Kinematics – describing motion | 运动学 – 描述运动

📚 Kinematics – describing motion | 运动学 – 描述运动

Kinematics is the branch of mechanics that deals with the description of motion without considering its causes. Understanding how to describe motion using quantities like displacement, velocity and acceleration – and interpreting motion through graphs – forms the foundation for all of A‑Level Physics. This article revisits the key concepts required for CIE A‑Level, from scalars and vectors to the analysis of projectile motion in two dimensions.

运动学是力学中描述物体运动而不考虑其成因的分支。掌握如何用位移、速度、加速度等物理量描述运动，以及通过图像解读运动，是A‑Level物理的基石。本文重新梳理CIE A‑Level所需的全部核心概念——从标量与矢量到二维抛体运动分析。

1. Scalars and Vectors in Motion | 运动中的标量与矢量

All kinematic quantities can be classified as either scalars or vectors. A scalar has only magnitude (size), while a vector has both magnitude and direction. In describing motion, direction is crucial for complete understanding.

所有运动学量都可以分为标量或矢量。标量仅有大小，矢量既有大小又有方向。在描述运动时，方向对于完整理解至关重要。

Scalar examples: distance, speed, time, mass.
标量例子：距离、速率、时间、质量。
Vector examples: displacement, velocity, acceleration, force.
矢量例子：位移、速度、加速度、力。

When solving problems, always assign a positive direction and treat vector components accordingly. A velocity of −5 m s⁻¹ means 5 m s⁻¹ in the negative direction.

解题时务必规定正方向并相应地处理矢量分量。速度为 −5 m s⁻¹ 表示沿负方向 5 m s⁻¹。

2. Distance and Displacement | 距离与位移

Distance is a scalar measuring the total path length travelled. Displacement is a vector drawn from the initial position to the final position in a straight line, together with the direction.

距离是标量，测量运动轨迹的总长度。位移是矢量，是从起点指向终点的有向直线段及其方向。

If a car drives 3 km east and then 4 km west, the total distance is 7 km, but the displacement is 1 km west. The distinction is vital when calculating average quantities.

若一辆汽车先向东行驶 3 km，再向西行驶 4 km，总距离为 7 km，而位移为向西 1 km。在计算平均量时这一区别极为重要。

3. Speed and Velocity | 速率与速度

Speed is the rate of change of distance. Velocity is the rate of change of displacement. Average speed = total distance ÷ total time. Average velocity = resultant displacement ÷ total time.

速率是距离的变化率。速度是位移的变化率。平均速率 = 总距离 ÷ 总时间。平均速度 = 合位移 ÷ 总时间。

average velocity, v_av = Δs / Δt

Instantaneous velocity is the velocity at a specific instant, found from the gradient of a displacement–time graph or from calculus. In A‑Level, we often use initial and final velocities with uniform acceleration.

瞬时速度是某一瞬间的速度，可由位移–时间图的斜率或微积分求得。在A‑Level中，我们常使用匀加速条件下的初速度和末速度。

4. Acceleration | 加速度

Acceleration is defined as the rate of change of velocity. It is a vector, so a change in the magnitude or direction of velocity results in acceleration. Deceleration is simply acceleration in the opposite direction to motion.

加速度定义为速度的变化率。它是矢量，因此速度大小或方向的改变都会产生加速度。减速只是与运动方向相反的加速度。

a = (v − u) / t

where u is initial velocity, v is final velocity, and t is time taken. The unit is m s⁻². Uniform acceleration means the velocity changes by equal amounts in equal time intervals.

其中 u 为初速度，v 为末速度，t 为所用时间。单位是 m s⁻²。匀加速意味着在相等时间内速度变化量相等。

5. Equations of Uniformly Accelerated Motion | 匀加速运动方程

For motion in a straight line with constant acceleration, four kinematic equations (suvat equations) link the five quantities s, u, v, a, t. These are derived from the definitions of velocity and acceleration.

