Time measured by a car-Special relativity

[PeroK]

I think that is a equivalent expression. In the situation in study, i think that t2-t1=x/v-a/v=(x-a)/v, and x2-x1=x-a.
Then, Δt'=γ((x-a)/v-v(x-a)/c^2)=γ((ct-a)/v-v(x-a)/c^2), where γ=1/sqrt(1-v^2/c^2).

Nevertheless, 0.9428s is not the correct answer.

 

[Paulo Figueiredo]

Nevertheless, 0.9428s is not the correct answer.

Maybe not. If you get another answer I will apreciate to know.
Thank you very much

 

[PeroK]

Maybe not. If you get another answer I will apreciate to know.
Thank you very much

Why don't you try finishing it off? I get:

$\Delta t' = t'_2 - t'_1 = \gamma((t_2 - t_1) - \frac{v(x_2-x_1)}{c^2}) = \gamma(1s - \frac{x_2}{3c})$

Note that I haven't used any numbers such 200,000 at all. Also, note that $x_2 = 1$ light-second $= 1c s$

 

[Paulo Figueiredo]

Why don't you try finishing it off? I get:

$\Delta t' = t'_2 - t'_1 = \gamma((t_2 - t_1) - \frac{v(x_2-x_1)}{c^2}) = \gamma(1s - \frac{x_2}{3c})$

Note that I haven't used any numbers such 200,000 at all. Also, note that $x_2 = 1$ light-second $= 1c s$

I think that the problem is the value of x1. x2=1c s -Okay-(it is approximately 300,000 Km). But x1=200,000 Km (2c/3 c s) ?

 

[PeroK]

I think that the problem is the value of x1. x2=1c s -Okay-(it is approximately 300,000 Km). But x1=200,000 Km (2c/3 c s) ?

Ah, okay. But $x_1 = 0$.

 

[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.