2. Kinematics in One Dimension

3. The Cheetah: A cat that is built for speed. Its strength and agility allow it to sustain a top speed of over 100 km/h. Such speeds can only be maintained for about ten seconds.

4. Objectives: After completing this lecture, you should be able to: Define and apply concepts of average and instantaneous velocity and acceleration.Define and apply concepts of average and instantaneous velocity and acceleration. Solve problems involving initial and final velocity, acceleration, displacement, and time.Solve problems involving initial and final velocity, acceleration, displacement, and time. Demonstrate your understanding of directions and signs for velocity, displacement, and acceleration.Demonstrate your understanding of directions and signs for velocity, displacement, and acceleration. Solve problems involving a free-falling body in a gravitational field.Solve problems involving a free-falling body in a gravitational field.

5. 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 d = 20 m Distance d is a scalar quantity (no direction): Contains magnitude only and consists of a number and a unit.

6. 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. A B Δs = 12 m, 20 o 

7. 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 Δx is from the origin to the final position: What is the distance traveled? d = 20 m !! 12 m,W Δx Δx = 4 m, W x 8 m,E x = +8x = -4

8. 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!

9. 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 s = = dtdt 20 m 4 s v s = 5 m/s Not direction dependent! A B d = 20 m Time t = 4 s

10. 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.) Direction required! A B s = 20 m Time t = 4 s Δx=12 m 20 o = 3 m/s at 20 0 N of E

11. Average Speed and Instantaneous Velocity The instantaneous velocity is the magn- itude and direction of the velocity 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

12. 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 m/s Direction does not matter! start s 1 = 200 m s 2 = 300 m

13. 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 = -100 m x o = 0 m; x = -100 m Direction of final displacement is to the left as shown. Average velocity: Note: Average velocity is directed to the west.

14. 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? 625 m 356 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:

15. 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

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

17. 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. - + - + +

18. Definition of Acceleration  An acceleration is the change in velocity per unit of time. (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.

19. 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

20. 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 = +8 m/s v 0 = +2 m/s t = 3 s Force

21. 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 (+)

22. 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 o = +8 m/s t = 4 s v = +20 m/s Force

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

24. 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 = -5 m/s Step 3. Label given info with + and - signs. E

25. Example 4 (Cont.): Wagon moving east at 20 m/s encounters a head-wind, causing it to change directions. Five seconds later, it is traveling west at 5 m/s. What is the average acceleration? Choose the eastward direction as positive. Initial velocity, v o = +20 m/s, east (+) Final velocity, v f = -5 m/s, west (-) The change in velocity,  v = v f - v 0  v = (-5 m/s) - (+20 m/s) = -25 m/s Choose the eastward direction as positive. Initial velocity, v o = +20 m/s, east (+) Final velocity, v f = -5 m/s, west (-) The change in velocity,  v = v f - v 0  v = (-5 m/s) - (+20 m/s) = -25 m/s

26. Example 4: (Continued) = = = = -25 m/s 5 s 5 s = - 5 m/s 2 Acceleration is directed to left, west (same as F). + Force v o = +20 m/s v = -5 m/s E  v = (-5 m/s) - (+20 m/s) = -25 m/s

27. Signs for Displacement Time t = 0 at point A. What are the signs (+ or -) of displacement at B, C, and D? At B, x is positive, right of origin At C, x is positive, right of origin At D, x is negative, left of origin + Force v o = +20 m/s v f = -5 m/s E a = - 5 m/s 2 A B C D

28. Signs for Velocity What are the signs (+ or -) of velocity at points B, C, and D?  At B, v is zero - no sign needed.  At C, v is positive on way out and negative on the way back.  At D, v is negative, moving to left. + Force v o = +20 m/s v f = -5 m/s E a = - 5 m/s 2 A B C D x = 0

29. What are the signs (+ or -) of acceleration at points B, C, and D?  The force is constant and always directed to left, so acceleration does not change.  At B, C, and D, a = -5 m/s, negative at all points. Signs for Acceleration + Force v o = +20 m/s v f = -5 m/s E a = - 5 m/s 2 A B C D

30. Definitions Average velocity: Average acceleration:

31. Graphical Analysis slope: velocity: acceleration:

32. x, (m) Position vs time graph (velocity)

33. v, (m/s) velocity vs time graph (acceleration)

34. Graphical Analysis xx tttt x2x2x2x2 x1x1x1x1 t2t2t2t2 t1t1t1t1 xx tt Time slope Displacement, x Average Velocity: Instantaneous Velocity:

35. Uniform Acceleration in One Dimension: Motion is along a straight line (horizontal, vertical or slanted).Motion is along a straight line (horizontal, vertical or slanted). Changes in motion result from a CONSTANT force producing uniform acceleration.Changes in motion result from a CONSTANT force producing uniform acceleration. The velocity of an object is changing by a constant amount in a given time interval.The velocity of an object is changing by a constant amount in a given time interval. The moving object is treated as though it were a point particle.The moving object is treated as though it were a point particle.

