For example, a particle with coordinates (x, y) relative to an origin O has a position vector r = x i + y j. As the particle moves, r changes with time, giving a vector function r(t). This approach unifies direction handling and simplifies analysis of curved paths.

例如，一个相对于原点 O 坐标为 (x, y) 的质点具有位置向量 r = x i + y j。随着质点运动，r 随时间变化，构成向量函数 r(t)。这种方法统一了方向处理，简化了曲线路径的分析。

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2. Vector Representation of Motion | 运动的向量表示

In the Edexcel specification, you will frequently see vectors written as r = xi + yj or as a column vector. Bold typeface or underlined letters indicate vector quantities. The magnitude of a vector, representing distance from origin for position or speed for velocity, is found using Pythagoras’ theorem: |r| = √(x² + y²).

在Edexcel大纲中，你会经常看到向量写作 r = xi + yj 或列向量形式。粗体或带下划线的字母表示向量。向量的大小——对位置而言表示到原点的距离，对速度而言表示速率——通过勾股定理求得：|r| = √(x² + y²)。

It is crucial to distinguish between displacement and distance, velocity and speed, when working in 2D. A particle travelling in a circle at constant speed has velocity changing due to direction, hence acceleration is present. Using vectors, acceleration points towards the centre of the circle.

在二维问题中，区分位移与路程、速度与速率至关重要。一个以恒定速率做圆周运动的质点，由于方向改变，速度在变化，因此存在加速度。用向量表示时，加速度指向圆心。

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3. Position, Velocity and Acceleration Vectors | 位置、速度和加速度向量

If the position vector of a particle is given as a function of time, r(t) = x(t) i + y(t) j, then the velocity vector v is the first derivative with respect to time: v = dr/dt = (dx/dt) i + (dy/dt) j. The acceleration vector a is the derivative of velocity: a = dv/dt = d²r/dt².

如果质点的位置向量表示为时间的函数 r(t) = x(t) i + y(t) j，那么速度向量 v 是对时间的一阶导数：v = dr/dt = (dx/dt) i + (dy/dt) j。加速度向量 a 是速度的导数：a = dv/dt = d²r/dt²。

In many exam questions, you will be given r(t) as a polynomial expression, such as r = (2t³ − t) i + (5t² + 3) j. You are expected to differentiate or integrate to find velocity and acceleration, or vice versa. Constant acceleration problems can be simplified using vector SUVAT, as explored next.

在许多考试题目中，你会得到 r(t) 的多项式表达式，例如 r = (2t³ − t) i + (5t² + 3) j。要求你通过微分或积分求出速度和加速度，或者反过来求解。匀加速问题可以用向量SUVAT来简化，接下来将进行探讨。

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4. SUVAT Equations in Vector Form | 向量形式的匀加速运动方程

For motion with constant acceleration in two dimensions, the standard SUVAT equations can be applied to each vector component. Using u for initial velocity, v for final velocity, a for acceleration, s for displacement, and t for time, the equations hold with vector addition.

对于二维匀加速运动，标准的SUVAT方程可以应用于每个向量分量。用 u 表示初速度，v 表示末速度，a 表示加速度，s 表示位移，t 表示时间，这些方程通过向量加法成立。

v = u + a t

s = u t + ½ a t²

s = ½ (u + v) t

v² = u² + 2 a·s (where the dot product may be used for magnitude: |v|² = |u|² + 2 a·s)

In practice, you often resolve horizontally and vertically. For a projectile under gravity, the horizontal acceleration is 0, and the vertical acceleration is −g j (taking upwards as positive). Thus, the horizontal motion is uniform, while the vertical motion is uniformly accelerated.

实际操作中，你通常分别处理水平和竖直分量。对于重力作用下的抛体，水平加速度为 0，竖直加速度为 −g j（设向上为正）。因此，水平运动是匀速的，而竖直运动是匀加速的。

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5. Projectile Motion: Horizontal and Vertical Components | 抛体运动：水平与垂直分量

Consider a particle projected from the origin with speed u at an angle θ to the horizontal. The initial velocity can be resolved as:

考虑一个质点从原点以速度 u 与水平方向成 θ 角抛出。初速度可分解为：

u_x = u cos θ ; u_y = u sin θ

Assuming the only force acting is gravity (neglecting air resistance), the horizontal acceleration a_x = 0, and the vertical acceleration a_y = −g (where g ≈ 9.8 m s⁻²). The equations of motion at time t become:

假设仅受重力作用（忽略空气阻力），水平加速度 a_x = 0，竖直加速度 a_y = −g（g ≈ 9.8 m s⁻²）。t 时刻的运动方程变为：

x = u cos θ × t

y = u sin θ × t − ½ g t²

This separation is a powerful tool. All projectile problems in Edexcel can be solved by writing these two independent equations and linking them through the time of flight.

