Reference Frames and Relative Motion · 参考系与相对运动
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| relative/ˈrelətɪv/ | 相对 | xiāng duì |
| reference frame/ˈrefrəns freɪm/ | 参考系 | cān kǎo xì |
| resultant/rɪˈzʌltənt/ | 合矢量 | hé shǐ liàng |
You are sitting still — at 30 km/s
- Right now you feel motionless. Yet Earth carries you around the Sun at about $30\ \tfrac{\text{km}}{\text{s}}$.
- Both are true. "At rest" is only meaningful relative to something.
- Motion is always measured from a chosen viewpoint — a reference frame 参考系.
- Change the frame and the same object can be still, slow, or racing.
你正静静坐着——以 30 km/s
- 此刻你感觉一动不动。然而地球正带着你以约 $30\ \tfrac{\text{km}}{\text{s}}$ 绕太阳飞行。
- 两者都对。"静止"只有相对于某物时才有意义。
- 运动总是从一个选定的视角来测量——一个参考系。
- 换个参考系,同一个物体可以是静止、缓慢、或飞驰。
What a reference frame is
- A reference frame is a viewpoint with its own origin and axes for measuring position.
- The ground, a moving train, a flowing river — each is a valid frame.
- A velocity only means something once you say "relative to which frame".
- Physics laws work in any frame moving at constant velocity (an inertial frame).
什么是参考系
- 参考系是一个视角,有自己的原点和坐标轴来测量位置。
- 地面、行驶的火车、流动的河水——每一个都是有效的参考系。
- 速度只有在你说明"相对于哪个参考系"之后才有意义。
- 物理定律在任何匀速运动的参考系(惯性系)中都成立。
Why can we say you are both "at rest" and "moving at $30\ \tfrac{\text{km}}{\text{s}}$" at once? · 为什么我们可以同时说你“静止”和“以$30\ \tfrac{\text{km}}{\text{s}}$的速度运动”？
Relative to the ground you are at rest; relative to the Sun you move with the Earth. Velocity is frame-dependent. · 相对于地面你是静止的；相对于太阳你随地球一起运动。速度是依赖于参考系的。
Select all · 所有 of these that can serve as a reference frame. · 选择所有可以作为参考系的选项。
Any physical object with an origin and axes can be a frame. A bare number is not a viewpoint. · 任何具有原点和坐标轴的物理物体都可以作为参考系。单纯的数字不是观察视角。
Adding velocities on a line
- A passenger walks at $+2\ \tfrac{\text{m}}{\text{s}}$ inside a train that moves at $+30\ \tfrac{\text{m}}{\text{s}}$.
- To the ground, their velocity is $30 + 2 = 32\ \tfrac{\text{m}}{\text{s}}$.
- Walk toward the back ($-2$) and the ground sees $30 - 2 = 28\ \tfrac{\text{m}}{\text{s}}$.
- In one dimension, relative 相对 velocities just add as signed numbers.
在一条线上相加速度
- 一名乘客在一列以 $+30\ \tfrac{\text{m}}{\text{s}}$ 行驶的火车内部以 $+2\ \tfrac{\text{m}}{\text{s}}$ 行走。
- 相对地面,他的速度是 $30 + 2 = 32\ \tfrac{\text{m}}{\text{s}}$。
- 若朝车尾走($-2$),地面看到的是 $30 - 2 = 28\ \tfrac{\text{m}}{\text{s}}$。
- 在一维中,相对速度就当作带符号的数直接相加。
A passenger walks at $+2\ \tfrac{\text{m}}{\text{s}}$ toward the front of a train moving at $+30\ \tfrac{\text{m}}{\text{s}}$. What is the passenger's velocity relative to the ground, in $\tfrac{\text{m}}{\text{s}}$? · 一名乘客以$+2\ \tfrac{\text{m}}{\text{s}}$走向以$+30\ \tfrac{\text{m}}{\text{s}}$行驶的车厢前端。相对于地面，乘客的速度是多少$\tfrac{\text{m}}{\text{s}}$？
Same line, same direction: $30 + 2 = 32\ \tfrac{\text{m}}{\text{s}}$. · 同一直线，同一方向：$30 + 2 = 32\ \tfrac{\text{m}}{\text{s}}$。
The same passenger now walks toward the back · 反向 at $2\ \tfrac{\text{m}}{\text{s}}$. Velocity relative to the ground, in $\tfrac{\text{m}}{\text{s}}$? · 同一乘客现在向后走，速度为 $2\ \tfrac{\text{m}}{\text{s}}$。相对于地面的速度是多少，单位为 $\tfrac{\text{m}}{\text{s}}$？
Now the walk is negative: $30 + (-2) = 28\ \tfrac{\text{m}}{\text{s}}$. · 现在行走方向为负：$30 + (-2) = 28\ \tfrac{\text{m}}{\text{s}}$。
Adding velocities in two dimensions
- A boat points straight across a river; the current sweeps it downstream.
