Motion in Two Dimensions AP Physics C Mrs. Coyle

Position Vector r=xî +yĵ +zk

Displacement Δr = r f – r i Δr=(x 2 –x 1 )î +(y 2 -y 1 )ĵ +(z 2 -z 1 )k –Displacement is independent of the path taken.

Displacement

Example 1 The initial position of a boat was r 1 =1î +4ĵ -2k and the final was r 2 =-3î +7ĵ +2k. Find the displacement from r 1 to r 2.

Instantaneous Velocity v = dr   dt Instantaneous speed is the magnitude of instantaneous velocity, |v|.

The direction of the instantaneous velocity is tangent to the path and in the direction of the motion.

Average Velocity Average Velocity= = Displacement = r Time t The direction of v avg is the same as the direction of the displacement. v avge is independent of the path taken. Why?

Average Acceleration Average Acceleration= Change in Velocity Time a avge is a vector in the direction of v a = v t

Instantaneous Acceleration a = dv   dt a = d ( dr ) = d 2 r dt dt dt 2

Kinematics Equations The same as in one dimensional motion. Treat each dimension separately because of the independence of vector quantities. So one for example the x axis can have constant v and the y axis can have constant a.

Example 2 The position vector for a particle has components x = at + b and y = ct 2 + d, where a = 1.00 m/s, b = 1.00 m, c = m/s 2, and d = 1.00 m. (a)Calculate the average velocity during the time interval from t = 2.00 s to t = 4.00 s. (b)Determine the velocity and the speed at t = 2.00 s. Answer: (a) (1.00i j) m/s; (b) v = (1.00 m/s)i + (0.500 m/s)j, |v| = 1.12 m/s

Example 3 A puck moves on an x-y plane. Its position is given by: x=-2t 2 +3t-1 and y= 4t 2 -3t Determine whether the x and y components of the acceleration are constant and whether the acceleration is constant.