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in the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?
A. velocity at position x, velocity at time t=0, position at time t=0
B. velocity at position x, velocity at position x=0, position x, and the original position
C. original velocity, velocity at position x=0, position x, and the original position
D. original velocity, zero velocity, original position, and the zero position

Respuesta :

B. velocity at position x, velocity at position x=0, position x, and the original position

In the equation

[tex]v_{x}^{2}[/tex] = [tex]v_{ox}^{2}[/tex] +2 a x (x - x₀)

[tex]v_{x}[/tex] = velocity at position "x"

[tex]v_{ox}[/tex] = velocity at position "x = 0 "

x = final position

[tex]x_{o}[/tex] = initial position of the object at the start of the motion

Explanation:

The equation of motion of an object is given by :

[tex]v_x^2=v_{ox}^2+2ax(x-x_o)[/tex]

Where

[tex]v_x[/tex] is velocity of a particle at position x

[tex]v_{ox}[/tex] is the velocity at position x = 0

x is the position of an object

[tex]x_o[/tex] is position at t = 0

So, the correct option is (b) "velocity at position x, velocity at position x=0, position x, and the original position". Hence, this is the required solution.                

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