# Integrating motion equation to derive displacement

• annamal

#### 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

##\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##?

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
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?

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
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.

• Steve4Physics
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##

##\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.

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

• kuruman
• • 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.