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)?
 A: 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 $\overrightarrow{a}=-\dfrac{v^{2}}{r}\overrightarrow{e}_{r}$
A: There is still not enough information to answer the question.  The initial speed is specified, but not the direction.  Given the direction of travel, you can find the center of the circle.  Then (as in my last comment) the angular velocity is V/R and you can apply the same trig functions.
A: I don't know what you mean by R.  If you start at the origin, $v_x=v\cos(\theta)$ and $v_y=v\sin(\theta)$, $x=v_xt$ and $y=v_yt$.
A: 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
$$R=\frac{V^2}{\tan(\theta)G}$$
