Question

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

Answer 1

Step-by-step 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