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PHYS_2014_Lecture_3

# PHYS_2014_Lecture_3 - Fall 2010 Oklahoma State University...

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Unformatted text preview: Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 3, Slide 1 Lecture 3: Motion in 2 and 3 Dimensions average speed: Δ the change in position Δ the change in time x v t = = average velocity: or x x v v t t Δ Δ = = Δ Δ a a a a instantaneous velocity: lim t x dx v t dt → Δ = = Δ a a a average acceleration: or v v a a t t Δ Δ = = Δ Δ a a a a instantaneous acceleration: 2 2 lim t v dv d s a t dt dt → Δ = = = Δ a a a a Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 3, Slide 2 Derivation of the four key Kinematic Equations for Motion with Constant Acceleration f f f f v v v a t t t v v a t at v v − Δ = = Δ − − = = − a a a a a a a a a a In the equation for average acceleration if we set t = 0, we get: f v v at = + a a a 1. Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 3, Slide 3 The average velocity can be thought of as being the average between the initial velocity and the final velocity: 2 f v v v + = a a a If the position of an object at t = 0 is , the position at a later time, t , will be: x a x x vt = + a a a Substituting our expression for into this equation for yields: v a x a ( ) 1 2 f x x v v t = + + a a a a 2. Derivation of the four key Kinematic Equations for Motion with Constant Acceleration Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 3, Slide 4 We can substitute our first equation, , into our second equation, , to get: f v v at = + a a a ( ) 1 2 f x x v v t = + + a a a a ( ) 1 2 x x v at v t = + + + a a a a a 2 1 2 x x v t at = + + a a a a or 3. Derivation of the four key Kinematic Equations for Motion with Constant Acceleration Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 3, Slide 5 Rewriting our first equation, , as , and substituting for t in our second equation, , yields: f v v at = + a a a f v v t a − = a a a ( ) 1 2 f x x v v t = + + a a a a ( ) 1 2 f f v v x x v v a − = + + a a a a a a a or 4. ( ) ( )( ) 2 f f a x x v v v v − = + − a a a a a a a ( ) 2 2 2 f v v a x x = + − a a a a a Derivation of the four key Kinematic Equations for Motion with Constant Acceleration Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 3, Slide 6 These are four Kinematic equations that describe motion with constant acceleration : f v v at = + a a a ( ) 1 2 f x x v v t = + + a a a a 2 1 2 x x v t at = + + a a a a ( ) 2 2 2 f v v a x x = + − a a a a a 1. 2. 3. 4. Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 3, Slide 7 Example: I throw the billiard ball directly upward with an initial velocity of 10 m/s. What is the maximum height the billiard ball will reach? When the ball reaches its maximum height, its velocity will be zero, . We already know that . We set up our coordinate system so that and + is upwards....
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PHYS_2014_Lecture_3 - Fall 2010 Oklahoma State University...

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