1. Chapter 3 - Kinematics: Acceleration Bruce Chittenden

2. The notion of acceleration emerges when a change in velocity is combined with the time during which the change occurs. Acceleration

3. Average Acceleration

4. 3.1 Average Acceleration The average acceleration (a av ) of a body is defined as the ratio of the change in its velocity over the time elapsed in the process Average acceleration = change in velocity time elapsed a AV = = v t v f – v i t f – t i

5. Average Acceleration

7. Acceleration and Decreasing Velocity

9. Uniform Acceleration Notice how the distances traveled in equal time intervals increases, if fact those distances go as the odd integers 1, 3, 5, 7,...

10. Centripetal Acceleration Whenever an object speeds up or slows down, its velocity changes size and it has an acceleration along, or tangent to, the path traveled. Whenever an object moves in a curve, its velocity changes direction and it has an acceleration perpendicular to the path traveled.

11. Average Acceleration a AV = = v t v f – v i t f – t i Usually we will take the direction in which the object first moves to be positive, v i > 0. With such a sign convention Average Acceleration can be written a scalar values.

12. Preferred SI Units for Acceleration The preferred SI units of acceleration are m/s 2, and is read meters per second squared although (m/s) / s is more informative. An object changes speed by so many meters per second every second.

13. 3.2 Instantaneous Acceleration a = lim v t t0 a = = v t v f – v i t f – t i

14. 3.3 Constant Acceleration Without Calculus, it is appropriate to stay with situations in which a is constant and therefore equal to a av. a = = v t v f – v i t f – t i

15. 3.4 Mean Speed v av = ½ (v i + v f ) A constant tangential acceleration produces a uniformly changing speed, and so the average value is midway between the initial and final speeds. Thus, Mean Speed is ½ the initial speed + the final speed and is given by the equation

16. Mean Speed A body that accelerates uniformly from v i to v f over a certain straight- line distance will cover exactly the same distance in the same time traveling at a fixed speed of v av known as the mean speed.

17. Uniformly Accelerating For a body that is uniformly accelerating, the mean speed, v av, is the height of the midpoint of the straight-line speed- time graph.

18. Mean Speed Theorem v i t + ½(v f - v i )t = (v i - ½ v i + ½ v f )t = ½(v i +v f )t The area under the sloping straight line representing uniform acceleration corresponds to the distance traveled. The area in question has two distinct pieces and can be thought of as the sum of the areas of the rectangle (base t times height v i ) and the triangle (one-half the base t time the altitude v f – v i ). The total area is then Setting this equal to s, the scalar value of the displacement, we arrive at the Mean Speed Theorem: s = ½ (v i + v f )

19. 3.5 The Equations of Constant Acceleration There are five motion variables (v i, v f, a T, s, t) and five equations that are traditionally used to interrelate them. These constant a equations derive from the basic definitions of speed and acceleration, and so there are actually only two independent equations and two variables that can be solved for. The five equations are developed primarily as a convenience in problem solving.

20. 1.Initial Velocity, v i 2.Final Velocity, v f 3.Acceleration, a T 4.Displacement, s 5.Time, t Constant Acceleration Variables

21. Constant Acceleration Equations Acceleration Average Speed Mean Speed Theorem Displacement Independent of Time v f = v i + at v av = ½(v i + v f ) s = v av t = ½(v i + v f )t s = v i t + ½a T t 2 v f 2 = v i 2 + 2a T s

22. Displacement

23. saTaT vfvf vivi t ?2.0 m/s 2 6.0 m/s8.0 s Displacement s = v i t + ½ a T t 2 = (6.0m/s)(8.0s) + ½ (2.0m/s 2 ) (8.0s) 2 = +110m

24. Displacement

25. saTaT vfvf vivi t ?31 m/s 2 62 m/s0 m/s Displacement v f 2 = v i 2 + 2a T s s = (v f 2 – v i 2 ) / 2a T = ((62m/s) 2 - (0m/s) 2 ) / 2(31m/s 2 ) = +62m

26. Free-Fall On Earth all objects falling through a vacuum accelerate downward at the same fairly constant rate, regardless of their weight.

27. 3.6 Air Drag When an object falls through the air, moving ever more rapidly, the resistance to its motion, the drag, also increases. The greater the rate of descent, the greater the resistance, as more air must be pushed aside per second. Finally, a balance is reached where no further increase in falling speed can occur. This point is called the terminal speed, and it is dependent on the shape, surface, and weight of the object.

28. 3.7 Acceleration Due to Gravity The Free-Fall acceleration, due as it is to gravity, is represented by its own symbol, g. There are some small variations in g over the planet, but we can generally take g to be a constant equal to its average value of 9.80665 m/s 2 or 9.81 m/s 2 (i.e. 32.2 ft/s 2 ).

