1. I. Distance vs. Displacement Kinematics: The study of motion without regard to the cause (i.e. – force). Position: Your location at a specific time vector quantity (magnitude & direction) change in position WITH direction Final position minus initial position Distance: scalar quantity (magnitude only) total length of path traveled by object

2. Think about this… A person runs around a track once. She starts and stops in the same position. What was the distance she covered and what was her displacement? Distance = 400 m (total length traveled) Displacement = 0 m (starts and stops in the same position)

3. Example: Joe has a date with Jill. Her house is 5 km east of his, but he has to do some things before he picks her up. He first heads 7 km N of his house to get flowers. He then travels 5 km east to pick up the movie tickets. Then he heads to her house. N E

4. Mr. Sicienski bikes 15 km east then 30 km north (a) what is the distance he traveled? (b) what is the magnitude of his displacement?

5. II. Speed vs. Velocity Vector Quantity Rate of change of displacement or more simply Displacement divided by time Average Velocity:

6. II. Speed vs. Velocity scalar Quantity Rate of change of distance or more simply distance divided by time Average Speed:

7. II. Speed vs. Velocity 1. If Joe’s drive in example 1 above takes him 30 min., determine his average speed and average velocity.

8. II. Speed vs. Velocity  Velocity at a particular instant in time. Instantaneous Velocity Instantaneous Speed  Speed at a particular instant in time.

9. II. Speed vs. Velocity 1. Is it possible to travel at a constant speed, but not a constant velocity? Explain Questions: Yes, you can maintain a constant speed (60 mph), but change your direction, therefore changing your velocity

10. II. Speed vs. Velocity Questions: 2. A car travels 10 km due north, then 8 km due east in 30 minutes. Find the car’s a) distance, b) displacement, c) average speed, d) average velocity A)Distance = 10 km +8 km = 18 km B)Displacement = 12.8 km 38.7 degrees (E of N) C)0.6 km/min D)0.43 km/min 38.7 degrees (E of N)

11. III. Acceleration - vector quantity - rate of change in velocity (change in magnitude OR direction)

12. Think about this… Is the object accelerating in the following cases: Yes. There is a change in the magnitude of the velocity (60 mph  20 mph) - A car slams on its brakes and goes from 60 mph to 20 mph - A car goes around a circular track at a constant 15 mph Yes. It moves at a constant speed, but changes its direction, therefore, changing its velocity

13. Acceleration Equation (see reference tables) a = acceleration  V= Change in velocity (v f - v i final velocity/speed –initial velocity/speed ) t = time final velocity/speed –initial velocity/speed ) t = time Units: m/s/s or m/s 2 Means every second the velocity changes by that magnitude

14. Part B: Free Fall Definition: acceleration due to gravity is the acceleration of an object in free fall that results from the influence on Earth’s force of gravity 9.81 m/s 2 (toward center of earth) (see reference tables) = -9.81 m/s 2 ( - = down) NEGLECTING AIR RESISTANCE, ALL OBJECTS FALL WITH THE SAME ACCELERATION (REGARDLESS OF MASS)

15. Cool videos Mythbusters - Penny of the Empire State (4 min) World’s Largest Vacuum Chamber (5 min) Feather vs. Hammer on the Moon (1 min)

16. t = 0 s, v = 0 m/s 1 s, v = 9.81 m/s 2 s, v = 19.6 m/s 3 s, v = 29.4 m/s 4 s, v = 39.2 m/s 5 s, v = 49.1 m/s a = 9.81 m/s/s the entire fall

17. Tossing a Ball up in the Air t = 0 s, v = +40 m/s → t = 1 s, v = +30 m/s → t = 2 s, v = +20 m/s → t = 3 s, v = +10 m/s → t = 4 s, v = 0 m/s → ← t = 5s, v = -10 m/s ← t = 6s, v = -20 m/s ← t = 7s, v = -30 m/s ← t = 8s, v = -40 m/s

18. Tossing a Ball up in the Air 0 m/s (comes to a rest) time to fall back to original position speed at original position on the way down

19. The equations of motion The equations of motion can be used when an object is accelerating at a constant rate There are four equations relating five quantities u initial velocity, v final velocity, s displacement, a acceleration, t time SUVAT equations

20. The four equations 1This is a re-arrangement of 2This says displacement = average velocity x time 3With zero acceleration, this becomes displacement = velocity * time 4Useful when you don’t know the time

21. Important Note! All quantities are vectors. These equations are normally done in one dimension, so a negative result means displacement/ velocity/ acceleration in the opposite direction.

22. How to Complete a Physics Problem: 1. List Givens 2. List what you want to find 3. Find correct equation and manipulate it to solve for desired variable 4. Plug in numbers with units

23. Example 1 Mr Rayner is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms -1 and his decceleration is 2 ms -2 how long will it take for the car to come to rest?

24. Example 1 Mr Rayner is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms -1 and his decceleration is 2 ms -2 how long will it take for the car to come to rest? What does the question tell us. Write it out.

