LectureNotes03

# V v i f r r r δ j ˆ t v i ˆ t v t v a y x avg δ

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v v i f r r r - Δ j ˆ t v i ˆ t v t v a y x avg Δ Δ + Δ Δ = Δ Δ r r is parallel to v r Δ a avg r Instantaneous Acceleration j ˆ a i ˆ a dt v d t v Lim 0 t ) a ( Lim 0 t a y x avg + = Δ Δ Δ = Δ r r r r If the solution is known find: ) t ( r r v x (t), v y (t) by differentiating x(t), y(t) a x (t), a y (t) by differentiating v x (t), v y (t) is NOT tangent to the path a r a x IS tangent to a graph of v x (t) a y IS tangent to a graph of v y (t) a r a x affects v x and x, but not v y or y Similarly for a y Rate of change of velocity

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Copyright R. Janow – Spring 2012 Summary – 3D Kinematics Formulas k ˆ z j ˆ y i ˆ x r + + r k ˆ t z j ˆ t y i ˆ t x t r v avg Δ Δ + Δ Δ + Δ Δ = Δ Δ r r k ˆ v j ˆ v i ˆ v dt r d ) v ( Lim 0 t v z y x avg + + = Δ r r r dt dz v dt dy v dt dx v z y x where k ˆ a j ˆ a i ˆ a dt v d ) a ( Lim 0 t a z y x avg + + = Δ s s r dt dv a dt dv a dt dv a z z y y x x where Definitions: Same for z Easiest to use in Cartesian coordinates – The x, y, z, dimensions move independently - Choose a x , a y , a z and initial conditions independently Cartesian Scalar Form - Constant Acceleration t a v v x x0 xf + = 2 x 2 1 x0 0 f t a t v x x + + = ) x (x 2a v v f x 2 x0 2 xf 0 - + = t ] v v x x xf x0 [ 2 1 0 f + + = t a v v y y0 yf + = 2 y 2 1 y0 0 f t a t v y y + + = ) y (y 2a v v f y 2 y0 2 yf 0 - + = t ] v v y y yf y0 [ 2 1 0 f + + = Vector Form - Motion with Constant Acceleration - 2D or 3D t ) v v ( r r 0 f 2 1 0 f r r v v + = - o f 0 f t t t t a v v - + = r r r 2 2 1 0 0 f t a t v r r r r v r + + = r a 2 v v v v 2 0 f f 2 f r o r v o v Δ + = PARABOLAS
Copyright R. Janow – Spring 2012 Kinematic Equations in 2D: Graphical Representation t a v (t) v i f r r r + = 2 2 1 i i f t a t v r (t) r r r v r + + = v x v y ) (constants conditions initial are a r v ally" parametric " motions y and x links t t t i i i f r r v -

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Copyright R. Janow – Spring 2012 Projectile Motion: Motion of a particle under constant, downward gravitation only Assume: Free fall along y direction (up/down) with horizontal motion as well a y = constant = - g, a x , a z = 0 Velocity in x-z direction is constant. Trajectory (a parabola) lies in a plane. Can choose it to be x-y. Motion is 2D, not 3D. Usually pick initial location at origin: x 0 = 0, y 0 = 0 at t=0 Initial conditions: square4 v 0 has x and y components v x0 = v 0 cos( θ ), v y0 = v 0 sin( θ ) square4 v x is constant, no drag or non-gravitation forces usually square4 If v x = 0, motion is strictly vertical, range = 0 x y path 0 v r v 0x v 0y θ Equations: v (t) v x0 x = t v x x(t) x0 0 + = v (t) v 2 x0 2 x = No acceleration along x gt v (t) v y0 y - = 2 2 1 y0 0 gt t v y y(t) - + = ] y [y(t) 2g v (t) v 2 y0 2 y 0 - - = Acceleration = - g along y
Copyright R. Janow – Spring 2012 Trajectory of a Projectile: y as a function of x Launch Range R (return to launch altitude) v y at E = -v yi Maximum Height t v x(t) xi = 2 2 1 yi gt t v y(t) - = xi x v ) t ( v = gt v (t) v yi y - = v R

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Copyright R. Janow – Spring 2012 Example: Show Projectile Trajectory is a Parabola (in x) Initial Conditions: x y path i v r v ix v iy θ i 0 = i x 0 = i y ) cos( v v i i xi θ = ) sin( v v i i yi θ = Eliminate time from Kinematics Equations: t v x(t) xi = v x(t) t
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• Fall '09
• DARILPEDIGO
• Acceleration, Velocity, Yi, Copyright R. Janow

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