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3 Motion in Two & Three Dimensions Displacement, velocity, acceleration Case 1: Projectile Motion Case 2: Circular Motion Hk: 51, 55, 69, 77, 85, 91.
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Position & Displacement Vectors
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Velocity Vectors
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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
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Example:
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Acceleration Vectors
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Direction of Acceleration Direction of a = direction of velocity change (by definition) Examples: rounding a corner, bungee jumper, cannonball (Tipler), Projectile (29, 30 below)
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Projectile Motion begins when projecting force ends ends when object hits something gravity alone acts on object
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Horizontal V Constant
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Two Dimensional Motion (constant acceleration)
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Range vs. Angle
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Example 1: Calculate Range (R) v o = 6.00m/s o = 30° x o = 0, y o = 1.6m; x = R, y = 0
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Example 1 (cont.) Step 1
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Quadratic Equation
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Example 1 (cont.) End of Step 1
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Example 1 (cont.) Step 2 (a x = 0) “Range” = 4.96m End of Example
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Circular Motion Uniform Non-uniform Acceleration of Circular Motion
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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
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Tangential Acceleration Direction = forward along path (speed increasing) Direction = backward along path (speed decreasing)
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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) + 3.0 m/s/s (L) a(total) = 5.0 m/s/s
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Summary Two dimensional velocity, acceleration Projectile motion (downward pointing acceleration) Circular Motion (acceleration in any direction within plane of motion)
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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 m/s @ 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)
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PM Example 2: v o = 6.00m/s o = 0° x o = 0, y o = 1.6m; x = R, y = 0
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PM Example 2 (cont.) Step 1
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PM Example 2 (cont.) Step 2 (a x = 0) “Range” = 3.43m End of Step 2
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PM Example 2: Speed at Impact
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v1 1. v1 and v2 are located on trajectory. a
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Q1. Givenlocate these on the trajectory and form v.
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Velocity in Two Dimensions v avg // r instantaneous “v” is limit of “v avg ” as t 0
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Acceleration in Two Dimensions a avg // v instantaneous “a” is limit of “a avg ” as t 0
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Displacement in Two Dimensions roro r rr
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v1 1. v1 and v2 are located on trajectory. a
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Ex. If v1(0.00s) = 12m/s, +60° and v2(0.65s) = 7.223 @ +33.83°, find a ave.
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Q1. Givenlocate these on the trajectory and form v.
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Q2. If v3(1.15s) = 6.06m/s, -8.32° and v4(1.60s) = 7.997, -41.389°, write the coordinate-forms of these vectors and calculate the average acceleration. v3 v4 vv a
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