2D - What is the next point given this info.

[stvrbbns]

In a two-dimensional environment:
Given:

• a starting point (x1,y1)
• an initial velocity (speed and 2D heading/direction)
• a constant rate of turning (yaw, since there is no roll or pitch in only two dimensions)
• a constant acceleration of the object from itself (not like gravity where the direction of the acceleration is relatively constant - "down" - but like a rocket that is propelling itself)
• a time change t1 to t2

What is the formula for the final point (x2,y2)? What kind of physics problem do I have on my hands? Is this just a variation on kinematics with variable acceleration (varying direction of acceleration instead of amount), or is it qualitatively different?

There is no sliding.

(turning without accelerating example) If it is headed north/up from (0,0) and turns 90 degrees to the right at a speed of pi/10 per second, then after 10 seconds it will be at (2,2).

I just guessed at thread title prefix thread level. Let me know if it is incorrect and I can try to change it.

 

[stvrbbns]

Intuitively I think that this should result in a path which might be visualized as a quarter of an ellipse and I was trying to work off of http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations and http://www.numericana.com/answer/ellipse.htm but I'm not sure this is the correct direction/method.

 

[jbriggs444]

Intuitively I think that this should result in a path which might be visualized as a quarter of an ellipse and I was trying to work off of http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations and http://www.numericana.com/answer/ellipse.htm but I'm not sure this is the correct direction/method.

Note that the acceleration is independent of velocity. So the first trick you should automatically try is to transform the velocity away and assume that starting velocity is always zero.

Now note that acceleration is cyclical. The "rocket motor" spins at a constant rate regardless. This is consistent with uniform circular motion. So the resulting motion will be the superposition of uniform circular motion on constant velocity motion. So the first trick was slightly wrong. One needs to transform the initial velocity away, leaving just the appropriate tangential velocity.

At that point, solving for position as a function of time is trivial.