2-D Vector Equations have the same form as 1-D Kinematics. f i i

Transcription

1 2-D Vector Equations have the same form as 1-D Kinematics v = v + at f i 1 r = r + v t+ at f i i 2 2

2 2-D Vector Equations have the same form as 1-D Kinematics v = viˆ+ v ˆj f x y = ( v + ati ) ˆ+ ( v + at) ˆj xi x yi y Similarly, f = ( v iˆ+ v ˆj ) + ( a tiˆ+ a tj ˆ) v = v + at xi yi x y i 1 r = r + v t + at f i i 2 2

3 Vector Drivers Heather in her Corvette accelerates at the rate of (3.00iˆ 2.00 ˆjm ) / s while Jill in her Jaguar accelerates at ˆ 2 (1.00i jm ) / s. They both start from rest at the origin of an xy coordinate system. After 5.00 s, find What is Heather's acceleration relative to Jill? the velocity of each driver and Heather's speed with respect to Jill. The displacement vector for each driver and how far apart are they, and draw it all. What is Heather's acceleration relative to Jill? ˆ 2

4 Relative Velocity Two observers moving relative to each other generally do not agree on the outcome of an experiment For example, observers A and B below see different paths for the ball and measure different velocities: Velocity of ball relative to observer B v = v + v bb ba AB Velocity of ball relative to observer A Velocity of A relative to observer B

5 Co-Linear Motion Just add or subtract the magnitudes of vectors! v = v + v PG PT TG Notice how the inner subscripts cancel!

6 2D Relative Velocity The boat can travel 2.50 m/s relative to the river. The river current flows at 1.00 m/s relative to the Earth. What is the total velocity of the boat relative to the Earth (shore = Earth)? v = v + v be br re 2 2 vbe = (2.5 m/ s) + (1 m/ s) = 2.69 m/ s 1 m/ s 1 1m/s θ = tan = m/s 2.50 m/ s

7 Relative Velocity Again The boat can travel 10 m/s relative to the river. The river current flows at 5.00 m/s relative to the shore. If the boat wants to travel straight across, what must be his heading? What is its total speed? v = v + v be br re From the triangle: 1 5 m/ s θ = sin = m/ s 2 2 v = (10 m/ s) (5 m/ s) = 8.66 m/ s be v = v + v br be re 10 m/ s 5 m/ s

8 Train Rain A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 5.00 m/s relative to the ground. When the train moves at a constant velocity, the rain drops make an angle of 25 degrees when theymove past the window, as the drawing shows. How fast is the train moving? We know: vrt = vrg + vgt = v v RG TG v RT 25 vrg We want v TG : v TG v TG = v RG tan 25 = 5 m/ stan 25 = 2.33 m/ s

9 Previous problems all involved right triangles.how do you solve if you don t have right triangle relationship between relative velocities?

10 Using ijk vector component addition A ferry boat is traveling in a direction 35.1 degrees north of east with a speed of 5.12 m/s relative to the water. A passenger is walking with a velocity of 2.71m/s due east relative to the boat. What is the velocity of the passenger with respect to the water? Determine the angle relative to due east. v BW θ PW v PB v PW vpw = vpb + vbw = [5.12 cos(35.1) iˆ+ 5.12sin(35.1) ˆj] m/ s+ (2.71 iˆ) m/ s = (6.90iˆ ˆjm ) / s 2 2 vpw = (6.9) + (2.94) m/ s = 7.50 m/ s θ PW = tan ( ) =

11 Using Law of Sines and Cosines A ferry boat is traveling in a direction 35.1 degrees north of east with a speed of 5.12 m/s relative to the water. A passenger is walking with a velocity of 2.71m/s due east relative to the boat. What is the velocity of the passenger with respect to the water? Determine the angle relative to due east. vpw = vpb + vbw v BW α θ PW 35.1 v PB v PW v = v + v 2v v cos PW BW PB BW PB vpw = ( (5.12)(2.71) cos144.9 ) / m s vpw = 7.50 m/ s of cosines: 2 cos 20 = + c a b ab Low Law of sines: sinα sin β sinγ = = a b c α =11.99 θ PW = = 23.1 sinα sin144.9 = θ = 23.1 PW

12 Which Way is Best???? Using Law of Sines and Cosines A ferry boat is traveling in a direction 35.1 degrees north of east with a speed of 5.12 m/s relative to the water. A passenger is walking with a velocity of 2.71m/s due east relative to the boat. What is the velocity of the passenger with respect to the water? Determine the angle relative to due east. vpw = vpb + vbw v BW 35.1 v PB v PW Depending on the givens, COMPONENTS IS EASIER!!! AND it works for more than three vectors!!!!

