Calculating the instantaneous position of an object constant speed and varying direction Given an initial position for an object that's moving at constant speed but with a direction that varies in a linear fashion over time, how can I calculate the instantaneous position of the object?
For example, if an object is moving a 2 km/h, with an initial position of (0,0), an initial direction of 0° and a final direction of 90° over a  period of 1 hour, how can I get it's position at any point during that hour?
Please bear in mind that I'm a programmer, not a mathematician! With that in mind, I have no idea what to tag this question with....
 A: Assuming you mean that the angle goes from start angle to final angle at a constant rate, the object traces out a circular arc.
You can prove this by writing out the differential equations and solving them.  I won't give a proof, however, because it doesn't seem like you are necessarily looking for one.  Instead, I'll just give some general details about the arc traced by the object.
Suppose $\theta_0$ is the initial angle, $\theta_1$ is the final angle, $T$ is the total time between the two angles, and $V$ is the constant velocity.  Then:


*

*The angle at any given time $t$ is $\theta(t) = \theta_0 + \frac{t}{T} (\theta_1 - \theta_0)$.

*Assuming the angles are in radians, the radius of the circle traced out is $R = \frac{VT}{\theta_1 - \theta_0}$.
This follows from that the distance traveled is $R(\theta_1 - \theta_0)$, so the velocity must be $V = \frac{R\theta_1 - \theta_0}{T}$; then just solve for $R$.

*If the object starts at $(0,0)$, the center of the circle is at $(R\cos(\theta_0 + 90^\circ), R\sin(\theta_0 + 90^\circ))$.  (Adding $90^\circ$ to the angle $\theta_0$ gives us the direction from the object to the circle's center.)
This simplifies to $(-R\sin \theta_0, R \cos \theta_0)$.

*The displacement from the center of the circle to the object at any time is $(-R \cos(\theta + 90^\circ), - R \sin(\theta +90^\circ))$, where as above $\theta$ depends on $t$.  This simplifies to $(R \sin \theta, - R \cos \theta)$.
To get the position of the object (in $(x,y)$ coordinates) at a general time $t$ just add the center of the circle and the displacement together:
$$
\Big(x(t), y(t)\Big) = ( R\sin \theta - R\sin \theta_0, -R\cos \theta + R\cos \theta_0 )
$$
You can of course factor out the $R$ if you want.
Clarification
Regarding the limit as $\theta_1 - \theta_0 \to 0$, consider the $x$-coordinate:
$$
R (\sin \theta - \sin \theta_0) = VT \frac{\sin \theta - \sin \theta_0}{\theta_1 - \theta_0}
= VT \frac{\sin \left((\theta_0 + \frac{t}{T} (\theta_1 - \theta_0)\right) - \sin \theta_0}{\theta_1 - \theta_0}
$$
Write $\theta_1 - \theta_0 = \Delta$.
The limit as $\Delta$ goes to zero may be found by recognizing the definition of the derivative:
$$
x(t) = \lim_{\Delta \to 0} VT \frac{\sin \left((\theta_0 + \frac{t}{T} \Delta\right) - \sin \theta_0}{\Delta} = VT \left( \frac{t}{T} \cos(\theta_0) \right) = Vt\cos\theta_0
$$
Similarly one may compute $y(t) = Vt\sin\theta_0$.   Hence the object is just traveling in a straight line in direction $\theta_0$ with velocity $V$ (as expected).
A: We are looking for $d(t)=(d_x(t),d_y(t))$, and we will need an expression for $v(t)=(v_x(t),v_y(t))$ to find it. If we consider the direction of the velocity with respect to the x-axis (call the angle $\theta$) we find that $\tan(\theta)=\frac{v_y}{v_x}$, or 
$$v_y =v_x\tan (\theta)$$
We are given that the direction of the velocity changes linearly so
$$\frac{d\theta}{d t}=c$$
for some constant $c$, giving as a solution $\theta = ct + \theta_0$. Here $\theta_0$ is the direction of the initial velocity. Now if we use this expression in our expression for $v_y$ we find:
$$v_y = v_x \tan(ct+\theta_0)$$
We also know that the magnitude of the velocity is equal to $V$ at all times. This magnitude is given by:
$$\sqrt{v_x^2+v_y^2} = \sqrt{ v_x^2(1+\tan^2(ct+\theta_0))    }=\frac{v_x}{\cos(ct)+\theta_0}=V$$
Here we use $\sqrt{ (1+\tan^2(ct+\theta_0)} =\frac{1}{\cos(ct+\theta_0)}$ which is easily deduced from $sin^2+cos^2=1$. 
Now we have:
$$v_x=\cos(ct+\theta_0)V$$
and thus 
$$v_y = v_x \tan(ct+\theta_0)=\sin(ct+\theta_0)V$$
We can find $c$ from $\theta = ct+\theta_0$. For this we will need the direction of the velocity, $\theta_1$ at some $t_1$. these must be specified before hand or we cannot solve for $c$. Then we have, after rearranging, $c=\frac{\theta_1-\theta_0}{t_1}$. With $V$ assumed constant we now have $v(t)=(v_x(t),v_y(t))$:
$$v(t)= \left(V\cos \left(\frac{\theta_1-\theta_0}{t_1}\cdot t +\theta_0 \right) , V\sin \left(\frac{\theta_1-\theta_0}{t_1}\cdot t + \theta_0\right)\right)$$
to find the displacement we integrate the velocity from the initial time to some time $t$:
$$d(t) - d(0) = \int\limits_{0}^{t} \left(V\cos \left(\frac{\theta_1-\theta_0}{t_1}\cdot s  + \theta_0 \right) , V\sin \left(\frac{\theta_1-\theta_0}{t_1}\cdot s + \theta_0\right)\right)ds = V \left[\frac{t_1}{\theta_1-\theta_0}\sin \left(\frac{\theta_1-\theta_0}{t_1}\cdot t  + \theta_0\right) , -\frac{t_1}{\theta_1-\theta_0}\cos \left(\frac{\theta_1-\theta_0}{t_1}\cdot t  + \theta_0\right)\right]_{t=0}^{t=t}$$
Here we used the fact that $d(0)=(0,0)$. Finally, this equals:
$$d(t) = \frac{V t_1}{\theta_1-\theta_0} \left(       \sin\left(  \frac{\theta_1-\theta_0}{t_1}\cdot t  + \theta_0 \right) - \sin(\theta_0),  -\cos\left(  \frac{\theta_1-\theta_0}{t_1}\cdot t  + \theta_0 \right) +\cos(\theta_0) \right)$$
Just as a side note, in your example $\theta_0 = 0$, $\theta_1=\frac{\pi}{2}$, $t_1=1$ and $V=2$. When you use these in the final equation for the displacement you will find a much easier expression.
