- #1
leroyjenkens
- 616
- 49
Homework Statement
Ok I have a moving person (primed) going 50 m/s in the positive x direction, and I have someone stationary (unprimed) observing them.
At t = 0, the moving person is at x(0) = 100m
Write an equation for the object’s position as a function of time x(t)
seen by the observer in the unprimed (stationary) reference frame.
And it says the only unknown in my expression should be the variable t.
So to be clear, the stationary reference frame is unprimed. The moving reference frame is primed.
Homework Equations
[tex]x'=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
[tex]x=\frac{x'-vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
The Attempt at a Solution
First of all, the wording of this confuses me. From what I've been reading in the book, the first equation is for the stationary (unprimed) observer. And the primed (moving) reference frame is the second equation.
But the question asks me to write an equation for x(t) for the unprimed observer. Shouldn't that be x'(t)?
Also, since the primed reference frame is moving with the person who is moving in the primed reference frame, does that mean their position is always 0? From their perspective, it's the unprimed observer who we consider stationary who is moving.
So I get...
[tex]0=\frac{100-50t}{\sqrt{1-\frac{50^2}{(3x10^8)^2}}}[/tex]
So I replaced all the variables I knew, which was x, given in the original problem as 100. And v, given as 50. And the assumed x'(t) of 0. Is that right? I think I'm way off and not even close to understanding what I'm being asked to do here.
Thanks