Mechanics- Relative Velocity HELP

Hi! I have the solution to these questions but unsure as to how it was worked out, why were the velocities of A and B added instead of subtracted to find the relative velocity?

The velocity of A relative to B denoted by $v_{AB}$ is found by $v_B-v_A$ - likewise for $r_{AB}$

Since $v_A=0{\bf i}-25{\bf j}$ and from your diagram, you have $v_{BA}=30 {\bf i} -25 {\bf j}$

Then since at $t=0$ you have $r_A=0 {\bf i } + 610{\bf j }$ and $r_B= 610 {\bf i } + 0{\bf j }$ you have $r_{BA}=-610 {\bf i } + 610{\bf j }$

So at time $t$ you have $r_{BA}=(-610 {\bf i } + 610{\bf j })+(30t {\bf i} -25t {\bf j})=(-610+30t) {\bf i}+(610-25t) {\bf j}$

So really the ${\bf i}$ component is incorrect there by a multiple -1, but of course the correct answer is achieved as the multiple of -1 goes to +1 when squared as does +1 squared if it were correctly written down.

(610 - 30t) and (610 - 25t) are displacements from the origin, not velocities.

The straight line distance between the two cars is found by Pythagoras: d^2 = x^2 + y^2, where all of those distances are functions of t.

The rest of the working consists of minimising d^2 (which is as good as minimising the modulus of d) by differentiating with respect to t and setting the differential equal to zero.

There is no addition or subtraction of velocities involved here.