Unit B01 – Motion in One
Dimension
[UNAUTHORIZED COPYING OR USE OF ANY PART
OF ANY ONE OF THESE SLIDES IS ILLEGAL.]
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.
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.
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.
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?
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.
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
Section 2-1: Reference Frames and Displacement
Example: Consider the coordinate axes below. Each box represents 1
meter.
y
Fred’s position is Point A, which is
at (x = 3, y = –3). Mark this point.
x
A
Section 2-1: Reference Frames and Displacement
Example: Consider the coordinate axes below. Each box represents 1
meter.
y
Fred walks 5 meters to the left, 5
B
x
A
meters up, and 5 meters to the
right, arriving at point B. Mark
point B and Fred’s path on the
diagram.
A. (2, 2)
B. (–2, 2)
C. (–2, –3)
D. (–3, –2)
E. (2, 3)
F. (3, 2)
Section
ReferenceD.
Frames
and Displacement
A. 5 2-1:
meters
5 meters
North
B. 10 the
meters
E. axes
10 meters
West
Example: Consider
coordinate
below. Each
box represents 1
meter.
C. 15 meters F. 15 meters North
y
What distance did Fred travel?
15 m
B
x
A
What is the distance between
points A and B?
5m
What is Fred’s displacement from
his original position?
5 m North
Section 2-2: Average Velocity
Give a word-equation for average speed.

distance traveled 
average speed  
time elapsed 
Give a word-equation for average velocity.

net displaceme nt 
average velocity  
time elapsed 
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).
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.
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.
v
“Change” is represented by the letter “delta” ()
“Change” is determined by subtracting:
(value) = (final value) – (initial value)
Section 2-2: Average Velocity
Example: Consider the coordinate axes below. Each box represents 1
meter.
y
Frank’s position is Point C, which
is at (x = 4, y = 4). Mark this
point.
C
x
Example:
meter.
A.Section
(4, –4)
D. (–5,
4)
2-2: Average
Velocity
B.
(–4,
4)
E.
(5,
–4)
Consider the coordinate axes below. Each box represents 1
C. (–5, –4) F. (–4, –4)
y
D
Frank walks down at a speed of 2
m/s for 4 seconds.
C
x
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.
Example:
meter.
A. 8Section
meters
D. 8 Velocity
meters North
2-2: Average
B.
9
meters
E.
9
meters
West
Consider the coordinate axes below. Each box represents 1
C. 25 meters F. 25 meters West
y
D
During this time, what distance did
Frank travel?
C
25 m
x
What is the distance between
points C and D?
9m
What is Frank’s displacement from
his original position?
9 m West
Example:
meter.
A. 0.45
m/s
D. 0.45
m/s West
Section
2-2: Average
Velocity
B.
1.25
m/s
E.
1.25
m/s
West
Consider the coordinate axes below. Each box represents 1
C. 1.67 m/s F. 1.67 m/s West
y
D
What is Frank’s average speed
during this time?
C
(Avg. Speed) = (Distance)/(Time)
x
(Avg. Speed) = (25)/(20)
(Avg. Speed) = 1.25 m/s
Example:
meter.
A. 0.45
m/s
D. 0.45
m/s West
Section
2-2: Average
Velocity
B.
1.25
m/s
E.
1.25
m/s
West
Consider the coordinate axes below. Each box represents 1
C. 1.67 m/s F. 1.67 m/s West
y
D
C
What is Frank’s average velocity
during this time?
(Avg. Vel.) = (Displacement)/(Time)
x (Avg. Velocity) = (9)/(20)
(Avg. Velocity) = 0.45 m/s West
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.
A. 5 seconds
B. 10 seconds
C. 15 seconds
D. 20 seconds
E. 25 seconds
F. 30 seconds
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.

distance traveled 
average speed  
time elapsed 
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.
Section 2-2: Average Velocity
Would this amount of wait-time have also halved his average velocity?

net displaceme nt 
average velocity  
time elapsed 
…so this is also cut in half.
Answer: YES.
This was doubled…
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.
Section 2-3: Instantaneous Velocity
What does a speedometer in a car measure?
A. Average Speed
B. Average Velocity
C. Instantaneous Speed
D. Instantaneous Velocity
Instantaneous Speed – It tells you your rate of motion right now, but
does not give you any information about direction.
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.
Section 2-4: Acceleration
Write a word-equation for acceleration.

