IB Physics: Kinematics Fundamentals and Description | 运动学基础与描述

📚 IB Physics: Kinematics Fundamentals and Description | 运动学基础与描述

Kinematics is the branch of physics that describes motion without considering its causes. In this article, we will explore the fundamental concepts of displacement, velocity, acceleration, and the mathematical tools used to describe motion in one and two dimensions.

运动学是物理学中描述运动而不考虑其成因的分支。在本文中,我们将探讨位移、速度、加速度等基本概念,以及用于描述一维和二维运动的数学工具。


1. Reference Frames and Position | 参考系与位置

A reference frame is a coordinate system in which the position of an object is measured. The position of an object is defined relative to a chosen origin and is a vector quantity, often denoted by \(\mathbf{r}\) or simply by a coordinate \(x\). In IB Physics, we usually work with a fixed Earth-based frame unless otherwise stated.

参考系是一个用于测量物体位置的坐标系。物体的位置是相对于选定的原点定义的,它是一个矢量量,通常用 \(\mathbf{r}\) 或坐标 \(x\) 表示。在 IB 物理中,除非另有说明,我们通常使用以地球为参照的固定参考系。

For one-dimensional motion along a straight line, position is described by a single coordinate. For example, if a particle is 3 m to the right of the origin, we write \(x = +3\ \text{m}\).

对于沿直线的一维运动,位置由单一坐标描述。例如,若一个质点位于原点右侧 3 m 处,我们写作 \(x = +3\ \text{m}\)。


2. Distance and Displacement | 路程与位移

Distance is a scalar quantity that measures the total length of the path travelled, regardless of direction. Displacement, on the other hand, is a vector quantity that measures the change in position from the initial point to the final point.

路程是标量,它度量的是所经过路径的总长度,与方向无关。位移则是矢量,它度量的是从起点到终点的位置变化。

Mathematically, displacement is calculated as \(\Delta x = x_f – x_i\), where \(x_f\) is the final position and \(x_i\) is the initial position. If an object returns to its starting point, its displacement is zero, even though the distance travelled may be large.

数学上,位移计算公式为 \(\Delta x = x_f – x_i\),其中 \(x_f\) 是末位置,\(x_i\) 是初位置。如果物体回到出发点,其位移为零,即使走过的路程可能很大。

  • Distance is always positive and never decreases with time.

    路程始终为正,且不会随时间减少。

  • Displacement can be positive, negative, or zero depending on direction.

    位移可以为正、负或零,这取决于方向。


3. Speed and Velocity | 速率与速度

Speed is the scalar rate at which distance is covered, defined as \(v = \frac{\text{distance}}{\text{time}}\). Velocity is the vector rate of change of displacement, defined as \(v = \frac{\Delta x}{\Delta t}\).

速率是路程被覆盖的标量率,定义为 \(v = \frac{\text{路程}}{\text{时间}}\)。速度是位移变化的矢量率,定义为 \(v = \frac{\Delta x}{\Delta t}\)。

For example, a car driving 100 km north in 2 hours has an average speed of 50 km/h and an average velocity of 50 km/h north. If it returns to the starting point in another 2 hours, the average speed is 50 km/h, but the average velocity over the whole journey is zero because the total displacement is zero.

例如,一辆汽车向北行驶 100 km,用时 2 小时,其平均速率为 50 km/h,平均速度为 50 km/h 方向向北。如果它再用 2 小时返回出发点,全程的平均速率为 50 km/h,但由于总位移为零,全程平均速度为零。

Instantaneous velocity is the velocity at a particular instant, obtained by taking the limit \(\Delta t \to 0\). On a position-time graph, the instantaneous velocity is the slope of the tangent line at that point.

瞬时速度是某一时刻的速度,通过取极限 \(\Delta t \to 0\) 获得。在位置-时间图像上,瞬时速度等于该点切线的斜率。


4. Acceleration | 加速度

Acceleration is the rate of change of velocity with respect to time. As a vector quantity, it is defined by \(a = \frac{\Delta v}{\Delta t}\). The SI unit of acceleration is metres per second squared (\(\text{m/s}^2\)).

