1. Chapter 2: Motion in One Dimension

2. How to solve any physics problem:
DUFAS
Diagram or draw a picture
Units and variables labeled
Formulas written and ready to use
Algebra shown with numbers and units
Solution boxed with correct units and sig. figs.

3. There are three ways to describe motion.
Displacement
Velocity
Acceleration

4. Displacement and Velocity

5. First off, to describe motion, we must specify position.

6. But to specify position, we need to relate it to another known location
You are here
You want to be here

7. For motion in one dimension, it is convenient to use the x-axis.
From this concept, we can create a number line.

8. +

9. -

10. Displacement
Displacement of an object - the change of position of the object
NOTE: displacement is not always equal to the distance traveled.

11. Displacement (x) x = xf – xi Where x = displacement
x = xf – xi
Where x = displacement
xf = final position xi = initial position

12. Example
What is the displacement if a car moves from an initial position of 10 m to a final position of 80 m?
x =80 m – 10 m = 70 m

13. Example
What is the displacement if a car moves from an initial position of 80 m to a final position of 20 m?
x =20 m – 80 m = -60 m
Note: the negative sign indicates the direction of displacement.

14. Distance and Displacement
Every morning you drive 4 miles from home to NPHS and then come back home at night using the same route. What is the best statement describing your daily trip? (Assume you do not move far while at NPHS)
                                                   
Your displacement is 0, distance traveled is 8 mi
Your displacement is 8 mi, distance traveled is 0
Your displacement is 0, distance traveled is 0
Your displacement is 8 mi, distance traveled is 8 mi

15. VECTORS
HIDDEN!
A VECTOR is a physical quantity that has both magnitude and direction (displacement, velocity, acceleration)

16. VELOCITY – change in displacement over time.
Average velocity :

17. Units for velocity
The SI unit for velocity is m/s; however, a velocity can be given in other units such as:
Km/sec or km/hr or mi/hr
When calculating the velocity, be sure all your units match. If they don’t, you need to convert.

18. Example: It takes 6.0 hours to drive to San Francisco from Newbury Park if driving at 65 mi/hr. How far is the trip?
Newbury Park
San Fransisco
t = 6.0 hrs
v = 65 mi/hrs
Dx = ?

19. Example: The United States is 4,300 km wide
Example: The United States is 4,300 km wide. How long (in hours) would it take to drive across the country if someone were to drive at a steady 22 m/s the whole way?
Dx = 4,300 km
v = 22 m/s
t = ?

20. Speed vs. Velocity
Speed is a SCALAR meaning it is only a number, direction doesn’t matter.
Speed = (total distance)/(time)
Velocity is a VECTOR meaning it has a number (magnitude) and direction. Direction matters.
Velocity = (displacement)/(time)

21. Acceleration
Acceleration – rate of change in velocity over time.

22. Units of Acceleration

23. Example: A car is initially coasting up hill at 8. 5 m/s. 4
Example: A car is initially coasting up hill at 8.5 m/s seconds later, the car is now rolling backwards at 2.6 m/s down the same hill. What is the car’s acceleration during this time period?
- Vf
+ Vo
Vo = 8.5 m/s
t = 4.75 sec
Vf = m/s

24. Another way to describe motion is to graph displacement and velocity as a function of time it

25. Plug motion sensor into one of the ports at the top (line the groove up)
GLX Time!
A position vs time graph should automatically open
The “Play” button lets you start and stop data taking.
There is no need to erase your graph. Just hit “Play” again to take more data and overwrite your old graph.

26. We can represent a trip on the train track with a position vs
We can represent a trip on the train track with a position vs. time graph.
HIDDEN!!
Position (m)
Time (s)

27. (1)Create this graph on your GLX using your motion sensor.
(2) Describe your motion in your notes.
In this case, displacement stays constant with time…there is no movement.
Position (m)
Time (s)

28. Position (m) Time (sec)
(1)Create this graph on your GLX using your motion sensor.
(2) Describe your motion in your notes.
Here the object has a low constant velocity in the beginning, and then it changes to a higher constant velocity.
Position (m)
Time (sec)

29. (1)Create this graph on your GLX using your motion sensor.
(2) Describe your motion in your notes.
This time the object travels away from the starting point at constant velocity, stops for some amount of time and then travels back to its starting point at constant velocity.
Position (m)
Time (s)

30. (1)Create this graph on your GLX using your motion sensor.
(2) Describe your motion in your notes.
Good luck on this one!
Velocity is not constant. Velocity (slope) is increasing at a constant rate.
Position (m)
Time (s)

31. SPEED vs. VELOCITY
Both describe how fast the position is changing with respect to time.
Speed is a SCALAR quantity. It indicates an amount (magnitude), but not direction.
Velocity is a VECTOR quantity. It indicates both an amount (magnitude) and a direction.

