How Does the Displacement Equation Calculate Position Changes in Physics?

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The displacement equation in physics, represented as x = x0 + v0t + (1/2)at^2, calculates the position of an object over time, incorporating initial position, velocity, and acceleration. The terms "displacement 1" and "displacement 2" refer to different instances or conditions of the same object's movement rather than separate bodies. Kinematics focuses on the motion of an object rather than its physical dimensions, clarifying that displacement is not related to the object's length. Therefore, displacement is defined as the change in position, not the size of the object itself. Understanding these distinctions is crucial for accurately applying kinematic equations in physics.
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We can have the velocity as v1 or v2 , time also, but In some cases i see it in displacement for e.x
##x=x_0+v_0 t + \frac{1}{2} at^2##, how can it be displacement 1 , displacement 2, can some one explain? Is it the displacement of the body 1, and displacement of the body 2 or what?
 
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-Physician said:
We can have the velocity as v1 or v2 , time also, but In some cases i see it in displacement for e.x
##x=x_0+v_0 t + \frac{1}{2} at^2##, how can it be displacement 1 , displacement 2, can some one explain? Is it the displacement of the body 1, and displacement of the body 2 or what?
In that version of the kinematic equation, x0 is the initial position (at time t = 0), v0 the initial velocity, and x is the position at time t.
 
Wouldn't that be the length of the body, like if the length of the body is 5m and distance 10m, the displacement would be 5m?
 
-Physician said:
Wouldn't that be the length of the body, like if the length of the body is 5m and distance 10m, the displacement would be 5m?
No. Kinematics describes the movement of a body, not its dimensions.
 
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