Constant turn radius with Δ airspeed and bank Can someone explain how speed impacts rate of turn and and degree of bank required?

I'm trying to understand this practice problem but can't get the concept: Say you have 2 vehicles travelling at different speeds and carrying out a turn at the same radius. The example states that the faster vehicle will have to bank at a shallower angle and will have a lower rate of turn. 

In my mind, the faster vehicle would have to maintain a steeper bank angle to compensate for the higher relative speed, otherwise their radius would increase. If you're moving faster, and banking at a lower angle, wouldn't that mean you're travelling at greater forward distance and less lateral distance thus increasing you're radius?
You are right on all accounts.

In a steady-state turn, from circular motion, we have, with $F_c$ being centripetal force, $V$ being airspeed in zero wind, $R$ being the radius of turn, and $m$ being the aircraft mass:

$$F_c=m\frac{V^2}{R}$$

The centripetal acceleration in a coordinated turn is created from the banked lift vector ($L$), therefore, with $\phi$ being the bank angle:

$$L\sin\phi=m\frac{V^2}{R}$$

Therefore, the radius of turn is:

$$R=\frac{V^2}{g\tan\phi}$$

and the rate of turn is:

$$\dot{\psi}=\frac{g\tan\phi}{V}$$

As speed increases, you would need an ever larger bank angle in order to achieve the same rate of turn. Of course, there's a limit to how much lift you can get out of the wing at a given speed, especially at higher Mach when transonic buffet becomes the limiting factor.