对于匀加速直线运动，四个运动学方程（suvat 方程）将五个物理量 s、u、v、a、t 联系起来。它们源自速度与加速度的定义。

v = u + a t
s = u t + ½ a t²
v² = u² + 2 a s
s = ½ (u + v) t

s is displacement in metres, u initial velocity in m s⁻¹, v final velocity in m s⁻¹, a acceleration in m s⁻², t time in seconds. Remember to use consistent sign conventions.

s 为位移（米），u 为初速度 (m s⁻¹)，v 为末速度 (m s⁻¹)，a 为加速度 (m s⁻²)，t 为时间（秒）。注意保持一致的符号约定。

Select the equation that does not contain the unknown quantity you are not asked for. For example, if you need to find s and you are not asked for v, use s = u t + ½ a t².

选择不包含题目未要求的未知量的方程。例如，若需求 s 且不要求 v，则使用 s = u t + ½ a t²。

6. Free Fall under Gravity | 重力作用下的自由落体

A body falling freely near the Earth’s surface experiences a constant downward acceleration due to gravity, g, typically 9.81 m s⁻². The kinematic equations apply directly by setting a = g or a = −g depending on the chosen positive direction.

在地球表面附近自由下落的物体受到恒定的向下重力加速度 g，通常为 9.81 m s⁻²。根据选定的正方向，设 a = g 或 a = −g，可直接套用运动学方程。

If upward is taken as positive, a = −9.81 m s⁻². A ball thrown upward moves upward while slowing down, stops momentarily, and then accelerates downward. At the highest point, v = 0 but a = −9.81 m s⁻², never zero.

若向上为正，则 a = −9.81 m s⁻²。向上抛出的球向上运动时减速，瞬间停止后向下加速。在最高点处 v = 0，但 a = −9.81 m s⁻²，永远不为零。

7. Displacement–Time Graphs | 位移–时间图

A displacement–time graph (s‑t graph) plots displacement on the vertical axis and time on the horizontal. The gradient of the graph gives the instantaneous velocity.

位移–时间图（s‑t 图）纵轴为位移，横轴为时间。图线的斜率表示瞬时速度。

A straight, sloping line → constant velocity.
倾斜直线 → 匀速运动。
A horizontal line → stationary (zero velocity).
水平线 → 静止（速度为零）。
A curved line → changing velocity (acceleration).
曲线 → 变速运动（加速度存在）。

The sign of the gradient indicates direction. A decreasing gradient means deceleration or acceleration in the negative direction, depending on context.

斜率的正负表示方向。斜率减小可能意味着减速或沿负方向加速，需结合具体情境判断。

8. Velocity–Time Graphs | 速度–时间图

A velocity–time graph (v‑t graph) is extremely powerful: its gradient equals acceleration, and the area under the graph between two times represents the displacement during that interval.

速度–时间图（v‑t 图）功能极强：其斜率等于加速度，图线与时间轴之间所围面积表示该时间段内的位移。

Straight sloping line → constant acceleration.
倾斜直线 → 匀加速。
Horizontal line → constant velocity (zero acceleration).
水平线 → 匀速（零加速度）。
Area above time axis → positive displacement; area below → negative displacement.
时间轴上方面积 → 正位移；下方面积 → 负位移。

To calculate total displacement, add areas algebraically. To find total distance, add the absolute values of all areas.

计算总位移时，代数相加各面积；计算总距离则取所有面积的绝对值之和。

9. Acceleration–Time Graphs | 加速度–时间图

Acceleration–time graphs (a‑t graphs) show how acceleration varies with time. The area under the curve between two times gives the change in velocity over that interval.

加速度–时间图（a‑t 图）展示加速度随时间的变化。图线在时间区间内的面积等于该时段内的速度变化量。

A constant positive acceleration is shown by a horizontal line above the time axis. A sudden jump to zero area means velocity no longer changes. Sudden changes in acceleration are common in exam graphs, e.g., when engines or brakes are applied.

