objectives

1. objectives Homework: read 3.1 & 3.2 Exercise: 3.3, 3.5, 3.7 Position, velocity, acceleration vectors

2. Position vector

3. (x2 – x1)i (y2 – y1)j (z2 – z1)k vav = + + ∆t ∆t ∆t Average velocity vector • During a time interval t the particle moves from P1 , where its position vector is r1, to P2, where its position vector is r2. One dimension

4. (2D) Instantaneous velocity

5. dr v = dt

6. Example 3.1: calculating average and instantaneous velocity

8. Test your understanding • In which of these situations would the average velocity vector vav over an interval be equal to the instantaneous velocity v at the end of interval: • A body moving along a curved path at constant speed • A body moving along a curved path and speeding up • A body moving along a straight line at constant speed • A body moving along a straight line and speeding up.

9. 2b 3c t = example If r = bt2i+ ct3j Where b and c are positive constants, when does the velocity vector make an angle of 45.0o with the x- and y-axes? v = dr/dt = 2bt i + 3ct2 j = 1

10. acceleration

11. (v2x – v1x)i (v2y – y1y)j (v2z – v1z)k aav = + + ∆t ∆t ∆t One dimension

12. d2x dvx dvy dvz d2y d2z a = i + j + k a = i + j + k dt dt dt dt dt dt ay tanθ = ax |a| = √ax2 + ay2 + az2 Instantaneous Acceleration

13. All about acceleration • Equal to time rate of change of velocity • ≠ 0 if velocity changes in magnitude or direction. • It does not have same direction as velocity vector • Acceleration vector lies on concave side of curved path.

14. Example 3.2 • Given: • Find the components of the average acceleration in the interval from t = 0.0 s to t = 2.0 s. • Find the instantaneous acceleration at t = 2.0 s, its magnitude and its direction.

15. Parallel and perpendicular components of acceleration • The acceleration vector a for a particle can describe changes in the particle’s speed, its direction of motion, or both. • The component of acceleration parallel to a particle’s path (parallel to the velocity) tells us about changes in the particle's speed. • The acceleration component perpendicular to the path (perpendicular to the velocity) tells us about changes in the particle’s direction of motion.

17. The effect of acceleration’s directions • When acceleration is perpendicular to particle’s velocity: velocity’s magnitude does not change, only its direction changes, particle moves in a curved path at constant speed. • When acceleration is parallel to particle’s velocity: velocity’s magnitude changes only, its direction remains the same, particle moves in a straight line with changing speed.

18. In most cases, the particle’s speed and direction change. The acceleration a has both parallel and perpendicular components. Increasing speed decreasing speed

20. Example 3.3 • Given: • Find the parallel and perpendicular components of the instantaneous acceleration at t = 2.0 s

21. Example 3.4 • A skier moves along a ski-jump ramp as shown in the figure. The rap is straight from point A to point C onward. The skier picks up speed as she moves downhill from point A to point E. draw the direction of the acceleration vector at points B, D, E and F.

22. Test your understanding 3.2 • A sled travels over the crest of a snow-covered hill. The sled slows down as it climbs up one side of the hill and gains speed as it descends on the other side. Which of the vectors (1 though 9) in the figure correctly shows the direction of the sled’s acceleration at the crest? Choice 9 is that the acceleration is zero.)

23. 3.6

24. a = 0

25. objective • Projectile motion • Homework: Read 3.3; Exercise: 3.9, 3.11, 3.13

26. A projectile is any body that is given an initial velocity and then follows a path determined entirely by the effects of gravitational acceleration and air resistance. Projectile Motion

27. Projectiles move in TWO dimensions • Since a projectile moves in 2-dimensions, it therefore has 2 components just like a resultant vector. • Horizontal and Vertical The path of a projectile is called a trajectory

28. θ Horizontal Component • Velocity is constant: vx0 = vi0cosθ • Acceleration: ax = 0 • Displacement x = x0 + vx0∙t In other words, the horizontal velocity is CONSTANT. BUT WHY? Gravity DOES NOT work horizontally to increase or decrease the velocity.

29. θ Vertical Component • Acceleration: ay = -g • Velocity: vy = vy0 – gt = vi0sinθ - gt • Displacement: y = y0 + vy0∙t - ½ gt2

30. If air resistance is negligible, the trajectory of a projectile is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. The path of the projectile is parabolic.

31. The magnitude of the position: The magnitude of the velocity: The direction of the velocity: • If at t = 0, x0 = y0 = 0 then we can find the x, y coordinates and the x, y velocity at time t: In y direction: vy = v0sinθ - gt y = (v0sinθ)t – ½ gt2 vy2 = (v0sinθ)2– 2gy In x direction: vx = v0cosθt x = (v0cosθ)t

32. Air resistance isn’t always negligible.

33. Example 3.5 Let’s consider again the skier in example 3.4. What is her acceleration at points G, H, and I after she flies off the ramp? Neglect air resistance. The acceleration at points G, H, I are the same: ax = 0; ay = -9.8 m/s2

34. Example 3.6:a body projected horizontally • A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. find the motorcycle’s position,distance and velocity from the edge of the cliff, and velocity after 0.50 s.

35. Example: 3.7 – height and range of a projectile I – a batted baseball

36. objective • Projectile • Homework: 3.15, 3.17, 3.19

37. Example 3.8 • For a projectile launched with speed v0 at initial angle α0 (between 0o and 90o), derive general expressions for the maximum height h and horizontal range R. For a given v0, what value of α0 gives maximum height? What value gives maximum horizontal range?

38. You toss a ball from your window 8.0 m above the ground. When the ball leaves your hand, it is moving at 10.0 m/s at an angle of 20o below the horizontal. How far horizontally from your window will the ball hit the ground? Ignore air resistance.

39. A monkey escapes from the zoo and climbs a tree. After failing to entice the monkey down, the zookeeper fires a tranquilizer dart directly at the monkey. The clever monkey lets go at the same instant the dart leaves the gun barrel, intending to land on the ground and escape. Show that the dart always hits the monkey, regardless of the dart’s muzzle velocity (provided that it gets to the monkey before he hits the ground).

40. tanα0 = h / d

41. Check your understanding 3.3 • In example 3.10, suppose the tranquilizer dart has a relatively low muzzle velocity so that the dart reaches a maximum height at a point before striking the monkey. When the dart is at point P, will the monkey be • At point A (higher than P) • At point B (at the same height as P) • At point C (lower than P)?

42. objective • Motion in a circle • Homework: 3.29, 3.31, 3.33

43. 3.4 motion in a circle • When a particle moves along a curved path, the direction of its velocity is tangent to the curve, and its acceleration is pointing toward the concave side. Increasing speed

44. Decreasing speed

45. Constant speed – uniform circular motion

46. There is no component of acceleration parallel (tangent) to the path; the acceleration vector is perpendicular (normal) to the path and hence directed inward toward the center of the circular path.

47. The subscript “rad” is a reminder that the direction of the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

48. Uniform circular motion vs. projectile motion • the magnitude of acceleration is constant at all times. • the direction of the direction of the acceleration in projectile always points down; • the magnitude of acceleration is constant at all times. • the direction of acceleration changes continuously - always points toward the center of the circle.

49. Example 3.11

50. Example 3.12