Position-time and velocity-time graphs Uniform motion problems algebra Acceleration and displacement

Transcription

1 Position-time and velocity-time graphs Uniform motion problems algebra Acceleration and displacement

2 Topics: The kinematics of motion in one dimension: graphing and calculations Problem-solving strategies Free fall Sample question: Horses can run much much faster than humans, but if the length of the course is right, a human can beat a horse in a race. When, and why, can a man outrun a horse?

3 Speed, Velocity, & Acceleration Speed (v) how fast you go (scalar) Velocity (v) how fast and which way; the rate at which position changes (vector) Average speed ( v ) distance / time Acceleration (a) how fast you speed up, slow down, or change direction; the rate at which velocity changes

4 During your 8 mi. trip, which took 15 min., your speedometer displays your instantaneous speed, which varies throughout the trip. Your average speed is 32 mi/hr. Your average velocity is 32 mi/hr in a SE direction. At any point in time, your velocity vector points tangent to your path. The faster you go, the longer your velocity vector.

5 Acceleration Acceleration how fast you speed up, slow down, or change direction; it s the rate at which velocity changes. Two examples: t (s) v (mph) t (s) v (m/s) a = +2 mph / s a = -3 m / s s = -3 m / s 2

6 Velocity & Acceleration Sign Chart V E L O C I T Y A C C E L E R A T I O N Moving forward; Moving backward; Speeding up Slowing down - Moving forward; Moving backward; Slowing down Speeding up

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8 Displacement is a change of position in a certain direction, not the total distance traveled

9 Motion Takes place over time Depends upon the frame of reference What is frame of reference? How do we choose one? Once we decide upon a frame of reference, it remains fixed for the problem

10 Keep in mind Displacement is not always equal to the distance traveled! Displacement can be positive or negative!

11 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: Δx = 4 m, W What is the distance traveled? d = 20 m!! Δx 8 m,e x = -4 x = m,w x

12 Displacement When things start moving, the length of a straight line drawn from the object s initial position to it s final position is it s displacement In one dimension x x x f i Change in position along x-axis = (final position on x-axis) (initial position on x-axis)

13 Motion in One Dimension Same concepts apply to any axis! y y y f i Change in position along y-axis = (final position on y-axis) (initial position on y-axis)

14 POSITIVE AND NEGATIVE QUANTITIES ARE JUST DIRECTIONS IN MOTION PROBLEMS!

15 Positiontime graphs the relationship between the shape of a p-t graph and the motion of the object.

16 consider a car moving with a constant, rightward (+) velocity of +10 m/s. If the position-time data for the car were graphed, it would look like the graph at the right.

17 Now consider a car moving with a rightward (+), changing velocity - that is, a car that is moving rightward but speeding up or accelerating. If the position-time data for such a car were graphed, the graph would look like the graph at the right.

18 . Position-time graphs for the two types of motion - constant velocity and changing velocity (acceleration). Constant Velocity Positive Velocity Positive Velocity Changing Velocity (acceleration)

19 Motion diagram (student walking to school) Provide a reasonable explanation for the motion diagram. Draw a position-time graph. (time on the x-axis)

20 Motion diagram (student walking to school) Table of data Graph

21 Interpreting position-time graphs Starting at t = 0: Position away from the origin in + direction has + velocity Position away from the origin in direction has a velocity No change in position 0 velocity

22 x p-t graph B Describe this motion. A t C A Starts at home (origin) and goes forward slowly B Not moving (position remains constant as time progresses) C Turns around and goes in the other direction quickly, passing up home

23 E x A C B THS D W Practice problems 1-3, pg. 85

24 Uniform motion (constant velocity) Equal displacements occur during successive equal time intervals. the slope of the line on a position-time graph reveals useful information about the velocity of the object.

