---
# System prepended metadata

title: 2020.4.1(3)
tags: [Kinematics, Mechanics]

---

---
tags : Mechanics,Kinematics
title : 2020.4.1(3)
---
# Mechanics(3)
## §1.1 Kinematics in one-dimensional space
### 1.1.5 Deceleration Movement
Equation of motion when an object catches deceleration proportional to
velocity *v* by fluid is

$\large
\frac{dv}{dt} = \eta v$

The equation given as a derivative relation such this  is called as **differentional equation**.
Above equation has only an first derivative $\frac{dv}{dt}$ , so we regard it as a first differentional equation.
We can solve it by **separation of variables**.
:::success
#### Remark .Differentional Equation and Separation of Variables
Ceremony : $\large \frac{dx}{dt} =f(x)g(x)$ 
Devide both sides the function f(x),

$\large \frac{1}{f(x)} \frac{dx}{dt} = g(t)$

Integrate t~0~ to t~1~

$\large \int_{t0}^{t1} \frac{1}{f(x)} \frac{dx}{dt} dt = \int_{t0}^{t1} g(t) dt$

Convert variable *t* to *x* because *x* = *x*(*t*)

$\large dx = \frac{dx}{dt} dt$

Therefore,
<font color="red">
$\huge \int_{x_(t0)}^{x_(t1)} \frac{dx}{f(x)} =\int_{t0}^{t1} g(t) dt$
</font>

This matches the integration of the equation which dx/dt transferd dt to right side ,

$\large \frac{dx}{f(x)} = g(t)dt$
:::