加速度恒为正时，图线是时间轴上方的一条水平线。面积突变为零意味着速度不再变化。加速度的突变在考题图像中很常见，例如发动机或制动器介入时。

Be sure to relate the a‑t graph to the v‑t and s‑t graphs: the gradient of v‑t gives a; the area of a‑t gives Δv.

务必联系 a‑t 图与 v‑t、s‑t 图：v‑t 的斜率得出 a；a‑t 的面积得出 Δv。

10. Projectile Motion | 抛体运动

A projectile is an object moving under the influence of gravity alone, after being launched. Projectile motion is analysed by treating horizontal and vertical components independently, because the horizontal acceleration is zero and the vertical acceleration is g (downward).

抛体是抛出后仅受重力影响的运动物体。抛体运动通过独立分析水平与竖直分量来处理，因为水平方向加速度为零，竖直方向加速度为向下的 g。

Horizontal component: constant velocity u_x = u cos θ.
水平分量：匀速，u_x = u cos θ。
Vertical component: constant acceleration a_y = −g (if upward is positive). Initial vertical velocity u_y = u sin θ.
竖直分量：匀加速 a_y = −g（取向上为正），初速竖直分量 u_y = u sin θ。

The time of flight is determined entirely by the vertical motion. The horizontal range is simply the constant horizontal velocity multiplied by the time of flight.

飞行时间完全由竖直运动决定。水平射程就是不变的水平速度乘以飞行时间。

11. Analysing Projectile Motion Components | 分析抛体运动分量

For a projectile launched from ground level at speed u and angle θ to the horizontal:

对于从地面以速度 u、仰角 θ 发射的抛体：

Time to reach highest point: t_peak = u sin θ / g

Total time of flight: t_total = 2 u sin θ / g

Maximum height: H = (u² sin² θ) / (2g)

Horizontal range: R = (u² sin 2θ) / g

All these results follow from applying the suvat equations to the vertical motion and using the horizontal constant-velocity relationship.

所有这些结果都是将 suvat 方程应用于竖直运动，并结合水平匀速关系推导而出的。

Note that the maximum range for a given launch speed occurs when θ = 45°, because sin 2θ is maximised at 1. Complementary angles (e.g., 30° and 60°) give the same range.

注意，给定发射速度下的最大射程出现在 θ = 45°，因为此时 sin 2θ 取最大值 1。互余角（如 30° 和 60°）产生相同的射程。

12. Applying Kinematic Equations: A Worked Example | 运动学方程应用实例

A common exam question: A ball is thrown vertically upward with speed 20 m s⁻¹ from a height of 1.5 m above the ground. Find the total time before it hits the ground. (Take g = 9.81 m s⁻².)

常见考题：一球以 20 m s⁻¹ 的速度从离地 1.5 m 高处竖直向上抛出。求球落地前经过的总时间。（取 g = 9.81 m s⁻²。）

Solution: Set upward as positive. Given u = +20 m s⁻¹, a = −9.81 m s⁻², displacement to ground s = −1.5 m (since ground is below launch point). Use s = u t + ½ a t².

解答：设向上为正。已知 u = +20 m s⁻¹，a = −9.81 m s⁻²，落地位移 s = −1.5 m（地面在抛出点下方）。使用 s = u t + ½ a t²。

−1.5 = 20 t + ½ (−9.81) t² → 4.905 t² − 20 t − 1.5 = 0. Solve the quadratic: t ≈ 4.15 s (discard the negative root). This type of multi‑step problem tests sign conventions and selection of the correct equation.

−1.5 = 20 t + ½ (−9.81) t² → 4.905 t² − 20 t − 1.5 = 0。解二次方程得 t ≈ 4.15 s（舍去负根）。这类多步骤问题考查符号约定与正确选用方程的能力。

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Kinematics – describing motion | 运动学 – 描述运动