1. Unit B01 – Motion in One Dimension
AP Physics

2. Section 2-1: Reference Frames and Displacement
Explain what “frame of reference” is in terms a junior-high student could understand.
A frame of reference is like a point of view. Specifically, it is the point of view of an observer making measurements and describing what is going on.

3. Section 2-1: Reference Frames and Displacement
Explain what “frame of reference” is in terms a junior-high student could understand.
Example: Adam (red) says the lightning bolt struck 2 mi north and 2 mi west of him. Betty (blue) says the lightning struck 4 mi south and 3 mi west. Both give a different position for the lightning bolt. Who is right?
Both observed the same thing, but made different measurements based on their different frames of reference.

4. Section 2-1: Reference Frames and Displacement
Explain the difference between “distance” and “displacement”.
Distance – The length of the path an object travels.
Displacement – Distance straight from start to end, with direction specified.

5. Section 2-1: Reference Frames and Displacement
Explain the difference between “distance” and “displacement”.
What is Adam’s (red) distance from the lightning strike?
What is Adam’s displacement from the lightning strike?
What is Betty’s (blue) distance from the lightning strike?
What is Betty’s displacement from the lightning strike?
What is the lightning’s displacement from Adam?
What is Adam’s distance from Betty?

6. Section 2-1: Reference Frames and Displacement
Explain what a vector is.
A vector is any measured quantity that has both a magnitude (amount) and a direction.
Explain what a scalar is.
A scalar is any measured quantity that has only a magnitude (amount) but no meaningful direction.

7. Section 2-1: Reference Frames and Displacement
What letter is the symbol for displacement used in mathematical equations? x
Distance and displacement are measured in what units?
Meters

8. Section 2-1: Reference Frames and Displacement
Example: Consider the coordinate axes below. Each box represents 1 meter. y x
Fred’s position is Point A, which is at (x = 3, y = –3). Mark this point. A

9. Section 2-1: Reference Frames and Displacement
Example: Consider the coordinate axes below. Each box represents 1 meter. y x
Fred walks 5 meters to the left, 5 meters up, and 5 meters to the right, arriving at point B. Mark point B and Fred’s path on the diagram. B
(2, 2)
(–2, 2)
(–2, –3)
(–3, –2)
(2, 3)
(3, 2) A

10. Section 2-1: Reference Frames and Displacement
Example: Consider the coordinate axes below. Each box represents 1 meter. y x
What distance did Fred travel?
15 m B
What is the distance between points A and B?
5 m A
What is Fred’s displacement from his original position?
5 meters
10 meters
15 meters
5 meters North
10 meters West
15 meters North
5 m North

11. Section 2-2: Average Velocity
Give a word-equation for average speed.
Give a word-equation for average velocity.

12. Section 2-2: Average Velocity
What is the difference between speed and velocity?
Speed is a scalar—it only has a magnitude (such as 60 mph).
Velocity is a vector—it has magnitude and direction (such as 60 mph north).

13. Section 2-2: Average Velocity
Which is a true statement?
(A) Average speed can be greater than average velocity.
(B) Average velocity can be greater than average speed.
The first one.
A person travels 10 m east, then 5 m west in 10 seconds.
His displacement is 5 m (east), but the distance he traveled is 15 m.
This makes his average velocity (5 m east)/(10 sec) = 0.5 m/s east.
His average speed is (15 m)/(10 sec) = 1.5 m/s.

14. Section 2-2: Average Velocity
The symbol for time is t. Time is measured in seconds.
The symbol for velocity is v. Velocity is measured in meters/second.
“Average” is indicated by putting a straight bar over the symbol v.
“Change” is represented by the letter “delta” ()
“Change” is determined by subtracting:
(value) = (final value) – (initial value)

15. Section 2-2: Average Velocity
Example: Consider the coordinate axes below. Each box represents 1 meter. y x
Frank’s position is Point C, which is at (x = 4, y = 4). Mark this point. C

16. Section 2-2: Average Velocity
Example: Consider the coordinate axes below. Each box represents 1 meter. y x
Frank walks down at a speed of 2 m/s for 4 seconds.
He then immediately walks left 3 m/s for 3 seconds.
He waits for 5 seconds, and then walks 1 m/s upward for 8 seconds and stops at point D.
Plot point C, point D, and Frank’s path. D C
(4, –4)
(–4, 4)
(–5, –4)
(–5, 4)
(5, –4)
(–4, –4)

17. Section 2-2: Average Velocity
Example: Consider the coordinate axes below. Each box represents 1 meter. y x
During this time, what distance did Frank travel? D C
25 m
What is the distance between points C and D?
9 m
What is Frank’s displacement from his original position?
8 meters
9 meters
25 meters
8 meters North
9 meters West
25 meters West
9 m West

