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In the inertial system S, an event is observed to take place at point A on the x-axis and 10−6S later another event takes place at point B, 900 m further down. Find the magnitude and direction of the velocity of S with respect to S in which these two events appear simultaneous

User Joe Grund
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The magnitude of the velocity of S with respect to S' is 0.866 times the speed of light, and it is in the negative x-direction (i.e., opposite to the direction of motion of S').

Step-by-step explanation:

The problem can be solved using the concept of relativity of simultaneity and the Lorentz transformation equations. Let's assume that the inertial system S is moving with a velocity v with respect to S', where the two events are simultaneous.

Let's first find out the time difference between the two events as observed in S. According to S, the two events occur at different times because they are separated by a distance of 900m. Let's call this time difference Δt.

Δt = 900m / c = 3 × 10^−6 S

where c is the speed of light.

Now, let's apply the Lorentz transformation equations to relate the time difference Δt in S to the time difference Δt' in S':

Δt' = γ(Δt - vΔx/c^2)

where Δx is the distance between the two events as measured in S, and γ is the Lorentz factor given by:

γ = 1 / sqrt(1 - v^2/c^2)

Since the two events are simultaneous in S', we have Δt' = 0. Also, Δx = 900m, and Δt = 3 × 10^−6 S.

Solving for v, we get:

v = Δx / (γΔt)

Substituting the values of Δx, Δt, and γ, we get:

v = 0.866c

User Mhvelplund
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