29. Acceleration Due to Gravity

30. 3.8 Straight Up & Down All equations of uniformly accelerated motion apply, as is, with a T = g. In that case g is treated as just another acceleration, and like a T, its numerical value can be positive or negative. On the way up a projectile has negative acceleration. On the way down it has a positive acceleration. In any event, in free-fall the acceleration is always g and it’s always straight downward regardless of the motion.

31. A Falling Stone A stone is dropped from the top of a tall building. After 3.00s of free fall, what is the displacement y of the stone? Straight Down

32. saTaT vfvf vivi t ?-9.80 m/s 2 0 m/s3.00 s Straight Down s = v i t + ½ a T t 2 = (0m/s)(3.00s) + ½ (-9.80m/s 2 ) (3.0s) 2 = -44.1 m

33. How High Does it Go? The referee tosses the coin up with an initial speed of 5.00m/s. In the absence if air resistance, how high does the coin go above its point of release? Straight Up

34. saTaT vfvf vivi t ?-9.80 m/s 2 0 m/s+5.00 m/s v f 2 = v i 2 + 2a T s s = (v f 2 – v i 2 ) / 2a T = ((0m/s) 2 – (5.0m/s) 2 ) / 2 (-9.80m/s 2 ) = 1.28m

35. Two-Dimensional Motion: Projectiles Consider an object moving in two dimensions. At any moment it velocity vector can be resolved into two perpendicular components. The motion unfolds as if it were composed of these two separate motions superimposed: v = v x + v y

36. Two Dimensional Motion

38. The x part of the motion occurs exactly as it would if the y part did not occur at all, and vice versa. Two Dimensional Motion

39. Example 1 A Moving Spacecraft In the x direction, the spacecraft has an initial velocity component of +22 m/s and an acceleration of +24 m/s 2. In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s 2. Find (a) x and v x, (b) y and v y, and (c) the final velocity of the spacecraft at time 7.0 s. Two Dimensional Motion

40. Reasoning Strategy 1. Make a drawing. 2. Decide which directions are to be called positive (+) and negative (-). 3. Write down the values that are given for any of the five kinematic variables associated with each direction. 4. Verify that the information contains values for at least three of the kinematic variables. Do this for x and y. Select the appropriate equation. 5. When the motion is divided into segments, remember that the final velocity of one segment is the initial velocity for the next. 6. Keep in mind that there may be two possible answers to a kinematics problem. Two Dimensional Motion

41. Example 1 A Moving Spacecraft In the x direction, the spacecraft has an initial velocity component of +22 m/s and an acceleration of +24 m/s 2. In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s 2. Find (a) x and v x, (b) y and v y, and (c) the final velocity of the spacecraft at time 7.0 s. xaxax vxvx v ox t ?+24.0 m/s 2 ?+22 m/s7.0 s yayay vyvy v oy t ?+12.0 m/s 2 ?+14 m/s7.0 s Two Dimensional Motion

42. xaxax vxvx v ox t ?+24.0 m/s 2 ?+22 m/s7.0 s Two Dimensional Motion

43. yayay vyvy v oy t ?+12.0 m/s 2 ?+14 m/s7.0 s Two Dimensional Motion

46. Projectiles The faster each cannonball is fired, the farther it will go, but all fall at the same rate and hit the water after the same flight time. The horizontal speed v ix, is constant, provided air friction is negligible, the vertical speed increases at the rate g.

47. Under the influence of gravity alone, an object near the surface of the Earth will accelerate downwards at 9.80 m/s 2. Projectile Motion

48. Example 3 A Falling Care Package The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. Determine the time required for the care package to hit the ground.

49. Example 4 The Velocity of the Care Package What are the magnitude and direction of the final velocity of the care package? Projectile Motion

50. yayay vyvy v oy t -1050 m-9.80 m/s 2 0 m/s? Projectile Motion

51. yayay vyvy v oy t -1050 m-9.80 m/s 2 ?0 m/s14.6 s Projectile Motion

52. Ballistic Flight For unpowered Ballistic Flight the motion occurs as if it were two independent motions, one vertical with initial scalar velocity of v iy = v i sin θ the other horizontal, with an initial scalar velocity of v ix = v i cos θ The displacement vector s drawn from the point of launch to the projectile at any moment has a scalar horizontal component of s x = v ix t because a x = 0 and a scalar vertical component of s y = v iy + ½ gt 2 because a y = g.

53. Ballistic Flight

54. The Height of a Kickoff A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, determine the maximum height that the ball attains. Projectile Motion: Height

55. v o c os Projectile Motion: Height

56. yayay vyvy v oy t ?-9.80 m/s 2 014 m/s Projectile Motion: Height

57. The Time of Flight of a Kickoff What is the time of flight between kickoff and landing? Projectile Motion: Time

58. yayay vyvy v oy t 0-9.80 m/s 2 14 m/s? Projectile Motion: Time

59. yayay vyvy v oy t 0-9.80 m/s 2 14 m/s? Projectile Motion: Time

60. The Range of a Kickoff Calculate the range R of the projectile. Projectile Motion: Distance

61. Actual Projectile Data