Time measured by a car-Special relativity

[Paulo Figueiredo]

Ah, okay. But $x_1 = 0$.

Let's see: O and A are initially in the same place (t=t'=x=x'=0). Correct? Then A moves with a velocity c/3. It was as if O moves farway from A at that velocity. The event 1 is the event concerning at t1=2 sec. Then x1 (distance between O and A, measured by O) is 200,000 Km. I don´think that x1=0. This value was not the value when the event 1 had occured.

 

[PeroK]

Let's see: O and A are initially in the same place (t=t'=x=x'=0). Correct? Then A moves with a velocity c/3. It was as if O moves farway from A at that velocity. The event 1 is the event concerning at t1=2 sec. Then x1 (distance between O and A, measured by O) is 200,000 Km. I don´think that x1=0. This value was not the value when the event 1 had occured.

Event 1 is O emitting the signal. $x_1$ is the location of the event, not the position of A. So, $x_1 = 0$. This is in O's frame. You could use the Lorentz Transformation to get the coordinate of Event 1 in A's frame, which would be $x'_1 = \gamma(x_1 - vt_1) = -\gamma vt_1$.

 

[Paulo Figueiredo]

Event 1 is O emitting the signal. $x_1$ is the location of the event, not the position of A. So, $x_1 = 0$. This is in O's frame. You could use the Lorentz Transformation to get the coordinate of Event 1 in A's frame, which would be $x'_1 = \gamma(x_1 - vt_1) = -\gamma vt_1$.

You are right. Now I understand my error.
Thank you very much.