1. Physics: Principles with Applications, 6th edition
Lecture PowerPoints
Chapter 2
Physics: Principles with Applications, 6th edition
Giancoli
© 2005 Pearson Prentice Hall
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2. Describing Motion: Kinematics in One Dimension
Chapter 2
Describing Motion: Kinematics in One Dimension

3. 2-1 Reference Frames and Displacement
Any measurement of position, distance, or speed must be made with respect to a reference frame.
For example, if you are sitting on a train and someone walks down the aisle, their speed with respect to the train is a few miles per hour, at most. Their speed with respect to the ground is much higher.

4. 2-1 Reference Frames and Displacement
We make a distinction between distance and displacement.
Displacement (blue line) is how far the object is from its starting point, regardless of how it got there.
Distance traveled (dashed line) is measured along the actual path.

5. 2-1 Reference Frames and Displacement
The displacement is written:
Left:
Displacement is positive.
Right:
Displacement is negative.

6. 2-2 Average Velocity
Speed: how far an object travels in a given time interval
(2-1)
Velocity includes directional information:

7. How far can a cyclist travel in 2
How far can a cyclist travel in 2.5 h along a straight road if her average velocity is 18km/hr?
Ex 2.1 p34
Ex 2.2 p 37
Ex 2.3 p 41
CW : p 57 #16, 17, 18,

8. 2-3 Instantaneous Velocity
The instantaneous velocity is the average velocity, in the limit as the time interval becomes infinitesimally short.
(2-3)
These graphs show (a) constant velocity and (b) varying velocity.

9. 2-4 Acceleration
Acceleration is the rate of change of velocity.

10. 2-4 Acceleration
Acceleration is a vector, although in one-dimensional motion we only need the sign.
The previous image shows positive acceleration; here is negative acceleration:

11. 2-4 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. The magnitude of the velocity is decreasing.

12. 2-4 Acceleration
The instantaneous acceleration is the average acceleration, in the limit as the time interval becomes infinitesimally short.
(2-5)

13. A car accelerates along a straight road from rest to 75 km/h in 5. 0 s
A car accelerates along a straight road from rest to 75 km/h in 5.0 s. What is the magnitude of its average acceleration?

14. If the velocity of an object is zero, does it mean that the acceleration is zero?
If the acceleration is zero, does it mean that the velocity is zero?

15. a = ?
a = ?

16. 2-5 Motion at Constant Acceleration
The average velocity of an object during a time interval t is
The acceleration, assumed constant, is

17. 2-5 Motion at Constant Acceleration
In addition, as the velocity is increasing at a constant rate, we know that
Combining these last three equations, we find:
(2-8)
(2-9)

18. 2-5 Motion at Constant Acceleration
We can also combine these equations so as to eliminate t:
We now have all the equations we need to solve constant-acceleration problems.
(2-10)
(2-11a)
(2-11b)
(2-11c)
(2-11d)

19. Example
Ex 2.5 p 44,
Ex 2.6 p 46
CW : p 59 # 37 m/s/s,#39
P 60 #53,63

20. 2-7 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.

21. 2-7 Falling Objects
In the absence of air resistance, all objects fall with the same acceleration, although this may be hard to tell by testing in an environment where there is air resistance.

22. 2-7 Falling Objects
The acceleration due to gravity at the Earth’s surface is approximately 9.80 m/s2.

23. Figure 2-22 Tossing an object up
An object thrown into the air leaves the thrower's hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2-12, 2–13, 2–14, and 2–15.

24. Design your own experiment
Practice Problems
Ex 2.9 p 52
Do : Reaction Time
Design your own experiment

25. Graphs of (a) y vs. t (b) v vs. t for a ball thrown upward.

26. 2-8 Graphical Analysis of Linear Motion
This is a graph of x vs. t for an object moving with constant velocity. The velocity is the slope of the x-t curve.

