Vectors and Motion in Two Dimensions · 二维空间中的矢量与运动
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| component/kəmˈpəʊnənt/ | 分量 | fèn liàng |
| resultant/rɪˈzʌltənt/ | 合矢量 | hé shǐ liàng |
| projectile/prəˈdʒektaɪl/ | 抛体 | pāo tǐ |
A plane flies northeast — two motions in one
- One arrow points "northeast", but the plane is really moving east and north at once.
- Almost no real motion is purely left–right or up–down.
- The trick of two-dimensional kinematics: split each vector into perpendicular pieces.
- Handle the pieces separately, then put them back together.
飞机朝东北飞——一个箭头,两个运动
- 一个箭头指向"东北",但飞机其实同时在向东和向北移动。
- 几乎没有哪个真实的运动是纯粹的左右或上下。
- 二维运动学的诀窍:把每个矢量拆成相互垂直的两部分。
- 分别处理这两部分,再把它们合起来。
Resolving into components
- Any vector can be replaced by two perpendicular components 分量 — one along $x$, one along $y$.
- Together the two components have exactly the same effect as the original vector.
- Choose axes that make the problem easy (often horizontal and vertical).
- The components carry signs, so "left" and "down" are just negative values.
分解为分量
- 任何矢量都可以用两个相互垂直的分量代替——一个沿 $x$,一个沿 $y$。
- 这两个分量合在一起,效果与原矢量完全相同。
- 选择让问题变简单的坐标轴(通常是水平和竖直)。
- 分量带符号,所以"向左""向下"就是负值。
The component formulas
- For a vector $\vec A$ of magnitude $A$ at angle $\theta$ above the $x$-axis:
- $A_x = A\cos\theta$ and $A_y = A\sin\theta$.
- These come straight from the right triangle the vector makes with its components.
- Going back: $A = \sqrt{A_x^2 + A_y^2}$ and $\theta = \tan^{-1}\!\big(\tfrac{A_y}{A_x}\big)$.
分量公式
- 对于一个大小为 $A$、与 $x$ 轴成 $\theta$ 角的矢量 $\vec A$:
- $A_x = A\cos\theta$,$A_y = A\sin\theta$。
- 它们直接来自矢量与其分量所构成的直角三角形。
- 反过来:$A = \sqrt{A_x^2 + A_y^2}$,$\theta = \tan^{-1}\!\big(\tfrac{A_y}{A_x}\big)$。
A velocity of $10\ \tfrac{\text{m}}{\text{s}}$ points $37^\circ$ above the horizontal. Its horizontal component is $10\cos 37^\circ$. With $\cos 37^\circ = 0.8$, what is $v_x$ in · 入 $\tfrac{\text{m}}{\text{s}}$? · 一个$10\ \tfrac{\text{m}}{\text{s}}$的速度指向水平上方$37^\circ$度。其水平分量为$10\cos 37^\circ$。已知$\cos 37^\circ = 0.8$，求$v_x$在$\tfrac{\text{m}}{\text{s}}$中的值？
$v_x = A\cos\theta = 10 \times 0.8 = 8\ \tfrac{\text{m}}{\text{s}}$.
For the same velocity, the vertical component is $10\sin 37^\circ$. With $\sin 37^\circ = 0.6$, what is $v_y$ in · 入 $\tfrac{\text{m}}{\text{s}}$? · 对于相同的速度，垂直分量为$10\sin 37^\circ$。已知$\sin 37^\circ = 0.6$，求$v_y$在$\tfrac{\text{m}}{\text{s}}$中的值？
$v_y = A\sin\theta = 10 \times 0.6 = 6\ \tfrac{\text{m}}{\text{s}}$.
A vector has components $A_x = 3$ and $A_y = 4$. What is its magnitude? · 一个矢量的分量为$A_x = 3$和$A_y = 4$。它的大小是多少？
$A = \sqrt{A_x^2 + A_y^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. Not $3+4=7$ — the components are perpendicular. · $A = \sqrt{A_x^2 + A_y^2} = \sqrt{9 + 16} = \sqrt{25} = 5$。而不是$3+4=7$——因为分量是相互垂直的。
The horizontal component of a vector $A$ at angle $\theta$ is $A$ ____ $\theta$. · 矢量的水平分量 $A$ 在角度 $\theta$ 处为 $A$ ____ $\theta$.
