Question

Two events are observed in a frame of reference S to occur at the same space point, the second occurring 1.80 s after the first. In a frame S′ moving relative to S, the second event is observed to occur 2.05 s after the first. What is the difference between the positions of the two events as measured in S^?

Answer 1

Answer:

The difference between the positions of the two events as measured in = 3.53 *10^8 m/s

Explanation:

As we know -

$\Delta x = -\gamma \mu\Delta t$

Here,

$\Delta x$ is the difference between the positions of the two events as measured in S^

$\gamma$ $= \frac{1}{\sqrt{1-\frac{\mu^2}{c^2} } }$

And

$\mu$ = 0.547 c

Substituting the given values in above equation, we get -

$\Delta x = (0.547 c)*\frac{1}{\sqrt{1-\frac{\mu^2}{c^2} } }*2.15\\\Delta x = (0.547 c)*\frac{1}{\sqrt{1-\frac{(0.547 c)^2}{c^2} } }*2.15\\\Delta x = (0.547 *3*10^8)*\frac{1}{\sqrt{(1-\(0.547 )^2 } }*2.15\\\Delta x = 3.53 *10^8$meter per second