Further Kinematics | 进阶运动学

📚 Further Kinematics | 进阶运动学

In Edexcel A Level Mathematics, Further Kinematics extends the kinematics you met at AS level into two dimensions using vectors, variable acceleration and projectile motion. You will use calculus to move between displacement, velocity and acceleration, both in one dimension and in i-j form, and you will apply these ideas to model projectiles under gravity.

在 Edexcel A Level 数学中，进阶运动学将你在 AS 阶段学过的运动学推广到二维，使用向量、变加速度和抛体运动。你将运用微积分在位移、速度和加速度之间转换，包括一维形式和 i-j 向量形式，并将这些方法应用于重力作用下的抛体模型。

1. Vector Kinematics Basics | 向量运动学基础

In further kinematics, position, velocity and acceleration are often written as vectors. The position vector r gives the displacement from a fixed origin; the velocity vector v is the rate of change of position; and the acceleration vector a is the rate of change of velocity.

在进阶运动学中，位置、速度和加速度通常用向量表示。位置向量 r 给出质点相对固定原点的位移；速度向量 v 是位置的变化率；加速度向量 a 是速度的变化率。

If a particle moves so that its position vector is r = x(t) i + y(t) j, then each component is differentiated independently to find velocity and acceleration.

如果质点运动的位置向量为 r = x(t) i + y(t) j，那么每一个分量都可以独立求导，从而得到速度和加速度。

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

a = dv/dt = (d²x/dt²) i + (d²y/dt²) j

The speed of the particle is the magnitude of the velocity vector, and the direction of motion is the angle between the velocity vector and the positive i-direction.

质点的速率是速度向量的大小，运动方向是速度向量与正 i 方向之间的夹角。

2. Differentiation with Vectors | 向量的微分

When the position vector is given as a function of time, differentiating each component gives the velocity vector. Differentiating the velocity vector then gives the acceleration vector.

当位置向量表示为时间的函数时，对每个分量求导即可得到速度向量。再对速度向量求导，就可以得到加速度向量。

For example, if r = (3t² + 1) i + (t³ − 2t) j, then v = 6t i + (3t² − 2) j and a = 6 i + 6t j. The coefficients of i and j are treated as separate scalar functions.

例如，如果 r = (3t² + 1) i + (t³ − 2t) j，那么 v = 6t i + (3t² − 2) j，a = 6 i + 6t j。i 和 j 的系数被当作独立的标量函数处理。

This process is often tested with questions asking for the velocity and acceleration at a particular time, so after differentiating you should substitute the given t value carefully.

这类过程经常在题目中出现，要求你求出某一时刻的速度和加速度，因此在求导之后应仔细代入给定的 t 值。

3. Integration with Vectors | 向量的积分

If the acceleration vector is known as a function of time, integration is used to recover the velocity and position vectors. The constants of integration are determined by initial conditions such as the velocity or position at t = 0.

如果已知加速度向量是时间的函数，则可以通过积分还原速度向量和位置向量。积分常数由初始条件确定，例如 t = 0 时的速度或位置。

The standard process is to integrate acceleration to get velocity, then integrate velocity to get position. Each component has its own constant, so you must use the given initial vector for each direction.

标准步骤是先将加速度积分得到速度，再将速度积分得到位置。每个分量都有自己的常数，因此你必须对每个方向使用给定的初始向量。

v = ∫ a dt + c, and r = ∫ v dt + d

Here c and d are constant vectors, often found by substituting the initial velocity vector u and initial position vector r₀.

这里的 c 和 d 是常向量，通常通过代入初始速度向量 u 和初始位置向量 r₀ 来求出。

4. Variable Acceleration in One Dimension | 一维变加速度

When acceleration is not constant, it is often given as a function of time, a = f(t). The velocity is found by integrating acceleration, and displacement is found by integrating velocity.

当加速度不是常量时，它通常表示为时间的函数 a = f(t)。速度通过对加速度积分求得，位移通过对速度积分求得。

Key exam skills include finding maximum velocity, the time when a particle changes direction, and the distance travelled. If velocity changes sign, distance travelled is not the same as displacement, so you must split the motion into intervals.

