No, they're spherical children in a vacuum.

You'll note that I already assumed that they were on a plane, not the surface of a sphere.

I'm trying to apply the most simple math possible and it seems to add up.

After one second, their distance is √(5² + 1²) = ~5.1 ft

After two seconds, their distance is √(10² + 2²) = ~10.2 ft

After three seconds, it's √(15² + 3²) = ~15.3 ft

As speed is the rate of change of distance over time, you can see it's a constant 5.1 ft/s. You're free to point out any error, but I don't think you need anything more than Pythagoras' theorem.

The question specifically asks for their seperation speed at 5s to ignore any initial change in their speed as they first need to accelerate, I'd assume.

I don't see why the distance between them isn't growing at a constant speed.

At any given time t seconds after separation, the boy is 5t north, and the girl is 1t east. The distance between them is defined by the square root of ((5t)^2 + (t)^2 ), or about 5.099t.

In other words, the distance between them is simply a function defined as 5.099t, whose first derivative with respect to time is just 5.099.

If you're at the south pole, would every direction count as north?

Sure, but there is a north say 30 ft away from the north pole.

Their distance is the hypotenuse of a triangle with sides 5t and t which will be root((5t)^2^ + t^2^). So the distance at time t of the ex lovers will be root(26) × t. You can basically grasp intuitively that the speed is indeed constant and equals to the root(26)=5.1 ft/sec. Technically you’d use the derivative power rule to drop the t and get the speed.

Velocity is not the difference between distances.