Can Velocity Reversal Alter Temporal Order in Special Relativity?

[Ayame17]

Homework Statement

a) Event A = [ct(A)=1, x(A)=5, y(A)=0, z(A)=0] precedes event B = [ct(B)=2, x(B)=1, y(B)=0, z(B)=0]. What is the nature of the interval (HAVE DONE THIS). Then, it is desired to transform to another inertial sysyem O', moving at velocity, vi, relative to O, where the temporal order of the two events are reversed ie t'(B)$$\leq$$t'(A). What condition on v does this requirement impose?

Homework Equations

N/A for second part I assume!

The Attempt at a Solution

Have worked out that the nature of the interval is "Spacelike". Also I know that to transform to another inertial system, the velocity normally becomes -vi. However, I don't know if the temporal order will mean that something different happens, and it just being negative seems too easy an answer! Any help would be appreciated.

 

[Galileo]

The Attempt at a Solution

Have worked out that the nature of the interval is "Spacelike". Also I know that to transform to another inertial system, the velocity normally becomes -vi.

If an object stands still in one inertial frame, then from an inertial frame moving relative to the first with velocity v, the object is seen to have velocity -v. Is that what you meant?
It isn't really relevant to this problem however.
You have been given the spacetime coordinates of two events in some inertial frame. (And you found that the events are spacelikely separated.)
The spacecoordinates of the events will be different in an inertial frame moving relative to the first. What is the relation between the coordinates in the two frames as a function of v?