13 Relative Velocity A plane is moving at 45m/s due north relative to the air, while its velocity relative to the ground is 38.0m/s, 20 degrees west of north. What is the velocity of the wind relative to due west? One Method: Use Law of Cosines: c = a + b ab c = + v = v + v plane/ ground plane/ wind wind / ground 2 cos (38)(45) cos 20 vw = 16 m/ s 1 38sin(20) m/ s θ = cos = m/ s

14 Projectile Motion

15 Galileo Challenged Aristotle Physics In a vacuum, all objects fall with the same acceleration due to gravity: 9.80 m/s 2, independent of their weight.

16 What Goes Up Must Come Down Someone standing at the edge of a cliff throws one ball straight up and one straight down at the same speed. Ignoring air resistance, which ball strikes the ground with the greatest speed?

17 Throwing up is Also Free Fall! Symmetry of G Field. Estimate : a= g ~ 10 m/ s 2 v = v + gt f 0 1 y = v0t + gt 2 2

18 Free Fall: Throwing Up What is the speed at the top of the path? ZERO! +y What is the acceleration at the top? a = m/s 2 What is the velocity at the same height on the way down? -30 m/s

19 QuickCheck 2.18 A ball is tossed up in the air. At its very highest point, the ball s instantaneous acceleration a y is A. Positive. B. Negative. C. Zero. +y Slide 2-96

20 QuickCheck 2.18 A ball is tossed up in the air. At its very highest point, the ball s instantaneous acceleration a y is A. Positive. B. Negative. C. Zero. +y Slide 2-97

21 Zero at the Top! Y component of velocity is zero at the top of the path in both cases! You know they have the y component of velocity is the same in both cases because they reached the same height!

22 Projectile Motion Same cannons, Same height. One dropped, One shot. Which hits the ground first? SAME! Both falling the same height! Horizontal speed doesn t affect vertical speed or the time to hit the ground! Only y determines time!

23 Projectile Motion Ignore Air Resistance! Most Important: X and Y components are INDEPENDENT of each other!

24 Projectile Motion: Vector Picture Motion with no acceleration 1 r = v t gt f i 2 2 = v iˆ+ ( v gt) ˆj v xi yi

25 Projectile Motion Problem Solving x and y directions are INDEPENDENT INDEPENDENT kinematics equations for x and y direction gravity affects only y direction solve for x with a = 0! Use a = g for y direction only! time for events to occur is the same for x and y directions time is the link between x and y!!!!

26 First the SIMPLE Case: Horizontal Launch (Ignore Air Resistance) The x-component doesn t change (no acceleration in x-direction.) The y-component changes (a = -g.)

27 Projectile Motion Horizontal and vertical components are independent of each other! Gravity acts in the vertical direction but not in the horizontal direction!! Speed in vertical direction speeds up! Speed in horizontal direction stays the same!

28 Projectile Motion No Change a = x 0 Actual path is a vector sum of horizontal and vertical motions. change a y = g

29 Relative Motion Horizontal and vertical components are independent of each other! Horizontal component remains unchanged without air resistance. Only the vertical component changes!

30 Dropping From Moving Frame Plane and Package An airplane traveling at a constant speed and height drops a care package. Ignoring air resistance, at the moment the package hits the ground, where is it relative to the plane? a) Behind the plane. b) Under the plane. c) In front of the plane. Any object dropped from a plane has the same initial velocity as the plane!

31 Care Package An airplane moves horizontally with constant velocity of 115 m/s at an altitude of 1050m and drops a care package as shown. How far from the release point does the package land? 1050m x =?

32 Care Package Strategy: Find the time the package drops to get the horizontal distance. The time to drop is just the free fall time!!! The horizontal displacement takes the same time as it takes the package to drop. x= vt x

33 Care Package Knowns : v = 0, v = 115 m/ s y0 x y = 1050 m, x=? y /, x 0 a = m s a = Strategy: Find time from y info to solve for x= vt x y y= vt 0 + at t= a t = 2 y y 2( 1050 m) 2 ( 9.8 m/ s ) t = 14.6s x= vt x = 115 m/ s(14.6 s) x= 1680m With what velocity does it hit the ground?

34 Knowns y0 : v = 0, v = 115 m/ s x y = 1050 m, x= 1680m /, x 0 ay = m s a = t 14.6s Care Package Strategy: Find final velocity in y direction and use it in: v y v= vx + vy, θ = tan vx v = v + at yf y0 y = 2 = 0 + ( 9.8 m/ s )(14.6 s) = 143 m/ s v= v + v 2 2 x y = (115 m/ s) + ( 143 m/ s) = 184 m/ s 2 2 v = (184 m/ s, 51.3 ) v 1 y m/ s θ = tan = tan vx 115 m/ s θ = 51.3

35 Projectile Motion Same cannons, Same height. One dropped, One shot. Which hits the ground first? SAME! Both falling the same height! Horizontal speed doesn t affect vertical speed or the time to hit the ground! Only y determines time!