32. SLOPE REVIEW

33. SLOPE REVIEW

34. SLOPE REVIEW

35. SLOPE REVIEW

36. SLOPE REVIEW

37. Position (m)
Rise
Run
Time (s)

38. VELOCITY
Velocity is represented by the SLOPE of the curve on a displacement vs. time graph.
In this class, a positive (+) slope indicates a forward direction and a negative (-) slope indicates a backwards direction (return).

40. Use the position graph to answer the following:
a. What is the object’s velocity from 10 – 15 seconds?
0 m/s (object is at rest)
b. What is the object’s velocity from 15 – 25 seconds?
= -4 m/s
c. What is the object’s velocity from 0 – 40 seconds?
= -1 m/s

41. Consider this trip…
Position (m)
Time (s)

42. How fast is the car going at this instant in time?
INSTANTANEOUS VELOCITY
How fast is the car going at this instant in time?
Position (m)
Time (s)

43. INSTANTANEOUS VELOCITY
The slope of a curve at a given point, is equal to the slope of a tangent line at that point.
Position (m)
Time (s)

44. INSTANTANEOUS VELOCITY
The slope of this line represents instantaneous velocity at the indicated point.
Position (m)
Time (s)

45. Hint: What is the meaning of the slope of the x(t) graph?
Describing Motion
The x(t) graph describes a 1-D motion of a train. What must be true about this motion?
Speeds up all the time
Slows down all the time
Speeds up part of the time
Slows down part of the time
Speeds up & then slows down
Slows down & then speeds up
Hint: What is the meaning of the slope of the x(t) graph?

46. Position vs. Time GraphsB t x A t1
The x(t) graph displays motions of two trains A and B on parallel tracks. Which statement is true?
At t1 both trains have the same velocity
At t1 both trains have the same speed
At t1 both trains have the same acceleration
Both trains have the same velocity sometime before t1
The trains never have the same velocity

47. End Position Graphs

49. Motion can be described with a velocity vs. time graph.

50. Velocity vs. Time 1 Velocity (m/s) Time (s) constant acceleration
constant velocity
Time (s)

51. Velocity vs. Time 2 + - Velocity (m/s) Time (s) + velocity 0 velocity
Time (s)
- velocity -

52. The Slope of Velocity vs. Time Graphs
Velocity (m/s)
Rise =  v
Run =  t
Time (s)

53. Use the velocity graph to answer the following:
a. What is the object’s velocity from 4 – 7 seconds?
3 m/s
b. What is the object’s acceleration from 4 – 7 seconds?
0 m/s2 (constant velocity)
c. What is the object’s acceleration from 2 – 4 seconds?
= 1 m/s2

54. Consider a trip…
A train travels at 5 miles per hour for 1 hour. What is its displacement after 1 hour?
Displacement can also be determined by finding the area under the curve of a velocity vs. time graph.

55. Using a graph
Velocity (m/s)
5 mi/hr
1 hour
Time (hrs)

56. Find the “area under the curve” (AUC).
Velocity (m/s)
5 mi/hr
1 hour
Time (s)
Area = l x w = 1 hour x 5 mi/hr = 5 miles.

57. Graphical analysis summary
Displacement vs. Time graph:
Slope = velocity
Velocity vs. Time graph:
Slope = acceleration
AUC = displacement

58. Draw a “v vs. t” graph for:
A blue car moving at a constant speed of 10 m/s passes a red car that is at rest. This occurs at a stoplight the moment that the light turns green.
The red car accelerates from rest at 4 m/s/s for three seconds and then maintains a constant speed for 9 s. The blue car maintains a constant speed of 10 m/s for the entire 12 seconds.

59. The answer

60. QUICK QUIZ 2.3
Parts (a), (b), and (c) of the figure below represent three graphs of the velocities of different objects moving in straight-line paths as functions of time. The possible accelerations of each object as functions of time are shown in parts (d), (e), and (f). Match each velocity-time graph with the acceleration-time graph that best describes the motion.