加速度是速度随时间的变化率。作为矢量量,其定义为 \(a = \frac{\Delta v}{\Delta t}\)。加速度的国际单位是米每二次方秒(\(\text{m/s}^2\))。

Acceleration can be positive, negative, or zero. A negative acceleration does not always mean the object is slowing down; it depends on the direction of velocity. For instance, if a car moving in the negative direction speeds up, its acceleration is also negative.

加速度可以为正、负或零。负加速度并不总是意味着物体在减速,这取决于速度的方向。例如,如果一辆汽车沿负方向加速行驶,其加速度也是负的。

When velocity is constant, acceleration is zero. When acceleration is constant, we say the motion is uniformly accelerated.

当速度恒定时,加速度为零。当加速度恒定时,我们称之为匀变速运动。


5. Uniformly Accelerated Motion: The SUVAT Equations | 匀变速运动:SUVAT 方程

For an object moving with constant acceleration, there are four kinematic equations that relate displacement \(s\), initial velocity \(u\), final velocity \(v\), acceleration \(a\), and time \(t\). These are often called the SUVAT equations.

对于做匀变速直线运动的物体,有四个运动学方程,将位移 \(s\)、初速度 \(u\)、末速度 \(v\)、加速度 \(a\) 和时间 \(t\) 联系起来。它们通常被称为 SUVAT 方程。

v = u + at

s = ut + ½at²

v² = u² + 2as

s = ½(u + v)t

These equations are only valid under constant acceleration. You must choose the equation that contains the unknown quantity and three known quantities.

这些方程仅在加速度恒定时成立。你必须选择包含未知量以及三个已知量的方程。

For example, a ball is thrown upward with an initial velocity of 20 m/s. What is its displacement after 3 s? Taking upward as positive, we have \(u = 20\ \text{m/s}\), \(a = -9.8\ \text{m/s}^2\), \(t = 3\ \text{s}\). Using \(s = ut + ½at²\), we get \(s = 20 \times 3 + ½ \times (-9.8) \times 3^2 = 60 – 44.1 = 15.9\ \text{m}\).

例如,一个小球以 20 m/s 的初速度竖直上抛。3 s 后它的位移是多少?取向上为正,则 \(u = 20\ \text{m/s}\),\(a = -9.8\ \text{m/s}^2\),\(t = 3\ \text{s}\)。使用 \(s = ut + ½at²\),得 \(s = 20 \times 3 + ½ \times (-9.8) \times 3^2 = 60 – 44.1 = 15.9\ \text{m}\)。


6. Free Fall and Gravitational Acceleration | 自由落体与重力加速度

Free fall is the motion of an object under the influence of gravity alone. Near the Earth’s surface, the acceleration due to gravity is approximately \(g = 9.8\ \text{m/s}^2\), directed downward.

自由落体是物体仅受重力作用下的运动。在地球表面附近,重力加速度约为 \(g = 9.8\ \text{m/s}^2\),方向竖直向下。

In the absence of air resistance, all objects fall with the same acceleration regardless of their mass. This principle was famously demonstrated by Galileo and later by astronauts on the Moon.

在忽略空气阻力的情况下,所有物体无论质量大小,都以相同的加速度下落。这一原理由伽利略著名地证实,后来也由宇航员在月球上演示。

When solving free-fall problems, it is crucial to choose a consistent sign convention. If the positive direction is upward, then \(a = -g\). The SUVAT equations apply with \(a = -g\).

在解决自由落体问题时,选择一致的符号约定至关重要。若取向上为正方向,则 \(a = -g\)。SUVAT 方程在 \(a = -g\) 的情况下依然适用。

A useful result: an object thrown upward with speed \(u\) reaches its maximum height when \(v = 0\). The time to reach maximum height is \(t = u/g\), and the maximum height is \(H = u^2/(2g)\).

