The equations of motion · 运动学方程
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| free fall/friː fɔːl/ | 自由落体 | zì yóu luò tǐ |
| equations of motion/ɪˈkweɪʒnz ɒv ˈməʊʃn/ | 运动方程 | yùn dòng fāng chéng |
| constant acceleration/ˈkɒnstənt əkˌseləˈreɪʃn/ | 匀加速 | yún jiā sù |
| symmetric/sɪˈmetrɪk/ | 对称 | duì chèn |
| light gate/laɪt ɡeɪt/ | 光电门 | guāng diàn mén |
A hammer and a feather
- On the Moon, an astronaut dropped a hammer and a feather together — they landed at the same time.
- With no air, free fall 自由落体 is the same for every mass.
- To handle motion like this, we use the equations of motion 运动方程.
锤子和羽毛
- 在月球上,一位宇航员同时放下一把锤子和一根羽毛——它们同时落地。
- 没有空气时,自由落体对每个质量都一样。
- 要处理这样的运动,我们用 运动学方程(equations of motion)。
Five symbols
- Straight-line motion with constant acceleration 匀加速 uses five symbols.
- $u$ start velocity · $v$ final velocity · $a$ acceleration · $s$ displacement · $t$ time.
A speedometer shows speed: the distance travelled per unit time
五个符号
- 匀加速(恒定加速度)直线运动用五个符号。
- $u$ 初速度 · $v$ 末速度 · $a$ 加速度 · $s$ 位移 · $t$ 时间。
速度计显示速率:单位时间内走过的距离
The equations of motion · 运动学方程
v = u + at
The suvat results come straight from the v–t graph: gradient = a, area = displacement. · SUVAT 各式直接来自 v–t 图:斜率 = a,面积 = 位移。
The four equations
- $v = u + at$
- $s = ut + \tfrac{1}{2}at^{2}$
- $s = \tfrac{1}{2}(u + v)t$
- $v^{2} = u^{2} + 2as$
Acceleration–time graph derived from the same motion
四个方程
- $v = u + at$
- $s = ut + \tfrac{1}{2}at^{2}$
- $s = \tfrac{1}{2}(u + v)t$
- $v^{2} = u^{2} + 2as$
由同一运动导出的加速度–时间图
When may the equations of motion be used? Select all · 所有 that apply. · 什么时候可以使用运动学方程?选出所有适用的。
Air resistance makes the acceleration change, so only graph methods work there. A curved velocity-time graph is the visible warning. · 空气阻力使加速度改变,所以那里只有图解法管用。弯曲的速度-时间图就是看得见的警告。
Choosing the right one
- Each equation uses four of the five symbols.
- Write down what you know and what you want, then pick the equation with exactly those.
Displacement–time graph for uniform acceleration — the slope at any point equals the instantaneous velocity
选对方程
- 每个方程用到五个符号中的 四个。
- 写下你 已知 的和 想求 的,再选恰好含这些符号的方程。
匀加速的位移–时间图——任意一点的斜率等于瞬时速度
Each SUVAT equation leaves out one symbol. Match each to the one it does not · 不 contain. · 每个 SUVAT 方程都漏掉一个符号。把每个方程与它 不含 的那个符号配对。
Spotting the missing symbol is how you choose: pick the equation that has your four known/wanted symbols and leaves out the one you neither know nor want. · 找出缺失的符号正是选择方程的方法:选含有你那四个已知/想求符号、并漏掉你既不知道也不想求的那个符号的方程。
A car starts at $u = 2\ \dfrac{\text{m}}{\text{s}}$ and accelerates at $a = 3\ \dfrac{\text{m}}{\text{s}^2}$ for $t = 4\ \text{s}$. Find $v$. · 一辆车以 $u = 2\ \dfrac{\text{m}}{\text{s}}$ 开始,以 $a = 3\ \dfrac{\text{m}}{\text{s}^2}$ 加速 $t = 4\ \text{s}$。求 $v$。
Use $v = u + at = 2 + 3 \times 4 = 14\ \dfrac{\text{m}}{\text{s}}$. · 用 $v = u + at = 2 + 3 \times 4 = 14\ \dfrac{\text{m}}{\text{s}}$。
Put the method for choosing a suvat equation in order. · 把选择运动学方程的方法按顺序排列。
Picking the equation that is missing the quantity you neither have nor want saves solving two equations together. · 挑那个不含你既没有又不想要的量的方程,省掉联立两个方程。
Match each equation to the quantity it does not contain. · 把每个方程与它不含的那个量配对。
Knowing which quantity each equation leaves out is what makes the choice instant. · 知道每个方程漏掉了哪个量,选择就变得立刻可做。
Where they come from
- $v = u + at$ is the gradient of a velocity–time line ($a = \tfrac{v-u}{t}$).
- $s = \tfrac{1}{2}(u+v)t$ is the area under it (a trapezium).
它们从哪来
- $v = u + at$ 是速度–时间直线的 斜率($a = \tfrac{v-u}{t}$)。
- $s = \tfrac{1}{2}(u+v)t$ 是它下方的 面积(一个梯形)。
Which equation is just the gradient of a velocity–time graph? · 哪个方程就是速度–时间图的 斜率?
