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Published byHorace Godfrey Howard Modified over 8 years ago
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Why Study Motion First? Position Time
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Types of Motion Accelerated motion.Speed or direction or both change.
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Speed
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Some Common Speeds 8,000 mi 93,000,000 mi
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The Ultimate Speed 8,000 mi 93,000,000 mi
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Vector versus Scalar 10 mi0 mi7 mi
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Vector Components a ayay axax a ayay azaz axax j k i x y
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Vectors (cont) x y a ayay axax θ
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Addition of Vectors b c a θcθc b a θaθa b a c c θbθb b a
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Subtraction of Vectors b a c = a - b c -b a c = a + (- b) d = b - a d = b + (- a) -a b d
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Subtraction of Vectors (cont) r2r2 r1r1 ∆r = r 2 – r 1 ∆r = r 2 + (- r 1 ) -r 2 r1r1 ∆rr2r2 r1r1
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Multiplication of Vectors Dot Product b a c = a. b = ab cos θ b a a b cosθ b a b sin θ d
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Constant Velocity xt v t
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Constant Linear Acceleration x t v t a t
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Constant Linear Acceleration Equations t v v0v0 A1A1
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Constant Acceleration Equations 2 v t v0v0
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Constant Acceleration Graphs
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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?
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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
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Acceleration Due To Gravity y = distance traveled Up is +Down isv o = initial velocity g = acceleration due to gravity
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Acceleration Due To Gravity
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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
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Graph for Sample Problem 1 v0v0 -v 0 v (m/s) t (s) 20 (s) -20 (s) 2 s4 s 0
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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
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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
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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
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Projectile Motion 1
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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
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Sample Problem 3 x = ___ m t = ___ sv ox = 36.0 m/s y = - 6.00 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?
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Sample Problem 4 v ox = ___ mt = ___ s x = 28.0 my = - 2.00 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.
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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
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Range of a Projectile vovo R
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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
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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
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Circular Motion v = v f - v i vivi R vfvf vfvf -v i vv aa a = v 2 /R a = (4 2 /T 2 )R
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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
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Rolling Wheel
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Sample Problem 8 R =.525 m v ox R Top View v ox y x Side View T = 1.20 s y = -2.20 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.
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Satellite Motion x y x y vsvs v s 18,000 mph v e 25,000 mph
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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
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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
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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?
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