# Integrating motion equation to derive displacement

• annamal
In summary: The ##x_i## is the constant that results from the first integration. Don't replace it with something else.I have taken differential equations already, so feel free to explain it to me in those terms as wellYes, 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}{
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?

PeroK
annamal said:
##\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##?

kuruman said:
What is ##x_i##?
##x_i## is Initial position. I swapped the ##x_i## and ##x_f## positions in the integral

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

PeroK
kuruman said:
And does the initial position change as the object is in motion?
No, the initial position is constant

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

annamal said:
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.

nasu
kuruman said:
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
Steve4Physics said:
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.

Steve4Physics said:
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.

Steve4Physics
kuruman said:
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##

annamal said:
##\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.
annamal said:
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.

PeroK and Steve4Physics
in the Classical Physics Forum (which is where this thread started before being moved).
* @kuruman, @Steve4Physics

kuruman

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

Steve4Physics
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.

## 1. What is the purpose of integrating motion equations to derive displacement?

The purpose of integrating motion equations is to determine the displacement of an object over a certain period of time. This is useful in understanding the position, velocity, and acceleration of the object and how they change over time.

## 2. What are the basic equations used for integrating motion to derive displacement?

The basic equations used for integrating motion to derive displacement are the position equation (x = x0 + v0t + 1/2at^2), the velocity equation (v = v0 + at), and the acceleration equation (a = (vf - v0)/t).

## 3. How is integration used to derive displacement from these equations?

Integration is a mathematical process that involves finding the area under a curve. In the context of motion equations, integrating the acceleration equation yields the velocity equation, and integrating the velocity equation yields the position equation. This allows us to determine the displacement of an object by finding the area under the velocity curve.

## 4. What are the key steps in integrating motion equations to derive displacement?

The key steps in integrating motion equations to derive displacement are: 1) determining the initial conditions (position, velocity, and acceleration) of the object, 2) integrating the acceleration equation to find the velocity equation, 3) integrating the velocity equation to find the position equation, and 4) using the position equation to determine the displacement of the object at a specific time.

## 5. What are some real-world applications of integrating motion equations to derive displacement?

Integrating motion equations to derive displacement is used in various fields such as physics, engineering, and astronomy. It is used to analyze the motion of objects in space, design structures and machines, and predict the trajectory of projectiles. It is also used in sports science to understand the movement of athletes and in medicine to track the motion of body parts during physical therapy.

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