设 S'相对S系的速度为 v，则
t1'=(t1-vx1/c²)/√[1-(v/c)²]
t2'=(t2-vx2/c²)/√[1-(v/c)²]
所以 （t2'-t1') = [(t2-t1) -v(x2-x1)]/√[1-(v/c)²]
由题意 x1=x2 t2-t1= 4 t2'-t1'=5
所以 √[1-(v/c)²] = 4/5 v=3c/5
x1'=(x1-vt1)/√[1-(v/c)²]
x2'=(x2-vt2)/√[1-(v/c)²]
则 (x2'-x1')= [(x2-x1)-v(t2-t1)]/√[1-(v/c)²] = -3c
即 |△x'|= 3c=9X10^8 m