Given a radius and velocity calculate position of an aircraft banking to make a turn I have a radius, R, for an aircraft traveling at velocity, V. If we start at point, (X,Y), what is the position of the point at time, t.

For example:
The aircraft is at point (0,0) and traveling at 250 knots and initiates a turn with a bank angle, phi, of 5 degrees. Assume that the aircraft can instantaneously rotate to the five degree bank. The equation for the turn radius, R where g is the acceleration due to gravity (9.81) is:
\begin{equation}
\text{R} = \frac{V^2}{\text{g} \tan{\phi}}
\end{equation}

For this example, R = 10.4 nautical miles. Where is the aircraft at t = 2 if the aircraft is traveling at a heading of 90 degrees (straight along the y axis)?
See banked turn in  aeronautics



In a motion along a circular path centered at $(0,0)$ (see figure below) with constant angular velocity $\omega$ the position of a point $P(x,y)$ (in Cartesian coordinates and $P(r,\varphi)$ in cylindrical coordinates) starting at $(r,0)$ at $t=0$ is given by the law of the uniform circular motion:

$$x=r\cos(\omega t)$$

$$y=r\sin(\omega t),$$

where $r$ is the radius of the circumference and $\omega=\dfrac{d\varphi}{dt}$. The magnitude of the tangential velocity is $v=\omega r$ and the radial velocity is zero. Thus $\omega=\dfrac{v}{r}$.

$$x=r\cos\left(\frac{vt}{r}\right)$$

$$y=r\sin\left(\frac{vt}{r}\right)$$ 

In your case $r=R$, $v=V$ (it is assumed that the aircraft mantains the speed $V$) and the centre of the circumference is $(X,Y)=(-R,0)$. If you make the change of coordinates $X=x-R$ and $Y=y$, you get the following $X,Y$ coordinates as a function of time $t$:

$$X=R\left(\cos\left(\frac{V}{R}t\right)-1\right)$$

$$Y=R\sin\left(\frac{V}{R}t\right)$$

In this $XY$-coordinate system the motion starts at $(X,Y)=(0,0)$ ($t=0$).

I assumed that the direction of the turn is to the left. If it is to the right, then one has a symmetric motion with respect to the $y-$axis:

$$X=R\left(-\cos\left(\frac{V}{R}t\right)+1\right)$$

$$Y=R\sin\left(\frac{V}{R}t\right).$$



The numerical result is obtained for $t=2$ and the other given data. 

Added: For a different starting position $\left( X_{0},Y_{0}\right) $ at $t=0$ we have to incorporate these values in the motion equations. Integrating $%
\omega =\dfrac{d\varphi }{dt}$, we obtain $\varphi =\omega t+C$, where $C$ is
a constant. Then

$$x=r\cos \left( \omega t+C\right) ,y=r\sin \left( \omega t+C\right) $$

$$x_{0}=r\cos \left( C\right) ,y_{0}=r\sin \left( C\right) $$

and

$$\dfrac{y_{0}}{x_{0}}=\dfrac{\sin \left( C\right) }{\cos \left( C\right) }%
=\tan \left( C\right) ,C=\arctan \left( \dfrac{y_{0}}{x_{0}}\right) $$

$$x=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right)
,y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right) .$$

Since

$$X=x-r=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right)
\right) -r,Y=y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}%
\right) \right) $$

we get

$$X_{0}=x_{0}-r,Y_{0}=y_{0},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+r}$$

$$r=R,\omega =\dfrac{v}{r}=\dfrac{V}{R},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+R}.$$

Finally, we obtain the equations of the motion of the aircraft in the $X,Y$-plane:

$$X=x-R=R\left( \cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) -1\right) $$

$$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right),$$

where $\left( X_{0},Y_{0}\right) $ is the position of the aircraft at $t=0$.

If the direction of the turn is to the right, then the motion is symmetric with respect to the $y-$axis:

$$X=-x+R=R\left( -\cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) +1\right) $$

$$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right).$$

Note: I changed the notation; in the question the position of the aircraft
at $t=0$ is $\left( X,Y\right) $.



The velocity vector is $\overrightarrow{v}=v\overrightarrow{e}_{\varphi }$ and the acceleration vector $\overrightarrow{a}=-\dfrac{v^{2}}{r}\overrightarrow{e}_{r}$
Use the following:


$L$ = lift
$R$ = radius
$\theta$ = bank angle
$M$ = mass
$C$ = centrifugal force
$G$ = gravity


Then

$$L\cos(\theta)=MG\\L\sin(\theta)=C=\frac{MV^2}{R}$$

and so

$$\tan(\theta)=\cfrac{\frac{MV^2}{R}}{MG}=\frac{V^2}{RG}$$

Ergo

$$R=\frac{V^2}{\tan(\theta)G}$$