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Chabot College Physics 4A Spring 2015 Chapter 2 Motion Along a Straight Line!

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1 Chabot College Physics 4A Spring 2015 Chapter 2 Motion Along a Straight Line!

2 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!) time Position x(t)

3 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! time Velocity v(t)

4 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 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 Introduction A bungee jumper speeds up during the first part of his fall and then slows to a halt.bungee jumper Model his position in time with function x(t) Model his velocity in time v(t) Model his acceleration in time a(t)

7 2-1 Position, Displacement, and Average Velocity For this chapter, we restrict motion in three ways: 1.We consider motion along a straight line only 2.We discuss only the motion itself, not the forces that cause it 3.We consider the moving object to be a particle © 2014 John Wiley & Sons, Inc. All rights reserved.

8 2-1 Position, Displacement, and Average Velocity A particle is either: A point-like object (such as an electron) Or an object that moves such that each part travels in the same direction at the same rate (no rotation or stretching) © 2014 John Wiley & Sons, Inc. All rights reserved.

9 2-1 Position, Displacement, and Average Velocity Position is measured relative to a reference point: o The origin, or zero point, of an axis © 2014 John Wiley & Sons, Inc. All rights reserved. Origin x = 0 x at time t1

10 2-1 Position, Displacement, and Average Velocity Position is still measured relative to the same reference point: o The origin, or zero point, of an axis © 2014 John Wiley & Sons, Inc. All rights reserved. Origin x = 0 x at time t2

11 2-1 Position, Displacement, and Average Velocity Position has a sign: o Positive direction is in the direction of increasing numbers o Negative direction is opposite the positive © 2014 John Wiley & Sons, Inc. All rights reserved.

12 2-1 Position, Displacement, and Average Velocity A change in position is called displacement o ∆x is the change in x o Always (final position) – (initial position) © 2014 John Wiley & Sons, Inc. All rights reserved.

13 Displacement is written: SIGN matters! Direction matters! It is a VECTOR!! Vectors have THREE things… Magnitude Direction Units Displacement Displacement could be: 3 m North, or – 5 cm on the x axis, or +83 miles NW

14 Displacement is written: SIGN matters! Direction matters! It is a VECTOR!! <= Positive displacement Displacement is negative => Displacement

15 2-1 Position, Displacement, and Average Velocity © 2014 John Wiley & Sons, Inc. All rights reserved. Examples A particle moves... o From x = 5 m to x = 12 m? o ∆x = 7 m (positive direction)

16 2-1 Position, Displacement, and Average Velocity © 2014 John Wiley & Sons, Inc. All rights reserved. Examples A particle moves... o From x = 5 m to x = 12 m: ∆x = 7 m (positive direction) o From x = 5 m to x = 1 m? o ∆x = -4 m (negative direction)

17 2-1 Position, Displacement, and Average Velocity © 2014 John Wiley & Sons, Inc. All rights reserved. Examples A particle moves... o From x = 5 m to x = 12 m: ∆x = 7 m (positive direction) o From x = 5 m to x = 1 m: ∆x = -4 m (negative direction) o From x = 5 m to x = 200 m to x = 5 m: o ∆x = 0 m !! The actual distance covered is irrelevant

18 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.

19 Displacement vs. Distance Q: You make a round trip to the store 1 mile away. What distance do you travel? What is your displacement?

20 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?

21 2-1 Position, Displacement, and Average Velocity © 2014 John Wiley & Sons, Inc. All rights reserved. Answer: pairs (b) and (c) (b) -7 m – -3 m = -4 m (c) -3 m – 7 m = -10 m

22 Speed is the ratio of 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

23 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 Always think about your answer …. Is it reasonable?? VECTOR!

24 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 Not with traffic is 60 mph even possible!

25 2-1 Position, Displacement, and Average Velocity Average velocity is the ratio of: o A displacement, ∆x o To the time interval in which the displacement occurred, ∆t © 2014 John Wiley & Sons, Inc. All rights reserved. Average velocity has units of (distance) / (time) o Meters per second, m/s

26 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!

27 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

28 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

29 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

30 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 but it is a scalar! No direction needed Example of Average SPEED

31 Negative velocity??? Average x-velocity is negative during a time interval if particle moves in negative x- direction for that time interval.

32 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?

33 Displacement, time, and average velocity 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

34 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? “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.

35 Displacement, time, and average velocity 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?

36 A position-time graph x(t) vs t 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.

37 A position-time graph x(t) vs t

38 2-2 Instantaneous Velocity and Speed The graph shows the position and velocity of an elevator cab over time. The slope of x(t), and so also the velocity v, is zero from 0 to 1 s, and from 9s on. During the interval bc, the slope is constant and nonzero, so the cab moves with constant velocity (4 m/s). Example

39 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

40 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

41 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 time X(t)

42 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

43 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

44 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

45 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

46 2-2 Instantaneous Velocity and Speed © 2014 John Wiley & Sons, Inc. All rights reserved.