25 The importance of Slope The slope of the line on a position-time graph reveals useful information about the velocity of the object. Whatever characteristics the velocity has, the slope will exhibit the same (and vice versa). If the velocity is constant, then the slope is constant (i.e., a straight line). If the velocity is changing, then the slope is changing (i.e., a curved line). If the velocity is positive, then the slope is positive (i.e., moving upwards and to the right). This very principle can be extended to any motion

26 The object represented by the graph on the right is traveling faster than the object represented by the graph on the left. Slow, Rightward(+) Constant Velocity Fast, Rightward(+) Constant Velocity

27 Consider the two graphs below. Both graphs show plotted points forming a curved line. The curved line of changing slope is a sign of accelerated motion (i.e., changing velocity). Applying the principle of slope to the graph on the left, the object depicted by the graph is moving with a negative velocity (since the slope is negative ). The object is starting with a small velocity (small slope) and finishes with a large velocity. This object is moving in the negative direction and speeding up. Negative (-) Velocity Slow to Fast Leftward (-) Velocity Fast to Slow The graph on the right also depicts negative velocity (since there is a negative slope). The object begins with a high velocity (the slope is initially large) and finishes with a small velocity (since the slope becomes smaller). So this object is moving in the negative direction and slowing down.

28 Applying algebra to uniform motion The slope of the line on a position-time graph is equal to the velocity of the object. If the object is moving with a velocity of +2 m/s, then the slope of the line will be +2 m/s.

29 Slope = rise/run For p-t graph rise = d run = t slope of the line = ave. velocity V ave = d = d 1 d 0 t t 1 t 0 ave. velocity is d NOT d* t t *instantaneous speed

30 A v a = 2 - (-2) = 4 m = 10 m/s m B A s book, pg.86 B v a = -4 - (-2) = -2 m = -5 m/s

31 For Uniform motion the derivative: V ave = d = d 1 d 0 t t 1 t 0 When t 0 = 0, it can be eliminated from the problem V ave = d 1 d 0 t 1 d 1 = V ave (t 1 ) + d 0 V ave = v because constant velocity so no need for ave. d = vt + d 0 d 0 represents position at t = 0

32 An airplane is moving at a uniform velocity of 75 m/s from a starting position of 40 m. Find the position of the airplane after 2.5 seconds. d = vt + d 0 d = 40 m + (75 m/s)(2.5 s) d = 230 m practice problems 9-12, pg. 89

33 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.

34 graphing velocity

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

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

37 Average Speed and Instantaneous Velocity The average speed depends ONLY on the distance traveled and the time required. s = 20 m C A Time t = 4 s B The instantaneous velocity is the magnitude and direction of the velocity at a particular instant. (v at point C)

38 Example: 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: s 2 = 300 m s 1 = 200 m start Average speed total path time 500 m 60 s Avg. speed= 8 m/s Direction does not matter!

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

40 Graphing velocity: uniform motion (constant velocity) is represented by a horizontal line on a v-t graph. B A C D Increasing speed?

41 V (m/s) + 0 Constant speed? Increasing speed? Backwards and increasing speed? Backwards at a constant speed? Decreasing speed?

42 ? What does the intersection of 2 lines on a v-t graph tell us? Note: v-t graphs give NO information about position.

43 Checking Understanding Here is a motion diagram of a car moving along a straight stretch of road: Which of the following velocity-versus-time graphs matches this motion diagram? A. B. C. D.

44 A graph of position versus time for a basketball player moving down the court appears like so: Which of the following velocity graphs matches the above position graph? A. B. C. D. Slide 2-15

45 A graph of velocity versus time for a hockey puck shot into a goal appears like so: Which of the following position graphs matches the above velocity graph? A. B. C. D.

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48 slope slope displacement vs. time velocity vs. time acceleration vs. time Area underneath = Δd Area underneath = Δv

49 slope slope displacement vs. time velocity vs. time acceleration vs. time Area underneath = Δd Area underneath = Δv p-t v-t a-t time (s) time (s) time (s) area: none m/s x s = m = Δd m/s 2 x s = m/s = Δv slope: m/s = v Rate at which displacement changes m/s/s = a Rate at which velocity changes

50 Acceleration The rate of change of velocity

51 Think about this... What are three ways to change the velocity of a car? Accelerate Decelerate Change direction

52 average acceleration is the rate of change in velocity between t 0 and t 1 a = v t acceleration = change in velocity change in time acceleration (m/s 2 ) = (v f ) - (v i ) time If a car moves at a constant velocity, then its acceleration is zero

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54 Checking Understanding These four motion diagrams show the motion of a particle along the x-axis. Which motion diagrams correspond to a positive acceleration? Which motion diagrams correspond to a negative acceleration? Slide 2-22