一个有用的结论:以速率 \(u\) 竖直上抛的物体,当 \(v = 0\) 时达到最大高度。到达最大高度的时间为 \(t = u/g\),最大高度为 \(H = u^2/(2g)\)。


7. Motion Graphs: Position-Time, Velocity-Time, Acceleration-Time | 运动图像:位置-时间、速度-时间、加速度-时间

Graphs provide a visual representation of motion and allow us to extract kinematic information quickly.

图像提供了运动的直观表示,使我们能够快速提取运动学信息。

Position-time (s-t) graph: The slope of the s-t graph gives the velocity. A straight line means constant velocity; a curve means changing velocity.

位置-时间(s-t)图像: s-t 图像的斜率给出速度。直线代表匀速运动;曲线代表变速运动。

Velocity-time (v-t) graph: The slope of the v-t graph gives the acceleration. The area under the v-t graph gives the displacement.

速度-时间(v-t)图像: v-t 图像的斜率给出加速度。v-t 图像下方的面积给出位移。

Acceleration-time (a-t) graph: The area under the a-t graph gives the change in velocity.

加速度-时间(a-t)图像: a-t 图像下方的面积给出速度的变化量。

Graph | 图像 Slope | 斜率 Area | 面积
s-t Velocity | 速度 Not used | 不常用
v-t Acceleration | 加速度 Displacement | 位移
a-t Jerk (not in IB) | 加加速度(IB不要求) Change in velocity | 速度变化量

8. Interpreting Graph Features | 解读图像特征

When interpreting motion graphs, pay attention to intercepts, turning points, and intervals where the graph crosses the time axis.

在解读运动图像时,需要注意截距、拐点以及图像与时间轴相交的区间。

On an s-t graph, a horizontal segment means the object is at rest. A segment with positive slope means moving in the positive direction; a segment with negative slope means moving in the negative direction. The y-intercept gives the initial position.

在 s-t 图像中,水平线段表示物体静止。斜率为正的线段表示沿正方向运动;斜率为负的线段表示沿负方向运动。y 轴截距给出初始位置。

On a v-t graph, a horizontal segment means constant velocity, which implies zero acceleration. The y-intercept gives the initial velocity. If the graph crosses the time axis, the object changes direction.

在 v-t 图像中,水平线段表示匀速运动,即加速度为零。y 轴截距给出初速度。如果图像与时间轴相交,物体改变运动方向。

For the area under a v-t graph, regions above the time axis contribute positive displacement, and regions below contribute negative displacement. The net displacement is the algebraic sum.

对于 v-t 图像下方的面积,时间轴上方的区域贡献正的位移,时间轴下方的区域贡献负的位移。净位移是它们的代数和。


9. Relative Motion | 相对运动

Relative velocity describes the velocity of one object as observed from another moving object. If object A has velocity \(v_A\) and object B has velocity \(v_B\), both measured in the same reference frame, then the velocity of A relative to B is \(v_{AB} = v_A – v_B\).

相对速度描述的是从一个运动物体上观察另一个物体时的速度。如果物体 A 的速度为 \(v_A\),物体 B 的速度为 \(v_B\),两者在同一参考系中测量,则 A 相对 B 的速度为 \(v_{AB} = v_A – v_B\)。

For example, two cars moving in opposite directions on a straight road, one at +30 m/s and the other at -20 m/s, have a relative speed of 50 m/s. If they move in the same direction, the relative speed is 10 m/s.

例如,两辆汽车在直道上相向而行,一辆速度为 +30 m/s,另一辆为 -20 m/s,它们的相对速度为 50 m/s。若同向行驶,则相对速度为 10 m/s。

Relative motion problems often involve boats crossing rivers or aircraft flying in wind. The key is to use vector addition and choose a consistent reference frame.

相对运动问题通常涉及船过河或飞机在风中飞行。关键在于使用矢量加法并选择一致的参考系。


10. Two-Dimensional Motion: Projectile Motion Basics | 二维运动:抛体运动基础

When an object is launched at an angle to the horizontal, its motion can be separated into independent horizontal and vertical components. This is known as projectile motion.