18. Section 2-2: Average Velocity
Example: Consider the coordinate axes below. Each box represents 1 meter. y x
What is Frank’s average speed during this time? D C
(Avg. Speed) = (Distance)/(Time)
(Avg. Speed) = (25)/(20)
(Avg. Speed) = 1.25 m/s
0.45 m/s
1.25 m/s
1.67 m/s
0.45 m/s West
1.25 m/s West
1.67 m/s West

19. Section 2-2: Average Velocity
Example: Consider the coordinate axes below. Each box represents 1 meter. y x
What is Frank’s average velocity during this time? D C
(Avg. Vel.) = (Displacement)/(Time)
(Avg. Velocity) = (9)/(20)
(Avg. Velocity) = 0.45 m/s West
0.45 m/s
1.25 m/s
1.67 m/s
0.45 m/s West
1.25 m/s West
1.67 m/s West

20. Section 2-2: Average Velocity
Originally, Frank waited for 5 seconds (below point D). If he wanted to make his average speed half as much, how long should he have waited? Explain your reasoning.
5 seconds
10 seconds
15 seconds
20 seconds
25 seconds
30 seconds

21. Section 2-2: Average Velocity
Originally, Frank waited for 5 seconds (below point D). If he wanted to make his average speed half as much, how long should he have waited? Explain your reasoning.
In order to cut this in half…
…double the denominator.
The total time from before was 20 seconds.
To cut the original avg. speed in half, the total time needs to be 40 sec.
This adds 20 seconds to the original wait time of 5 seconds.
New wait time: 25 seconds.

22. Section 2-2: Average Velocity
Would this amount of wait-time have also halved his average velocity?
…so this is also cut in half.
This was doubled…
Answer: YES.

23. Section 2-3: Instantaneous Velocity
Explain what “instantaneous velocity” is in the simplest terms possible. Make sure you point out the difference between this and “average velocity”.
“Instantaneous velocity” is the rate at which an object moves at a particular point in time.
“Average velocity” is the rate at which an object moves over a particular interval of time.

24. Section 2-3: Instantaneous Velocity
What does a speedometer in a car measure?
Average Speed
Average Velocity
Instantaneous Speed
Instantaneous Velocity
Instantaneous Speed – It tells you your rate of motion right now, but does not give you any information about direction.

25. Section 2-3: Instantaneous Velocity
What two instruments in a car can be used to measure the car’s average speed during a long car trip? Explain how to use these instruments to calculate average speed.
Odometer (mileage gauge) – Measures the distance of the car trip.
Clock– Measures the time of the car trip.
Take the total distance traveled and divide by the total time of the trip to get the average speed over a long car trip.

26. Section 2-4: Acceleration
Write a word-equation for acceleration.
The symbol for acceleration is a. The units are m/s2.

27. Section 2-4: Acceleration
Explain where the “square” in the units for acceleration comes from.
These units are m/s
These units are s
Plug in the units of m/s and s to get:

28. Section 2-4: Acceleration
Explain the difference between velocity and acceleration using the words “rate of change of”.
Velocity – Rate of change of an object’s position. (Or: the rate at which an object’s position changes.)
6 m/s means that the object gains 6 meters of position every second.
Acceleration – Rate of change of an object’s velocity. (Or: the rate at which an object’s velocity changes.)
6 m/s2 means that the object gains 6 m/s of speed every second.

29. Section 2-4: Acceleration
Acceleration is a vector. Explain how to determine the direction of acceleration.
If an object is speeding up, then acceleration points in the same direction as the object’s velocity.
If an object is slowing down, then acceleration points in the direction opposite the object’s velocity.

30. Section 2-4: Acceleration
Example: Each diagram shows a car’s position at one second intervals as it travels along a line in the same direction. The solid [white] car is the final position at time = 5 seconds. For each diagram, make a graph of position vs. time, and draw an arrow indicating the direction of velocity and an arrow for acceleration: 5 10 15 20 25 x t 5 10 15 20 25 1 2 3 4
Velocity
Acceleration
(A) 
(B) 
(C)
(D)
(E)
(F)

31. Section 2-4: Acceleration
Example: Each diagram shows a car’s position at one second intervals as it travels along a line in the same direction. The solid [white] car is the final position at time = 5 seconds. For each diagram, make a graph of position vs. time, and draw an arrow indicating the direction of velocity and an arrow for acceleration: 5 10 15 20 25 x t 5 10 15 20 25 1 2 3 4
Velocity
Acceleration
(A) 
(B) 
(C)
(D)
(E)
(F)