关键的考试技能包括求最大速度、质点改变方向的时刻以及运动路程。如果速度改变符号，路程与位移并不相同，因此你必须把运动分成若干区间。

v = ∫ a dt, x = ∫ v dt

Initial conditions such as v(0) and x(0) are needed to determine the constants of integration, and they should be written down before substituting values.

求解积分常数时需要 v(0) 和 x(0) 等初始条件，在代入数值之前应先把这些条件写清楚。

5. Variable Acceleration in Two Dimensions | 二维变加速度

The same integration and differentiation ideas extend to two dimensions. If the acceleration vector has components that depend on time, integrate the i-component and j-component separately.

同样的积分和微分思想可以推广到二维。如果加速度向量的分量依赖于时间，则分别对 i 分量和 j 分量进行积分。

For example, if a = 2t i + 3 j, and the particle starts at rest at the origin, then v = t² i + 3t j and r = (t³/3) i + (3t²/2) j.

例如，如果 a = 2t i + 3 j，且质点从原点静止开始运动，则 v = t² i + 3t j，r = (t³/3) i + (3t²/2) j。

When the components are integrated, the resulting velocity and position vectors describe curved motion in the plane. You can still find the speed by calculating the magnitude of v.

当分量被积分后，得到的速度和位置向量描述的是平面上的曲线运动。你仍然可以通过计算 v 的大小来求速率。

6. Constant Acceleration with Vectors | 矢量形式的匀加速运动

If acceleration is constant in both magnitude and direction, the familiar SUVAT equations can be used in vector form. Each equation combines vectors, but the same structure applies.

如果加速度在大小和方向上都是恒定的，那么熟悉的 SUVAT 方程可以写成向量形式。每个方程都结合向量，但结构保持不变。

For constant acceleration a, with initial velocity u and displacement r from the starting point, the main vector equations are:

对于恒定加速度 a、初始速度 u 和相对起点的位移 r，主要的向量方程如下：

v = u + a t

r = u t + ½ a t²

r = ½ (u + v) t

In calculations, write each vector in i-j form and then work separately with the i-components and j-components to avoid mixing horizontal and vertical motion.

在计算中，将每个向量写成 i-j 形式，然后分别处理 i 分量和 j 分量，避免混合水平和竖直运动。

7. Projectile Motion: Modelling Assumptions | 抛体运动：建模假设

Projectile motion is an application of constant acceleration in two dimensions. The standard model assumes that the projectile is a particle, air resistance is negligible, and the only force acting is weight.

抛体运动是二维恒定加速度的一个应用。标准模型假设抛体是质点、空气阻力可以忽略，并且只受重力作用。

Under these assumptions, the horizontal component of velocity is constant, and the vertical motion has constant downward acceleration g, usually taken as 9.8 m s⁻² in Edexcel questions.

在这些假设下，速度的水平分量保持不变，竖直方向具有恒定的向下加速度 g，Edexcel 题目中通常取 g = 9.8 m s⁻²。

The motion is typically modelled from a point O with initial speed u at an angle θ above the horizontal. The horizontal and vertical components are then u cos θ and u sin θ.

运动通常从点 O 开始建立模型，初速度为 u，方向与水平面成 θ 角。水平分量为 u cos θ，竖直分量为 u sin θ。

8. Projectile Key Equations | 抛体运动的关键方程

Taking the initial position as the origin, the displacement components at time t are obtained by using constant velocity horizontally and constant acceleration vertically.

以初始位置为原点，在 t 时刻的位移分量可以通过水平方向的匀速运动和竖直方向的匀加速运动计算。

x = u t cos θ

y = u t sin θ − ½ g t²

The velocity components at time t are:

在 t 时刻的速度分量为：

vₓ = u cos θ

v_y = u sin θ − g t

These four equations are the starting point for finding the path equation, the time of flight, the range and the greatest height. Always choose upward as positive for y, so the initial vertical velocity is positive and gravity is negative.

