Motion in 2 Dimensions | 二维运动

📚 Motion in 2 Dimensions | 二维运动

In Edexcel A-Level Mechanics, motion in two dimensions is described using vector quantities. You will use i-j notation, differentiate and integrate vector functions of time, and apply constant acceleration equations separately to horizontal and vertical components. Projectile motion is a central application.

在 Edexcel A-Level 力学中，二维运动用矢量来描述。你将使用 i-j 表示法，对时间的矢量函数进行求导和积分，并将匀加速运动方程分别应用于水平与竖直分量。抛体运动是核心应用。

1. Vector Notation and Components | 矢量表示法与分量

In two dimensions, a vector is written in terms of perpendicular unit vectors i and j. By convention, i points horizontally to the right and j points vertically upwards. For example, a position vector can be written as r = x i + y j, where x and y are scalar components.

在二维空间中，矢量用互相垂直的单位矢量 i 和 j 表示。通常规定 i 水平向右，j 竖直向上。例如，位置矢量可写作 r = x i + y j，其中 x 与 y 是标量分量。

Vectors are added or subtracted component by component. If a = a₁ i + a₂ j and b = b₁ i + b₂ j, then:

矢量按分量逐项相加或相减。若 a = a₁ i + a₂ j，b = b₁ i + b₂ j，则：

a + b = (a₁ + b₁) i + (a₂ + b₂) j

You can also multiply a vector by a scalar, which scales each component. This is essential when using equations such as F = ma in vector form.

你也可以将矢量乘以标量，这会按比例缩放每个分量。这在矢量形式的方程 F = ma 中非常关键。

2. Position and Displacement | 位置与位移

For a particle moving in a plane, its position vector at time t is a function of t: r(t) = x(t) i + y(t) j. The coordinates x(t) and y(t) are ordinary scalar functions.

对于在平面内运动的质点，它在时刻 t 的位置矢量是 t 的函数：r(t) = x(t) i + y(t) j。坐标 x(t) 与 y(t) 是普通的标量函数。

Displacement is the change in position. If the position changes from r₁ to r₂, the displacement is:

位移是位置的变化量。若位置从 r₁ 变为 r₂，则位移为：

Δr = r₂ – r₁

Distance is the length of the path actually travelled, while displacement is a vector from the starting point to the final point. The magnitude of displacement is found using Pythagoras: |Δr| = √((Δx)² + (Δy)²).

路程是实际经过路径的长度，而位移是从起点指向终点的矢量。位移的大小用勾股定理求出：|Δr| = √((Δx)² + (Δy)²)。

3. Velocity and Speed | 速度与速率

Velocity is the rate of change of position with respect to time. If r(t) = x(t) i + y(t) j, then differentiating each component gives:

速度是位置关于时间的变化率。若 r(t) = x(t) i + y(t) j，则对每个分量求导得到：

v = dr/dt = (dx/dt) i + (dy/dt) j

Velocity is a vector: it has both magnitude and direction. Speed is the magnitude of velocity and is a scalar:

速度是矢量：既有大小又有方向。速率是速度的大小，是标量：

speed = |v| = √(v₁² + v₂²)

The direction of velocity is usually given as an angle θ from the positive i-direction, where tan θ = v₂ / v₁. You must consider the quadrant when finding θ because the inverse tangent function alone can be ambiguous.

速度的方向通常用与正 i 方向的夹角 θ 表示，其中 tan θ = v₂ / v₁。求 θ 时需要考虑象限，因为仅用反正切函数可能会产生歧义。

4. Acceleration and Vector Differentiation | 加速度与矢量求导

Acceleration is the rate of change of velocity with respect to time. It is the first derivative of velocity and the second derivative of position:

加速度是速度关于时间的变化率。它是速度的一阶导数，也是位置的二阶导数：

a = dv/dt = d²r/dt²

In component form, if v = v₁ i + v₂ j, then a = (dv₁/dt) i + (dv₂/dt) j. A particle accelerates if the magnitude of velocity changes, if the direction of motion changes, or both.

用分量形式表示，若 v = v₁ i + v₂ j，则 a = (dv₁/dt) i + (dv₂/dt) j。当速度大小改变、方向改变或两者同时改变时，质点都具有加速度。

For example, a particle moving at constant speed around a circle still has acceleration because its direction is continuously changing. In A-Level questions, you will usually differentiate given position or velocity vectors component by component.

例如，以恒定速率做圆周运动的质点仍然具有加速度，因为其方向在不断变化。在 A-Level 考题中，你通常需要对给出的位置或速度矢量逐分量求导。

5. Integrating Acceleration in 2D | 二维加速度积分

If acceleration is given as a function of time, you can integrate each component separately to find velocity, then integrate again to find position. Do not forget the constant of integration, which is found using initial conditions.