32. Section 2-4: Acceleration
Example: Each diagram shows a car’s position at one second intervals as it travels along a line in the same direction. The solid [white] car is the final position at time = 5 seconds. For each diagram, make a graph of position vs. time, and draw an arrow indicating the direction of velocity and an arrow for acceleration: 5 10 15 20 25 x t 5 10 15 20 25 1 2 3 4
Velocity
Acceleration
(A) 
(B) 
(C)
(D)
(E)
(F)

33. Section 2-4: Acceleration
Example: Each diagram shows a car’s position at one second intervals as it travels along a line in the same direction. The solid [white] car is the final position at time = 5 seconds. For each diagram, make a graph of position vs. time, and draw an arrow indicating the direction of velocity and an arrow for acceleration: 5 10 15 20 25 x t 5 10 15 20 25 1 2 3 4
Velocity
Acceleration
(A) 
(B) 
(C)
(D)
(E)
(F)

34. Section 2-5: Motion at Constant Acceleration
Write the equation for velocity as a function of time for constant acceleration.
Write the equation for average velocity under constant acceleration.

35. Section 2-5: Motion at Constant Acceleration
Write the equation for position as a function of time for constant acceleration.
Write the equation that relates velocity, acceleration, and position (the “no time” equation).

36. Section 2-5: Motion at Constant Acceleration
Identify every symbol that you used in the four equations above.
x = Position
x0 = Initial Position
v = Velocity
v0 = Initial Velocity
a = Acceleration
t = Time

37. Section 2-5: Motion at Constant Acceleration

38. Section 2-6: Solving Problems
Example: A runner finishes a 100-meter dash in 8 seconds. Determine his average speed. G:
x = 100 m, t = 8 sec U:
v = ? E:
SS:
(100) = v(8)
v = 12.5 m/s

39. Section 2-6: Solving Problems
Example: A car can go from zero to 30 m/s in 2 seconds. Determine the car’s average acceleration. G:
v0 = 0, v = 30 m/s, t = 2 sec U:
a = ? E:
SS:
(30) = a(2)
a = 15 m/s2

40. Section 2-6: Solving Problems
Example: Another car can go from zero to 30 m/s in 3 s. How far does the car go while it accelerates? G:
v0 = 0, v = 30 m/s, t = 3 sec U:
x = ? E:
Avg. speed = average of start & end speeds
SS:
x = (15)(3)
x = 45 m

41. Section 2-6: Solving Problems
Example: The radius of Earth’s orbit is 1.5  1011 m. How fast does Earth move as it orbits the Sun?
r = 1.5  1011 m G:
t = 1 year = 365 days
= seconds U:
v = ? E:
x = circumference = 2r
SS:
2(1.5  1011) = v( )
v = m/s

42. Section 2-7: Falling Objects
What type of motion is free-fall?
Constant Position
Constant Velocity
Constant Acceleration
Changing Acceleration
Constant Acceleration

43. Section 2-7: Falling Objects
What does the  symbol mean in the top paragraph on page 33 that states “d  t2”? How is that different than the = symbol?
The  symbols says “is proportional to”. It means that two symbols have the same ratio all the time.
“d  t2” means that d/t2 = (the same number all the time).
When accelerating, d = ½at2, so d/t2 = ½a (the same number all the time).
If d = t2, then d/t2 = 1 all the time.
For a circle, A = r2. This means that A/r2 =  (the same number all the time), so A  r2

44. Section 2-7: Falling Objects
Explain Figure 2-17 on page 33. Why does the paper fall differently than the ball in one case, but not in the other?
The paper’s surface area contributes to the force of air resistance on it. When the paper is balled up, air resistance is greatly reduced and it falls at the same acceleration as the ball and everything else in free-fall.

45. Section 2-7: Falling Objects
What is the value of the acceleration of gravity on Earth’s surface? What symbol do we give this number?
The acceleration of gravity on earth is 9.8 m/s2. We give this value the special symbol g.
When a calculator is not available to us (or sometimes when it is just convenient), we use g = 10 m/s2.