这四个方程是求轨迹方程、飞行时间、射程和最大高度的出发点。始终选择向上为 y 的正方向，因此初始竖直速度为正，重力为负。

9. Path Equation, Time of Flight, Range and Greatest Height | 轨迹方程、飞行时间、射程和最大高度

The path of a projectile can be found by eliminating t from the x and y equations. This gives a quadratic equation for y in terms of x, showing that the trajectory is a parabola.

通过从 x 和 y 方程中消去 t，可以得到抛体的轨迹方程。这是 y 关于 x 的二次方程，表明轨迹是抛物线。

y = x tan θ − g x² / (2 u² cos² θ)

The time of flight T is found by setting y = 0 for a projectile that returns to its original vertical level.

对于回到原来竖直高度的抛体，令 y = 0 即可求出飞行时间 T。

T = 2 u sin θ / g

The horizontal range R is the value of x when t = T, and the greatest height H occurs when the vertical component of velocity is zero.

水平射程 R 是 t = T 时的 x 值；最大高度 H 出现在速度竖直分量为零的时刻。

R = u² sin 2θ / g

H = u² sin² θ / (2 g)

These standard results are valid only when the launch point and landing point are at the same vertical height. If the projectile is thrown from a height above the ground, you must return to the original equations and solve the relevant quadratic.

这些标准结果仅在抛出点和落地点处于相同竖直高度时成立。如果抛体从某一高度抛出，你必须回到原始方程并求解相应的二次方程。

10. Worked Example | 实例分析

A ball is projected from ground level with speed 25 m s⁻¹ at an angle of 40° above the horizontal. Take g = 9.8 m s⁻². Find the time of flight, the horizontal range and the greatest height.

一个球从地面以 25 m s⁻¹ 的速度、与水平面成 40° 角抛出。取 g = 9.8 m s⁻²。求飞行时间、水平射程和最大高度。

First, u sin θ = 25 sin 40° ≈ 16.07 m s⁻¹, so the time of flight is T = 2 u sin θ / g = 2 × 16.07 / 9.8 ≈ 3.28 s.

首先，u sin θ = 25 sin 40° ≈ 16.07 m s⁻¹，因此飞行时间为 T = 2 u sin θ / g = 2 × 16.07 / 9.8 ≈ 3.28 s。

The range is R = u² sin 2θ / g = 625 sin 80° / 9.8 ≈ 62.8 m, and the greatest height is H = u² sin² θ / (2g) = 625 sin² 40° / 19.6 ≈ 13.2 m.

水平射程为 R = u² sin 2θ / g = 625 sin 80° / 9.8 ≈ 62.8 m，最大高度为 H = u² sin² θ / (2g) = 625 sin² 40° / 19.6 ≈ 13.2 m。

If the ball were projected from a height, for example from a cliff, you would set y = −h at the landing point and solve the resulting quadratic in t before calculating the range.

如果球是从某一高度抛出，例如从悬崖上，你应令落地点的 y = −h，然后求解由此得到的关于 t 的二次方程，再计算射程。

11. Exam Tips and Common Errors | 考试提示与常见错误

In exam questions, draw a clear diagram showing the initial velocity, its horizontal and vertical components, and the positive direction. This helps avoid sign errors, especially with upward and downward motion.

在考试题中，画出清晰的示意图，标出初速度、水平与竖直分量以及正方向。这有助于避免符号错误，尤其是在涉及向上和向下运动时。

Do not use the range or time-of-flight formula when the launch point and landing point have different vertical heights. Instead, use the full equations x = u t cos θ and y = u t sin θ − ½ g t² and substitute the given landing condition.

当抛出点和落地点不在同一竖直高度时，不要直接使用射程公式或飞行时间公式。应使用完整方程 x = u t cos θ 和 y = u t sin θ − ½ g t²，并代入给定的落地条件。

When integrating vectors, always add the constant vector and use the initial conditions for both components. A common mistake is to forget that the constant has an i and a j part.

对向量积分时，一定要加上常向量，并对两个分量都使用初始条件。常见错误是忘记常数包含 i 分量和 j 分量。

Finally, remember that distance travelled is not the same as displacement when the velocity changes sign

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Further Kinematics | 进阶运动学