36. Average velocity for constant a: setting t o = 0 combining both equations: For constant acceleration:

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

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

39. Example: (Cont.) + Step 3. Step 3. List given; find information with signs. Given: v o = 400 ft/s- initial velocity of airplane v = 0 - final velocity after traveling Δx = +300 ft v = 0 - final velocity after traveling Δx = +300 ft Find: a = ?- acceleration of airplane Δx = 300 ft v o = 400 ft/s v = 0

40. Step 4. a t Step 4. Select equation that contains a and not t. v 2 - v o 2 = 2aΔx 0 a = = -v o 2 2x -(400 ft/s) 2 2(300 ft) a = - 300 ft/s 2 a = - 300 ft/s 2 Why is the acceleration negative? Because Force is in a negative direction which means that the airplane slows down Given: v o = +400 ft/s v = 0 v = 0 Δx = +300 ft

41. Example 5: A ball 5.0 m from the bottom of an incline is traveling initially at 8.0 m/s. Four seconds (4.0 s) later, it is traveling down the incline at 2.0 m/s. How far is it from the bottom at that instant? 5.0 m ΔxΔx 8.0 m/s -2.0 m/s t = 4.0 s + Given: d = 5.0 m Given: d = 5.0 m- distance from initial position of the ball v o = 8.0 m/s velocity v o = 8.0 m/s - initial velocity v = -2.0 m/s - final velocity after t = 4.0 s Find:x = ?- distance from the bottom of the incline

42. x = 17.0 m Given: d = 5.0 m v o = 8.0 m/s velocity v o = 8.0 m/s - initial velocity v = -2.0 m/s - final velocity after t = 4.0 s Find:x = ?- distance from the bottom of the incline Solution: where

43. 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!

44. 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

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

47. The Signs of Displacement Displacement is positive (+) or negative (-) based on LOCATION.Displacement is positive (+) or negative (-) based on LOCATION. The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION. 2 m -1 m -2 m

48. Acceleration Produced by Force Acceleration is (+) or (-) based on direction of force (NOT based on v).Acceleration is (+) or (-) based on direction of force (NOT based on v). A push or pull (force) is necessary to change velocity, thus the sign of a is same as sign of F. F a (-) a (-) F a (+) More will be said later on the relationship between F and a.

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

50. Acceleration Due to Gravity Every object on the earth experiences a common force: the force due to gravity.Every object on the earth experiences a common force: the force due to gravity. This force is always directed toward the center of the earth (downward).This force is always directed toward the center of the earth (downward). The acceleration due to gravity is relatively constant near the Earth’s surface.The acceleration due to gravity is relatively constant near the Earth’s surface. Every object on the earth experiences a common force: the force due to gravity.Every object on the earth experiences a common force: the force due to gravity. This force is always directed toward the center of the earth (downward).This force is always directed toward the center of the earth (downward). The acceleration due to gravity is relatively constant near the Earth’s surface.The acceleration due to gravity is relatively constant near the Earth’s surface. Earth Wg

51. Gravitational Acceleration In a vacuum, all objects fall with same acceleration.In a vacuum, all objects fall with same acceleration. Equations for constant acceleration apply as usual.Equations for constant acceleration apply as usual. Near the Earth’s surface:Near the Earth’s surface: In a vacuum, all objects fall with same acceleration.In a vacuum, all objects fall with same acceleration. Equations for constant acceleration apply as usual.Equations for constant acceleration apply as usual. Near the Earth’s surface:Near the Earth’s surface: a = g = -9.80 m/s 2 or -32 ft/s 2 Directed downward (usually negative). Directed downward (usually negative).

52. Experimental Determination of Gravitational Acceleration. The apparatus consists of a device which measures the time required for a ball to fall a given distance. Suppose the height is 1.20 m and the drop time is recorded as 0.650 s. What is the acceleration due to gravity? y tttt

53. Experimental Determination of Gravity (y 0 = 0; y = -1.20 m) y tttt y = -1.20 m; t = 0.495 s Acceleration of Gravity: Acceleration a is negative because force W is negative. + W

54. Sign Convention: A Ball Thrown Vertically Upward Velocity is positive (+) or negative (-) based on direction of motion.Velocity is positive (+) or negative (-) based on direction of motion. Displacement is positive (+) or negative (-) based on LOCATION. Displacement is positive (+) or negative (-) based on LOCATION. Release Point UP = + Acceleration is (+) or (-) based on direction of force (weight).Acceleration is (+) or (-) based on direction of force (weight). y = 0 y = + y = 0 Negative y = - Negative v = + v = 0 v = - Negative v= - Negative a = -

55. Same Problem Solving Strategy Except a = g:  Draw and label sketch of problem.  Indicate + direction and force direction.  List givens and state what is to be found. Given: ____, _____, a = - 9.8 m/s 2 Find: ____, _____  Select equation containing one and not the other of the unknown quantities, and solve for the unknown.

56. Example 7: A ball is thrown vertically upward with an initial velocity of 30.0 m/s. What are its position and velocity after 2.00 s, 4.00 s, and 7.00 s? Find also the maximum height attained a = g + v o = +30.0 m/s Given: a = -Δ9.8 m/s 2 v o = 30.0 m/s v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s t = 2.00 s; 4.00 s; 7.00 s Find: Δy = ? – displacement Δy = ? – displacement v = ? - final velocity v = ? - final velocity After those three “times” Δy = ? – maximum height Δy = ? – maximum height

57. Given: a = -9.8 m/s 2 ; v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s t = 2.00 s; 4.00 s; 7.00 sSolutions: For t = 2.00 s: For t = 4.00 s: For t = 7.00 s:

58. Given: a = -9.8 m/s 2 ; v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s t = 2.00 s; 4.00 s; 7.00 sSolutions: For t = 2.00 s: For t = 4.00 s: For t = 7.00 s:

59. Given: a = -9.8 m/s 2 ; v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s t = 2.00 s; 4.00 s; 7.00 sSolutions: For maximum height, v = 0 (the ball stops at maximum height):

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

61. Experiment 10 Uniformly Accelerated Motion (Acceleration due to Gravity) 39 (06A)

62. Summary of Formulas DerivedFormulas Derived Formulas: For Constant Acceleration Only

63. Summary: Procedure  Draw and label sketch of problem.  Indicate + direction and force direction.  List givens and state what is to be found. Given: ____, _____, ______ Find: ____, _____  Select equation containing one and not the other of the unknown quantities, and solve for the unknown.