PHYS2001_Ch. 3

# PHYS2001_Ch. 3 - Ch 3 Kinematics in 2D Now let’s consider...

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Unformatted text preview: Ch. 3 Kinematics in 2D Now let’s 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 let’s 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 θ = 90 o 2-D Projectile motion Launch angle θ < 90 o θ v o v o θ For the 2-D projectile motion case, we can break v o down into its x- and y- components. y x v o θ v ox v oy θ sin o o v v y = θ cos o o v v x = So here’s 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!!! Let’s look at the velocity components of a projectile as it travels along its trajectory: θ 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 θ 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 let’s consider...

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