如果加速度作为时间的函数给出，你可以分别对每个分量积分求得速度，再积分一次求得位置。不要忘记积分常数，它们需要利用初始条件确定。

Example: a = 6t i + 2 j, with v(0) = 3i – 4j and r(0) = 0.

示例：a = 6t i + 2 j，且 v(0) = 3i – 4j，r(0) = 0。

Integrating acceleration gives v = 3t² i + 2t j + C. Using v(0) = 3i – 4j gives C = 3i – 4j, so:

对加速度积分得 v = 3t² i + 2t j + C。利用 v(0) = 3i – 4j 得 C = 3i – 4j，因此：

v = (3t² + 3) i + (2t – 4) j

Integrating velocity gives r = (t³ + 3t) i + (t² – 4t) j + D. Since r(0) = 0, we have D = 0. The position vector is therefore r = (t³ + 3t) i + (t² – 4t) j.

对速度积分得 r = (t³ + 3t) i + (t² – 4t) j + D。由于 r(0) = 0，得 D = 0。因此位置矢量为 r = (t³ + 3t) i + (t² – 4t) j。

6. Constant Acceleration Equations in Vector Form | 矢量形式的匀加速方程

When acceleration is constant, the usual SUVAT equations can be applied to each component independently. In vector form, with initial velocity u and constant acceleration a:

当加速度恒定时，通常的 SUVAT 方程可以分别应用于每个分量。用矢量形式表示，设初速度为 u，恒定加速度为 a：

v = u + a t

r = r₀ + u t + ½ a t²

Here r₀ is the initial position vector. In component form, if u = u₁ i + u₂ j and a = a₁ i + a₂ j, these become:

这里 r₀ 是初始位置矢量。用分量形式表示，若 u = u₁ i + u₂ j，a = a₁ i + a₂ j，可得：

v₁ = u₁ + a₁ t, v₂ = u₂ + a₂ t

x = x₀ + u₁ t + ½ a₁ t², y = y₀ + u₂ t + ½ a₂ t²

A common mistake is to treat the two components as if they interact. In motion with constant acceleration, the horizontal and vertical components are independent, so you solve them separately but the time t is the same for both.

常见错误是把两个分量当成相互影响。在匀加速运动中，水平与竖直分量是相互独立的，因此可以分别求解，但时间 t 对两个分量相同。

7. Projectile Motion: Component Setup | 抛体运动：分量建立

Projectile motion assumes that the only acceleration is gravity, acting vertically downwards. Taking i horizontal and j vertically upwards, the acceleration is a = -g j, so a₁ = 0 and a₂ = -g.

抛体运动假设唯一加速度是竖直向下的重力。取 i 水平、j 竖直向上，则加速度 a = -g j，因此 a₁ = 0，a₂ = -g。

If a projectile is launched with speed u at an angle θ above the horizontal, the initial velocity components are:

若抛体以速度 u 沿与水平方向成 θ 角斜向上发射，则初速度分量为：

u₁ = u cos θ, u₂ = u sin θ

Since horizontal acceleration is zero, horizontal velocity is constant. The vertical motion has constant acceleration -g. Therefore the position components are:

由于水平加速度为零，水平速度保持不变。竖直方向具有恒定加速度 -g。因此位置分量为：

x = u cos θ × t

y = u sin θ × t – ½ g t²

Use g = 9.8 m s⁻² unless told otherwise. Always state a consistent positive direction and stick to it throughout the problem.

除非题目另有说明，通常取 g = 9.8 m s⁻²。必须始终明确一个正方向，并在整道题中保持一致。

8. Time of Flight, Range and Maximum Height | 飞行时间、射程与最大高度

For a projectile launched from and landing on the same horizontal level, the highest point occurs when the vertical velocity is zero: v₂ = u sin θ – g t = 0. Hence the time to highest point is:

对于在同一水平高度发射和落地的抛体，最高点出现在竖直速度为零时：v₂ = u sin θ – g t = 0。因此到达最高点的时间为：

t = u sin θ / g

The total time of flight is twice this value, because the upward and downward sections are symmetric:

总飞行时间是该值的两倍，因为上升段与下降段是对称的：

T = 2u sin θ / g

The range R is the horizontal distance travelled in the total flight time. Since horizontal velocity is constant:

射程 R 是总飞行时间内的水平距离。由于水平速度恒定：

R = u cos θ × T = (u² sin 2θ) / g

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Motion in 2 Dimensions | 二维运动