In the equation x = x₀ + vt, 'x₀' means what?

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Discussion Overview

The discussion revolves around the meaning of 'x₀' in the equation x = x₀ + vt, focusing on its interpretation as the initial position of an object in the context of motion equations. Participants explore its implications in terms of displacement and position within a coordinate system.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that 'x₀' represents the original or starting value of position.
  • Others clarify that 'x₀' denotes the initial position of an object, particularly in a standard Cartesian coordinate system.
  • There is a discussion about the relationship between 'x', 'x₀', and displacement, with some participants noting that displacement is represented by Δx, which is the difference between 'x' and 'x₀'.
  • Some participants express confusion about the final position 'x' and seek clarification on how it relates to 'x₀'.
  • One participant mentions that the equation applies to constant velocity along a single axis and that the final position depends on both the initial position and the distance traveled over time.
  • There are requests for diagrams to better understand the initial and final positions in relation to the equation.
  • A later post introduces the idea that the direction of motion can be indicated by the sign in front of the velocity term.

Areas of Agreement / Disagreement

Participants generally agree that 'x₀' represents the initial position, but there is some disagreement regarding the interpretation of displacement and the final position 'x'. The discussion remains unresolved regarding the specifics of how to visualize these concepts.

Contextual Notes

Some participants express uncertainty about the definitions of terms and the relationships between initial and final positions, indicating a need for clearer explanations or visual aids.

Indranil
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In the equation x = x₀ + vt, 'x₀' means what?
 
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The subscript "0" pretty much always means the original/starting value.
 
russ_watters said:
The subscript "0" pretty much always means the original/starting value.
As I know x denotes 'displacement'. If x = 5 meters east, it means it's final position is 5 meters east but 'x₀' denotes what? is the object stationary in this case?
 
Indranil said:
In the equation x = x₀ + vt, 'x₀' means what?

It means the initial position of an object.

Indranil said:
As I know x denotes 'displacement'. If x = 5 meters east, it means it's final position is 5 meters east but 'x₀' denotes what? is the object stationary in this case?

If you write, ##x_0 = 5## then that would mean that the initial position of the object is, in a standard cartesian coordinate system (xy-plane), located at 5 units to the right of the origin.
 
Indranil said:
As I know x denotes 'displacement'. If x = 5 meters east, it means it's final position is 5 meters east but 'x₀' denotes what? is the object stationary in this case?
If ##x_0 = 5##, this means that the object in question starts out at 5 units in the positive direction (usually eastwards or to the right) from the origin (where x=0). This is the initial position.

The ##x## value being spit out by the function gives you the final position of the object. This can be any new number.

However ##x## is not the displacement. The displacement would be ##\Delta x##, pronounced Delta ##x##. ##\Delta x## is the difference between ##x_0## and ##x##.
 
lekh2003 said:
If ##x_0 = 5##, this means that the object in question starts out at 5 units in the positive direction (usually eastwards or to the right) from the origin (where x=0). This is the initial position.

The ##x## value being spit out by the function gives you the final position of the object. This can be any new number.

However ##x## is not the displacement. The displacement would be ##\Delta x##, pronounced Delta ##x##. ##\Delta x## is the difference between ##x_0## and ##x##.
Ok, then what would be the final position ( x ) from the concept above? what would be the final position value? I knew that the initial position is 0 and the final position is 5. Am I correct? Please explain.
 
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Indranil said:
In the equation x = x₀ + vt, 'x₀' means what?
x0 is the position at t = 0: the initial position.
 
Doc Al said:
x0 is the position at t = 0: the initial position.
Could you please provide me with a diagram so that I can clear my confusion? it's my humble request.
 
Indranil said:
Could you please provide me with a diagram so that I can clear my confusion? it's my humble request.
I'm not sure what sort of diagram you're looking for. The equation ##x = x_0 + vt## applies to constant velocity along a single axis (in this case the x axis). It gives you the final position along that axis after some time "t" passes. The final position depends on where you started (given by x0) and the distance you traveled in that time (given by vt).
 
  • #10
Doc Al said:
I'm not sure what sort of diagram you're looking for. The equation ##x = x_0 + vt## applies to constant velocity along a single axis (in this case the x axis). It gives you the final position along that axis after some time "t" passes. The final position depends on where you started (given by x0) and the distance you traveled in that time (given by vt).
Please provide me with any 'initial and final position' diagram. It would be very useful to me.
 
  • #11
Indranil said:
Please provide me with any 'initial and final position' diagram. It would be very useful to me.
Code:
x axis: -----0--------x0-----------------------xf----
             ^        ^                        ^
             Origin   Start here               End here
 
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  • #12
upload_2018-7-23_13-8-15.png

How about this?
 

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  • #13
Indranil said:
In the equation x = x₀ + vt, 'x₀' means what?

These are the coordinate equations for rectilinear uniform motion. x₀ is the coordinate of the starting point of motion. And plus or minus shows the direction of motion. In one case, when plus, it moves toward plus infinity, and when minus, it moves toward minus infinity.
v- velocit
t-unit of time
 
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  • #14
This thread is 5 years old and the OP is no longer at PhysicsForums. Your post does not add anything to what has been said before.

Thread locked.
 

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