Physics 203 College Physics I Fall 2012

1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 3 Motion in 2 Dimensions – Part 2

2. Today’s Topics • Motion Concepts in 2 Dimensions • We will use vectors to define displacement and velocity in 2 dimensions. • Constant Acceleration: Projectile Motion

3. 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. Question 4 from the Quiz… → • Vector A has a magnitude of 10 and a direction angle θ = 60omeasured counter-clockwise from the +x axis. What are the magnitude and direction angle of the vector – 2A? • A. – 20, 60o • B. 20, 240o • C. 20, – 30o • D. – 20, 240o • E. – 20, – 30o → → A y θ = 60o x

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

6. Two Ships • Suppose one ship (A) is 20o E of N from a port, at a distance of 2300 m. A second ship (B) is 50o W of N, 1100 m from the port. • Sketch the two position vectors of the ships and the displacement vector from ship Ato ship B. • Calculate the magnitude and direction of the displacement vector from ship Ato ship B.

7. Ships’ Position Vectors Cartesian coordinate system Ship 1 = A Magnitude A = 2300 m Ship 2 = B Magnitude B = 1100 m N y 2300 m → 20o → 50o A 1100 m → W E → x 0 B S

8. 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!

9. Motion and Displacements • When an object moves, the vector rwhich describes its position changes with time. → t = 0 s t = 1 s t = 2 s t = 3 s t = 4 s t = 5 s t = 6 s t = 7 s t = 8 s → r O

10. Motion and Displacements • The change in position from start to finish is the displacement vector • Δr=rf–ri. • The displacement only depends on the initial and final points, not the path. → → rf → → → Δ r → ri O

11. 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/ Δt → → Δ r → v → →

12. Average Velocity • The average velocity always points in the same direction as the displacement. → Δ r → v

13. 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 s t = 1 s t = 2 s t = 3 s t = 4 s t = 5 s t = 6 s t = 7 s t = 8 s → r O

14. 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 _

15. 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|

16. Constant Velocity • The velocity is constant when the object moves in a straight line at a constant speed. → r t = 0 s t = 1 s t = 2 s t = 3 s t = 4 s t = 5 s O

17. 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.

18. 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.

19. 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/s2) • Acceleration vector g points downward with magnitude g. →

20. Path of a Projectile

21. 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.

22. Projectile Motion • A projectile is shot from the edge of a cliff as shown with an initial speed v0 = 65.0 m/s. The height of the cliff is h = 125 m, and the angle with horizontal is q = 37.0o. a) What velocity does it have just before it hits the ground? (magnitude and direction) 0 x y (b) How long does it take to hit the ground?

23. Football • A football is kicked from the 30 yard line with an initial angle of 35o. 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.) y → v0 h θ x d

24. Football • Given: d = 120 ft, h = 10 ft, θ = 35o. • Horizontal: d = v0x t with v0x = v0cosθ • Vertical: h = v0y t – ½ g t2 with v0y = v0 sin θ y → v0 h θ x d

25. Football • Substitute t = d/v0x into h = v0y t – ½ g t2: • h = d (v0y/v0x) – ½ g d2/v0x2 • = d tan θ – ½ g (d2/v02) sec2θ • Simplify: • 2 cos2θ (d tan θ – h) = ½g d2/v02 • d/v0 = 2cos θ √(d tan θ – h)/g = 1.638 √ 7.55 s2 • 120 ft/v0 = 4.50 s • v0 = 120 ft / 4.50 s = 26.7 ft/s