Show My Homework: Relativity

a) $t2A = X_A/V = 200/(0.6*c) = 1.11*10^{-6} sec$

b) in the moving reference system the rest length are seen contracted.. Thus the intervals of time becomes bigger.

$\Delta(t_B) = t2A/\sqrt{(1-V^2/c^2)} =1.11*10^{-6}/\sqrt{(1-0.6^2)} =1.3875*10^{-6} sec$

c) $XB1 =XA1=0 m$

$XB2 = XA2/\sqrt{(1-v^2/c^2)} = 200/\sqrt{(1-0.6^2)} =160 m$

(The length in moving reference system are seen contracted.)

d) proper time is by definition

$\tau = \sqrt{(\Delta(t)^2 – \Delta(x)^2/c^2)} =$

$= \sqrt {[(1.1*10^{-6})^2 -(200)^2/c^2]} =8.7496*10^{-7} seconds$

e) the length of the ruler for a B observer is

The length in moving reference system are seen contracted.

$LB = LA/\sqrt{(1-v^2/c^2)} =200/\sqrt{(1-0.6^2)} =160 m$

f) The proper length is:

$L = c*\tau = 3*10^8*8.7496*10^{-7} =262.488 m$

$L = (c^2*\Delta(t)^2 -\Delta(x)^2) =262.488 m$