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Microsoft PowerPoint - Knight Ch 4 - kinematics in TWo Dimendions - new

# Microsoft PowerPoint - Knight Ch 4 - kinematics in TWo Dimendions - new

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1 i f Δ r r r = t Δ Δ = r v dt d t lim 0 t r r v = Δ Δ = Δ Kinematics in Two Dimensions – Chapter 4 ____________ j ˆ v i ˆ v j ˆ dt dy i ˆ dt dx dt r d v y x + = + = = r r j ˆ y i ˆ x r + = r Position and velocity are expressed in terms of both horizontal and ____________components. 2D motion with Constant Acceleration ____________ ____________ Relating Position to Velocity 2 y 2 x y x v v v sin v dt dy v cos v dt dx v + = θ = = θ = = ____________ ____________ Acceleration Vector dt d t lim 0 t v v a = Δ Δ = Δ

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2 Acceleration Vectors j dt dv i dt dv a y x ˆ ˆ + = r ____________ Projectile (2D) Motion A ____________ is any body that is given an initial velocity and then follows a path determined entirely by the effects of __________________ acceleration and _________resistance. The path followed by a projectile is called its _________ . In projectile motion, the ________ acceleration is due to gravity! This means there is ________ acceleration in the “x” direction! Therefore, x and y velocity components are ___________ of one another! Projectile (2D) Motion What does this mean mathematically? a x = _________ a y = _________ Projectile motion, no air resistance 6.14 Projectile motion equations t x Δt v x x ix i f + = For a projectile with motion in the x and y direction, ( ) t y Δt g 2 1 Δt v y y 2 iy i f + = t v const v v ix fx = = t v gΔΔ v v iy fy =
3 x y CAR BALL cosΘ gt 2 1 t v x x 2 ox o + + = cosΘ gt 2 1 t v x x 2 ox o + + = 0 y = sinΘ gt 2 1 t v y 2 oy = The equations in x are _____________, therefore, when the ball returns to the x axis, it will ___________________________. θ ____________ ____________ ______ Maximum Height What is the maximum height y (v=0, t=?) attained by the projectile? At the maximum height 0 sin = = gt v v i y θ Plug t maxH into position equation 2 2 1 sin gt t v y y i i + = θ g v t i H θ sin max = g v H i 2 sin 2 2 θ = 2 sin 2 1 sin sin 0 + = g v g g v v H i i i θ θ θ ____________ _____ ________ Maximum Range What is the maximum range x(t = t tot ) attained by the projectile? Plug t TOTAL into position equation θ θ cos sin 2 = g v v R i i t TOTAL = _____________ Identity: 2 cos θ sin θ = sin 2 θ θ cos t v x x i i + = g v R i θ 2 sin 2 = ________ Playing Darts Playing darts, one player throws a dart towards the bulls eye located 8.00 ft (2.44 m) away. If the velocity of the dart when it leaves his hands parallel to the floor is 5.00 m/s, how far below the bulls eye will it hit the target? v o x y 2 2 2 1 sin 2 1 cos t a t v y y t a t v x x y i i x i i + + = + + = θ θ 2 y ox t a 2 1 y and t v x = = Solving the x-t equation for t and plug into y-t equation solving for y 2 2 1 = = ix y ix v x a y and v x t ( ) m s m m s m y 17 . 1 / 00 . 5 44 . 2 / 80 . 9 2 1 2 2 = = ____ ____________

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4 In Speed , a bus is equipped with a bomb that will explode if the speed of the bus falls below 50 mph (22 m/s). The police route the bus to a freeway under construction – ie no traffic. SOMEHOW, they overlooked a 50 ft (15 m) gap in a bridge. IF the bus slows down it will kill all aboard, so they decide to jump the gap. The missing segment lies on an effectively level stretch of elevated freeway, 50 ft (15 m) above the ground, yet somehow they manage to jump the gap at 67 mph (30 m/s) and land safely on the other side.
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Microsoft PowerPoint - Knight Ch 4 - kinematics in TWo Dimendions - new

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