Use the relativistic coordinate transformation

(x, y, z, t) − (x′, y′, z′, t′) shown and given
below where the latter frame S′, (x′, y′, z′, t′),
has a velocity 2.69813 × 108 m/s in the
positive direction relative to the frame S,
(x, y, z, t).

For convenience, assume that the origins of the two reference frames overlap at time
t = t′ = 0.

what is the observed time interval T = t2 − t1 in the stationary reference frame given T′ = t′2 − t′1 in the moving reference frame?
The speed of light is 2.99792 × 108 m/s and the time measured in the primed frame is 52 s

there were some other details for this problem but the above is the question..what is the time interval...

so far i've got the speed of light, velocity, i found B= 0.900000667, Y=2.294164587, and t'=52 seconds

and now i'm stuck as to how to plug them in to find the answer. please help thanks

To find the observed time interval (T) in the stationary reference frame, given the time interval in the moving reference frame (T'), we can use the relativistic time dilation equation:

T = γ * T'

Where T is the observed time interval in the stationary reference frame, T' is the time interval in the moving reference frame, and γ (gamma) is the Lorentz factor, which depends on the velocity.

First, let's calculate γ using the velocity provided:
v = 2.69813 × 10^8 m/s (velocity)

The Lorentz factor γ can be calculated using the formula:

γ = 1 / sqrt(1 - (v^2 / c^2))

Where c is the speed of light, given as 2.99792 × 10^8 m/s.

Substituting the values, we get:

γ = 1 / sqrt(1 - (2.69813 × 10^8)^2 / (2.99792 × 10^8)^2)

Now we can calculate γ using the above equation.

Once we have γ, we can calculate T using the equation mentioned earlier.

T = γ * T'

Substituting the values of γ and T', we have:

T = γ * 52 seconds

Now, you can calculate the value of T by substituting the calculated γ value and T' value into the equation.