47 2-2 Instantaneous Velocity and Speed © 2014 John Wiley & Sons, Inc. All rights reserved. Answers: (a) Situations 1 and 4 (zero) (b) Situations 2 and 3

48 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… acceleration = 2.10 m/s 2 in the +x direction Acceleration

49 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

50 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

51 Acceleration Acceleration = the rate of change of velocity.

52 (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?

53 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

54 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

55 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

56 Finding velocity on an x-t graph At any point on an x-t graph, instantaneous x-velocity is equal to slope of tangent to curve at that point.

57 Motion diagrams A motion diagram shows position of a particle at various instants, and arrows represent its velocity at each instant.

58 Motion diagrams A motion diagram shows position of a particle at various instants, and arrows represent its velocity at each instant.

59 Average acceleration Acceleration describes the rate of change of velocity with time. The average x-acceleration is a av-x =  v x /  t.

60 Finding acceleration on a v x -t graph Use  x vs. t graph to find instantaneous acceleration and average acceleration.

61 Instantaneous acceleration The instantaneous acceleration is a x = dv x /dt. Q 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 acceleration in the first second? What was its average acceleration in the first 4 seconds?

62 An x-t graph and a motion diagram

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64 A v x -t graph and motion diagram

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66 Motion with constant acceleration For a particle with constant acceleration, the velocity changes at the same rate throughout the motion.

67 Acceleration given x(t) A particle is moving in a straight line with position 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) instantaneous acceleration as function of time.

68 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)

69 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)

70 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

71 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

72 2-4 Constant Acceleration In many cases acceleration is constant, or nearly so. For these cases, 5 special equations can be used. Note that constant acceleration means a velocity with a constant slope, and a position with varying slope (unless a = 0).

73 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

74 2-4 Constant Acceleration First basic equation o When the acceleration is constant, the average and instantaneous accelerations are equal o Rewrite Eq. 2-7 and rearrange © 2014 John Wiley & Sons, Inc. All rights reserved. Eq. (2-11) This equation reduces to v = v 0 for t = 0 Its derivative yields the definition of a, dv/dt

75 2-4 Constant Acceleration Second basic equation © 2014 John Wiley & Sons, Inc. All rights reserved. Eq. (2-15) Eq. (2-12) o Average = ((initial) + (final)) / 2: Eq. (2-13) Eq. (2-14) o Substitute 2-14 into 2-12

76 2-4 Constant Acceleration These are enough to solve any constant acceleration problem o Solve as simultaneous equations Additional useful forms: © 2014 John Wiley & Sons, Inc. All rights reserved. Eq. (2-18) Eq. (2-17) Eq. (2-16)

77 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

78 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 ! x = x 0 + v x t – 1/2a x t 2

79 2-4 Constant Acceleration Table 2-1 shows 5 key equations & the quantities missing from each.

80 2-4 Constant Acceleration © 2014 John Wiley & Sons, Inc. All rights reserved. Answer: Situations 1 (a = 0) and 4.

81 A motorcycle with constant acceleration Follow Example 2.4 for an accelerating motorcycle.

82 Two bodies with different (constant) accelerations A motorist is moving at a constant speed of 15m/s (in a 25 mph zone, through a school crosswalk) passes a police officer on a motorcycle, who is stopped. The officer immediately accelerates at a constant 3.0 m/s 2 in pursuit, and overtakes the motorist. How far are they from the point where the motorist first passed the officer? How much time elapsed?

83 Two bodies with different accelerations A motorist is moving at a constant speed of 15m/s (in a 25 mph zone, through a school crosswalk) passes a police officer on a motorcycle, who is stopped. The officer immediately accelerates at a constant 3.0 m/s 2 in pursuit, and overtakes the motorist.

84 Two bodies with different accelerations Two different initial/final velocities and accelerations TIME and displacement are linked!

85 Two bodies with different accelerations Two different initial/final velocities and accelerations TIME and displacement are linked!

86 Two bodies with different accelerations (part 2!) A motorist is moving at a constant speed of 15m/s (in a 25 mph zone, through a school crosswalk) passes a police officer on a motorcycle, who is stopped. The officer waits 3 seconds, and then accelerates at a constant 3.0 m/s 2 in pursuit, and overtakes the motorist. tptp tmtm

87 Two bodies with different accelerations Two different initial/final velocities and accelerations & times? TIME and displacement are STILL linked, but times aren’t equal… tmtm Officer tptp Displacement (m) Motorist Time (s)

88 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.

89 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.

90 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

91 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. Example: Falling from a tower.

92 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. Example: Falling from a tower.

93 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 t = 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.

94 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

95 Example: Ball thrown up! A person standing on top of a building 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

96 Up-and-down motion in free fall An object is in free fall even when it is moving upward.

97 Is the acceleration zero at the highest point? The vertical velocity, but not the acceleration, is zero at the highest point.

98 Is the acceleration zero at the highest point? The vertical velocity, but not the acceleration, is zero at the highest point.

99 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).

100 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

101 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.