61. One way to look at the motion of an object is by using a motion map.

62. Position and direction
Imagine a toy car traveling along a piece of paper and dropping a dot of ink at a given time interval (say 1 drop every second). It could produce a trail that looks like this:
Position and direction
of object at a given
instant in time
Same time interval
Between dots
Time starts
at zero
distance

63. Position and direction
If the car produced the following motion map, how long did it take the car to travel the length of the paper?
distance
Same time interval
Between dots
Position and direction
of object at a given
instant in time
Time starts
at zero
1 sec
2 sec
3 sec
4 sec
5 sec
6 sec
7 sec
8 sec
8 seconds total

64. Position and direction
Describe the motion of the car? Stopped, constant velocity, accelerating or decelerating?
distance
Same time interval
Between dots
Position and direction
of object at a given
instant in time
Time starts
at zero
Constant velocity
How do you know?

65. Graphing Examples

67. a. Draw the motion map for the following:
Object accelerates for 3 seconds.
Then travels at a constant velocity for 2 seconds
Then decelerates for 3 seconds
Stops for 2 seconds
Then returns to the start in 4 seconds at a constant velocity.
b. Sketch the velocity graph for the above motion.

68. Velocity with Constant Accelerationvo
time t
time

69. vf vo t (p. 35) velocity time Rise = v = vf - vo
Run = t = (t – 0) = t t
time
(p. 35)

70. Break the area into two parts…vf
velocity
Find the AUC… vi
time t
Break the area into two parts…
Area = (l x w) + (½b x h)
x = vit + ½ t(vf-vi)

71. AUC = x = vot + ½ t(vf-vo)
velocity
Find the AUC… vi
time t
AUC = x = vot + ½ t(vf-vo)
x = vot + ½ vft – ½ vot distribution
x = ½ vot + ½ vft simplification
x = ½ (vo+vf)t p.(35)

72. We now know… 1) vf = vo + at and 2) x = ½ (vo + vf)t
By substitution of equation #1 into #2
x = ½ [vo + (at +vo)]t substitution
x = ½ (2vo + at)t combining terms
x = vot + ½ at2
(p. 36)

73. 1) vf = vo + at 2) x = ½ (vo + vf)t (p. 36)
Solve Eq #1 for t & substitute into Eq #2…
1) vf = vo + at
2) x = ½ (vo + vf)t
(p. 36)

74. 5 Parameters of Motion a = acceleration 2. x = displacement m
3. vf = final velocity
4. vi = initial velocity
5. t = time sec

75. vf = vo + at x = ½(vo + vf)t x = vot + ½a(t)2 vf2 = vo2 + 2ax
To solve a constant acceleration problem, you must know, or be able to find, three of the five parameters.
Then use the following equations to solve for the other two:
vf = vo + at
x = ½(vo + vf)t
x = vot + ½a(t)2
vf2 = vo2 + 2ax

76. Example… vf = vi + at a) vi = +100m/s vf= 0 m/s a = -5.0 m/s2 t = 20s
Example: A jet plane lands with a velocity of 100 m/sec and can slow down (-acceleration) at a maximum rate of –5.0 m/s2. Find (a) the time required for the plane to come to rest, and (b) the minimum size of the runway.
Example…
vf = vi + at
a) vi = +100m/s
vf= 0 m/s
0m/s = 100m/s + (-5m/s2)t
a = -5.0 m/s2
t = 20s
b) Solve for x
vf2 = vi2 + 2ax
0m/s = (100m/s)2 + 2(- 5m/s2) x
x = 1000 m

77. x = (20m/s)(6s) + ½ (-1m/s2)(6s)2
Example: A train is traveling down a straight track at 20 m/sec when the engineer applies the brakes, resulting in an acceleration of –1m/sec2 as long as the train is in motion. How far does the train travel in the first 6 seconds after the breaks are applied?
vi = 20 m/s
x = vit + ½ at2
a = -1 m/sec2
x = (20m/s)(6s) + ½ (-1m/s2)(6s)2
t = 6 sec
x = 120m – 18m = 102 m
x = ?
x = 100 m (sig. figs!)

78. vf2 = vi2 + 2ax vf2 = 0 + 2(5m/s2)(30.5m) vf = 17.5 m/s
Example: A racing car starting from rest accelerates at a rate of 5.00 m/s2. What is the velocity of the car after it has traveled 100. ft?
vi = 0 m/s
x = 100 ft= 30.5 m
a = 5.0 m/sec2
vf = ?
vf2 = vi2 + 2ax
vf2 = 0 + 2(5m/s2)(30.5m)
vf = 17.5 m/s