36 Question g = 10 m/ s 2 The ball is thrown horizontally at 20 m/s. About how long does it take to hit the ground? 0 1 y = v yit + gt 2 2 t 2 y = = 1s g d How far does it travel in the horizontal direction? m x= vt= 20 1s = 20m xi s

37 Question g = 10 m/ s 2 The ball is thrown horizontally at 30 m/s. About how long does it take to hit the ground? 1 y = v yit + gt 2 2 t 2 y = = 1s g Only y determines time! d How far does it travel in the horizontal direction? m x= vt= 30 1s = 30m xi s

38 Question g = 10 m/ s 2 The ball is thrown horizontally at 100 m/s. How long does it take to hit the ground? 1 Second!! How far does it travel in the horizontal direction? m x= vt= 100 1s = 100m xi s

39 Curvature of Earth Curvature of the Earth: Every 8000 m, the Earth curves by 5 meters!

40 Curvature of Earth If you threw the ball at 8000 m/s off the surface of the Earth (and there were no buildings or mountains in the way) how far would it travel in the vertical and horizontal directions in 1 second?

41 Curvature of Earth If you threw the ball at 8000 m/s off the surface of the Earth (and there were no buildings or mountains in the way) how far would it travel in the vertical and horizontal directions in 1 second? ( )( ) horizontal : x = v t = 8000 m / s 1s = 8000m x 1 2 ( ) vertical : y = gt = 5t = 5 1s = 5m

42 Curvature of Earth If you threw the ball at 8000 m/s off the surface of the Earth (and there were no buildings or mountains in the way) how far would it travel in the vertical and horizontal directions in 1 second? ( )( ) horizontal : x = v t = 8000 m / s 1s = 8000m x 1 2 ( ) vertical : y = gt = 5t = 5 1s = 5m

43 Curvature of Earth If you threw the ball at 8000 m/s off the surface of the Earth (and there were no buildings or mountains in the way) how far would it travel in the vertical and horizontal directions in 1 second? Does the ball ever hit the Earth????

44 Orbital Velocity If you can throw a ball at 8000m/s, the Earth curves away from it so that the ball continually falls in free fall around the Earth it is in orbit around the Earth! Ignoring air resistance. Above the atmosphere

45 Projectile Motion IS Orbital Motion The Earth is in the way!

46 Orbital Motion & Escape Velocity 8km/s: Circular orbit Between 8 & 11.2 km/s: Elliptical orbit 11.2 km/s: Escape Earth 42.5 km/s: Escape Solar System!

47 Projectiles Launched at an Angle: The simple case: y=0

48 Projectile Motion A place kicker kicks a football at an angle of 40 degrees above the horizontal with an initial speed of 22 m/s. Ignore air resistance and find the total time of flight, the maximum height and the range the ball attains.

49 Maximum range is achieved at a launch angle of 45!

50 Symmetry in the Projectile Range sin 2θ is symmetric about 45 Range Equation: R = 2 sin 2 v i i g θ

51 Projectile Motion Launched at an Angle (Ignore Air Resistance) The x-component doesn t change (no acceleration in x-direction.) The y-component changes (a = -g.)

52 Same rock, same speed, same angle. Which rock hits the water first? a) Rock 1 b) Rock 2 c) same Which rock hits the water with the greatest speed? a) Rock 1 b) Rock 2 c) same

53 Same rock, same speed, same angle. Which rock hits the water first? a) Rock 1 b) Rock 2 c) same Which rock hits the water with the greatest speed? a) Rock 1 b) Rock 2 c) same Spatial Symmetry In G Field!

54 How Tall is the Building? A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53degrees above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?

55 Harder Horizontal Launch Problems: Hitting an incline! Distance traveled is given by the trajectory but the net displacement is the diagonal!

56 A skier leaves the ramp of a ski jump with a velocity of 10.0 m/s, 15.0 above the horizontal, as shown. The slope is inclined at 50.0, and air resistance is negligible. Find the distance from the ramp to where the jumper lands and the time of flight. m x= v x0 t dcos 50 = 10 cos15 t s 1 2 y= vy0t+ at y 2 m m 2 d sin 50 = 10 sin15 t 4.9 t 2 s s m cos15 d = 10 t = 15.03t d s cos50 y m m 2 (15.03 t)sin 50 = 10 sin15 t 4.9 t 2 s s x = m t 2 t = 2.88s s x = 27.8 m, d = 43.2 m

57 HW Problem : A dive bomber has a velocity of 280 m/s at an angle below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.25 km. Find the angle. HINT: You don t know time so set up two equations for y and x, eliminate t and use the displacement values given to solve for θ.