这种分离是一个强大的工具。Edexcel 中所有的抛体问题都可以通过写出这两个独立方程，并通过飞行时间将它们联系起来求解。

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6. Trajectory Equation Derivation | 轨迹方程推导

To find the Cartesian equation of the path, eliminate t from the parametric equations. From x = u cos θ t, we have t = x / (u cos θ). Substituting into the y-equation yields:

为求出轨迹的直角坐标方程，需从参数方程中消去 t。由 x = u cos θ t 可得 t = x / (u cos θ)。代入 y 的方程得到：

y = x tan θ − (g / (2 u² cos² θ)) x²

This is a quadratic in x, indicating that the trajectory is parabolic. The coefficient of x² is always negative (for standard projection above horizontal), so the parabola opens downwards. This equation is particularly useful for determining whether a projectile clears a wall or hits a target at given coordinates.

这是一个关于 x 的二次方程，表明轨迹是抛物线。x² 的系数始终为负（对于标准的地面以上抛射），因此抛物线开口向下。该方程在判断抛体是否越过墙壁或击中给定坐标的目标时特别有用。

You might also be asked to find the maximum height by locating the vertex of this parabola, or by using vertical motion equations: v_y² = u_y² − 2g y, setting final vertical velocity zero at the peak.

你可能还需要通过寻找抛物线的顶点来求最大高度，或者使用竖直运动方程：v_y² = u_y² − 2g y，令最高点的竖直末速度为零。

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7. Time of Flight, Range, Maximum Height | 飞行时间、射程、最大高度

For a projectile launched from and landing on the same horizontal level, the time of flight T is found by setting y = 0 (return to original level) and solving T > 0:

对于从同一水平面发射和着陆的抛体，飞行时间 T 可通过令 y = 0（返回原高度）并求解 T > 0 得到：

T = (2 u sin θ) / g

The horizontal range R is the distance travelled in that time: R = u cos θ × T = (u² sin 2θ) / g. This shows that maximum range for a given speed is achieved when θ = 45°.

水平射程 R 是该时间内经过的距离：R = u cos θ × T = (u² sin 2θ) / g。这表明在给定初速下，θ = 45° 时射程最大。

Maximum height H occurs when v_y = 0, using v_y² = u_y² − 2g H:

最大高度 H 发生在 v_y = 0 时，利用 v_y² = u_y² − 2g H 得到：

H = (u² sin² θ) / (2g)

These standard results should be memorised or, better, derived quickly from the SUVAT equations in any problem. Edexcel often sets questions where launch and landing heights differ, so understanding the underlying equations is more important than rigidly applying formulas.

这些标准结果应当熟记，或者更好的是，在任何问题中根据SUVAT方程快速推导出来。Edexcel 经常设置发射与着陆高度不同的题目，因此理解底层方程比死记公式更为重要。

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8. Projection from a Height | 从高处抛出的抛体

If a particle is projected from a point above the landing level, the vertical displacement y is not zero at landing. For example, if launched from a cliff of height h, the time of flight is found by solving y = −h (if downwards is negative). The quadratic equation −½ g t² + (u sin θ) t + h = 0 gives the time. Only the positive root is relevant.

如果质点从高于落地平面的点抛出，落地时竖直位移 y 不为零。例如，从高 h 的悬崖上发射，通过解方程 y = −h（设向下为负）可求得飞行时间。二次方程 −½ g t² + (u sin θ) t + h = 0 给出时间，只需取正根。

Similarly, range is found by substituting this time into the horizontal equation x = u cos θ t. The impact speed is obtained from the vector sum of horizontal and vertical velocity components at that instant, often using energy considerations or v = u + a t for vertical motion.

类似地，将此时间代入水平方程 x = u cos θ t 可求得射程。撞击速率可由该瞬时的水平和竖直速度分量通过向量和求得，通常利用能量原理或竖直运动的 v = u + a t。

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9. Impact Velocity and Angle | 碰撞速度和角度

To find the velocity with which a projectile hits the ground, calculate the horizontal component (which remains constant) and the vertical component at the time of impact. The impact speed is √(v_x² + v_y²). The direction of the velocity relative to the horizontal is given by θ = tan⁻¹(|v_y| / v_x), often asked as the angle to the horizontal or vertical.

为求抛体撞击地面时的速度，需计算水平分量（保持不变）和撞击时刻的竖直分量。撞击速率为 √(v_x² + v_y²)。速度方向与水平面的夹角为 θ = tan⁻¹(|v_y| / v_x)，常被问及与水平面或竖直方向的夹角。

If the landing height is the same as launch, the speed at impact equals the initial speed (by conservation of energy), but the direction is mirrored. Edexcel questions may require you to find the angle with vector i or j, so pay attention to signs.

