Lecture 5: Vectors & Motion in 2 Dimensions

Questions of Yesterday 2) I drop ball A and it hits the ground at t 1. I throw ball B horizontally (v 0y = 0) and it hits the ground at t 2. Which is correct? a) t 1 < t 2 b) t 1 > t 2 c) t 1 = t 2

Questions of Yesterday 1) Can a vector A have a component greater than its magnitude A? a) YES b) NO 2) What are the signs of the x- and y-components of A + B in this figure? a) (x,y) = (+,+) b) (+,-) c) (-,+) d) (-,-) A B

Displacement in 2 Dimensions Position vectors no longer accounted for by + and - Displacement = change in position vector of object =  r 1 Dimension  x = x f - x i  r = r f - r i 2 Dimensions x (m) y (m) riri rfrf rr object path

Velocity in 2 Dimensions Average Velocity r f - r i t f - t i rtrt v av == rtrt v = lim  t -> 0 Instantaneous Velocity x (m) y (m) riri rfrf rr object path titi tftf

Acceleration in 2 Dimensions v f - v i t f - t i vtvt a av == Instantaneous Acceleration vtvt a = lim  t -> 0 Average Acceleration v x (m/s) v y (m/s) vivi vfvf vv object’s instant. velocity titi tftf

Acceleration in 2 Dimensions v f - v i t f - t i vtvt a av == If a car is going North at a constant speed and makes a left turn while maintaining its constant speed and then continues West at the same speed… does the car accelerate during this trip? Average Acceleration A runner is running on a circular track at constant speed? Is she accelerating?

Acceleration in 2 Dimensions v f - v i t f - t i vtvt a av == Velocity is a vector with both magnitude & direction, so… An object can accelerate by either changing its SPEED or changing its DIRECTION Average Acceleration

Projectile Motion Motion in 2 Dimensions under constant gravitational acceleration Horizontal component of velocity is constant over entire path! v x = v 0x No acceleration in horizontal direction

Projectile Motion Motion in 2 Dimensions under constant gravitational acceleration Vertical component of velocity constantly changing due to gravitational acceleration in -y direction v 0y --> 0 -> -v 0y

t= 1 s t= 2 s t= 3 s t= 4 s t= 5 s Projectile Motion Horizontal and Vertical motions are completely independent of each other!! Motion in one direction has NO EFFECT on motion in the other direction Initial velocity in horizontal direction no gravity Initial velocity in horizontal direction with gravity Initial velocity in vertical direction with gravity

Important Features of Projectile Motion Acceleration is ALWAYS m/s 2 in the vertical direction Parabolic motion is symmetric v f = -v 0 At the top of the trajectory: t = 1/2 of total time x = 1/2 of total horizontal range t= 1 s t= 2 s t= 3 s t= 4 s t= 5 s Total time of trajectory is independent of horizontal motion

At what point in the object’s trajectory is the speed a minimum? What about velocity? t= 1 s t= 2 s t= 3 s t= 4 s t= 5 s Important Features of Projectile Motion

2D Motion under Constant Acceleration Because x and y motions are independent… we can apply 1D equations for constant acceleration motion separately to each x- and y- direction But…. v 0 has both x- and y-components Need to separate v 0 into x- and y- components  v 0x = v 0 cos  v 0y = v 0 sin 

2D Motion under Constant Acceleration Recall equations for 1D motion under constant acceleration v = v 0 + at  x = v 0 t + 1/2at 2 v 2 = v a  x 2D motion equivalent to superposition of two independent motions in the x- and y-directions v x = v 0x + a x t  x = v 0x t + 1/2a x t 2 v x 2 = v 0x 2 + 2a x  x Horizontal Motion v y = v 0y + a y t  y = v 0y t + 1/2a y t 2 v y 2 = v 0y 2 + 2a y  y Vertical Motion

Horizontal Motion of Projectile v x = v 0x + a x t  x = v 0x t + 1/2a x t 2 v x 2 = v 0x 2 + 2a x  x v 0x = v 0 cos  a x = 0 Time is still determined by y-direction motion! t= 1 s t= 2 s t= 3 s t= 4 s t= 5 s v x = v 0 cos  = constant  x = v 0x t = (v 0 cos  t

Vertical Motion of Projectile v y = v 0y + a y t  y = v 0y t + 1/2a y t 2 v y 2 = v 0y 2 + 2a y  y v 0y = v 0 sin  a y = g = m/s 2 v y = v 0 sin  + gt  y = (v 0 sin  t + 1/2gt 2 v y 2 = (v 0 sin  2 + 2g  y Pay attention to sign convention! 

Equations for Motion of Projectile v y = v 0 sin  + gt  y = (v 0 sin  t + 1/2gt 2 v y 2 = (v 0 sin  2 + 2g  y v x = v 0 cos  = constant  x = v 0x t = (v 0 cos  t v = (v x 2 + v y 2 ) 1/2 tan  = v y /v x  = tan -1 (v y /v x ) -90 <  < 90 Vertical Component Motion Horizontal Component Motion Combined 2D Motion

Problem #1 A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below this line if it has been traveling for 1 s? For 2 s? Does your answer depend on the angle at which the projectile is launched? What about the speed? t= 1 s t= 2 s t= 3 s t= 4 s t= 5 s

Questions of the Day 2) Two projectiles are thrown with the same initial speed, one at an angle  with respect to the ground and the other at an angle 90 o - . Both projectiles strike the ground at the same distance from the projection point. Are both projectiles in the air for the same length of time? a) YES b) NO 1) A heavy crate is dropped from a high-flying airplane as it flies directly over your shiny new car? Will your car get totaled? a) YES b) NO