Motion in Two or Three Dimensions (Physics)

I actually finished reading chapter 3, "Motion in Two or Three Dimensions," of Young and Freedman's University Physics quite a long time ago.  I try to read two pages of either this physics or a calculus text a day.  So since I read this chapter I have read the first chapter of Stewart's sixth edition calculus text and am almost done with chapter 4 of Young and Freedman.  So I want to catch up with the old before I finish chapter 4.

One observation I've made is that the broad strokes of physics and chemistry are really not too difficult to explain.  The Devil is in the details of application.  As always, I'm tempted to write something to boil things down without the burden to be technically correct.

This chapter's review list:

• The main difference between motion in 2 and 3 dimensions from one dimension is simply that one must now use vectors.  This is just same old same old, dividing out the x, y, and perhaps z components of displacement, velocity, or acceleration, whether as an average or as an instantaneous derivative.
• Projectile motion is 2 dimensional motion.  Using trig, the initial velocity x component in projectile motion will be v_0x=v₀cosθ while the initial y velocity component will be v_0y=v₀sinθ.
• The x velocity component is constant because of Newton's first law, Galileo's law of inertia, so the distance, x=(v₀cosθ)t very straightforwardly (distance=velocity x time).
• The y velocity component must deal with the constant negative acceleration of gravity, -g, which yields the slightly more involved formula for y=(v₀sinθ)t-½gt², based on the formula, d=vt + ½at²
• Motion in a circle is also 2 dimensional.  Although I don't think converting from radians to degrees and visa versa is a big deal, the reasons for radians hasn't quite clicked with me.  In circular motion with a constant speed, the acceleration always points inward toward the center of the circle.  This "centripetal acceleration" is a=v²/R
•  Since speed is distance/time, and in uniform circular motion, the distance around is the circumference, which equals πD (diameter, which is also 2R, radius) = 2πR, the velocity thus equals 2πR/T.
• Accordingly, if we substitute 2πR/T for v in the equation above, we get a=4π²R/T²
• In non-uniform motion, we have to take the derivative of the velocity to get the component of acceleration that is tangent to the circle, and the vector sum of the total acceleration will not point directly toward the center.  It will point either ahead or behind, depending on whether we have acceleration or deceleration.
• The final section of the chapter had to do with velocity relative to different reference frames.  So the velocity of a point in frame B in relation to frame A is a vector sum--the velocity of point P in relation to B "plus" the velocity of frame B in relation to frame A.