102 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) How long does it take the ball to reach the base of the cliff? (b) What is the total distance traveled by the ball? Ignore air resistance (likely to be significant, so our result is an approximation).

103 Variable Acceleration; Integral Calculus Deriving the kinematic equations through integration: For constant acceleration,

104 Variable Acceleration; Integral Calculus Then: For constant acceleration,

105 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?

106 Graphical Analysis and Numerical Integration The total displacement of an object can be described as the area under the v-t curve:

107 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.

108 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.

109 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

110 Motion with changing acceleration

111 2-3 Acceleration Acceleration is a vector quantity: o Positive sign means in the positive coordinate direction o Negative sign means the opposite o Units of (distance) / (time squared)

112 2-3 Acceleration Note: accelerations can be expressed in units of g © 2014 John Wiley & Sons, Inc. All rights reserved. Eq. (2-10)

113 2-3 Acceleration © 2014 John Wiley & Sons, Inc. All rights reserved. Example If a car with velocity v = -25 m/s is braked to a stop in 5.0 s, then a = + 5.0 m/s 2. Acceleration is positive, but speed has decreased.

114 2-3 Acceleration © 2014 John Wiley & Sons, Inc. All rights reserved.

115 2-3 Acceleration © 2014 John Wiley & Sons, Inc. All rights reserved. Answers: (a) + (b) - (c) - (d) +

116 2-3 Acceleration The graph shows the velocity and acceleration of an elevator cab over time. When acceleration is 0 (e.g. interval bc) velocity is constant. When acceleration is positive (ab) upward velocity increases. When acceleration is negative (cd) upward velocity decreases. Steeper slope of the velocity- time graph indicates a larger magnitude of acceleration: the cab stops in half the time it takes to get up to speed. Example

117 2-5 Free-Fall Acceleration 2.16 Identify that if a particle is in free flight (whether upward or downward) and if we can neglect the effects of air on its motion, the particle has a constant downward acceleration with a magnitude g that we take to be 9.8m/s 2. 2.17 Apply the constant acceleration equations (Table 2-1) to free-fall motion. © 2014 John Wiley & Sons, Inc. All rights reserved. Learning Objectives Figure 2-12

118 2-5 Free-Fall Acceleration Free-fall acceleration is the rate at which an object accelerates downward in the absence of air resistance o Varies with latitude and elevation o Written as g, standard value of 9.8 m/s 2 o Independent of the properties of the object (mass, density, shape, see Figure 2-12) The equations of motion in Table 2-1 apply to objects in free-fall near Earth's surface o In vertical flight (along the y axis) o Where air resistance can be neglected © 2014 John Wiley & Sons, Inc. All rights reserved.

119 2-5 Free-Fall Acceleration The free-fall acceleration is downward (-y direction) o Value -g in the constant acceleration equations © 2014 John Wiley & Sons, Inc. All rights reserved. Answers: (a) The sign is positive (the ball moves upward); (b) The sign is negative (the ball moves downward); (c) The ball's acceleration is always -9.8 m/s 2 at all points along its trajectory

120 2-6 Graphical Integration in Motion Analysis 2.18 Determine a particle's change in velocity by graphical integration on a graph of acceleration versus time. 2.19 Determine a particle's change in position by graphical integration on a graph of velocity versus time. © 2014 John Wiley & Sons, Inc. All rights reserved.

121 2-6 Graphical Integration in Motion Analysis Integrating acceleration: o Given a graph of an object's acceleration a versus time t, we can integrate to find velocity The Fundamental Theorem of Calculus gives: © 2014 John Wiley & Sons, Inc. All rights reserved. Eq. (2-27)

122 2-6 Graphical Integration in Motion Analysis © 2014 John Wiley & Sons, Inc. All rights reserved. The definite integral can be evaluated from a graph: Eq. (2-28)

123 Example 2-6 Graphical Integration in Motion Analysis The graph shows the acceleration of a person's head and torso in a whiplash incident. To calculate the torso speed at t = 0.110 s (assuming an initial speed of 0), find the area under the pink curve: area A = 0 area B = 0.5 (0.060 s) (50 m/s 2 ) = 1.5 m/s area C = (0.010 s) (50 m/s 2 ) = 0.50 m/s total area = 2.0 m/s © 2014 John Wiley & Sons, Inc. All rights reserved.

124 Position Relative to origin Positive and negative directions Displacement Change in position (vector) Average Speed Distance traveled / time Average Velocity Displacement / time (vector) Eq. (2-1) Eq. (2-2) Eq. (2-3) 2 Summary © 2014 John Wiley & Sons, Inc. All rights reserved.

125 Instantaneous Velocity At a moment in time Speed is its magnitude Average Acceleration Ratio of change in velocity to change in time Constant Acceleration Includes free-fall, where a = -g along the vertical axis Instantaneous Acceleration First derivative of velocity Second derivative of position Eq. (2-4) Eq. (2-7) Eq. (2-8) Tab. (2-1) 2 Summary © 2014 John Wiley & Sons, Inc. All rights reserved.


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