Step 1: Picture the motion.
$A$ goes north, $B$ goes east, so their paths make a right angle. The gap between them is the hypotenuse of a right triangle whose legs are how far each has gone.
Step 2: Find the distance each has covered by 2 PM.
They start at noon, so by 2 PM, $2$ hours have passed. $A$: $60 \times 2 = 120$ km north. $B$: $80 \times 2 = 160$ km east.
Step 3: Find the gap between them.
By Pythagoras, the separation is $\sqrt{120^2 + 160^2} = \sqrt{14400 + 25600} = \sqrt{40000} = 200$ km.
Step 4: Relate the rates instead of differentiating from scratch.
Let $y$ be the north distance and $x$ the east distance, and $s$ the gap, with $s^2 = x^2 + y^2$. The rate of separation is $\frac{ds}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{s}$.
Step 5: Plug in the numbers.
Here $x = 160$, $\frac{dx}{dt} = 80$, $y = 120$, $\frac{dy}{dt} = 60$, and $s = 200$. So $\frac{ds}{dt} = \frac{160(80) + 120(60)}{200} = \frac{12800 + 7200}{200} = \frac{20000}{200}$.
Step 6: Get the answer.
That equals $100$ km/hr. The two riders are pulling apart at $100$ km/hr. \[ \boxed{100\ \text{km/hr}} \]