1. Copyright © 2015 Pearson Education Inc Modified SH 8/15. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Chapter 2 Motion Along a Straight Line

2. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Goals for Chapter 2 Describe straight-line motion in terms of velocity and acceleration Distinguish between average and instantaneous velocity and acceleration Interpret graphs –position versus time, x(t) versus t (slope = velocity!)

3. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Goals for Chapter 2 Describe straight-line motion in terms of velocity and acceleration Distinguish between average and instantaneous velocity and acceleration Interpret graphs –velocity versus time, v(t) versus t Slope = acceleration!

4. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Goals for Chapter 2 Understand straight-line motion with constant acceleration Examine freely falling bodies Analyze straight-line motion when the acceleration is not constant

5. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Introduction Kinematics is the study of motion. Displacement, velocity and acceleration are important physical quantities. A bungee jumper speeds up during the first part of his fall and then slows to a halt.bungee jumper

6. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Displacement vs. Distance Displacement (blue line) = how far the object is from its starting point, regardless of path Distance traveled (dashed line) is measured along the actual path.

7. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Displacement vs. Distance Q: You make a round trip to the store 1 mile away. What distance do you travel? What is your displacement?

8. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Displacement vs. Distance Q: You walk 70 meters across the campus, hear a friend call from behind, and walk 30 meters back the way you came to meet her. What distance do you travel? What is your displacement?

9. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Displacement is written: SIGN matters! Direction matters! It is a VECTOR!! <= Positive displacement Displacement is negative => Displacement vs. Distance

10. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Speed is how far an object travels in a given time interval (in any direction) Speed vs. Velocity Ex: Go 10 miles to Chabot in 30 minutes Average speed = 10 mi / 0.5 hr = 20 mph

11. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Velocity includes directional information: Speed vs. Velocity Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes Average velocity = 20 mi / 0.333 hr = 60 mph NORTH VECTOR!

12. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Speed vs. Velocity Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes Average velocity = 20 mi / 0.333 hr = 60 mph NORTH

13. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Velocity includes directional information: Speed vs. Velocity Ex: Go 10 miles on 880 Northbound to Chabot in 30 minutes Average velocity = 10 mi / 0.5 hr = 20 mph NORTH VECTOR!

14. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Speed is a SCALAR 60 miles/hour, 88 ft/sec, 27 meters/sec Velocity is a VECTOR 60 mph North 88 ft/sec East 27 m/s @ azimuth of 173 degrees Speed vs. Velocity

15. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Position of runner as a function of time is plotted as moving along the x axis of a coordinate system. During a 3.00-s time interval, a runner’s position changes from x 1 = 50.0 m to x 2 = 30.5 m What was the runner’s average velocity? Example of Average Velocity

16. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 During a 3.00-sec interval, runner’s position changes from x 1 = 50.0 m to x 2 = 30.5 m What was the runner’s average velocity? V avg = (30.5 - 50.0) meters/3.00 sec = -6.5 m/s in the x direction. The answer must have value 1, units 2, & DIRECTION 3 Example of Average Velocity Note!  x = FINAL – INITIAL position

17. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 During a 3.00-s time interval, the runner’s position changes from x 1 = 50.0 m to x 2 = 30.5 m. What was the runner’s average speed? S avg = |30.5-50.0| meters/3.00 sec = 6.5 m/s The answer must have value & units Example of Average SPEED

18. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Negative velocity??? Average x-velocity is negative during a time interval if particle moves in negative x-direction for that time interval.

19. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Displacement, time, and average velocity A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds?

20. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Displacement, time, and average velocity—Figure 2.1 QA racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds? Solution Method: 1.What do you know? What do you need to find? What are the units? What might be a reasonable estimate? 2.DRAW it! Visualize what is happening. Create a coordinate system, label the drawing with everything. 3.Find what you need from what you know

21. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Displacement, time, and average velocity—Figure 2.1 A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds? “starts from rest” = initial velocity = 0 car moves along straight (say along an x-axis) –has coordinate x = 0 at t=0 seconds –has coordinate x=+19 meters at t =1 second –Has coordinate x=+277 meters at t = 1+3 = 4 seconds.

22. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Displacement, time, and average velocity—Figure 2.1 QA racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds?

23. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A position-time graph—Figure 2.3 A position-time graph (an “x-t” graph) shows the particle’s position x as a function of time t. Average x-velocity is related to the slope of an x-t graph.

24. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Instantaneous speed is the average speed in the limit as the time interval becomes infinitesimally short. Ideally, a speedometer would measure instantaneous speed; in fact, it measures average speed, but over a very short time interval. Note: It doesn’t measure direction! Instantaneous Speed

25. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Instantaneous velocity is the average velocity in the limit as the time interval becomes infinitesimally short. Velocity is a vector; you must include direction! V = 27 m/s west… Instantaneous Speed

26. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Instantaneous velocity The instantaneous velocity is the velocity at a specific instant of time or specific point along the path and is given by v x = dx/dt.

27. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Instantaneous velocity The instantaneous velocity is the velocity at a specific instant of time or specific point along the path and is given by v x = dx/dt. The average speed is not the magnitude of the average velocity!

28. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A jet engine moves along an experimental track (the x axis) as shown. Its position as a function of time is given by the equation x = At 2 + B where A = 2.10 m/s 2 and B = 2.80 m. Instantaneous Velocity Example

29. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A jet engine’s position as a function of time is x = At 2 + B, where A = 2.10 m/s 2 and B = 2.80 m. (a)Determine the displacement of the engine during the time interval from t 1 = 3.00 s to t 2 = 5.00 s. (b)Determine the average velocity during this time interval. (c)Determine the magnitude of the instantaneous velocity at t = 5.00 s. Instantaneous Velocity Example

30. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A jet engine’s position as a function of time is x = At 2 + B, where A = 2.10 m/s 2 and B = 2.80 m. (a)Determine the displacement of the engine during the time interval from t 1 = 3.00 s to t 2 = 5.00 s. @ t = 3.00 s x 1 = 21.7 @ t = 5.00 s x 2 = 55.3 x 2 – x 1 = 33.6 meters in +x direction Instantaneous Velocity Example

31. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A jet engine’s position as a function of time is x = At 2 + B, where A = 2.10 m/s 2 and B = 2.80 m. b) Determine the average velocity during this time interval. Vavg = 33.6 m/ 2.00 sec = 16.8 m/s in the + x direction Instantaneous Velocity Example

32. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A jet engine’s position as a function of time is x = At 2 + B, where A = 2.10 m/s 2 and B = 2.80 m. (c) Determine the magnitude of the instantaneous velocity at t = 5.00 s |v| = dx/dt @ t = 5.00 seconds = 21.0 m/s Instantaneous Velocity Example

33. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Acceleration = the rate of change of velocity. Units: meters/sec/sec or m/s^2 or m/s 2 or ft/s 2 Since velocity is a vector, acceleration is ALSO a vector, so direction is crucial… A = 2.10 m/s 2 in the +x direction Acceleration

34. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A car accelerates along a straight road from rest to 90 km/h in 5.0 s. What is the magnitude of its average acceleration? KEY WORDS: “straight road” = assume constant acceleration “from rest” = starts at 0 km/h Acceleration Example

35. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A car accelerates along a straight road from rest to 90 km/h in 5.0 s What is the magnitude of its average acceleration? |a| = (90 km/hr – 0 km/hr)/5.0 sec = 18 km/h/sec along road better – convert to more reasonable units 90 km/hr = 90 x 10 3 m/hr x 1hr/3600 s = 25 m/s So |a| = 5.0 m/s 2(note – magnitude only is requested) Acceleration Example

36. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Acceleration Acceleration = the rate of change of velocity.

37. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 (a) If the velocity of an object is zero, does it mean that the acceleration is zero? (b) If the acceleration is zero, does it mean that the velocity is zero? Think of some examples. Acceleration vs. Velocity?

38. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 An automobile is moving to the right along a straight highway. Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v 1 = 15.0 m/s, and it takes 5.0 s to slow down to v 2 = 5.0 m/s, what was the car’s average acceleration? Acceleration Example

39. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 An automobile is moving to the right along a straight highway. Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v 1 = 15.0 m/s, and it takes 5.0 s to slow down to v 2 = 5.0 m/s, what was the car’s average acceleration? Acceleration Example

40. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A semantic difference between negative acceleration and deceleration: “Negative” acceleration is acceleration in the negative direction (defined by coordinate system). “Deceleration” occurs when the acceleration is opposite in direction to the velocity. Acceleration Example

41. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Finding velocity on an x-t graph At any point on an x-t graph, the instantaneous x- velocity is equal to the slope of the tangent to the curve at that point.

42. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Motion diagrams A motion diagram shows position of a particle at various instants, and arrows represent its velocity at each instant.

43. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Average acceleration Acceleration describes the rate of change of velocity with time. The average x-acceleration is a av-x =  v x /  t.

44. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Instantaneous acceleration The instantaneous acceleration is a x = dv x /dt. Consider an accelerating racing car:

45. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Average Acceleration QA racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average acceleration in the first second? What was its average acceleration in the first 4 seconds?

46. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Finding acceleration on a v x -t graph Use Velocity vs. Time (  x - t) graph to find instantaneous acceleration & average acceleration.

47. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A v x -t graph and a motion diagram Figure 2.13 shows the v x -t graph and the motion diagram for a particle.

48. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 An x-t graph and a motion diagram Figure 2.14 shows the x-t graph and the motion diagram for a particle.

49. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Motion with constant acceleration—Figures 2.15 and 2.17 For a particle with constant acceleration, the velocity changes at the same rate throughout the motion.

50. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Acceleration given x(t) A particle is moving in a straight line with its position is given by x = (2.10 m/s 2 ) t 2 + (2.80 m). Calculate (a) its average acceleration during the interval from t 1 = 3.00 s to t 2 = 5.00 s, & (b) its instantaneous acceleration as a function of time.

51. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A particle is moving in a straight line with its position is given by x = (2.10 m/s 2 ) t 2 + (2.80 m). Calculate (a) its average acceleration during the interval from t 1 = 3.00 s to t 2 = 5.00 s V = dx/dt = (4.2 m/s) t V 1 = 12.6 m/s V 2 = 21 m/s  v/  t = 8.4 m/s/2.0 s = 4.2 m/s 2 Acceleration given x(t)

52. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A particle is moving in a straight line with its position is given by x = (2.10 m/s 2 ) t 2 + (2.80 m). Calculate (b) its instantaneous acceleration as a function of time. Acceleration given x(t)

53. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Graph shows Velocity as a function of time for two cars accelerating from 0 to 100 km/h in a time of 10.0 s Compare (a) the average acceleration; (b) instantaneous acceleration; and (c) total distance traveled for the two cars. Analyzing acceleration

54. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Velocity as a function of time for two cars accelerating from 0 to 100 km/h in a time of 10.0 s Compare (a) the average acceleration; (b) instantaneous acceleration; and (c) total distance traveled for the two cars. Same final speed in time => Same average acceleration But Car A accelerates faster… Analyzing acceleration

55. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Constant Acceleration Equations FIVE key variables:  x displacement v initial, v final, a cceleration t ime FIVE key equations:   x = ½ (v i +v f )t   x = v i t + ½ at 2   x = v f t – ½ at 2  v f = v i + at  v f 2 = v i 2 + 2a  x

56. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 The equations of motion with constant acceleration Initial velocity, final velocity, acceleration, time Displacement (x – x0), initial velocity, time, acceleration Initial velocity, final velocity, acceleration, displacement Displacement, initial velocity, final velocity, time Displacement (x – x0), final velocity, time, acceleration Equation of MotionVariables Present x = x 0 + v x t – 1/2a x t 2

57. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 The equations of motion with constant acceleration Initial velocity, final velocity, acceleration, time Displacement (x – x0), initial velocity, time, acceleration Initial velocity, final velocity, acceleration, displacement Displacement, initial velocity, final velocity, time Equation of MotionFind 3 of 4, solve for 4 th ! Initial velocity, final velocity, acceleration, time Displacement (x – x0), initial velocity, time, acceleration Initial velocity, final velocity, acceleration, displacement Displacement, initial velocity, final velocity, time Displacement (x – x0), final velocity, time, acceleration x = x 0 + v x t – 1/2a x t 2

58. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 A motorcycle with constant acceleration What is position and velocity at t = 2.0 sec? Where is he when speed = 25 m/s?

59. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Two bodies with different accelerations (Ex. 2.5) A motorist traveling at a constant 15m/s passes school crossing where speed limit is 20 mph. Just as motorist passes the sign a police officer stopped on a motorcycle starts off in pursuit with acceleration 3.0 m/s 2.

60. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Two bodies with different accelerations (Ex. 2.5) WHEN does officer catch up to the car? How FAST is officer going at that time? How FAR has each travelled?

61. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Two bodies with different accelerations (Ex. 2.5) Two different initial/final velocities and accelerations Two different equations of position in time x(t) –Xm(t) for the motorist gives his position as f(time) –Xp(t) for the police officer gives *his* position But…time & displacement are linked!

62. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Two bodies with different accelerations (Ex. 2.5) Two different initial/final velocities and accelerations TIME and displacement are linked!

63. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Freely falling bodies Free fall is the motion of an object under the influence of only gravity. In the figure, a strobe light flashes with equal time intervals between flashes. The velocity change is the same in each time interval, so the acceleration is constant.

64. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Freely Falling Objects In the absence of air resistance, all objects fall with the same acceleration, although this may be tricky to tell by testing in an environment where there is air resistance.

65. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 The acceleration due to gravity at the Earth’s surface is approximately 9.80 m/s 2. At a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration. Freely Falling Objects

66. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Example: Falling from a tower. Suppose that a ball is dropped ( v 0 = 0) from a tower 70.0 m high. How far will it have fallen after a time t 1 = 1.00 s, t 2 = 2.00 s, and t 3 = 3.00 s? Ignore air resistance. Freely Falling Objects

67. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Example: Thrown down from a tower Suppose a ball is thrown downward with an initial velocity of 3.00 m/s, instead of being dropped. (a) What then would be its position after 1.00 s and 2.00 s? (b) What is its speed after 1.00 s and 2.00 s? Compare with the speeds of a dropped ball.

68. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Example: Ball thrown up! A person throws a ball upward into the air with an initial velocity of 15.0 m/s Calculate (a) how high it goes, & (b) how long the ball is in the air before it comes back to the hand. Ignore air resistance. Freely Falling Objects

69. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Up-and-down motion in free fall An object is in free fall even when it is moving upward.

70. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Is the acceleration zero at the highest point?—Figure 2.25 The vertical velocity, but not the acceleration, is zero at the highest point.

71. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Ball thrown upward (cont.) Consider again a ball thrown upward, & calculate (a) how much time it takes for the ball to reach the maximum height, (b) the velocity of the ball when it returns to the thrower’s hand (point C).

72. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Give examples to show the error in these two common misconceptions: (1) that acceleration and velocity are always in the same direction (2) that an object thrown upward has zero acceleration at the highest point. Freely Falling Objects

73. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 The quadratic formula For a ball thrown upward at an initial speed of 15.0 m/s, calculate at what time t the ball passes a point 8.00 m above the person’s hand.

74. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Ball thrown at edge of cliff A ball is thrown upward at a speed of 15.0 m/s by a person on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below. (a) Ignoring air resistance, how long does it take the ball to reach the base of the cliff? (b) What is the total distance traveled by the ball?

75. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Ball thrown at edge of cliff A ball is thrown upward at a speed of 15.0 m/s by a person on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below. Step 1: DRAW IT! Establish your coordinate system!

76. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Ball thrown at edge of cliff A ball is thrown upward at a speed of 15.0 m/s by a person on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below. Step 2: What do you know? V 0 = +15 m/s a = - 9.8 m/s/s  y = - 50 m

77. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Ball thrown at edge of cliff A ball is thrown upward at a speed of 15.0 m/s by a person on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below. Step 3: What do you need to find? Units? “How long does it take the ball to reach the base of the cliff?” TIME!

78. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Ball thrown at edge of cliff A ball is thrown upward at a speed of 15.0 m/s by a person on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below. Step 4: What equation relates what you need to what you know?  y = v 0 * t + ½ at 2 -50 = +15t - 4.9 t 2 Solve the quadratic to find… t = 5.07 sec OR – 2.01 sec ???

79. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Variable Acceleration; Integral Calculus Deriving the kinematic equations through integration: For constant acceleration,

80. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Variable Acceleration; Integral Calculus Then: For constant acceleration,

81. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Variable Acceleration; Integral Calculus Example: Integrating a time-varying acceleration. An experimental vehicle starts from rest ( v 0 = 0) at t = 0 and accelerates at a rate given by a = (7.00 m/s 3 ) t. What is (a) its velocity and (b) its displacement 2.00 s later?

82. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Graphical Analysis and Numerical Integration The total displacement of an object can be described as the area under the v-t curve:

83. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Graphical Analysis and Numerical Integration Similarly, the velocity may be written as the area under the a-t curve. However, if the velocity or acceleration is not integrable, or is known only graphically, numerical integration may be used instead.

84. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Example: Numerical integration An object starts from rest at t = 0 and accelerates at a rate a(t) = (8.00 m/s 4 ) t 2. Determine its velocity after 2.00 s using numerical methods.

85. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Velocity and position by integration The acceleration of a car is not always constant. The motion may be integrated over many small time intervals to give

86. Copyright © 2015 Pearson Education Inc. Modified SH 8/15 Motion with changing acceleration (Ex. 2.9) Look at the example where acceleration DOES vary in time. a x = 2.0 m/s 2 – (0.10 m/s 3 )t You cannot use the set equations - you must INTEGRATE