25. Example 1 Mr Rayner is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms -1 and his decceleration is 2 ms -2 how long will it take for the car to come to rest? u = 10 ms -1 v = 0 ms -1 a = -2 ms -2 t = ? s

26. Example 1 Mr Rayner is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms -1 and his decceleration is 2 ms -2 how long will it take for the car to come to rest? u = 10 ms -1 v = 0 ms -1 a = -2 ms -2 t = ? s Choose the equation that has these quantities in v = u + at

27. Example 1 Mr Rayner is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms -1 and his decceleration is 2 ms -2 how long will it take for the car to come to rest? u = 10 ms -1 v = 0 ms -1 a = -2 ms -2 t = ? s v = u + at 0 = 10 + -2t 2t = 10 t = 5 seconds

28. Example 2 Jack steps into the road, 30 metres from where Mr Rayner’s engine stops working. Mr Rayner does not see Jack. Will the car stop in time to miss hitting Jack?

29. Example 2 Jack steps into the road, 30 metres from where Mr Rayner’s engine stops working. Mr Rayner does not see Jack. Will the car stop in time to miss hitting Jack? What does the question tell us. Write it out.

30. Example 2 Jack steps into the road, 30 metres from where Mr Rayner’s engine stops working. Mr Rayner does not see Jack. Will the car stop in time to miss hitting Jack? u = 10 ms -1 v = 0 ms -1 a = -2 ms -2 t = 5 s s = ? m

31. Example 2 Jack steps into the road, 30 metres from where Mr Rayner’s engine stops working. Mr Rayner does not see Jack. Will the car stop in time to miss hitting Jack? u = 10 ms -1 v = 0 ms -1 a = -2 ms -2 t = 5 s s = ? m Choose the equation that has these quantities in v 2 = u 2 + 2as

32. Example 2 Jack steps into the road, 30 metres from where Mr Rayner’s engine stops working. Mr Rayner does not see Jack. Will the car stop in time to miss hitting Jack? u = 10 ms -1 v = 0 ms -1 a = -2 ms -2 t = 5 s s = ? m v 2 = u 2 + 2as 0 2 = 10 2 + 2x-2s 0 = 100 -4s 4s = 100 s = 25m, the car does not hit Jack. 

33. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1.

34. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. When is the velocity of the ball 12 m.s -1 ?

35. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. When is the velocity of the ball 12 m.s -1 ? u = 24 m.s -1 a = -9.8 m.s -2 v = 12 m.s -1 t = ?

36. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. When is the velocity of the ball 12 m.s -1 ? u = 24 m.s -1 a = -9.8 m.s -2 v = 12 m.s -1 t = ? v = u + at

37. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. When is the velocity of the ball 12 m.s -1 ? u = 24 m.s -1 a = -9.8 m.s -2 v = 12 m.s -1 v = u + at 12 = 24 + -9.8t -12 = -9.8t t = 12/9.8 = 1.2 seconds

38. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. When is the velocity of the ball -12 m.s -1 ?

39. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. When is the velocity of the ball -12 m.s -1 ? u = 24 m.s -1 a = -9.8 m.s -2 v = -12 m.s -1 t = ?

40. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. When is the velocity of the ball -12 m.s -1 ? u = 24 m.s -1 a = -9.8 m.s -2 v = -12 m.s -1 v = u + at

41. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. When is the velocity of the ball -12 m.s -1 ? u = 24 m.s -1 a = -9.8 m.s -2 v = -12 m.s -1 v = u + at -12 = 24 + -9.8t -36 = -9.8t t = 36/9.8 = 3.7 seconds

42. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the displacement of the ball at those times? (t = 1.2, 3.7)

43. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the displacement of the ball at those times? (t = 1.2, 3.7) t = 1.2, v = 12, a = -9.8, u = 24 s = ?

44. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the displacement of the ball at those times? (t = 1.2, 3.7) t = 1.2, v = 12, a = -9.8, u = 24 s = ? s = ut + ½at 2 = 24x1.2 + ½x-9.8x1.2 2 s = 28.8 – 7.056 = 21.7 m

45. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the displacement of the ball at those times? (t = 1.2, 3.7) t = 3.7, v = 12, a = -9.8, u = 24 s = ? s = ut + ½at 2 = 24x3.7 + ½x-9.8x3.7 2 s = 88.8 – 67.081 = 21.7 m (the same?!)

46. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the velocity of the ball 1.50 s after launch?

47. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the velocity of the ball 1.50 s after launch? u = 24, t = 1.50, a = -9.8, v = ?

48. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the velocity of the ball 1.50 s after launch? u = 24, t = 1.50, a = -9.8, v = ? v = u + at

49. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the velocity of the ball 1.50 s after launch? u = 24, t = 1.50, a = -9.8, v = ? v = u + at v = 24 + -9.8x1.50 = 9.3 m.s -1

50. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the maximum height reached by the ball?

51. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the maximum height reached by the ball? u = 24, a = -9.8, v = 0, s = ?

52. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the maximum height reached by the ball? u = 24, a = -9.8, v = 0, s = ? v 2 = u 2 + 2as 0 = 242 + 2x-9.8xs 0 = 242 -19.6s

53. Example 3 A ball is thrown upwards with a velocity of 24 m.s -1. What is the maximum height reached by the ball? u = 24, a = -9.8, v = 0, s = ? 0 = 242 -19.6s 19.6s = 242 s = 242/19.6 = 12.3 m

54. III. Motion Graphs for Acceleration due to Gravity

56. Acceleration (m/s 2 ) t (s) -9.81 m/s 2

57. Let’s do a shed load of questions Questions 1, 5, 6, 7, 8, 14, 16, 21, 26, 36