What's the difference between Δ and d in physics ?

[Peter25samaha]

Is the acceleration : a=Δv/Δt equivalent to a=dv/dt ??
Anx what's the difference between Δ and d and can i use Δ when i want ?
a=d²x / dt²
Can we write the same using Δ ?

 

[PeroK]

Is the acceleration : a=Δv/Δt equivalent to a=dv/dt ??
Anx what's the difference between Δ and d and can i use Δ when i want ?
a=^dx/d_t
Can we write the same using Δ ?

$\Delta$ usually represents a finite change in something. So, technically:

Average Acceleration = $\frac{\Delta v}{\Delta t}$

Where $\Delta t$ is some finite (not necessarily small) period of time.

$a = \frac{dv}{dt}$ denotes the derivative of velocity with respect to time. This gives the instantaneous acceleration.

The two are related by:

$\frac{dv}{dt} = \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}$

 

[Peter25samaha]

$\Delta$ usually represents a finite change in something. So, technically:

Average Acceleration = $\frac{\Delta v}{\Delta t}$Where $\Delta t$ is some finite (not necessarily small) period of time.
$a = \frac{dv}{dt}$ denotes the derivative of velocity with respect to time. This gives the instantaneous acceleration.
The two are related by:

$\frac{dv}{dt} = \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}$

a=d²x/dt² and here what does it mean the : d²x
Shoudn't be :dx² ?

 

[PeroK]

a=d²x/dt² and here what does it mean the : d²x
Shoudn't be :dx² ?

$\frac{d^2x}{dt^2} = \frac{d}{dt}(\frac{dx}{dt}) = \frac{dv}{dt}$

The notation is probably because you can view $\frac{d}{dt}$ as the differential operator, hence $\frac{d}{dt}\frac{d}{dt} = \frac{d^2}{dt^2}$ but I wouldn't read too much into the notation. Other notations for acceleration include:

$x''$ and $\ddot{x}$

 

[NascentOxygen]

Is the acceleration : a=Δv/Δt equivalent to a=dv/dt ??

These will have identical values if velocity vs. time graph is a straight line (around the region of interest), otherwise Δv/Δt is an approximation to the exact slope of the tangent, dv/dt.

 

[SteamKing]

a=d²x/dt² and here what does it mean the : d²x
Shoudn't be :dx² ?

The trouble you are having is this is calculus notation. If you haven't studied calculus yet, you're going to be understandably confused.

 

[Peter25samaha]

These will have identical values if velocity vs. time graph is a straight line (around the region of interest), otherwise Δv/Δt is an approximation to the exact slope of the tangent, dv/dt.

Yes and if i am using a 3D graph with x y and z i will put vector on each one and i have to write it like this : dv/dt right ?

 

[Peter25samaha]

$\frac{d^2x}{dt^2} = \frac{d}{dt}(\frac{dx}{dt}) = \frac{dv}{dt}$

The notation is probably because you can view $\frac{d}{dt}$ as the differential operator, hence $\frac{d}{dt}\frac{d}{dt} = \frac{d^2}{dt^2}$ but I wouldn't read too much into the notation. Other notations for acceleration include:

$x''$ and $\ddot{x}$

Okay but this :
a=d2x/dt2
Can be equal to delta squared if velocity and time graph is a straight light ?
a=Δ²x/dt²
Those two acceleration are equal ?

 

[SteamKing]

Okay but this :
a=d2x/dt2
Can be equal to delta squared if velocity and time graph is a straight light ?
a=Δ²x/dt²
Those two acceleration are equal ?

Generally, one doesn't mix Δ and d notation in the same expression.

 

[PeroK]

Okay but this :
a=d2x/dt2
Can be equal to delta squared if velocity and time graph is a straight light ?
a=Δ²x/dt²
Those two acceleration are equal ?

First $\Delta^2$ has no obvious meaning that I can see. $(\Delta x)^2$ has the obvious meaning.

Second, as Steam King points out, $\frac{\Delta}{d}$ is meaningless.

Third, you should be able to see for yourself why $a \ne \frac{(\Delta x)^2}{(\Delta t)^2}$

More generally, you seem to be just groping in the dark here. You seem to be struggling to understand what is going on with time and distance and motion. This material should make sense. Are you learning calculus?

 

[Peter25samaha]

First $\Delta^2$ has no obvious meaning that I can see. $(\Delta x)^2$ has the obvious meaning.

Second, as Steam King points out, $\frac{\Delta}{d}$ is meaningless.Third, you should be able to see for yourself why $a \ne \frac{(\Delta x)^2}{(\D elta t)^2}$

More generally, you seem to be just groping

in the dark here. You seem to be struggling to understand what is going on with time
and distance and motion. This material
should make sense. Are you learning calculus?

Yes i am learning but i only want to know when i use d and when i use Δ if i have to calculate the velocity or acceleration

 

[PeroK]

Yes i am learning but i only want to know when i use d and when i use Δ if i have to calculate the velocity or acceleration

Hopefully that was answered in post #2.

 

[Peter25samaha]

Hopefully that was answered in post #2.

Okay sorry i haven't seen this