Integrating motion equation to derive displacement

[annamal]

$\frac{dx}{dt} = \frac{dx_i}{dt} + \frac{d^2x}{dt^2}t$
Multiplying dt on both sides and integrating we have
$\int_{x_f}^{x_i} dx = \int_{0}^{v_i t} dx_i + \int_{0}^{at} dvt$
so $x_f - x_i = v_it + at^2$, which is not right

Where did I go wrong?

I understand that if we substitute a for $\frac{d^2x}{dt^2}$, integrating atdt = 0.5at, but what if we derived the equation for displacement by multiplying by dt?

 

[kuruman]

$\frac{dx}{dt} = \frac{dx_i}{dt} + \frac{d^2x}{dt^2}t$
Multiplying dt on both sides and integrating we have
$\int_{x_f}^{x_i} dx = \int_{0}^{v_i t} dx_i + \int_{0}^{at} dvt$
so $x_f - x_i = v_it + at^2$, which is not right

Where did I go wrong?

I understand that if we substitute a for $\frac{d^2x}{dt^2}$, integrating atdt = 0.5at, but what if we derived the equation for displacement by multiplying by dt?

What is $x_i$?

 

[annamal]

What is $x_i$?

$x_i$ is Initial position. I swapped the $x_i$ and $x_f$ positions in the integral

 

[kuruman]

And does the initial position change as the object is in motion?

 

[annamal]

And does the initial position change as the object is in motion?

No, the initial position is constant

 

[kuruman]

So what is $dx_i$ if $x_i$ is constant?

 

[Steve4Physics]

No, the initial position is constant

If $x_i$ is constant, what are the meanings of $\frac{dx_i}{dt}$ and $dx_i$ in your Post #1 equations?

It is not clear what question you are trying to answer. E.g. is this about an object with constant or variable acceleration? A complete and clear problem-statement is needed.

 

[Steve4Physics]

So what is $dx_i$ if $x_i$ is constant?

Wow, you beat me by a fraction of a second. I avoided replying for about an hour in case you replied. But then gave way to temptation!

 

[kuruman]

Wow, you beat me by a fraction of a second. I avoided replying for about an hour in case you replied. But then gave way to temptation!

I appreciate the consideration. I was out walking the dog.

It is not clear what question you are trying to answer. E.g. is this about an object with constant or variable acceleration? A complete and clear problem-statement is needed.

I think @annamal started with $v=v_i+at$, substituted $v=\frac{dx}{dt}$, $v_i=\frac{dx_i}{dt}$ and $a=\frac{d^2x}{dt^2}$ and now wants to know how to solve this equation for $x$.

To @annamal : It looks like you need a course in ordinary differential equations. For the time being, start with $\frac{dv}{dt}=a$ where a is constant. Integrate that to get the velocity as a function of time, $v(t)$. Don't forget the integration constant. Then integrate once more $\frac{dx}{dt}=v(t)$ to find $x(t)$. Again, don't forget the integration constant.

 

[annamal]

I appreciate the consideration. I was out walking the dog.I think @annamal started with $v=v_i+at$, substituted $v=\frac{dx}{dt}$, $v_i=\frac{dx_i}{dt}$ and $a=\frac{d^2x}{dt^2}$ and now wants to know how to solve this equation for $x$.

To @annamal : It looks like you need a course in ordinary differential equations. For the time being, start with $\frac{dv}{dt}=a$ where a is constant. Integrate that to get the velocity as a function of time, $v(t)$. Don't forget the integration constant. Then integrate once more $\frac{dx}{dt}=v(t)$ to find $x(t)$. Again, don't forget the integration constant.

I have taken differential equations already, so feel free to explain it to me in those terms as well

Yes, you're right in my intention, but I don't want to integrate in that way. I want to multiply dt throughout the equation and then integrate. so instead of saying $\frac{dv}{dt}dt$ = adt, we multiply dt so that adt = dv and then integrate. I am wondering why I can't derive the equation for position by multiplying dt throughout with a starting equation of $\frac{dx}{dt} = \frac{dx_i}{dt} + \frac{d^2x}{dt^2}t$. Multiplying dt and integrating:

$\int_{x_i}^{x_f} dx = \int_{0}^{v_i t} dx_i + \int_{0}^{at} dvt$

 

[kuruman]

$\int_{x_i}^{x_f} dx = \int_{0}^{v_i t} dx_i + \int_{0}^{at} dvt$

This equation is nonsense. To begin with, if $x_i$ is the initial position, it is a constant which means that $dx_i=0$. As $dvt$, it is meaningless unless you mean $d(vt)=vdt+tdv$ in which case you must write that.
Now for what you want to do.

I am wondering why I can't derive the equation for position by multiplying dt throughout with a starting equation of $\frac{dx}{dt} = \frac{dx_i}{dt} + \frac{d^2x}{dt^2}t$. Multiplying dt and integrating.

You can't because you are going around in a vicious circle. The equation
$$x=x_i+v_it+\frac{1}{2}at^2$$was the result of integrating twice the equation $$\frac{d^2x}{dt^2}=a=\mathbf{constant}.$$ You can't go back to the equation you derived and replace the constant $a$ with a function of time $\dfrac{d^2 x}{dt^2}$ and expect something sensible to emerge.

 

[SammyS]

For those^(*) helping with this thread, you might want to check the thread: https://www.physicsforums.com/threads/integrating-motion-equation-to-derive-displacement.1015857/
in the Classical Physics Forum (which is where this thread started before being moved).
* @kuruman, @Steve4Physics

 

[nasu]

Your link take you back to this thread. It is self-referencing. Or maybe your intention was to post it in the other thread?

 

[kuruman]

Your link take you back to this thread. It is self-referencing. Or maybe your intention was to post it in the other thread?

I believe @SammyS very cleverly provided a self-referencing link to match the self-referencing "derivation" by @annamal.

 

[SammyS]

I believe @SammyS very cleverly provided a self-referencing link to match the self-referencing "derivation" by @annamal.

No - not that clever today.
Sorry, I messed that up. Here's the correct link (maybe).
https://www.physicsforums.com/threads/deriving-kinematic-equation-for-position.1015909/unread

 

[kuruman]

No - not that clever today.
Sorry, I messed that up. Here's the correct link (maybe).
https://www.physicsforums.com/threads/deriving-kinematic-equation-for-position.1015909/unread

Too bad. I thought I was onto something. I wasn't aware of the other thread.

 

[PeroK]

To add to what has been said:$$v_i = v(t_i) \equiv v\rvert_{t=t_i} \equiv \frac{dx}{dt}\rvert_{t = t_i}$$In other words, $v_i$ is the velocity function of time evaluated at a specific time $t_i$. Likewise $x_i$ is the position function of time evaluated at $t_i$. It's not a function of $t$, so $\dfrac{dx_i}{dt}$ is meaningless.