如果着陆高度与发射高度相同，撞击速率等于初速（根据能量守恒），但方向镜像对称。Edexcel 题目可能要求求出与向量 i 或 j 的夹角，因此要注意符号。

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10. Relative Motion in 2D | 二维相对运动

Relative motion extends vector kinematics to two moving objects. The position of A relative to B is _Ar_B = r_A − r_B. The velocity of A relative to B is _Av_B = v_A − v_B. These vectors help answer questions like: When are two particles closest? Do they collide?

相对运动将向量运动学推广到两个运动物体。A 相对于 B 的位置是 _Ar_B = r_A − r_B。A 相对于 B 的速度是 _Av_B = v_A − v_B。这些向量有助于回答诸如两粒子何时最接近、是否相撞等问题。

To find if a collision occurs, set r_A(t) = r_B(t) and solve for t. If a positive solution exists and the objects are at the same point at that time, they collide. Shortest distance problems often involve minimising the magnitude of the relative position vector using calculus or vector geometry.

要判断是否发生碰撞，令 r_A(t) = r_B(t) 并求解 t。若存在正解且此时物体位于同一点，则发生碰撞。最短距离问题通常需要使用微积分或向量几何对相对位置向量的大小进行最小化。

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11. Worked Examples | 实例解析

Example 1: A ball is thrown from ground level with speed 20 m s⁻¹ at 30° to the horizontal. Find its range. (Take g = 9.8)

例1: 一球从地面以 20 m s⁻¹ 的速度与水平面成 30° 角抛出，求射程。（取 g = 9.8）

Solution: u_x = 20 cos30° ≈ 17.32, u_y = 20 sin30° = 10. Time of flight T = 2u_y/g = 20/9.8 ≈ 2.041 s. Range = u_x × T ≈ 17.32 × 2.041 ≈ 35.3 m. Alternatively, R = u² sin2θ / g = 400 sin60° / 9.8 ≈ 35.3 m.

解：u_x = 20 cos30° ≈ 17.32, u_y = 20 sin30° = 10。飞行时间 T = 2u_y/g = 20/9.8 ≈ 2.041 s。射程 = u_x × T ≈ 17.32 × 2.041 ≈ 35.3 m。或者，R = u² sin2θ / g = 400 sin60° / 9.8 ≈ 35.3 m。

Example 2: A particle has position vector r = (3t² − 2t) i + (t³ − 4) j. Find its velocity and acceleration at t = 2 s.

例2: 一质点的位置向量为 r = (3t² − 2t) i + (t³ − 4) j，求 t = 2 s 时的速度和加速度。

v = dr/dt = (6t − 2) i + 3t² j. At t=2: v = (10) i + 12 j. a = dv/dt = 6 i + 6t j. At t=2: a = 6 i + 12 j. Magnitudes can then be computed.

v = dr/dt = (6t − 2) i + 3t² j。t=2 时：v = 10 i + 12 j。a = dv/dt = 6 i + 6t j。t=2 时：a = 6 i + 12 j。然后可计算大小。

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12. Common Mistakes and Tips | 常见错误与技巧

Many students forget that the horizontal and vertical motions are independent only under gravity without air resistance. A frequent error is using the same time variable for horizontal and vertical equations without linking them correctly; remember, t is the same for both components.

许多学生忘记在无空气阻力且仅受重力作用时，水平和竖直运动是相互独立的。常见的错误是在水平和竖直方程中使用同一个时间变量却没有正确关联它们；请记住，t 对两个分量是相同的。

Another mistake is sign errors: if upwards is positive, acceleration is −g. The displacement y can be negative when landing below launch point. Use a consistent coordinate system. In relative motion, subtracting vectors incorrectly leads to wrong relative velocities; always use v_A − v_B for velocity of A relative to B.

另一个错误是符号出错：若设向上为正，加速度为 −g。当落点在发射点下方时，位移 y 可能为负。应使用一致的坐标系。在相对运动中，向量相减不正确会导致相对速度错误；始终使用 v_A − v_B 表示 A 相对于 B 的速度。

Lastly, practise drawing diagrams and clearly resolving vectors. For projectile problems, always write down u_x, u_y, and the relevant SUVAT. Derive equations rather than jumping to memorised formulas, as exam questions often modify initial conditions.

最后，多做画图练习，清晰地分解向量。对于抛体问题，始终写下 u_x、u_y 以及相关的 SUVAT。推导方程而不要直接套用背诵的公式，因为考试题目经常修改初始条件。

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