27. 2-8 Graphical Analysis of Linear Motion
On the left we have a graph of velocity vs. time for an object with varying velocity; on the right we have the resulting x vs. t curve. The instantaneous velocity is tangent to the curve at each point. What would an a vs. t graph look like?

28. The velocity of an automobile as a function of time, starting from a dead stop. The jumps in the curve represent gear shifts.

29. 2-8 Graphical Analysis of Linear Motion
The displacement, x, is the area beneath the v vs. t curve.

30. The shaded area represents the displacement during the time interval t = 2.0s to t = 6.0s.

31. PRACTICE PROBLEMS :
P 58 # 22 , #23
Provide students with 3 types of graphs

32. EXPERIMENTS Uniform Motion Uniform Acceleration Reaction Time
PROJECT : Mousetrap Car
EGG X PRESS mesa.ucop.edu
mesa day rules
mesa competition rules
HS – Egg Xpress

33. # 1 Notes 2-7 Falling Objects
27. 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.

34. 2-7 Falling Objects
28. In the absence of air resistance, all objects fall with the same acceleration, although this may be hard to tell by testing in an environment where there is air resistance.

35. 2-7 Falling Objects
28. The acceleration due to gravity at the Earth’s surface is approximately 9.80 m/s2.

36. FIGURE 2-22 TOSSING AN OBJECT UP
An object thrown into the air leaves the thrower's hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2-12, 2–13, 2–14, and 2–15. 36

37. FREE FALL Really, just a special case of one dimensional kinematics.
Uses the same equations.
But, since displacement is vertical, it is written as y (instead of x).
Acceleration is no longer a variable, it is a fixed value 'g'. 37

38. Free Fall
Free Fall: The only acceleration or force is due to a constant gravity.
29. All objects moving under the influence of
“only gravity” are said to be in free fall.
30. Free fall acceleration does not depend on the object’s original motion, or on the mass.
31. Acceleration is the one thing constant to all falling objects. It is constant for the entire time they are in the air. 38

39. Free Fall
32. Free Fall: The only acceleration or force is due to a constant gravity.
This means that the significant figures of force due to other forces like Air Resistance are InSignificant. 39

40. Freely Falling Objects
A freely falling object is any object moving under the influence of gravity alone.
Acceleration does not depend upon the initial motion of the object!!!!!!!
33. Examples of Freefall with different initial motions.
Dropped –released from rest
Thrown downward
Thrown upward
Throwing a ball Up or DOWN AFTER the ball leaves the hands.
Examples of NonFreefall or NOT FREEFALL.
Throwing a ball Up or DOWN BEFORE the ball leaves the hands, while the hands are still pushing on it is NOT freefall.
A rocket engine is still burning and pushing up on a rocket. 40

41. Freely Falling Objects
A freely falling object is any object moving under the influence of gravity alone.
Acceleration does not depend upon the initial motion of the object!!!!!!!
Examples of Freefall with different initial motions.
Dropped –released from rest
Thrown downward
Thrown upward
Throwing a ball Up or DOWN AFTER the ball leaves the hands.
So what does initial motion affect then???
It changes the vi in vf = vi + at.
It does not change the a. 41

42. Acceleration of Freely Falling Object
The acceleration of an object in free fall is directed downward, regardless of the initial motion
34.The magnitude of free fall acceleration is g= 9.80 m/s2 which is the number used in your textbooks, so use so that you match the black and white textbooks.
Only valid on the Earth’s surface at OUR ALTITUDE.
Not to be confused with g for grams
British textbooks use 9.81 m/s2 instead of 9.80 m/s2 like American textbooks. Why? 42

43. STRENGTH OF GRAVITY (.028 % less)
The value of 'g' is affected by altitude.
At the top of Mt. Everest, g = m/s2
(.028 % less) 43

44. Acceleration of Free Fall, cont.
35. We will neglect air resistance
36. Free fall motion is constantly accelerated motion in ONLY one dimension, which is y.
DO NOT USE ax = m/s2
Let upward be positive
37. Use the kinematic equations with ay= -g= m/s2 44

