Physics 小笔记:一,Displacement,前言等

前言

怎么会有这篇文章的?

每天/周写个文章来归纳一下所学的内容是一个好选择,此外的话也可以练习一下打字,和KaTeX语法等。

https://katex.org/docs/supported.html

还有的话,写的时候也能帮我重新记忆一遍,此外的话,未来拿来作什么项目也不是不行,2333

此系列不出意外会持续更新,粒度分为周和日,我尽量每周再集合一下,使其更加规整。

#2021/6/1

Displacement

Displacement is the difference between the final and initial position of a point trajectory (for instance, the center of mass of a moving object). The actual path covered to reach the final position is irrelevant.

Displacement have direction and is a vector.

Distence moved in Velcoty-Time Graph is the Area of the graph.

Formulas

\begin{align*}
v^{2}&=u^{2}+2as \\
v&=final\:velocity \\
u&=initial\:velocity \\
a&=acceleration \\
s&=displacement
\end{align*}

Also, v-u=at

怎么得出来的呢,这里需画个图

v-u的部分是此次加速过程中变化的量,可以使用acceleration * time算出

#2021/6/2

位移便是图像的面积,将其分为一个长方形和一个三角形计算

Area of the Gray Triangle

\begin{align*}
&(v-u)(t-0)\div2 \\
=&\;\cfrac{t(v-u)}{2} \\
Also, (v-u)=&\;at \\
Area_{triangle}=&\;\cfrac{at^2}{2}
\end{align*}

Area of the Pink Rectangle

\begin{align*}
&(u-0)(t-0) \\
=&\;ut \\
Area_{rectangle}=&\;ut
\end{align*}

Then plus to shapes together

\begin{align*}
s=&\;ut+\cfrac{at^2}{2}
\end{align*}

You can get a Formula withoutt

\begin{align*}
v-u=&at \\
t=&\;\cfrac{v-u}{a} \\
put\;t\;in\;\;\;s=&\;ut+\cfrac{at^2}{2} \\
s=&\;u\;\cfrac{v-u}{a}+\cfrac{a(\cfrac{v-u}{a})^2}{2} \\
s=&\;\cfrac{vu-u^2}{a}+\cfrac{a(\cfrac{v^2-2vu+u^2}{a^2})}{2} \\
2s=&\;2\cfrac{vu-u^2}{a}+\cfrac{v^2-2vu+u^2}{a} \\
2s=&\;\cfrac{2vu-2u^2+v^2-2vu+u^2}{a} \\
2s=&\;\cfrac{v^2-u^2}{a} \\
s=&\;\cfrac{v^2-u^2}{2a} \\
2as=&\;v^2-u^2 \\
v^2=&\;u^2+2as
\end{align*}

Summary

Displacement is the difference between the final and initial position.

\begin{align*}
v^{2}&=u^{2}+2as \\
s&=\;\cfrac{v^2-u^2}{2a} \\
s&=\;ut+\cfrac{at^2}{2} \\
v-u&=at \\
v&=final\:velocity \\
u&=initial\:velocity \\
a&=acceleration \\
s&=displacement
\end{align*}

Practice

  1. A car increases its speed from 25 m s-1 to 31 m s-1 with a uniform acceleration of 1.8 m s-2. How far does it travel while accelerating?

Solution

  1. How far 说明他在求Displacement
  2. 在脑海中想象一波,绘制一个图像
  3. 计算图像的面积
  4. 用逻辑估算验证一下算出来的对不对

差个t,不过他已经告诉你 v, u, a了

\begin{align*}
v-u=&at \\
31-21=&1.8t \\
t=&\cfrac{10}{3}
\end{align*}

都有了,算一算面积,s就得出来了

实际过程

\begin{align*}
(31-25)\div1.8=&\cfrac{10}{3} \\
\cfrac{10}{3}\times\cfrac{1}{2}(31-25)=&10 \\
\cfrac{10}{3}\times25=&\cfrac{250}{3} \\
\cfrac{250}{3}+10=&93.3\;(3.s.f)
\end{align*}

注释

\begin{align*}
v^{2}&=u^{2}+2as \\
v&=final\:velocity \\
u&=initial\:velocity \\
a&=acceleration \\
s&=displacement
\end{align*}

\begin{align*}
v-u=&at \\
t=&\;\cfrac{v-u}{a} \\
put\;t\;in\;\;\;s=&\;ut+\cfrac{at^2}{2} \\
s=&\;u\;\cfrac{v-u}{a}+\cfrac{a(\cfrac{v-u}{a})^2}{2} \\
s=&\;\cfrac{vu-u^2}{a}+\cfrac{a(\cfrac{v^2-2vu+u^2}{a^2})}{2} \\
2s=&\;2\cfrac{vu-u^2}{a}+\cfrac{v^2-2vu+u^2}{a} \\
2s=&\;\cfrac{2vu-2u^2+v^2-2vu+u^2}{a} \\
2s=&\;\cfrac{v^2-u^2}{a} \\
s=&\;\cfrac{v^2-u^2}{2a} \\
2as=&\;v^2-u^2 \\
v^2=&\;u^2+2as
\end{align*}

\begin{align*}
v^{2}&=u^{2}+2as \\
s&=\;\cfrac{v^2-u^2}{2a} \\
s&=\;ut+\cfrac{at^2}{2} \\
v-u&=at \\
v&=final\:velocity \\
u&=initial\:velocity \\
a&=acceleration \\
s&=displacement
\end{align*}

\begin{align*}
(31-25)\div1.8=&\cfrac{10}{3} \\
\cfrac{10}{3}\times\cfrac{1}{2}(31-25)=&10 \\
\cfrac{10}{3}\times25=&\cfrac{250}{3} \\
\cfrac{250}{3}+10=&93.3\;(3.s.f)
\end{align*}

这玩意写得我想死。。

果然记忆深刻,还有点想写没写,一次写不动了,周末发布总和吧。

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