- Its velocity over the ground is the vector sum of boat-velocity and water-velocity.
- The two combine tip-to-tail into a resultant 合矢量 — a diagonal path.
- The boat crosses and drifts at the same time; neither motion cancels the other.
在二维中相加速度
- 一条船径直指向对岸;水流把它冲向下游。
- 它相对地面的速度是船速与水速的矢量和。
- 两者首尾相接合成一个合矢量——一条斜向的路径。
- 船同时在横渡又在漂移;两个运动谁也不抵消谁。
Add the boat and current vectors · 将船速与水流速度矢量相加
Combine a straight-across velocity with a downstream current and read off the resultant. · 结合垂直渡河的流速与顺流的水流，读取合速度。
Across a river, velocities must be added as ____, not just as plain numbers. · 横渡河流时，速度必须作为____相加，而不仅仅是普通数字。
Perpendicular velocities combine tip-to-tail as vectors, giving a diagonal resultant. · 垂直速度作为矢量首尾相接合成，得到一个对角线方向的合矢量。
The subscript rule
- Write "velocity of A relative to B" as $\vec v_{A/B}$.
- They chain: $\vec v_{A/\text{ground}} = \vec v_{A/B} + \vec v_{B/\text{ground}}$.
- Swap the order and you negate it: $\vec v_{B/A} = -\,\vec v_{A/B}$.
- Line up the inner subscripts and the outer pair is your answer.
下标规则
- 把"A 相对于 B 的速度"写成 $\vec v_{A/B}$。
- 它们可以串联:$\vec v_{A/\text{ground}} = \vec v_{A/B} + \vec v_{B/\text{ground}}$。
- 交换顺序就取负:$\vec v_{B/A} = -\,\vec v_{A/B}$。
- 让内侧的下标对齐,外侧的一对就是你的答案。
The velocity of A relative to B is the exact opposite (negative) of the velocity of B relative to A. · A相对于B的速度恰好是B相对于A的速度的相反数（负值）。
Reversing the pair reverses the direction: $\vec v_{B/A} = -\vec v_{A/B}$. · 交换这对顺序会反转方向：$\vec v_{B/A} = -\vec v_{A/B}$。
Relative velocities add as vectors, not always as plain numbers. Only when the motions are along the same line can you just add or subtract the values. Across a river the answer needs vector addition, so the boat's ground speed is larger than its speed through the water.
相对速度按矢量相加,不总是按普通的数。只有当运动在同一条直线上时,才能直接把数值相加或相减。横渡河流时答案需要矢量相加,所以船相对地面的速率大于它相对水的速率。
A person walks at $1.5\ \tfrac{\text{m}}{\text{s}}$ toward the front of a train moving at $12\ \tfrac{\text{m}}{\text{s}}$.
- Same direction, one line: add the values.
- Velocity relative to the ground $= 12 + 1.5 = 13.5\ \tfrac{\text{m}}{\text{s}}$.
一个人在一列以 $12\ \tfrac{\text{m}}{\text{s}}$ 行驶的火车上朝车头以 $1.5\ \tfrac{\text{m}}{\text{s}}$ 行走。
- 同一方向、同一条线:把数值相加。
- 相对地面的速度 $= 12 + 1.5 = 13.5\ \tfrac{\text{m}}{\text{s}}$。
Motion is measured in a reference frame. To change frames, add velocities as vectors: $\vec v_{A/\text{ground}} = \vec v_{A/B} + \vec v_{B/\text{ground}}$, and reversing a pair negates it ($\vec v_{B/A} = -\vec v_{A/B}$). On one line they add as signed numbers; in two dimensions they add tip-to-tail.
运动是在参考系中测量的。要换参考系,就按矢量相加速度:$\vec v_{A/\text{ground}} = \vec v_{A/B} + \vec v_{B/\text{ground}}$,而交换一对下标就取负($\vec v_{B/A} = -\vec v_{A/B}$)。在一条线上按带符号的数相加;在二维中首尾相接相加。