Question

Observer О notes that two events are separated in space and time by 600 𝑚 and 8 × 10−7

𝑠. How fast

must an observer O' be moving relative to О in order that the events be simultaneous to O'?

Answer 1

d=600m

$t=8\times10^{-7}sec$

$O$ co-ordination System

$v=\frac{d}{t}$

$v=\frac{600}{8\times10^{-7}}=75\times10^{7}m/sec$

Now $O'$ co -ordination system

Similarly

d'=600m

$t'=8\times10^{-7}sec$

$v'=\frac{d'}{t'}$

$v'=\frac{600}{8\times10^{-7}}=75\times10^{7}m/sec$

$v=v'$