---
tags : Mechanics,Kinematics
title : 2020.4.1(2)
---
# Mechanics (2)
## §1.1 Kinematics in one-dimensional space
If we know how changing velocity between time 0 to *t* , we can write the position *x*(*t*) in time *t* by the time integration of velocity because the velocity *v* expresses the eime changing of the position.
For *v*(*t*') = dx(*t*')/d*t*' , we can lead it,
### $\int_0^t v(t') dt' = \int_0^t \frac{dx}{dt'} dt' =\int_{x_(0)}^{x_(t)} f(x) dx$ = *x*(*t*) - *x*(0)
Similarly as position , we can write velocity *v*(*t*) on time *t* , if we know the acceleration *a*(*t*) between time 0 to *t* and velocity *v*(0) ,
### *v*(*t*) = *v*(0) + $\int_0^t a(t') dt'$
Therefore, the position on time *t* is
### *x*(*t*) - *x*(0) = $\int_0^t v(t') dt' = \int_0^t \{ v(0) + \int_0^{t'} a(t'') dt''\} dt'$
Rewriting,
<font color="red">
$\huge
x(t) = x(0) + v(0)t + \int_0^t \{\int_0^{t'} a(t'') dt''\} dt'$

</font>

#### remark
First term is initial position, second term is moving distance of initial velocity, third term is modification of moving distance caused by acceleration.

If acceleration is constant, regard *a*(*t*) = $\alpha$,

$\large
x(t) = x(0) + V(0)t + \alpha \int_0^t t' dt' = x(0) + v(0)t + \frac{\alpha}{2}t^2$

This matches Galileo's one.