46. Section 2-7: Falling Objects
Example: A person throws a ball upward near the edge of a building as shown. Ignore air resistance. The first white dot is the position of the ball when it leaves the thrower’s hand. Each white dot after that represents the ball’s position every second after the ball leaves the thrower’s hand. In the table below, the time represents how many seconds after the ball leaves the thrower’s hand.
Time
(seconds)
Position x
(meters)
Speed
(m/s)
Velocity
Accel.
(m/s2) 1 2 3 4 5 20 20
–10 15 10
–10 10 20
–10 15 10
–10
–10 20
–20
–10
–25 30
–30
–10

47. Section 2-7: Falling Objects
Fill in the boxes representing the ball’s velocity at each time.
When the ball is at ____________________, we know for sure its velocity is ________________.
Each second, the ball’s velocity ______________________________________________________.
The speed is almost the same as velocity, except
__________________________________________________.
Fill in the ball’s acceleration at each time. How does the ball’s acceleration vary with time? ____________.
the highest height
zero
decreases by 10 m/s (has –10 m/s added to it every sec)
speed has no direction and is always positive
It doesn’t

48. Section 2-8: Graphical Analysis of Linear Motion
How can a curve be considered to have a slope?
We look at a point on the curve, and create a line tangent to the curve at that point. The slope of the tangent line is the same as the slope of the curve at that point. y x

49. Section 2-8: Graphical Analysis of Linear Motion
How can velocity at a certain time be found by looking at a position vs. time graph?
Velocity is the slope of a position vs. time graph. x t
Going slow in the positive direction
Going fast in the positive direction
Stop for an instant
Going fast in the negative direction

50. Section 2-8: Graphical Analysis of Linear Motion
How can the change in position between two times be found from a velocity vs. time graph?
On a velocity vs. time graph, the area between the graph and the horizontal axis gives the change in position of an object.
v (m/s)
t (s) 1 2 3 4 5 –1 –2 –3 –4 –5 6 7 8 9 10
This area (of 12) means the object went 12 m in these 5 sec.
This area of –6 means the object went 6 m
to the left in these 3 seconds.

51. Section 2-8: Graphical Analysis of Linear Motion
How can acceleration at a certain time be found by looking at a velocity vs. time graph?
Acceleration is the slope of a velocity vs. time graph.

52. Section 2-8: Graphical Analysis of Linear Motion
How can the change in velocity between two times be found from an acceleration vs. time graph?
On an acceleration vs. time graph, the area between the graph and the horizontal axis gives the change in velocity of an object.

53. Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following x vs. t graph.
x (m)
t (sec) 5 –5 10 20
Estimate the velocity at:
t = 4 ________ t = 10 ________ t = 12 ________
–2 m/s
0 m/s
1 m/s

54. Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following x vs. t graph.
x (m)
t (sec) 5 –5 10 20
During what time intervals is the velocity
positive? ______________________________
negative? ___________________________
zero? ___________________________
10 to 21 sec
0 to 8 sec
8 to 10 sec

55. Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following x vs. t graph.
x (m)
t (sec) 5 –5 10 20
During what time intervals is the acceleration
positive? ____________________________
negative? ___________________________
zero? ___________________________
4 to 8, 10 to 12, 14 to 16 sec
0 to 4, 12 to 14, 19 to 21 sec
8 to 10, 16 to 19 sec

56. Section 2-8: Graphical Analysis of Linear Motion
x (m)
t (sec) 5 –5 10 20
Velocity
Negative
Positive
Acceleration

57. Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following x vs. t graph.
x (m)
t (sec) 5 –5 10 20
What is the object’s average velocity between 0 and 10 seconds? _______________
What is the object’s acceleration between 0 and 4 seconds? _______________
–0.8 m/s
–1 m/s

58. Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following v vs. t graph.
v (m/s)
t (sec) 5 –5 10 20
What is the acceleration at:
t = 11 s? __________ t = 15 s? __________
–2 m/s2
0 m/s2

59. Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following v vs. t graph.
v (m/s)
t (sec) 5 –5 10 20
Area = 24
Area = –16
How far, and in which direction, did the object travel:
from t = 4 to t = 11 s? __________ from t = 13 to t = 18 s? __________
24 m right
16 m left

60. Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following v vs. t graph.
v (m/s)
t (sec) 5 –5 10 20
During what time intervals is the object
speeding up? ______________________________
slowing down? ______________________________
0 to 4, 11 to 13 seconds
9 to 11, 16 to 18 seconds

61. Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following v vs. t graph.
v (m/s)
t (sec) 5 –5 10 20
During what time intervals is the acceleration
positive? ______________________________
negative? ____________________________
zero? ____________________________
0 to 4, 16 to 18 seconds
9 to 13 seconds
5 to 9, 13 to 16 seconds

62. Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following v vs. t graph.
v (m/s)
t (sec) 5 –5 10 20
Area = –12
Area = –4
Area = –4
Area = 8
Area = 20
Area = 4
Net Area = 12
If the object is at x = –5 m at t = 0 s, where is the object at t = 20 s? __________
+7 m

