1. Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 3 Motion in 2 Dimensions – Part 2

2. Physics 203 – College Physics I Department of Physics – The Citadel Today’s Topics 1. Motion Concepts in 2 Dimensions We will use vectors to define displacement and velocity in 2 dimensions. 2. Constant Acceleration: Projectile Motion

3. Physics 203 – College Physics I Department of Physics – The Citadel Tuesday’s Assignment Read Ch. 4, sec. 1 – 5 (Newton’s Laws). A problem set on HW3 on Ch. 3 is due Tuesday. The first exam is scheduled for next Thursday. You do not need to memorize equations: the essential ones will be provided for the exam. We will finish Ch. 4 after the exam. The exam only covers chapters 1 – 3.

4. Physics 203 – College Physics I Department of Physics – The Citadel Question 4 from the Quiz… Vector A has a magnitude of 10 and a direction angle θ = 60 o measured counter-clockwise from the +x axis. What are the magnitude and direction angle of the vector – 2A? A. – 20, 60 o B. 20, 240 o C. 20, – 30 o D. – 20, 240 o E. – 20, – 30 o → → x y A → θ = 60 o

5. Physics 203 – College Physics I Department of Physics – The Citadel Question 4 from the Quiz… Vector A has a magnitude of 10 and a direction angle θ = 60 o measured counter-clockwise from the +x axis. What are the magnitude and direction angle of the vector – 2A? A. – 20, 60 o B. 20, 240 o C. 20, – 30 o D. – 20, 240 o E. – 20, – 30 o → → x y A → θ = 60 o – 2A → θ = 240 o 20 10

6. Physics 203 – College Physics I Department of Physics – The Citadel Two Ships Suppose one ship (A) is 20 o E of N from a port, at a distance of 2300 m. A second ship (B) is 50 o W of N, 1100 m from the port. Sketch the two position vectors of the ships and the displacement vector from ship A to ship B. Calculate the magnitude and direction of the displacement vector from ship A to ship B.

7. Physics 203 – College Physics I Department of Physics – The Citadel Ships’ Position Vectors x y N 0 EW S Cartesian coordinate system Ship 1 = A Magnitude A = 2300 m Ship 2 = B Magnitude B = 1100 m 20 o 2300 m A → → → 50 o 1100 m B →

8. Physics 203 – College Physics I Department of Physics – The Citadel Ships’ Cartesian Coordinates x y N 0 EW S A x = A sin 20 o = 787 m A y = A cos 20 o = 2161 m B x = – B sin 50 o = – 843 m B y = B cos 50 o = 707 m 20 o 2300 m A → 50 o 1100 m B →

9. Physics 203 – College Physics I Department of Physics – The Citadel Displacement Vector x y N 0 EW S The displacement vector from ship A to ship B is D = B – A. Components: D x = B x – A x = – 1630 m D y = B y – A y = – 1454 m A → → → → B → D →

10. Physics 203 – College Physics I Department of Physics – The Citadel Displacement Vector x y N 0 EW S Magnitude: Pythagorean theorem D = √D x 2 – D y 2 = 2184 m ≈ 2200 m Direction: φ = atan(D y /D x ) = 41.7 o ≈ 42 o South of West A → B → D → φ

11. Physics 203 – College Physics I Department of Physics – The Citadel Arctangent Caution x y N 0 EW S D x = – 1600 m D y = – 1500 m Direction: φ = tan –1 (D y /D x ) = 42 o South of West But the polar angle is θ = φ + 180 o = 222 o. D → φ θ

12. Physics 203 – College Physics I Department of Physics – The Citadel Motion in 2 Dimensions We want to describe how an object moves along a general path... where it is at any time how fast it moves and which direction. Displacement Velocity Acceleration All vectors!

13. Physics 203 – College Physics I Department of Physics – The Citadel Motion and Displacements When an object moves, the vector r which describes its position changes with time. → t = 0 st = 1 st = 2 st = 3 st = 4 st = 5 st = 6 st = 7 st = 8 s O r →

14. Physics 203 – College Physics I Department of Physics – The Citadel Motion and Displacements The change in position from start to finish is the displacement vector Δ r = r f – r i. The displacement only depends on the initial and final points, not the path. Δ r O → → →→ → → rfrf riri

15. Physics 203 – College Physics I Department of Physics – The Citadel Average Velocity The average velocity for the motion is the vector v defined to be the net displacement vector divided by the time: v = Δ r / Δ t Δ r v → → →→ →

16. Physics 203 – College Physics I Department of Physics – The Citadel Average Velocity The average velocity always points in the same direction as the displacement. Δ r v → →

17. Physics 203 – College Physics I Department of Physics – The Citadel Speed The rate at which the particle moves along the path is its instantaneous speed. Speed is a scalar, and is always positive. t = 0 st = 1 st = 2 st = 3 st = 4 st = 5 st = 6 st = 7 st = 8 s O r →

18. Physics 203 – College Physics I Department of Physics – The Citadel Average Speed The average speed means something different from average velocity: The average speed is the total distance traveled along the path divided by the time: v = l / t. l _

