3 Motion in Two & Three Dimensions Displacement, velocity, acceleration Case 1: Projectile Motion Case 2: Circular Motion Hk: 51, 55, 69, 77, 85, 91.

Position & Displacement Vectors

Velocity Vectors

Relative Velocity Examples: people-mover at airport airplane flying in wind passing velocity (difference in velocities) notation used: velocity “BA” = velocity of B with respect to A

Example:

Acceleration Vectors

Direction of Acceleration Direction of a = direction of velocity change (by definition) Examples: rounding a corner, bungee jumper, cannonball (Tipler), Projectile (29, 30 below)

Projectile Motion begins when projecting force ends ends when object hits something gravity alone acts on object

Horizontal V Constant

Two Dimensional Motion (constant acceleration)

Range vs. Angle

Example 1: Calculate Range (R) v o = 6.00m/s  o = 30° x o = 0, y o = 1.6m; x = R, y = 0

Example 1 (cont.) Step 1

Quadratic Equation

Example 1 (cont.) End of Step 1

Example 1 (cont.) Step 2 (a x = 0) “Range” = 4.96m End of Example

Circular Motion Uniform Non-uniform Acceleration of Circular Motion

18 Centripetal Acceleration Turning is an acceleration toward center of turn-radius and is called Centripetal Acceleration Centripetal is left/right direction a(centripetal) = v 2 /r (v = speed, r = radius of turn) Ex. V = 6m/s, r = 4m. a(centripetal) = 6^2/4 = 9 m/s/s

Tangential Acceleration Direction = forward along path (speed increasing) Direction = backward along path (speed decreasing)

Total Acceleration Total acceleration = tangential + centripetal = forward/backward + left/right a(total) = dv/dt (F/B) + v 2 /r (L/R) Ex. Accelerating out of a turn; 4.0 m/s/s (F) m/s/s (L) a(total) = 5.0 m/s/s

Summary Two dimensional velocity, acceleration Projectile motion (downward pointing acceleration) Circular Motion (acceleration in any direction within plane of motion)

Ex. A Plane has an air speed v pa = 75m/s. The wind has a velocity with respect to the ground of v ag = 8 330°. The plane’s path is due North relative to ground. a) Draw a vector diagram showing the relationship between the air speed and the ground speed. b) Find the ground speed and the compass heading of the plane. (similar situation)

PM Example 2: v o = 6.00m/s  o = 0° x o = 0, y o = 1.6m; x = R, y = 0

PM Example 2 (cont.) Step 1

PM Example 2 (cont.) Step 2 (a x = 0) “Range” = 3.43m End of Step 2

PM Example 2: Speed at Impact

v1 1. v1 and v2 are located on trajectory. a

Q1. Givenlocate these on the trajectory and form  v.

Velocity in Two Dimensions v avg //  r instantaneous “v” is limit of “v avg ” as  t  0

Acceleration in Two Dimensions a avg //  v instantaneous “a” is limit of “a avg ” as  t  0

Displacement in Two Dimensions roro r rr

v1 1. v1 and v2 are located on trajectory. a

Ex. If v1(0.00s) = 12m/s, +60° and v2(0.65s) = °, find a ave.

Q1. Givenlocate these on the trajectory and form  v.

Q2. If v3(1.15s) = 6.06m/s, -8.32° and v4(1.60s) = 7.997, °, write the coordinate-forms of these vectors and calculate the average acceleration. v3 v4 vv a