2. Chapter 2: One Dimension Motion 2.1 – 2.4, 2.5, 2.6 Position Distance and Displacement Speed and Velocity Acceleration Motion equations Problem-Solving 2.1 – 2.4, 2.5, 2.6 Position Distance and Displacement Speed and Velocity Acceleration Motion equations Problem-Solving

3. Objectives: Student is able 1.to express the motion od an object using narrative, mathematical, and graphical representations. 1.3.A.1.1: to express the motion od an object using narrative, mathematical, and graphical representations. 2.To analyze experimental data describing the motion of an object and to express the result using above representation. 2.3.A.1.3: To analyze experimental data describing the motion of an object and to express the result using above representation.

4. Learning Objective Define position, displacement, distance in a particular frame reference.Define position, displacement, distance in a particular frame reference. Distinguish between displacement and distanceDistinguish between displacement and distance Distinguish between speed and velocityDistinguish between speed and velocity Define acceleration and uniform or non-uniform motion.Define acceleration and uniform or non-uniform motion. Solve problems involving initial and final velocity, acceleration, displacement, and time.Solve problems involving initial and final velocity, acceleration, displacement, and time.

5. How to present a motion How to present a motion Motion can be described by Words Words Diagram, a graph, a picture Diagram, a graph, a picture equations equations

6. One dimension motion: Motion is along a straight line (horizontal, vertical or slanted).Motion is along a straight line (horizontal, vertical or slanted). The moving object is treated as though it were a point particle.The moving object is treated as though it were a point particle. Particle model – representing objectParticle model – representing object For Example: long distance runner, an airplane, and throwing a ball, etc

7. Picturing a Motion You are free to choose the origin and positive direction as you like, but once your choice is made, stick with it.

8. Picturing Motion The locations of your house, your friend’s house, and the grocery store in terms of a one- dimensional coordinate system.

9. Figure 2-3 One-dimensional motion along the x axis

11. What is a position? Location of an object in a particular time.Location of an object in a particular time. A teacher paces left and right while lecturing. Her position relative to Earth is given by x. The +2.0 m displacement of the teacher relative to Earth is represented by an arrow pointing to the right.

12. Distance and Displacement Distance is the length of the actual path taken by an object. Consider travel from point A to point B in diagram below: A B s = 20 m Distance s is a scalar quantity (no direction): Contains magnitude only and consists of a number and a unit. (20 m, 40 mi/h, 10 gal)

13. Distance and Displacement Displacement is the straight-line separation of two points in a specified direction. A vector quantity: Contains magnitude AND direction, a number, unit & angle. (12 m, 30 0 ; 8 km/h, N) A B D = 12 m, 20 o 

14. Distance and Displacement For motion along x or y axis, the displacement is determined by the x or y coordinate of its final position. Example: Consider a car that travels 8 m, E then 12 m, W.For motion along x or y axis, the displacement is determined by the x or y coordinate of its final position. Example: Consider a car that travels 8 m, E then 12 m, W. Net displacement D is from the origin to the final position: What is the distance traveled? 20 m !! 12 m,W D D = 4 m, W x 8 m,E x = +8x = -4

15. What is a displacement equation?

16. The Signs of Displacement Displacement is positive (+) or negative (-) based on LOCATION.Displacement is positive (+) or negative (-) based on LOCATION. 2 m -1 m -2 m The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION. Examples: The direction of motion does not matter!

17. It is usually convenient to consider motion upward or to the right as positive ( + ) and motion downward or to the left as negative ( − ).

18. Speed and velocity The motion of these racing snails can be described by their speeds and their velocities.

19. Definition of Speed Speed is the distance traveled per unit of time (a scalar quantity).Speed is the distance traveled per unit of time (a scalar quantity). v = = stst 20 m 4 s v = 5 m/s Not direction dependent! A B s = 20 m Time t = 4 s

20. Definition of Velocity Velocity is the displacement per unit of time. (A vector quantity.)Velocity is the displacement per unit of time. (A vector quantity.) v = 3 m/s at 20 0 N of E Direction required! A B s = 20 m Time t = 4 s D=12 m 20 o

21. Example 1. A runner runs 200 m, east, then changes direction and runs 300 m, west. If the entire trip takes 60 s, what is the average speed and what is the average velocity? Recall that average speed is a function only of total distance and total time: Total distance: s = 200 m + 300 m = 500 m Avg. speed 8.33 m/s Direction does not matter! start s 1 = 200 m s 2 = 300 m

22. Example 1 (Cont.) Now we find the average velocity, which is the net displacement divided by time. In this case, the direction matters. x o = 0 t = 60 s x 1 = +200 m x f = -100 m x 0 = 0 m; x f = -100 m Direction of final displacement is to the left as shown. Average velocity: Note: Average velocity is directed to the west.