55 Checking Understanding These four motion diagrams show the motion of a particle along the x-axis. Rank these motion diagrams by the magnitude of the acceleration. There may be ties. Slide 2-21

56 Checking Understanding These six motion diagrams show the motion of a particle along the x-axis. Rank the accelerations corresponding to these motion diagrams, from most positive to most negative. There may be ties. Slide 2-23

57 Acceleration There is a difference between negative acceleration and deceleration: Negative acceleration is acceleration in the negative direction as defined by the coordinate system. Deceleration occurs when the acceleration is opposite in direction to the velocity by W.H. Freeman and Company

58 Acceleration acceleration is the rate of change of velocity. a v t 2 2 v t 1 1 v t

59 Acceleration Acceleration is a vector, although in onedimensional motion we only need the sign. here is an example of deceleration: a = v 2 v 1 t 2008 by W.H. Freeman and Company

60 Helpful rule: If the sign of both the acceleration and the velocity are the same = speeding up If the signs of the acceleration and velocity are different = slowing down

61 THINK ABOUT THIS Why do highways have speed limits and not velocity limits?

62 or deriving for dum dums

63 Quick calculation: 1. A plane travels in a single direction on a runway. It reaches the end of the runway in 20 seconds and its velocity is 80 m/s. What is its acceleration? + 4 m/s 2 2. A skateboarder is traveling at 8 m/s. He slows and comes to a stop in 4 sec. What was the acceleration? - 2 m/s 2

64 3. A sailboat is moving at 12 m/s when a gust of wind changes its velocity to 18 m/s. The wind lasts 10 seconds. For that 10 sec, what was its acceleration? 0.6 m/s 2 4. A speed skater accelerates at 5 m/s 2. What will his speed be 4 seconds later? 20 m/s

65 At highway speeds, a particular automobile is capable of an acceleration of about 1.6 m/s 2. At this rate, how long does it take to accelerate from 80 km/h to 110 km/h?. The time can be found from the average acceleration, a v t 1m s 30 km h v 110 km h 80 km h 3.6 km h t 5.208s 5 s 2 2 a 1.6 m s 1.6 m s One more.

66 A chicken runs across the road at a speed of 12 m/s. If the road is 36 meters across, how long does it take for the chicken to cross the road? 3 s

67 A. Displacement with constant acceleration B. Velocity with constant acceleration C. Displacement with constant acceleration D. Final velocity after any displacement

68 Displacement with constant acceleration Displacement depends on acceleration, velocity and time. V avg = Δd Δt For an object moving with constant acceleration, the average velocity is equal to the average of the initial and final velocities. v-t v f Δd = V avg = v i + v f Δt 2 V (cm/s) v avg Multiply both sides by t 0 0 v i time (s) Δd = ½ (v i + v f ) Δt

69 Try one: A race car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and breaking system, and comes to rest 5.5 s later. Find how far the car moves while stopping.

70 Displacement with Constant Acceleration A car accelerates uniformly from rest to a speed of 23.7 km/h in 6.5 s. Find the distance the car travels during this time. 20 m

71 One more: When Maggie applies the brakes of her car, the car slows uniformly from m/s to 0.00 m/s in 2.50 s. How many meters before a stop sign must she apply the brakes in order to stop at the sign? 18.8 m

72 Velocity with Constant Acceleration a = v f -v i = v f -v i t f - t i t Rearrange the equation for acceleration to find v f Multiply by t and add the initial velocity to both sides v f = v i + a t Final velocity = initial velocity + (acceleration x time interval)

73 Displacement with Constant Acceleration Substitute this expression in for v f into the displacement with constant acceleration equation: d = 1 (v i + a t +v i ) t 2 Distribute t v f = v i + a t d = 1 (v f - v i ) t 2 d = 1 [2v i t + a( t) 2 ] 2 d = v i t + 1 a ( t) 2 2

74 Displacement with Constant Acceleration d = v i t + 1 a ( t) 2 2 This equation is useful not only for finding how far an object travels under constant acceleration but also for finding the distance required for an object to reach a certain speed or to come to a stop

75 Practice Problems: A boat with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s 2 for 3.6 s. Find the final speed and the displacement of the boat during this time. Hint: you need to solve using two equations 35.6 km/hr 28.9 m

76 displacement/velocity when time is unknown d = 1 (v f + v i ) t 2 2 d = (v f + v i ) t 2 d (v f + v i ) multiply by 2 = t You can find the final velocity of an accelerating object without knowing how long it has been accelerating. You can use this equation to derive another equation for finding final velocity after any displacement.