当物体以与水平方向成一定角度的初速度被抛出时,其运动可分解为相互独立的水平分量和竖直分量。这称为抛体运动。

The horizontal motion has zero acceleration (ignoring air resistance), so the horizontal velocity remains constant. The vertical motion has constant acceleration \(g\) downward, so the vertical velocity changes with time.

水平方向运动的加速度为零(忽略空气阻力),因此水平速度保持不变。竖直方向运动具有向下的恒定加速度 \(g\),因此竖直速度随时间变化。

Using the SUVAT equations, if the initial speed is \(u\) and the angle of projection is \(\theta\), the horizontal component is \(u_x = u \cos \theta\) and the vertical component is \(u_y = u \sin \theta\).

利用 SUVAT 方程,若初速度为 \(u\),抛射角为 \(\theta\),则水平分量为 \(u_x = u \cos \theta\),竖直分量为 \(u_y = u \sin \theta\)。

The time of flight, maximum height, and range can be derived from these components. For a level landing at the same height, the range is \(R = \frac{u^2 \sin 2\theta}{g}\).

飞行时间、最大高度和射程都可以由这些分量推导得出。对于落点与起点在同一高度的情况,射程为 \(R = \frac{u^2 \sin 2\theta}{g}\)。


11. Common Pitfalls and Exam Tips | 常见误区与考试技巧

  • Confusing distance and displacement. Always check whether the question asks for a scalar or vector quantity.

    混淆路程和位移。始终检查问题要求的是标量还是矢量。

  • Forgetting that velocity and acceleration have direction. Use a clear sign convention.

    忘记速度和加速度具有方向性。使用明确的符号约定。

  • Applying SUVAT equations when acceleration is not constant. These equations are only valid for constant acceleration.

    在加速度不恒定时使用 SUVAT 方程。这些方程仅对匀变速运动成立。

  • Misreading graphs: the slope of an s-t graph is velocity, not acceleration; the area under a v-t graph is displacement, not distance.

    误读图像:s-t 图像的斜率是速度,不是加速度;v-t 图像下方的面积是位移,不是路程。

  • For projectile motion, forgetting to treat horizontal and vertical components separately.

    对于抛体运动,忘记分别处理水平分量和竖直分量。

  • Not stating the direction of a vector answer. In IB exams, velocity and displacement answers require direction.

    不注明矢量答案的方向。在 IB 考试中,速度和位移答案需要方向。


12. Worked Example and Problem-Solving Strategy | 例题与解题策略

A ball is dropped from rest from a height of 45 m. How long does it take to reach the ground? What is its speed just before impact? Take \(g = 10\ \text{m/s}^2\) and ignore air resistance.

一个球从 45 m 高处由静止释放。它需要多长时间落到地面?落地前瞬间的速度是多少?取 \(g = 10\ \text{m/s}^2\),忽略空气阻力。

Using \(s = ut + ½at²\), with \(u = 0\), \(s = 45\ \text{m}\), and \(a = 10\ \text{m/s}^2\):

使用 \(s = ut + ½at²\),其中 \(u = 0\),\(s = 45\ \text{m}\),\(a = 10\ \text{m/s}^2\):

45 = 0 + ½ × 10 × t²

t² = 9, so t = 3 s

Then using \(v = u + at\), we have \(v = 0 + 10 \times 3 = 30\ \text{m/s}\) downward.

再利用 \(v = u + at\),得 \(v = 0 + 10 \times 3 = 30\ \text{m/s}\),方向向下。

A systematic approach to kinematics problems: 1) Identify the known and unknown quantities. 2) Choose a positive direction. 3) Select the SUVAT equation that fits. 4) Substitute and solve. 5) Check the reasonableness and include direction for vectors.

运动学问题的系统方法:1)识别已知量和未知量。2)选择正方向。3)选择合适的 SUVAT 方程。4)代入求解。5)检查合理性,并为矢量注明方向。


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