PHYS2001_Ch. 3

PHYS2001_Ch. 3 - Ch. 3 Kinematics in 2D Now lets consider...

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Unformatted text preview: Ch. 3 Kinematics in 2D Now lets consider the concepts of displacement , velocity , and acceleration in 2 dimensions. o f x x x- = t x x t x v o f- = = t v v t v a o f- = = 2 2 1 at t v x x o o f + + = a v v x o f 2 2 2- = 1-D o f r r r - = t r r t r v o f - = = t v v t v a o f - = = 2 2 1 t a t v r r o o f + + = a v v r o f 2 2 2- = 2-D In general, for 2D, displacements, velocities, and accelerations will have components in both the x- and y-direction. Concepts from Ch. 1 Thus, y x r r r + = y x v v v + = y x a a a + = 2 2 1 t a t v x x x o o f x + + = 2 2 1 t a t v y y y o o f y + + = We get 2 equations of motion under constant acceleration one for the x- direction, and one for the y-direction. Example : A car drives 60 o N of E at a constant speed of 35 m/s. How far east has the car traveled after 10 s? E, x N, y v = 35 m/s 60 o We can break the velocity down into its x- and y-components. v y v x Now I can calculate the values for v x and v y : o x v v 60 cos = o y v v 60 sin = m/s 5 . 17 ) m/s)( 35 ( 2 1 = = m/s 3 . 30 ) m/s)( 35 ( 2 3 = = For distance in the x-direction (east) we use v x : t v x x = m 175 s) m/s)(10 17.5 ( = = 3.3 Projectile Motion Now lets analyze the motion of projectiles launched into the air, but this time, the motion is not completely vertical 2-D projectile motion. Vertical motion Launch angle = 90 o 2-D Projectile motion Launch angle < 90 o v o v o For the 2-D projectile motion case, we can break v o down into its x- and y- components. y x v o v ox v oy sin o o v v y = cos o o v v x = So heres the most important thing to remember about projectile motion: The only acceleration we have to deal with is that due to gravity, and gravity only acts in the vertical ( y ) direction . and =- = x y a g a Remember, acceleration is defined as the change in velocity over time. So, if the acceleration in the x-direction is zero, the velocity of a projectile in the x- direction does not change!!! Lets look at the velocity components of a projectile as it travels along its trajectory: v o y x v x = v ox v y = v oy v v y < v oy v x = v ox v v = v x = v ox v y = 0! v v x = v ox-v y < v oy v o v y = -v oy v x = v ox Level Ground v o y x v x = v ox v y = v oy v v y < v oy v x = v ox v = v x = v ox v y = 0! v v x = v ox-v y < v oy v o v y = -v oy v x = v ox Level Ground Projectile Motion - Summary There is no acceleration in the horizontal ( x ) direction. The x-component of the velocity is the same everywhere along the trajectory. The velocity of the projectile at any point along the trajectory is the vector sum of its horizontal and vertical components....
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PHYS2001_Ch. 3 - Ch. 3 Kinematics in 2D Now lets consider...

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