45. Acceleration of Free Fall, cont.
38.
Use the kinematic equations with a= g= m/s2
REMEMBER: + means up, - means down
Write these equations on page:
Kinematics Equation Page (Choose 1 side or other.)
Y Axis
y = yi + vi t - ½ gt2
vf = vi - g t
vf2 = vi2 -2g Δy
g= 9.80 m/s2
(Because a = g)
Y Axis
y = yi + vi t + ½ at2
vf = vi + a t
vf2 = vi2 + 2a Δy
g= m/s2
(Because g = Neg) 45

46. #1 NOTES 1 # 9 CW Think, Solve , Small G, Big G
39. P 52 A boy on the bridge throws a stone vertically downward with an initial velocity of 14.7 m/s toward the river below. If the stone hits the water 2.0 sec later , what is the height of the bridge above the water?
# 9 CW Think, Solve , Small G, Big G
Follow up Exercise p 53

47. #10 CW : THINK, SOLVE, SMALL GROUP, BIG GROUP
AP PHYSICS WORKBOOK B –
# 8-9

48. Free Fall –an object dropped
If I drop a soccer ball off the roof of a very, very tall building, what is the velocity 8 seconds later?
What is the displacement?
Solve these in your powerpoint notes. 48

49. Free Fall –an object dropped
If I drop a soccer ball off the roof of a very, very tall building, what is the velocity 8 seconds later?
vf = vi + a t
v8 = (-9.80 m/s2)(8 s)
v8 = m
What is the displacement?
y8 = yi + vi t + ½ at2
y8 = (8)+½ (-9.80 m/s2)(8 s)2
y8 = -313 m
Vf = vi - g t
V8 = (9.80 m/s2)(8 s)
V8 = m
What is the displacement?
y8 = yi + vi t - ½ gt2
y8 = (8) - ½ (9.80 m/s2)(8 s)2
y8 = -313 m 49

50. Free Fall of an object thrown upward
At Top
vy = 0
ay = -g
Initial velocity is upward, so positive
The instantaneous velocity at the maximum height is zero
Vy=max = 0
ay= -g = m/s2 everywhere in the motion:
at the start, top, and end
And everywhere inbetween.
vi≠0
ay = -g 50

51. Free Fall -object thrown upward
At Top
vy = 0
ay = -g
A soccer ball is kicked directly upwards with a velocity of 8 m/s.
At what time will it’s velocity be zero?
At what time will its height peak?
What’s the highest it will go?
How much time will it spend in the air before it hits the ground?
With what velocity will it strike the ground?
Vi > 0
ay = -g 51

52. Free Fall -object thrown upward
A soccer ball is kicked directly upwards with a velocity of 8 m/s.
At what time will it’s velocity be zero?
Ans: vf = vi + a t  t=.816s = .8s
At what time will its height peak?
Ans: Same Question. Notice vy = 0
At the top.
What’s the highest it will go?
Ans: yf = yi + vi t + ½ at2  3.27m = 3m
Or use vf2 = vi2 + 2a Δy
How much time will it spend in the air before it hits the ground? Double 1st ans
t = 1.6s
With what velocity will it strike the ground? Discuss ball, bullets, arrows, and air resistance.
At Top
vy = 0
ay = -g
Vi > 0
ay = -g 52

53. thrown upwards, free fall only
The motion may be symmetrical IF it lands at the same height as thrown.
Then tup= tdown
Then vf = -vi
Break the motion into two parts
up and down 53

54. Free Fall –an object thrown downward
ay= -g= m/s2
Initial velocity ≠0
Since up is positive,
initial velocity will be negative.
Vo< 0
a = -g 54

