Chapter 4, Motion in 2 Dimensions

Position, Velocity, Acceleration Just as in 1d, in 2, object’s motion is completely known if it’s position, velocity, & acceleration are known. Position Vector  r –In terms of unit vectors discussed last time, for object at position (x,y) in x-y plane: r  x i + y j Object moving: r depends on time t: r = r(t) = x(t) i + y(t) j

Object moves from A (r i ) to B (r f ) in x-y plane: Displacement Vector  Δr = r f - r i If this happens in time Δt = t f - t i Average Velocity v avg  (Δr/Δt) Obviously, in the same direction as displacement. Independent of path between A & B

As Δt gets smaller & smaller, clearly, A & B get closer & closer together. Just as in 1d, we define the instantaneous velocity as:  velocity at any instant of time  average velocity over an infinitesimally short time Mathematically, instantaneous velocity: v = lim ∆t  0 [(∆r)/(∆t)] ≡ (dr/dt) lim ∆t  0  ratio (∆r)/(∆t) for smaller & smaller ∆t. Mathematicians call this a derivative.  Instantaneous velocity v ≡ time derivative of displacement r

Instantaneous velocity v = (dr/dt). Magnitude |v| of vector v ≡ speed. As motion progresses, speed & direction of v can both change. Object moves from A (v i ) to B (v f ) in x-y plane: Velocity Change  Δv = v f - v i This happens in time Δt = t f - t i Average Acceleration a avg  (Δv/Δt) As both speed & direction of v change, arbitrary path

As Δt gets smaller & smaller, clearly, A & B get closer & closer together. Just as in 1d, we define instantaneous acceleration as:  acceleration at any instant of time  average acceleration over infinitesimally short time Mathematically, instantaneous acceleration: a = lim ∆t  0 [(∆v)/(∆t)] ≡ (dv/dt) lim ∆t  0  ratio (∆v)/(∆t) for smaller & smaller ∆t. Mathematicians call this a derivative.  Instantaneous acceleration a ≡ time derivative of velocity v

2d Motion, Constant Acceleration Can show: Motion in the x-y plane can be treated as 2 independent motions in the x & y directions.  Motion in the x direction doesn’t affect the y motion & motion in the y direction doesn’t affect the x motion.

Object moves from A (r i,v i ), to B (r f,v f ), in x-y plane. Position changes with time: Acceleration a is constant, so, as in 1d, can write (vectors!):  r f = r i + v i t + (½)at 2 Velocity changes with time: Acceleration a is constant, so, as in 1d, can write (vectors!):  v f = v i + at

Acceleration a is constant, (vectors!):  r f = r i + v i t + (½)at 2, v f = v i + at Horizontal Motion: x f = x i + v xi t + (½)a x t 2, v xf = v xi + a x t Vertical Motion: y f = y yi + v yi t + (½)a y t 2, v yf = v yi + a y t

Projectile Motion

Equations to Use One dimensional, constant acceleration equations for x & y separately! x part: Acceleration a x = 0! y part: Acceleration a y = g (if take down as positive). Initial x & y components of velocity: v xi & v yi. x motion: v xf = v xi = constant. x f = x i + v xi t y motion: v yf = v yf + gt, y f = y i + v yi t + (½)g t 2 (v yf ) 2 = (v yi ) 2 + 2g (y f - y 0 )

Projectile Motion Simplest example: Ball rolls across table, to the edge & falls off edge to floor. Leaves table at time t = 0. Analyze y part of motion & x part of motion separately. y part of motion: Down is positive & origin is at table top: y i = 0. Initially, no y component of velocity: v yi = 0  v yf = gt, y f = (½)g t 2 x part of motion: Origin is at table top: x f = 0. No x component of acceleration(!): a x = 0. Initially x component of velocity is: v xf  v xf = v xi, x f = v xi t

Ball Rolls Across Table & Falls Off

Summary: Ball rolling across the table & falling. Vector velocity v has 2 components: v xf = v xi, v yf = gt Vector displacement D has 2 components: x f = v xf t, y f = (½)g t 2

Projectile Motion PHYSICS: y part of motion: v yf = gt, y f = (½)g t 2 SAME as free fall motion !!  An object projected horizontally will reach the ground at the same time as an object dropped vertically from the same point! (x & y motions are independent)

Projectile Motion General Case:

General Case: Take y positive upward & origin at the point where it is shot: x i = y i = 0 v xi = v i cosθ i, v yi = v i sinθ i Horizontal motion: NO ACCELERATION IN THE x DIRECTION! v xf = v xi, x f = v xi t Vertical motion: v yf = v yi - gt, y f = v yi t - (½)g t 2 (v yf ) 2 = (v yi ) 2 - 2gy f –If y is positive downward, the - signs become + signs. a x = 0, a y = -g = -9.8 m/s 2