The adjacent side of the right triangle is $A\cos\theta$, so $A_x = A\cos\theta$. · 直角三角形的邻边是$A\cos\theta$，因此$A_x = A\cos\theta$。
Adding vectors by components
- To add vectors, add the $x$-components and separately add the $y$-components.
- Never add the magnitudes directly — that only works if they point the same way.
- The sums $(\Sigma A_x,\ \Sigma A_y)$ are the components of the resultant 合矢量.
- Recombine with $\sqrt{\;}$ and $\tan^{-1}$ to get its magnitude and direction.
用分量相加矢量
- 要把矢量相加,就把 $x$ 分量相加,再单独把 $y$ 分量相加。
- 绝不要直接把大小相加——只有当它们方向相同时才成立。
- 这两个和 $(\Sigma A_x,\ \Sigma A_y)$ 就是合矢量的分量。
- 用 $\sqrt{\;}$ 和 $\tan^{-1}$ 重新组合,得到它的大小和方向。
Add two vectors by components · 通过分量相加两个矢量
Drag two vectors and watch the resultant — its components are the sums of the parts. · 拖动两个矢量并观察其合矢量——它的分量是各部分之和。
Order the steps to add two vectors using components, first step at the top. · 排序使用分量法添加两个矢量的步骤，第一步在最上面。
Resolve, add component-wise, then recombine magnitude and direction. · 分解，按分量相加，然后重新组合大小和方向。
Perpendicular motions are independent
- The $x$-motion and the $y$-motion evolve separately — neither affects the other.
- Gravity pulls only downward, so it changes $v_y$ but leaves $v_x$ alone.
- This independence is the whole secret of projectile 抛体 motion (Topic 2).
- Drop a ball and fire one horizontally at the same instant: they land together.
相互垂直的运动彼此独立
- $x$ 方向的运动和 $y$ 方向的运动各自独立演化——谁也不影响谁。
- 重力只向下拉,所以它改变 $v_y$,却让 $v_x$ 保持不变。
- 这种独立性正是抛体运动的全部秘密(主题 2)。
- 同一瞬间松开一个球、水平射出另一个球:它们会同时落地。
The horizontal and vertical parts of a two-dimensional motion are independent of each other. · 二维运动的水平和垂直部分是相互独立的。
Perpendicular components evolve separately — gravity changes $v_y$ but not $v_x$. This is the basis of projectile motion. · 垂直分量独立演化——重力改变$v_y$但不改变$v_x$。这是抛体运动的基础。
You cannot add two vectors by adding their sizes unless they point along the same line. A $3$-unit vector plus a perpendicular $4$-unit vector gives $5$, not $7$ — resolve into components (or use the triangle) first.
除非两个矢量沿同一条直线,否则你不能通过把大小相加来相加它们。一个 $3$ 单位的矢量加上一个垂直的 $4$ 单位矢量得到的是 $5$,不是 $7$——要先分解为分量(或用三角形)。
Resolve a velocity of $10\ \tfrac{\text{m}}{\text{s}}$ aimed at $37^\circ$ above the horizontal (take $\cos 37^\circ \approx 0.8$, $\sin 37^\circ \approx 0.6$).
- $v_x = 10\cos 37^\circ \approx 8\ \tfrac{\text{m}}{\text{s}}$.
- $v_y = 10\sin 37^\circ \approx 6\ \tfrac{\text{m}}{\text{s}}$.
Check: $\sqrt{8^2 + 6^2} = 10$. ✓
把一个 $10\ \tfrac{\text{m}}{\text{s}}$、与水平成 $37^\circ$ 的速度分解(取 $\cos 37^\circ \approx 0.8$,$\sin 37^\circ \approx 0.6$)。
- $v_x = 10\cos 37^\circ \approx 8\ \tfrac{\text{m}}{\text{s}}$。
- $v_y = 10\sin 37^\circ \approx 6\ \tfrac{\text{m}}{\text{s}}$。
验算:$\sqrt{8^2 + 6^2} = 10$。✓
Split a 2-D vector into perpendicular components: $A_x = A\cos\theta$, $A_y = A\sin\theta$. Add vectors by adding $x$- and $y$-components separately, then recombine with $A = \sqrt{A_x^2+A_y^2}$. Perpendicular motions are independent — the key to projectiles.
把二维矢量拆成相互垂直的分量:$A_x = A\cos\theta$、$A_y = A\sin\theta$。分别相加 $x$ 与 $y$ 分量来相加矢量,再用 $A = \sqrt{A_x^2+A_y^2}$ 重新组合。相互垂直的运动彼此独立——这是抛体运动的关键。