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Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures.

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Presentation on theme: "Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures."— Presentation transcript:

1 Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Chapter 2 Motion Along a Straight Line

2 Copyright © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. Displacement is written: SIGN matters! Direction matters! It is a VECTOR!! <= Positive displacement Displacement is negative => Displacement vs. Distance

10 Copyright © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. Velocity includes directional information: Speed vs. Velocity Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes Average velocity = 20 mi / hr = 60 mph NORTH VECTOR!

12 Copyright © 2012 Pearson Education Inc. Speed vs. Velocity Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes Average velocity = 20 mi / hr = 60 mph NORTH

13 Copyright © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 azimuth of 173 degrees Speed vs. Velocity

15 Copyright © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 = ( ) 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 © 2012 Pearson Education Inc. 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 = | | meters/3.00 sec = 6.5 m/s The answer must have value & units Example of Average SPEED

18 Copyright © 2012 Pearson Education Inc. Negative velocity??? Average x-velocity is negative during a time interval if particle moves in negative x-direction for that time interval.

19 Copyright © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. Instantaneous velocity — Figure 2.4 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 © 2012 Pearson Education Inc. Instantaneous velocity — Figure 2.4 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 t = 3.00 s x 1 = t = 5.00 s x 2 = 55.3 x 2 – x 1 = 33.6 meters in +x direction Instantaneous Velocity Example

31 Copyright © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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| = t = 5.00 seconds = 21.0 m/s Instantaneous Velocity Example

33 Copyright © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. Acceleration Acceleration = the rate of change of velocity.

37 Copyright © 2012 Pearson Education Inc. (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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. Motion diagrams A motion diagram shows position of a particle at various instants, and arrows represent its velocity at each instant.

43 Copyright © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. Instantaneous acceleration The instantaneous acceleration is a x = dv x /dt. Follow Example 2.3, which illustrates an accelerating racing car.

45 Copyright © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. Finding acceleration on a v x -t graph As shown in Figure 2.12, the  x -t graph may be used to find the instantaneous acceleration and the average acceleration.

47 Copyright © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. 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 © 2012 Pearson Education Inc. A motorcycle with constant acceleration Follow Example 2.4 for an accelerating motorcycle.

59 Copyright © 2012 Pearson Education Inc. Two bodies with different accelerations Two different initial/final velocities and accelerations TIME and displacement are linked!

60 Copyright © 2012 Pearson Education Inc. 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.

61 Copyright © 2012 Pearson Education Inc. 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.

62 Copyright © 2012 Pearson Education Inc. 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

63 Copyright © 2012 Pearson Education Inc. 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

64 Copyright © 2012 Pearson Education Inc. 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.

65 Copyright © 2012 Pearson Education Inc. 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

66 Copyright © 2012 Pearson Education Inc. Up-and-down motion in free fall An object is in free fall even when it is moving upward.

67 Copyright © 2012 Pearson Education Inc. Is the acceleration zero at the highest point?—Figure 2.25 The vertical velocity, but not the acceleration, is zero at the highest point.

68 Copyright © 2012 Pearson Education Inc. 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).

69 Copyright © 2012 Pearson Education Inc. 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

70 Copyright © 2012 Pearson Education Inc. 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.

71 Copyright © 2012 Pearson Education Inc. 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).

72 Copyright © 2012 Pearson Education Inc. Variable Acceleration; Integral Calculus Deriving the kinematic equations through integration: For constant acceleration,

73 Copyright © 2012 Pearson Education Inc. Variable Acceleration; Integral Calculus Then: For constant acceleration,

74 Copyright © 2012 Pearson Education Inc. 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?

75 Copyright © 2012 Pearson Education Inc. Graphical Analysis and Numerical Integration The total displacement of an object can be described as the area under the v-t curve:

76 Copyright © 2012 Pearson Education Inc. 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.

77 Copyright © 2012 Pearson Education Inc. 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.

78 Copyright © 2012 Pearson Education Inc. 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

79 Copyright © 2012 Pearson Education Inc. Motion with changing acceleration Follow Example 2.9. Figure 2.29 illustrates the motion graphically.


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