Solving the Equation: x = 1/2gt^2 + v0t + x0

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Homework Help Overview

The discussion revolves around the kinematics equation x = 1/2gt² + v₀t + x₀, which relates displacement, initial velocity, time, and acceleration due to gravity. Participants are exploring the meaning of the variables involved, particularly the constant g, and its implications in various contexts.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants inquire about the purpose of the equation and the meaning of the variable g. There are discussions on whether g can vary in different scenarios and locations, with some questioning the consistency of its value.

Discussion Status

Several participants have provided clarifications regarding the equation and the variable g, noting that it typically represents the acceleration due to gravity. There is an ongoing exploration of how g can differ based on location and context, with no explicit consensus reached on its variability.

Contextual Notes

Participants are considering the implications of using g in different contexts, such as varying altitudes and planetary conditions, while also acknowledging the standard value of g near the Earth's surface.

hamudi786
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help me out here please

what exactly is this equation :

x=1/2gt(squared)+v0t+x0


and what is g
 
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It's a kinematics equation relating the final and initial displacements, the initial velocity, time, and the acceleration of gravity, g.
 
Welcome to PF!

Hi hamudi! Welcome to PF! :smile:

It's [tex]x(t)\,=\,\frac{1}{2}gt^2\,+\,v_0t\,+\,x_0\,,[/tex]

and it's the constant-acceleration equation for the distance x reached after time t with constant acceleration g if you started (at time 0) at position x0 and with speed v0. :smile:

"g" usually means "gravity", by the way - if it was anything else, you'd probably use the letter "a" (for "acceleration", of course).
 
Yes, Tiny-tim is correct; a (acceleration) in your equation is g because it is probably looking to y-axis and of course acceleration in y-axis is downward g

when ever you get confused what is a parameter you can look to its dimension too, in here g has acceleration dimension. [L/(T^2)]
 
hey thanks ...that was a great help =)
 
wat exactly is the purpose of the equation
 
For physics problems relating those variables.
 
can the constant g change ... like for example in one testing of an object it is 9 and on the other object( a diffrent one) 10.6 ...is this possible
 
no g is always equal to -9.8 m/s^2. though the sign changes according to the orientation of the coordinate system
 
  • #10
It depends only on location. We have this value of g near the surface of the earth. Slightly different values in cities of different altitudes -- in Poughkeepsie, New York it happens to be 9.80665 -- but who cares about the way-out decimal places? It changes slightly while falling through different altitudes, if it's a very long drop. Very different values on other planets -- on Mars it's about 3.2 m/s^2.
 
  • #11
hamudi786 said:
can the constant g change ... like for example in one testing of an object it is 9 and on the other object( a diffrent one) 10.6 ...is this possible

Hi hamudi! :smile:

As mikelepore says, it changes in different locations.

But in the same place, it does stay the same. :smile:

(If it didn't, we wouldn't use the letter "g".)
 
  • #12
thx...well yeah that does make sense :smile:
 

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