23. Example 2. A sky diver jumps and falls for 600 m in 14 s. After chute opens, he falls another 400 m in 150 s. What is average speed for entire fall? 600 m 400 m 14 s 142 s A B Average speed is a function only of total distance traveled and the total time required. Total distance/ total time:

24. The Signs of Velocity First choose + direction; then v is positive if motion is with that direction, and negative if it is against that direction.  Velocity is positive (+) or negative (-) based on direction of motion. - + - + +

25. Examples of Speed Light = 3 x 10 8 m/s Orbit 2 x 10 4 m/s Jets = 300 m/s Car = 25 m/s

26. Speed Examples (Cont.) Runner = 10 m/s Snail = 0.001 m/s Glacier = 1 x 10 -5 m/s

27. Average Speed and Instantaneous Velocity The instantaneous velocity is the magn- itude and direction of the speed at a par- ticular instant. (v at point C)  The average speed depends ONLY on the distance traveled and the time required. A Bs = 20 m Time t = 4 s C

28. Average and Instantaneous v xx tttt x2x2x2x2 x1x1x1x1 t2t2t2t2 t1t1t1t1 xx tt Time slope Displacement, x Average Velocity: Instantaneous Velocity:

29. Quiz You and your dog go to for a walk to the park. On the way, your dog takes many side trips to chase squirrels or examine fire hydrants. When you arrive at the park, do you and your dog have the same displacement? 1.Yes 2.No Have you and your dog traveled the same distance?

30. Quiz (con) If the position of a car is zero, does its speed nave to be zero? 1.Yes 2.No 3.It depends on the position

31. Warm up problem: Average Round-Trip Speed A person travels from city A to city B with a speed of 40 mph and returns with a speed of 60 mph. What is his average round-trip speed? (A) 100 mph (B) 50 mph (C) 48 mph (D) 10 mph (E) None of theseA person travels from city A to city B with a speed of 40 mph and returns with a speed of 60 mph. What is his average round-trip speed? (A) 100 mph (B) 50 mph (C) 48 mph (D) 10 mph (E) None of these

32. Critical thinking question And The Winner Is... Two marbles roll along two horizontal tracks. One track has a dip, and the other has a bump of the same shape. Which marble wins?And The Winner Is... Two marbles roll along two horizontal tracks. One track has a dip, and the other has a bump of the same shape. Which marble wins?

33. Definition of Acceleration  An acceleration is the rate at which velocity changes (A vector quantity.)  A change in velocity requires the application of a push or pull (force). A formal treatment of force and acceleration will be given later. For now, you should know that: The direction of accel- eration is same as direction of force. The acceleration is proportional to the magnitude of the force.

34. Pulling the wagon with twice the force produces twice the acceleration and acceleration is in direction of force. Acceleration and Force F a 2F 2a

35. Why acceleration opposite direction in this situation? A subway train in Sao Paulo, Brazil, decelerates as it comes into a station. It is accelerating in a direction opposite to its direction of motion. (credit: Yusuke Kawasaki, Flickr)

36. Example of Acceleration The wind changes the speed of a boat from 2 m/s to 8 m/s in 3 s. Each second the speed changes by 2 m/s. Wind force is constant, thus acceleration is constant. + v f = +8 m/s v 0 = +2 m/s t = 3 s Force

37. Constant Acceleration Acceleration: Setting t o = 0 and solving for v, we have: Final velocity = initial velocity + change in velocity

38. The Signs of Acceleration Acceleration is positive (+) or negative (-) based on the direction of force.Acceleration is positive (+) or negative (-) based on the direction of force. Choose + direction first. Then acceleration a will have the same sign as that of the force F — regardless of the direction of velocity. F F + a (-) a (+)

39. Figure 2.14 a) a)This car is speeding up as it moves toward the right. It therefore has positive acceleration in our coordinate system. b. This car is slowing down as it moves toward the right. Therefore, it has negative acceleration in our coordinate system, because its acceleration is toward the left. The car is also decelerating: the direction of its acceleration is opposite to its direction of motion. Can you describe c and d?