77 Final Velocity after any Displacement substitute the expression for t into the final v equation 2 d (v f + v i ) = t v f = v i + a t v f = v i + a ( 2 d) (v f + v i ) Subtract v i from both sides v f -v i = a ( 2 x) (v f + v i ) Multiply both sides by (v f + v i ) to get velocities on the same side v f 2 - v i 2 = 2a x Solve for v f 2 v f 2 = v i 2 + 2a d

78 Final Velocity after any Displacement cont. When using this equation remember you MUST take the square root of the right side to find the final velocity!

79 practice v f 2 = v i 2 + 2a d A baby sitter pushing a stroller starts from rest and accelerates at a rate of m/s 2. What is the velocity of the stroller after it has traveled 4.75 m?? v f? v i = 0 m/s a = m/s 2 d = 4.75m v f 2 = (0 m/s) 2 + 2(0.500 m/s 2 )(4.75 m) x + v f 2 = 4.75m 2 /s 2 v f = 4.75m 2 /s 2 = m/s

80 v f 2 = v i 2 + 2a d 2. An electron is accelerated from rest in an accelerator at 4.5 X 10 7 m/s 2 over a distance of 95 km. Assuming constant acceleration, what is the electron s final velocity? watch your units! 9.2 x 10 5 m/s

81 2 part acceleration problems

82 Dinner at a Distance, Part I Chameleons catch insects with their tongues, which they can extend to great lengths at great speeds. A chameleon is aiming for an insect at a distance of 18 cm. The insect will sense the attack and move away 50 ms after it begins. In the first 50 ms, the chameleon s tongue accelerates at 250 m/s 2 for 20 ms, then travels at constant speed for the remaining 30 ms. Does its tongue reach the 18 cm extension needed to catch the insect during this time? Slide 2-24

83 Dinner at a Distance, Part II Cheetahs can run at incredible speeds, but they can t keep up these speeds for long. Suppose a cheetah has spotted a gazelle. In five long strides, the cheetah has reached its top speed of 27 m/s. At this instant, the gazelle, at a distance of 140 m from the running cheetah, notices the danger and heads directly away. The gazelle accelerates at 7.0 m/s² for 3.0 s, then continues running at a constant speed that is much less than the cheetah s speed. But the cheetah can only keep running for 15 s before it must break off the chase. Does the cheetah catch the gazelle, or does the gazelle escape? Slide 2-25

84 F A L L I N G O B J E C T S 2008 by W.H. Freeman and Company

85 What is my acceleration right now?

86 According to Galileo With no air resistance, all freely falling objects have the same constant acceleration. The Mechanical Universe

87 The magnitude of acceleration of a falling object: g= m/s/s

88 Galileo and Apollo Aug. 2, 1971, astronaut David Scott

89 Falling Objects Near the surface of the Earth, all objects experience approximately the same acceleration due to gravity. This is one of the most common examples of motion with constant acceleration by W.H. Freeman and Company

90 Consider the position of the free-falling object at regular time intervals, every 1 second. The fact that the distance which the ball travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. It is accelerating!

91 Falling Objects The acceleration due to gravity at the Earth s surface is approximately 9.80 m/s 2. Misconceptions about falling objects.

92 Because acceleration due to gravity is directed downward to the center of the earth, acceleration of a falling object is always negative - even when the object is thrown up into the air! m/s 2 + Based on the usual axis, downward direction is -.

93 A curved line on a position vs. time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration of g = m/s/s, you would expect that its position-time graph would be curved. A closer look at the position-time graph - the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object, the initial small slope indicates a small initial velocity and the final large slope indicates a large final velocity. Last, but not least, the negative slope of the line indicates a negative (i.e., downward) velocity.