19. Physics 203 – College Physics I Department of Physics – The Citadel Instantaneous Velocity The instantaneous velocity is a vector whose magnitude is the instantaneous speed, and whose direction is the direction of motion at that moment. v Instantaneous speed v = |v| →

20. Physics 203 – College Physics I Department of Physics – The Citadel Constant Velocity The velocity is constant when the object moves in a straight line at a constant speed. O t = 0 st = 1 st = 2 st = 3 st = 4 st = 5 s r →

21. Physics 203 – College Physics I Department of Physics – The Citadel Constant Velocity vs Constant Speed Both of our examples so far had constant speed, meaning that the particle follows the path at a uniform rate. But only the motion in a straight line had constant velocity. The velocity is constant only when the magnitude and direction are both fixed. If the velocity vector changes, the motion has acceleration. Acceleration is also a vector, but we will consider only special cases.

22. Physics 203 – College Physics I Department of Physics – The Citadel Acceleration The acceleration gives the rate of change of the velocity vector, and is also a vector. The acceleration is zero only when an object is at rest or moves in a straight line at a constant speed. If the acceleration is constant, that means the velocity only changes in the direction the acceleration vector points, and it changes at a constant rate.

23. Physics 203 – College Physics I Department of Physics – The Citadel Projectile Motion Projectile motion is the motion of an object under the influence of gravity alone. Constant velocity in the x direction, Constant acceleration in the y direction (g = 9.80 m/s 2 ) Acceleration vector g points downward with magnitude g. →

24. Physics 203 – College Physics I Department of Physics – The Citadel Path of a Projectile

25. Physics 203 – College Physics I Department of Physics – The Citadel Independence of Motions The vertical and horizontal motion are independent. The yellow ball is initially given a horizontal velocity, while the red one is just dropped. They reach the same height at every time, because they have identical vertical motion.

26. Physics 203 – College Physics I Department of Physics – The Citadel Projectile Motion A projectile is shot from the edge of a cliff as shown with an initial speed v 0 = 65.0 m/s. The height of the cliff is h = 125 m, and the angle with horizontal is  = 37.0 o. a) What velocity does it have just before it hits the ground? (magnitude and direction) y x 0 (b) How long does it take to hit the ground?

27. Physics 203 – College Physics I Department of Physics – The Citadel Projectile Motion (a) Strategy: Find the final velocity components. x direction: constant velocity, v x = v 0 x. y direction: constant acceleration, known distance: a = – g, Δ y = – h. v fy 2 = v 0y 2 + 2a Δ y = v 0y 2 + 2gh. y Given: h = 125 m v 0 = 65.0 m/s θ = 37.0 o x

28. Physics 203 – College Physics I Department of Physics – The Citadel Projectile Motion v x = v 0 cos 37 o always = 51.9 m/s v y 2 = (v 0 sin 37 o ) 2 + 2gh = (39.1 m/s) 2 + 2450 m 2 /s 2 = 3979 m 2 /s 2 v y = 63.1 m/s v = (v x 2 + v y 2 ) ½ = 81.7 m/s Angle: tan θ f = v y /v x = 63.1/51.9 θ f = 50.6 downward y x Given: h = 125 m v 0 = 65.0 m/s θ = 37.0 o

29. Physics 203 – College Physics I Department of Physics – The Citadel Projectile Motion (b) Use the vertical motion to find the time in the air. v 0y = –39.1 m/s. v y = 63.1 m/s.  Δ v y = 102.2 m/s = gt  g = 9.8 m/s 2  t = Δ v y / g = 10.4 s y x Given: h = 125 m v 0 = 65.0 m/s θ = 37.0 o

30. Physics 203 – College Physics I Department of Physics – The Citadel Football A football is kicked from the 30 yard line with an initial angle of 35 o. What is the minimum initial speed of a kick at this angle that can make it over the goal posts to score a field goal? (The distance of the kick is 40 yards = 120 ft, and the goal post bar is 10 ft high.) x y v0v0 d h → θ

31. Physics 203 – College Physics I Department of Physics – The Citadel Football Given: d = 120 ft, h = 10 ft, θ = 35 o. Horizontal: d = v 0x t with v 0x = v 0 cos θ Vertical: h = v 0y t – ½ g t 2 with v 0y = v 0 sin θ x y v0v0 d h → θ

32. Physics 203 – College Physics I Department of Physics – The Citadel Football Substitute t = d/v 0x into h = v 0y t – ½ g t 2 : h = d (v 0y /v 0x ) – ½ g d 2 /v 0x 2 = d tan θ – ½ g (d 2 /v 0 2 ) sec 2 θ Simplify: 2 cos 2 θ (d tan θ – h) = ½g d 2 /v 0 2 d/v 0 = 2cos θ √(d tan θ – h)/g = 1.638 √ 7.55 s 2 120 ft/v 0 = 4.50 s v 0 = 120 ft / 4.50 s = 26.7 ft/s