55. Free Fall --object thrown Downward
Same Equations, but Vo will be negative
A person fires a handgun straight downwards from a helicopter. The muzzle velocity of the gun is 300 m/s. The helicopter is 250 m above the ground.
With what velocity will the bullet hit the ground? 55

56. Free Fall --object thrown Downward
Same Equations, but Vo will be negative
A person fires a handgun straight downwards from a helicopter. The muzzle velocity of the gun is 300 m/s. The helicopter is 250 m above the ground.
With what velocity will the bullet hit the ground?
ANS: vf2 = vi2 + 2a Δy = -308 m/s
All vi ,a, Δy = negative
vf = choose neg sq.root. 56

57. Problem Solving tips Think about and understand the situation
GIVENS: List the given information. REMEMBER g = m/s2 and at the top of the arc, vy = 0 (y direction only.)
DIAGRAM: Make a quick drawing of the situation
FIND: List the unknown you are solving for.
EQUATIONS: List the equations used to solve.
CALCULATIONS: Show your work. Think about units
SOLUTION: List your answer.
Focus on the expected result
Think about what a reasonable answer should be 57

58. Example We throw a bowling ball upwards from the edge of the roof.
What happens at points B, C and D? 58

59. Example
Initial velocity at A is upward (+) and acceleration is -g(-9.80 m/s2)
At B, the velocity is 0 and the acceleration is -g(-9.80 m/s2)
At C, the velocity has the same magnitude as at A, but is in the opposite direction
The displacement is –50.0 m (it ends up 50.0 m below its starting point) 59

60. Problem Solving –
Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground? 60

61. Problem Solving –
Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground?
y = yi +vi t + ½ at2
v = vi + a t
vf2 = vi2 +2ay 61

62. Problem Solving –
Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground?
Just look at flea on the way up.
y is .440 m vi is what we’re looking for
a is -9.8 m/s2 vf is zero, since we’re looking at the top of his jump. 62

63. Problem Solving –
Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground?
Just look at flea on the way up.
y is .440 m vi is what we’re looking for
a is -9.8 m/s2 vf is zero, since we’re looking at the top of his jump.
vf2 = vi2 +2ay 63

64. Problem Solving –
Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground?
0 = vi (-9.80m/s2)(.440m)
Vf2 = vi2 +2ay 64

65. Problem Solving –
Question: If a flea can jump .440 m, what is the initial velocity as it leaves the ground?
Vf2 = vi a y
0 = vi (-9.80m/s2)(.440m)
= vi m2/s2 so
vi = + / m/s
Which makes sense, + or -? 65

66. yf = yi +vi t + ½ at2 vf = vi + a t vf2 = vi2 +2ay Problem Solving –
Question: If a flea can jump .440 m, how much time does he spend going up?
yf = yi +vi t + ½ at2
vf = vi + a t
vf2 = vi2 +2ay 66

67. Problem Solving –
Question: If a flea can jump .440 m, how much time does he spend going up?
a is -9.8 m/s2 vi = 2.94 m/s vf = 0
-9.8 m/s2 = ( m/s) / t
t = .3sec
vf = vi + a t 67

68. Problem Solving – What is the velocity when he lands?
Just look at the problem on the way down.
vf2 = 0 + (-9.80m/s2)(-.440m)
= m2/s2
vf = +/ m/s
Which makes sense, + or - ?
vf2 = vi2 +ay 68

69. PROBLEM SOLVING yf = yi +vi t + ½ at2 vf = vi + a t vf2 = vi2 +2ay
How much time does he spend in the air?
yf = yi +vi t + ½ at2
vf = vi + a t
vf2 = vi2 +2ay 69

70. PROBLEM SOLVING How much time does he spend in the air?
-2.94m/s = 2.94m/s + (-9.81m/s2)*t
-5.82m/s = -9.81m/s2*t
t = .6 sec
Twice the time it took to go up
vf = vi + a t 70

71. BENCHMARK :KINEMATICS
PRINCETON REVIEW P
FREE RESPONSE 1-2