94 Free Fall the velocity-time graph shows the object starts with a zero velocity (from rest) and finishes with a large, negative velocity (the object is moving in the negative direction and speeding up). An object moving in the negative direction and speeding up has a negative acceleration. Analysis of the slope of the graph is consistent with this moving with a constant acceleration of m/s 2.

95 Objects thrown in the air have a + velocity and acceleration. This means the object is slowing down. The object moves upward but with smaller and smaller velocity. At the top of its path, v has decreased and is momentarily 0. Now it moves downward with v and still a (this means the object is speeding up!)

96 If everything accelerates at the same rate, does that mean everything falls at the same rate? Even if they have a different weight? yes Even if they are different sizes? yes Even if they are different shapes? not if you include air resistance If you put it in a vacuum, then yes

97 Free Fall Acceleration Objects moving toward earth move faster and faster as they fall.

98 Acceleration 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). The acceleration due to gravity is relatively constant near the Earth s surface. g Earth W

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

100 Acceleration due to Gravity Near the surface of the Earth, all objects accelerate at the same rate (ignoring air resistance). a = -g = -9.8 m/s m/s 2 This acceleration vector is the same on the way up, at the top, and on the way down! Interpretation: Velocity decreases by 9.8 m/s each second, meaning velocity is becoming less positive or more negative. Less positive means slowing down while going up. More negative means speeding up while going down.

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

102 Given: a = -9.8 m/s 2 ; v o = 30.0 m/s Solutions: t = 2.00 s; 4.00 s; 7.00 s For t = 4.00 s: For t = 7.00 s:

103 Given: a = -9.8 m/s 2 ; v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s Solutions: For t = 4.00 s: For t = 7.00 s:

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

105 Can use equations for constant acceleration to solve problems. Example 1: A flower pot falls from a window 25.0 m above the sidewalk a. How fast is it moving when it hits the ground? b. How much time does someone on the ground have to get out of the way? 1. diagram the problem 2. Define what you know, and what you re looking for Δd = 25 m v i = 0 m/s a = -9.8 m/s 2?v f?? t? 25 m 3. Choose the equation(s) v f 2 = v i 2 + 2a y v f = v i + a t

106 Example 1: A flower pot falls from a window 25.0 m above the sidewalk a. How fast is it moving when it hits the ground? b. How much time does someone on the ground have to get out of the way? 4. Solve! a. v f 2 = v i 2 + 2a y v f 2 = 0 + 2(-9.8)(25) v f = 490 = 22 m/s 25 m b. v f = v i + a t t = v f /a = 22 m/s -9.8 m/s 2 = 2.25 s

107 Example 2: Jan sets a volleyball straight up, leaving her hands at 12.0 m/s. a. How high does it go? b. If, when the ball is on the way down, it is spiked at the same height at which it was set, how long was it in flight? c. How fast is it traveling when it is spiked? 1. diagram the problem 0 Place starting point of the ball at the origin (y i = 0 at t = 0) floor

108 2. Define what you know, and what you re looking for d i = 0 m v i = 12 m/s a = -9.8 m/s 2?v f?? t? 3. Choose the equation(s) v f 2 = v i 2 + 2a y v i = 12 m/s a = m/s solve! a. At the top of its flight, the ball has instantaneous velocity of 0. v f = 0 so: 0 = (-9.8 m/s) y = (-19.6)Δy Δy = 7.35 m floor

109 b. To analyze the rest of the problem: remember that the down half of the trip is a mirror image of the up half. So, if, while going up, the ball passes through a particular height at a particular velocity, on its way down it will pass through that height at the same speed (with its velocity directed down). This means that the up half of the trip takes the same time as the down half of the trip, so we could just figure out how long it takes to reach its max height, and then double that to get the total time. OR, you can do the math: d f = d i + v i t + ½ at 2 0 = t + ½(-9.8)t 2 0 = t t = 12/4.9 = 2.45 s 0 floor

110 c. The answer has to be 12 m/s down, because of the mirror-image relationship between the up half of the flight and the down half. OR, you can do the math: v f = v i + a t v f = (2.45s) = -12m/s

111 Practice! Reviewing Concepts Pg , 1, 2, 4, 6-9, 12, website the Physics Classroom Practice Problems pg. 109 Problems 27, 28, 34, 35, 40, 43, Questions like #49 59, 64-76

112 Lab Time!! Determining acceleration