The gradient is $a = \dfrac{v - u}{t}$, which rearranges to $v = u + at$. · 斜率是 $a = \dfrac{v - u}{t}$,整理后就是 $v = u + at$。
Pick a positive direction first and keep it. For a ball thrown up with "up" positive, $a = -g$ — gravity points the other way.
先选一个 正方向 并保持不变。对于向上抛的球,取“上”为正,则 $a = -g$——重力指向相反方向。
A ball is thrown straight up at 20 m/s. What is the greatest height it reaches, in metres? (g = 9.81 m/s^2) · 一个球以 20 m/s 竖直上抛。它达到的最大高度是多少米?(g = 9.81 m/s^2)
At the top v = 0, so v^2 = u^2 + 2as gives 0 = 400 - 2(9.81)s and s = 20.4 m. With up positive, a is negative for the whole flight. · 最高点 v = 0,所以 v^2 = u^2 + 2as 给出 0 = 400 - 2(9.81)s,s = 20.4 m。以向上为正时,整段飞行 a 都是负的。
Free fall
- Ignoring air, a falling object accelerates at $g \approx 9.81\ \dfrac{\text{m}}{\text{s}^2}$ downward — the same for every mass.
- Dropped from rest: $h = \tfrac{1}{2}gt^{2}$, $\;v = gt$, $\;v^{2} = 2gh$.
自由落体
- 忽略空气时,下落物体以 $g \approx 9.81\ \dfrac{\text{m}}{\text{s}^2}$ 向下加速——对每个质量都相同。
- 从静止释放:$h = \tfrac{1}{2}gt^{2}$,$\;v = gt$,$\;v^{2} = 2gh$。
With no air resistance, a heavy ball falls faster than a light one. · 没有空气阻力时,重的球比轻的球落得快。
No — free fall has the same acceleration $g$ for every mass. Air resistance is what makes a feather seem to fall slower. · 不是——自由落体对每个质量都有相同的加速度 $g$。是空气阻力让羽毛看起来落得慢。
A stone is dropped from rest and falls for $2.0\ \text{s}$. How far does it fall? (Use $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$.) · 一块石头从静止落下,下落 $2.0\ \text{s}$。它落了多远?(取 $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$。)
$h = \tfrac{1}{2}gt^{2} = \tfrac{1}{2} \times 9.81 \times 2.0^{2} \approx 19.6\ \text{m}$. · $h = \tfrac{1}{2}gt^{2} = \tfrac{1}{2} \times 9.81 \times 2.0^{2} \approx 19.6\ \text{m}$。
Thrown straight up
- The motion is symmetric 对称: it rises, stops, and falls back the same way.
- Greatest height $h = \dfrac{u^{2}}{2g}$; time to the top $t = \dfrac{u}{g}$; total time up and down $= \dfrac{2u}{g}$.
竖直上抛
- 运动是 对称的:上升、停住,再原路落回。
- 最大高度 $h = \dfrac{u^{2}}{2g}$;到顶点的时间 $t = \dfrac{u}{g}$;上下往返总时间 $= \dfrac{2u}{g}$。
A ball is thrown straight up at $20\ \dfrac{\text{m}}{\text{s}}$. How long until it reaches the top? (Use $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$.) · 一个球以 $20\ \dfrac{\text{m}}{\text{s}}$ 竖直上抛。多久到达最高点?(取 $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$。)
At the top $v = 0$, so $t = \dfrac{u}{g} = \dfrac{20}{9.81} \approx 2.0\ \text{s}$. · 在最高点 $v = 0$,所以 $t = \dfrac{u}{g} = \dfrac{20}{9.81} \approx 2.0\ \text{s}$。
Measuring $g$
- Drop an object a height $h$ and time the fall: $g = \dfrac{2h}{t^{2}}$.
- Plot $h$ against $t^{2}$ — the gradient is $\dfrac{g}{2}$. Light gates 光电门 remove reaction-time error.
测量 $g$
- 让物体下落高度 $h$ 并计时:$g = \dfrac{2h}{t^{2}}$。
- 画 $h$ 对 $t^{2}$ 的图——斜率是 $\dfrac{g}{2}$。光电门(light gates) 可消除反应时间误差。
You drop an object from several heights and plot $h$ against $t^{2}$. The gradient equals: · 你从几个高度释放物体,并画 $h$ 对 $t^{2}$ 的图。斜率等于:
From $h = \tfrac{1}{2}gt^{2}$, plotting $h$ against $t^{2}$ gives a straight line of gradient $\tfrac{1}{2}g$ — so $g$ is twice the gradient. · 由 $h = \tfrac{1}{2}gt^{2}$,画 $h$ 对 $t^{2}$ 得到一条斜率为 $\tfrac{1}{2}g$ 的直线——所以 $g$ 是斜率的两倍。
You've got it
- four SUVAT equations link $u, v, a, s, t$ — pick the one with the four symbols you have
- free fall: constant $g \approx 9.81\ \dfrac{\text{m}}{\text{s}^2}$ down, the same for every mass
- a vertical throw is symmetric: time up $= \dfrac{u}{g}$
你掌握了
- 四个 SUVAT 方程联系 $u, v, a, s, t$——选含有你已有的那四个符号的方程
- 自由落体:恒定 $g \approx 9.81\ \dfrac{\text{m}}{\text{s}^2}$ 向下,对每个质量都相同
- 竖直上抛是 对称的:上升时间 $= \dfrac{u}{g}$