Kinematics of motion in a straight line · 直线运动的运动学
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| acceleration/əkˌseləˈreɪʃn/ | 加速度 | jiā sù dù |
| kinematics/ˌkɪnɪˈmætɪks/ | 运动学 | yùn dòng xué |
| distance/ˈdɪstəns/ | 距离 | jù lí |
| speed/spiːd/ | 速率 | sù lǜ |
| displacement/dɪˈspleɪsmənt/ | 位移 | wèi yí |
| velocity/vəˈlɒsɪti/ | 速度 | sù dù |
| velocity–time graph/vəˈlɒsɪti taɪm ɡræf/ | 速度-时间图 | sù dù - shí jiān tú |
| gradient/ˈɡreɪdɪənt/ | 斜率 | xié lǜ |
| constant acceleration/ˈkɒnstənt əkˌseləˈreɪʃn/ | 匀加速 | yún jiā sù |
The language of motion
- How fast is a car going after 10 seconds of acceleration 加速度? How far has a ball fallen in 3 seconds?
- Kinematics 运动学 answers these questions with a small set of powerful equations — the suvat equations.
运动的语言
- 一辆车加速 10 秒后跑多快?一个球 3 秒内落了多远?
- 运动学(kinematics)用一小套强大的方程回答这些问题——suvat 方程。
Scalars, vectors and graphs
- Distance 距离 and speed 速率 are scalars; displacement 位移, velocity 速度 and acceleration are vectors.
- On a velocity–time graph 速度-时间图: area = displacement, gradient 斜率 = acceleration.
On a velocity–time graph, the gradient is the acceleration and the area underneath is the displacement.
- Differentiate to go displacement → velocity → acceleration; integrate to go back.
Reading a v–t graph. A straight line from $(0, 0)$ to $(4, 10)$: gradient $= \dfrac{10}{4} = 2.5\text{ m s}^{-2}$ (acceleration). Area $= \dfrac{1}{2} \times 4 \times 10 = 20\text{ m}$ (displacement).
标量、向量与图
- 距离(distance)和速率(speed)是标量;位移(displacement)、速度(velocity)和加速度(acceleration)是向量。
- 在一张速度-时间图(velocity–time graph)上:面积 = 位移,斜率 = 加速度。
在一张速度-时间图上,斜率是加速度,下面的面积是位移。
- 微分从 位移 → 速度 → 加速度;积分往回走。
读一张 v–t 图。 一条从 $(0, 0)$ 到 $(4, 10)$ 的直线:斜率 $= \dfrac{10}{4} = 2.5\text{ m s}^{-2}$(加速度)。面积 $= \dfrac{1}{2} \times 4 \times 10 = 20\text{ m}$(位移)。
Kinematics & the v–t graph · 运动学与 v–t 图
v = u + at
Constant acceleration: the gradient of the v–t line is a, and the area · 面积 is the displacement. · 恒定加速度:v–t 直线的斜率是 a,而面积是位移。
On a velocity–time graph, the area under the line gives the: · 在一张速度-时间图上,线下面的面积给出:
Area = displacement; the gradient gives the acceleration. · 面积 = 位移;斜率给出加速度。
The suvat equations
For constant acceleration 匀加速:
Constant acceleration only. The suvat equations assume acceleration doesn't change. If acceleration varies (e.g. air resistance), you need calculus instead.
On a velocity-time graph, area is displacement and gradient is acceleration
suvat 方程
对恒定加速度:
只适用于恒定加速度。 suvat 方程假设加速度不改变。如果加速度变化(例如空气阻力),你需要改用微积分。
在一张速度-时间图上,面积是位移,斜率是加速度
A car starts from rest and accelerates at 2.5 m/s² for 4 s. Using v = u + at, what is its speed (m/s)? · 一辆车从静止开始,以 2.5 m/s² 加速 4 秒。用 v = u + at,它的速率是多少(m/s)?
v = 0 + 2.5 × 4 = 10 m/s. · v = 0 + 2.5 × 4 = 10 m/s。
For the same car (u = 0, a = 2.5, t = 4), what distance does it travel? (s = ut + ½at²) · 对同一辆车(u = 0,a = 2.5,t = 4),它走多远?(s = ut + ½at²)
s = 0 + ½ × 2.5 × 4² = ½ × 2.5 × 16 = 20 m. · s = 0 + ½ × 2.5 × 4² = ½ × 2.5 × 16 = 20 m。
Using v² = u² + 2as with u = 0, a = 3, s = 24, find v. · 用 v² = u² + 2as,u = 0,a = 3,s = 24,求 v。
v² = 0 + 2(3)(24) = 144, so v = 12 m/s. · v² = 0 + 2(3)(24) = 144,所以 v = 12 m/s。
The suvat equations apply when acceleration is changing. · suvat 方程在加速度变化时适用。
The suvat equations assume constant acceleration. For variable acceleration, use calculus. · suvat 方程假设恒定的加速度。对变加速度,用微积分。
Worked example
- From rest ($u = 0$), $a = 2.5\text{ m s}^{-2}$, $t = 4\text{ s}$:
- $v = u + at = 0 + 2.5 \times 4 = 10\text{ m s}^{-1}$.
- $s = ut + \dfrac{1}{2}at^2 = 0 + \dfrac{1}{2}(2.5)(16) = 20\text{ m}$.
算例
- 从静止($u = 0$),$a = 2.5\text{ m s}^{-2}$,$t = 4\text{ s}$:
- $v = u + at = 0 + 2.5 \times 4 = 10\text{ m s}^{-1}$。
- $s = ut + \dfrac{1}{2}at^2 = 0 + \dfrac{1}{2}(2.5)(16) = 20\text{ m}$。
Variable acceleration
- When acceleration isn't constant, use calculus:
- $v = \dfrac{ds}{dt}$, $a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$.
- Integrate to go the other way: $s = \int v\,dt$, $v = \int a\,dt$.
- On a velocity–time graph the area is displacement; on a displacement–time graph the gradient is velocity. Differentiation and integration move between them; a negative acceleration is deceleration.
变加速度
- 当加速度不恒定时,用微积分:
- $v = \dfrac{ds}{dt}$,$a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$。
- 积分往另一个方向走:$s = \int v\,dt$,$v = \int a\,dt$。
- 速度–时间图(velocity–time graphs)的面积是位移;位移–时间图(displacement–time graphs)的斜率是速度。求导(differentiation)和积分(integration)在它们之间转换;负加速度是减速度(deceleration)。
To go from displacement to velocity, you: · 要从位移到速度,你:
Velocity is the derivative of displacement: v = ds/dt. · 速度是位移的导数:v = ds/dt。
You've got it
- v–t graph: area = displacement, gradient = acceleration
- suvat (constant a): $v = u+at$, $s = ut + \tfrac12 at^2$, $v^2 = u^2 + 2as$
- differentiate s→v→a; integrate to reverse
你掌握了
- v–t 图:面积 = 位移,斜率 = 加速度
- suvat(恒定 a):$v = u+at$,$s = ut + \tfrac12 at^2$,$v^2 = u^2 + 2as$
- 微分 s→v→a;积分反过来