40. Example 3 (No change in direction): A constant force changes the speed of a car from 8 m/s to 20 m/s in 4 s. What is average acceleration? Step 1. Draw a rough sketch. Step 2. Choose a positive direction (right). Step 3. Label given info with + and - signs. Step 4. Indicate direction of force F. + v 1 = +8 m/s t = 4 s v 2 = +20 m/s Force

41. Example 3 (Continued): What is average acceleration of car? Step 5. Recall definition of average acceleration. + v 1 = +8 m/s t = 4 s v 2 = +20 m/s Force

42. Average and Instantaneous a vv tt v2v2 v1v1 t2t2 t1t1 vv tt time slope

43. Example 4: A wagon moving east at 20 m/s encounters a very strong head-wind, causing it to change directions. After 5 s, it is traveling west at 5 m/s. What is the average acceleration? (Be careful of signs.) Step 1. Draw a rough sketch. + Force Step 2. Choose the eastward direction as positive. v o = +20 m/s v f = -5 m/s Step 3. Label given info with + and - signs. E

44. Quiz: Acceleration If the velocity of a car is non-zero, can the acceleration of the car be zero?If the velocity of a car is non-zero, can the acceleration of the car be zero? 1.Yes 2.No 3.Depends on the velocity

45. Review of Symbols and Units (x, x o ); meters (m)Displacement (x, x o ); meters (m) (v, v o ); meters per second (m/s)Velocity (v, v o ); meters per second (m/s) ( a ); meters per s 2 (m/s 2 )Acceleration ( a ); meters per s 2 (m/s 2 ) Time (t); seconds (s)Time (t); seconds (s) (x, x o ); meters (m)Displacement (x, x o ); meters (m) (v, v o ); meters per second (m/s)Velocity (v, v o ); meters per second (m/s) ( a ); meters per s 2 (m/s 2 )Acceleration ( a ); meters per s 2 (m/s 2 ) Time (t); seconds (s)Time (t); seconds (s) Review sign convention for each symbol

46. Velocity for constant a Average velocity: Setting t o = 0 and combining we have:

47. Formulas based on definitions: Derivedformulas Derived formulas: For constant acceleration only

48. Use of Initial Position x 0 in Problems. If you choose the origin of your x,y axes at the point of the initial position, you can set x 0 = 0, simplifying these equations. The x o term is very useful for studying problems involving motion of two bodies. 00 0 0

49. Example 5: A ball 5 m from the bottom of an incline is traveling initially at 8 m/s. Four seconds later, it is traveling down the incline at 2 m/s. How far is it from the bottom at that instant? x = x o + t v o + v f 2 = 5 m + (4 s) 8 m/s + (-2 m/s) 2 5 m x 8 m/s -2 m/s t = 4 s vovo vfvf + F Careful

50. x = 5 m + (4 s) 8 m/s - 2 m/s 2 x = 17 m x = 5 m + (4 s) 8 m/s + (-2 m/s) 2 5 m x 8 m/s -2 m/s t = 4 s vovo vfvf + F (Continued)

51. Acceleration in our Example a = -2.50 m/s 2 What is the meaning of negative sign for a ? 5 m x 8 m/s -2 m/s t = 4 s vovo v + F The force changing speed is down plane!

52. Example 6: A airplane flying initially at 400 ft/s lands on a carrier deck and stops in a distance of 300 ft. What is the acceleration? 300 ft +400 ft/s v o v = 0 + F Step 1. Draw and label sketch. F Step 2. Indicate + direction and F direction. X 0 = 0

53. Example: (Cont.) 300 ft +400 ft/s v o v = 0 + F Step 3. Step 3. List given; find information with signs. Given: v o = +400 ft/s v = 0 v = 0 x = +300 ft Find: a = ?; t = ? List t = ?, even though time was not asked for. X 0 = 0

54. Problem Solving Strategy:  Draw and label sketch of problem.  Indicate + direction  List givens and state what is to be found. Given: ____, _____, _____ (x,v,v o,t) Find: ____, _____  Select equation containing one and not the other of the unknown quantities, and solve for the unknown.