Question

In the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?

Answer 1

Explanation:

The equation of motion of an object is given by :

$v_x^2=v_{ox}^2+2ax(x-x_o)$

Where

$v_x$ is velocity of a particle at position x

$v_{ox}$ is the velocity at position x = 0

x is the position of an object

$x_o$ 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.                

Answer 2

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

In the equation

$v_{x}^{2}$ = $v_{ox}^{2}$ +2 a x (x - x₀)

$v_{x}$ = velocity at position "x"

$v_{ox}$ = velocity at position "x = 0 "

x = final position

$x_{o}$ = initial position of the object at the start of the motion