63. Slope = 1 Value = 1 v (m/s) t (sec) 5 –5 10 20 a (m/s2) t (sec) 1 –1–5 10 20
Slope = 1
a (m/s2)
t (sec) 1 –1 10 20 2 –2
Value = 1

64. Slope = 0 Value = 0 v (m/s) t (sec) 5 –5 10 20 a (m/s2) t (sec) 1 –1–5 10 20
Slope = 0
a (m/s2)
t (sec) 1 –1 10 20 2 –2
Value = 0

65. Slope = –2 Value = –2 v (m/s) t (sec) 5 –5 10 20 a (m/s2) t (sec) 1 –1–5 10 20
Slope = –2
a (m/s2)
t (sec) 1 –1 10 20 2 –2
Value = –2

66. Slope = 0 Value = 0 v (m/s) t (sec) 5 –5 10 20 a (m/s2) t (sec) 1 –1–5 10 20
Slope = 0
a (m/s2)
t (sec) 1 –1 10 20 2 –2
Value = 0

67. Slope = 2 Value = 2 v (m/s) t (sec) 5 –5 10 20 a (m/s2) t (sec) 1 –1–5 10 20
Slope = 2
a (m/s2)
t (sec) 1 –1 10 20 2 –2
Value = 2

68. Slope = 0 Value = 0 v (m/s) t (sec) 5 –5 10 20 a (m/s2) t (sec) 1 –1–5 10 20
Slope = 0
a (m/s2)
t (sec) 1 –1 10 20 2 –2
Value = 0

69. Area = 8 Value = 8 Now x =3 v (m/s) t (sec) 5 –5 10 20 x (m) t (sec)–5 10 20
Area = 8
x (m)
t (sec) 5 –5 10 20 15 25
Value = 8
Now x =3

70. Area = 20 Value = 20 Now x = 23 v (m/s) t (sec) 5 –5 10 20 x (m)–5 10 20
Area = 20
x (m)
t (sec) 5 –5 10 20 15 25
Value = 20
Now x = 23

71. Area = 4 Value = 4 Now x = 27 v (m/s) t (sec) 5 –5 10 20 x (m)–5 10 20
Area = 4
x (m)
t (sec) 5 –5 10 20 15 25
Value = 4
Now x = 27

72. Area = –4 Value = –4 Now x = 23 v (m/s) t (sec) 5 –5 10 20 x (m)–5 10 20
Area = –4
x (m)
t (sec) 5 –5 10 20 15 25
Value = –4
Now x = 23

73. Area = –12 Value = –12 Now x = 11 v (m/s) t (sec) 5 –5 10 20 x (m)–5 10 20
Area = –12
x (m)
t (sec) 5 –5 10 20 15 25
Value = –12
Now x = 11

74. Area = –4 Value = –4 Now x = 7 v (m/s) t (sec) 5 –5 10 20 x (m)–5 10 20
Area = –4
x (m)
t (sec) 5 –5 10 20 15 25
Value = –4
Now x = 7

75. Area = 0 Value = 0 Still x = 7 v (m/s) t (sec) 5 –5 10 20 x (m)–5 10 20
Area = 0
x (m)
t (sec) 5 –5 10 20 15 25
Value = 0
Still x = 7

76. Example: For each pair of graphs, use the given graph to make a sketch of the other graph. Dashed lines are for reference. x t v
SLOPE
SLOPE starts positive and
decreases to zero
SLOPE goes from zero into
the negatives.
VALUE goes from zero into
the negatives.
VALUE
VALUE starts positive and
decreases to zero

77. Example: For each pair of graphs, use the given graph to make a sketch of the other graph. Dashed lines are for reference. v t a
SLOPE starts negative and
goes to zero
SLOPE goes from zero into
the positives.
SLOPE is zero.
SLOPE
VALUE starts negative and
goes to zero
VALUE is zero.
VALUE
VALUE goes from zero into
the positives.

78. Example: For each pair of graphs, use the given graph to make a sketch of the other graph. Dashed lines are for reference. a t v
VALUE
VALUE starts at zero and goes positive
VALUE is a constant positive
VALUE goes from positive to zero
SLOPE starts at zero and goes positive
SLOPE
SLOPE is a constant positive
SLOPE goes from positive to zero

79. Example: For each pair of graphs, use the given graph to make a sketch of the other graph. Dashed lines are for reference. v t x
VALUE starts negative and goes to zero
VALUE goes from zero to negative
VALUE
VALUE goes from zero to a maximum positive
VALUE goes from maximum positive to zero
SLOPE starts negative and goes to zero
SLOPE goes from zero to a maximum positive
SLOPE
SLOPE goes from maximum positive to zero
SLOPE goes from zero to negative