Question

The animation in the simulation (linked below) is a simple depiction of the motion of an orange, which is treated as a particle. The simulation depicts two reference frames, which are labeled A and B. Run the animation now. The motion is from the perspective of an observer (perhaps named Alex) at rest in frame A. Two things are being observed from frame A: the orange, and frame B. Here are some ideas to keep in mind about the simulation:

For the default values of vBA,x=4 m/s and vBA,y=0, what are the values of xPA (the x coordinate of the orange (the particle) as observed by Alex in frame A) at t=0

 and at t=8 s?

Answer 1

The given parameters are:-

$v=4m/s,y=0,x=0$

as we Know speed of light, $c=3\times 10^8m/s$

As there are two cordinate frames-

From the alex point of view

There was a transfomation in the cordinate of orange.

According to lorentz transformation,

$x'=\dfrac{x+vt}{\sqrt{1-\dfrac{v^2}{c^2}}}$

At time t=0, The cordinate of x is-

$=\dfrac{0+4\times 0}{\sqrt{1-\dfrac{v^2}{c^2}}}=0$

at time t=8s, The cordinate of x is:-

$=\dfrac{0+4\times 8}{\sqrt{1-\dfrac{(4)^2}{(3\times 10^8)^2}}}$

As $4<<3\times 10^8$ So It can be neglected,

The cordinate of $x \text{ is } 32m.$