Let's analyze the scenario based on the information provided:
Clock A, B, and C are synchronized at the start (t = 0). C moves away from A with a velocity of 0.866c. When C arrives at B, clock B reads 2 seconds.
First, we need to determine the time it took for C to travel from A to B. We can use the time dilation equation to calculate the elapsed time experienced by C:
Δt' = Δt * √(1 - v^2/c^2)
Here, Δt' represents the time experienced by C, and Δt represents the time measured by the stationary observer (A and B).
Given that clock B reads 2 seconds when C arrives at B, we can equate Δt' and 2 seconds:
2 = Δt * √(1 - (0.866c)^2/c^2)
Solving for Δt, we find:
Δt = 2 / √(1 - 0.866^2)
Now, let's calculate the time dilation factor for C's velocity:
γ = 1 / √(1 - v^2/c^2)
Using v = 0.866c, we have:
γ = 1 / √(1 - 0.866^2)
To determine the elapsed time on clock C, we can multiply the time measured by the stationary observer (Δt) by the time dilation factor (γ):
Δt' = γ * Δt
Finally, let's calculate the values:
Δt = 2 / √(1 - 0.866^2) ≈ 4.898 seconds
γ = 1 / √(1 - 0.866^2) ≈ 2.0008
Δt' = γ * Δt ≈ 2.0008 * 4.898 ≈ 9.799 seconds
Therefore, clock A would read approximately 4.898 seconds when C arrives at B, and clock C would read approximately 9.799 seconds.
In this setup, the time dilation effect is evident in the difference between the elapsed times experienced by C and measured by the stationary observer at A and B. Due to C's high velocity (0.866c), time appears to pass more slowly for C relative to A and B. This is consistent with the principles of time dilation in special relativity, where the relative velocity between frames of reference causes time to dilate.