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Why Study Motion First? Position Time

Types of Motion Accelerated motion.Speed or direction or both change.

Speed

Some Common Speeds 8,000 mi 93,000,000 mi

The Ultimate Speed 8,000 mi 93,000,000 mi

Vector versus Scalar 10 mi0 mi7 mi

Vector Components a ayay axax  a ayay azaz axax j k i x y

Vectors (cont) x y a ayay axax θ

Addition of Vectors b c a θcθc b a θaθa b a c c θbθb b a

Subtraction of Vectors b a c = a - b c -b a c = a + (- b) d = b - a d = b + (- a) -a b d

Subtraction of Vectors (cont) r2r2 r1r1 ∆r = r 2 – r 1 ∆r = r 2 + (- r 1 ) -r 2 r1r1 ∆rr2r2 r1r1

Multiplication of Vectors Dot Product b a  c = a. b = ab cos θ b a  a b cosθ b a  b sin θ d

Constant Velocity xt v t

Constant Linear Acceleration x t v t a t

Constant Linear Acceleration Equations t v v0v0 A1A1

Constant Acceleration Equations 2 v t v0v0

Constant Acceleration Graphs

Truck and Car Problem A car is at rest at a red traffic light. Just as the light turns green the car is passed by a truck traveling at a constant velocity of 15.0 m/s. The car immediately accelerates at the constant rate of 3.0 m/s 2 in the same direction as the truck. Both the car and truck travel in the same straight line direction on a level road. a) At what time t 1 are the car and truck traveling at the same velocity? b) At what time t 2 does the car pass the truck? c) How far has the car traveled when it passes the truck? d) How fast is the car traveling when it passes the truck?

Truck and Car Problem v(m/s) t(s) Truck Car t1t1 t2t2 15 t 1 = truck & car at same speed t 2 = truck & car at same distance

Acceleration Due To Gravity y = distance traveled Up is +Down isv o = initial velocity g = acceleration due to gravity

Acceleration Due To Gravity

Sample Problem 1 What is the maximum height reached by the ball? v o = 20.0 m/s How long did it take for the ball to reach that height? g = 9.8 m/s 2 v0v0 -v 0 How long did the total flight last? y = 20.4 m t = 2.04 s t = 4.08 s

Graph for Sample Problem 1 v0v0 -v 0 v (m/s) t (s) 20 (s) -20 (s) 2 s4 s 0

Relative Motion v T/E v B/T v B/T = Velocity of the ball relative to the train. v T/E = Velocity of the train relative to the Earth. v B/E = Velocity of the ball relative to the Earth. v T/E v B/T v B/E  Top view If v B/T and v T/E are moving (in the plane of the Earth) at an angle 90 o with each other, then

Sample Problem 2 A boat crosses a river traveling at a speed of 10.0 m/s due east. The river is flowing at a speed of 3.0 m/s due south. The river is 200 m wide. a) What is the speed of the boat relative to the Earth? b) What is the value of  ? c) How long does it take to cross the river? d) How far downstream does the boat travel? e) At what angle  ’ should the boat move to land due east? v B/W v W/E v B/E  v B/W v W/E x  = 16.7 o

Sample Problem 2 (cont) v W/E c) How long does it take to cross the river? d) How far downstream does the boat travel? e) At what angle  ’ should the boat move to land due east? v B/W v B/E  v B/W v W/E x  ’ = 17.5 o ’’ v B/W v W/E v B/E

Projectile Motion 1

Projectile Motion 2 x = horizontal distance v x = horizontal velocity v ox = initial horizontal velocity y = vertical distance v y = vertical velocity v oy = initial vertical velocity Horizontal Vertical

Sample Problem 3 x = ___ m t = ___ sv ox = 36.0 m/s y = m t = 1.11 s v ox y x x = 40.0 m A ball is thrown horizontally with a velocity of 36.0 m/s over a level surface. It is at a height of 6.00 m when released. a) How long was the ball in flight? b) How far did the ball travel?

Sample Problem 4 v ox = ___ mt = ___ s x = 28.0 my = m t =.639 s v ox y x v ox = 43.8 m/s Is v ox > 45 m/s (90 mph)? A pitcher claims he can throw a baseball at a velocity greater than 90 mi/hr (45 m/s). He throws the ball horizontally over a level surface from a height of 2.00 m. The ball travels a horizontal distance of 28.0 m. Was the boy correct? The boy was not correct.

Sample Problem 5 x max = ? v ox = v o cos  y max = ? vovo  v ox v oy v oy = v o sin  v o = 20.0 m/s  = 30.0 o t total = ? y max = 5.10 m t total = 2.04 s x max = 35.3 m

Range of a Projectile vovo  R

Sample Problem 6 vovo  R a) What is the maximum range of battleship gun that can fire a shell at a velocity of 6.00 x 10 2 m/s? b) At what angle should the shell be fired to hit a target 10.0 mi away? 22.8 mi  = 13.0 o 90 o -  = 77.0 o

Sample Problem 7 vovo  x max v o = 35.0 m/s  = 30.0 o x max = 108 m y max x max, y max, t total y max = 15.6 m t total = 3.56 s

Circular Motion  v = v f - v i vivi R vfvf vfvf -v i vv aa a = v 2 /R a = (4  2 /T 2 )R

Relative Motion of a Rolling Wheel v A/T v C/T A C B v T/E v B/T Assume v T/E = 50 mph

Rolling Wheel

Sample Problem 8 R =.525 m v ox R Top View v ox y x Side View T = 1.20 s y = mx = ? m x = 1.84 m A girl whirls a stone above her head in a horizontal circle at a a distance 2.20 m above the ground. The period of rotation is 1.20 s. The stone is released and hits the ground after traveling a horizontal distance x. Find the value of x.

Satellite Motion x y x y vsvs v s  18,000 mph v e  25,000 mph

Relative Projectile Motion v T/E v xB/T y x y = -3.0 m v xB/T = 30.0 m/s v xT/E = 10.0 m/s t = ? S x B/E = ? M x B/T =? m t =.78 s x B/E = 31 m x B/T = 23 m

Rotational Speed vava R vbvb R R a b R a = R R b = ½ R Compare the speed of point a with the speed of point b. Tangential Speed v a = 2 v b Angular Speed ω a = ω b Tangential speed is measured in m/s Angular speed is measured in radians/s or degrees/s

Sample Problem 8 vava R vbvb R R a = 4.00 m R b = 2.00 m v a = ? m/s v b = ? m/s ω a = ? rad/s ω b = ? rad/s Tangential Speed v a = 1.00 m/s Angular Speed ω a =.251 rad/s v b =.50 m/s ω b =.251 rad/s A merry-go-round makes 1 complete revolution every 25.0 s. Child a sits at a distance R a = 4.00 m from the center and child b sits at a distance R b = 2.00 m from the center. What are the tangential and angular speeds of each child?