change in velocit y 
accelerati on  
time elapsed 
The symbol for acceleration is a. The units are m/s2.
Section 2-4: Acceleration
Explain where the “square” in the units for acceleration comes from.
These units are m/s

change in velocit y 
accelerati on  
time elapsed 
These units are s
Plug in the units of m/s and s to get:

m/s 
accelerati on  
s
 m  1   m 
accelerati on         2 
 s s  s 
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.
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.
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:
0
5
Velocity Acceleration
(A)
(B)
(C)
(D)
(E)
(F)










0
0
10
15
25
20
25
x
20
15
10
5
0
t
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:
0
5
Velocity Acceleration
(A)
(B)
(C)
(D)
(E)
(F)










0
0
10
15
25
20
25
x
20
15
10
5
0
t
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:
0
5
Velocity Acceleration
(A)
(B)
(C)
(D)
(E)
(F)










0
0
10
15
25
20
25
x
20
15
10
5
0
t
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:
0
5
Velocity Acceleration
(A)
(B)
(C)
(D)
(E)
(F)










0
0
10
15
25
20
25
x
20
15
10
5
0
t
Section 2-5: Motion at Constant Acceleration
Write the equation for velocity as a function of time for constant
acceleration.
v  at  v0
Write the equation for average velocity under constant acceleration.
x
v
t
v  v0
v
2
Section 2-5: Motion at Constant Acceleration
Write the equation for position as a function of time for constant
acceleration.
x  at  v0t  x0
1
2
2
Write the equation that relates velocity, acceleration, and position
(the “no time” equation).
v  2ax  x0   v0
2
2
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
Section 2-5: Motion at Constant Acceleration
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: x  v t
SS:
(100) = v(8)
v = 12.5 m/s
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: v  at  v0
SS:
(30) = a(2)
a = 15 m/s2
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: x  v t
SS:
x = (15)(3)
x = 45 m
Avg. speed = average of start & end speeds
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?
G:
r = 1.5  1011 m
t = 1 year = 365 days = 31536000 seconds
U:
v=?
E:
x  vt
SS:
2(1.5  1011) = v(31536000)
v = 29886 m/s
x = circumference = 2r
Section 2-7: Falling Objects
What type of motion is free-fall?
A. Constant Position
B. Constant Velocity
C. Constant Acceleration
D. Changing Acceleration
Constant Acceleration
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
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.
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.
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)
0
1
2
3
4
5
Position x
(meters)
Speed
(m/s)
Velocity
(m/s)
Accel.
(m/s2)
0
15
20
15
0
–25
20
10
0
10
20
30
20
10
0
–10
–20
–30
–10
–10
–10
–10
–10
–10
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
the highest height
zero
velocity is ________________.
Each second, the ball’s velocity
decreases by 10 m/s (has –10 m/s added to it every sec)
______________________________________________________.
The speed is almost the same as velocity, except
speed has no direction and is always positive
__________________________________________________.
Fill in the ball’s acceleration at each time. How does the ball’s
It doesn’t
acceleration vary with time? ____________.
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
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
Going slow in the positive direction
Going fast in the positive direction
Stop for an instant
Going fast in the negative direction
t
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)
5
4
3
2
1
0
–1
–2
–3
–4
–5
This area (of 12) means the object went 12 m in these 5 sec.
1 2 3 4 5 6 7 8 9 10
t (m/s)
This area of –6 means the object went 6 m
to the left in these 3 seconds.
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.
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.
Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following x vs. t graph.
x (m)
5
0
10
20
t (sec)
–5
Estimate the velocity at:
–2 m/s
t = 4 ________
0 m/s
t = 10 ________
1 m/s
t = 12 ________
Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following x vs. t graph.
x (m)
5
0
10
–5
During what time intervals is the velocity
10 to 21 sec
positive? ______________________________
0 to 8 sec
negative? ___________________________
zero? ___________________________
8 to 10 sec
20
t (sec)
Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following x vs. t graph.
x (m)
5
0
10
–5
During what time intervals is the acceleration
4 to 8, 10 to 12, 14 to 16 sec
positive? ____________________________
0 to 4, 12 to 14, 19 to 21 sec
negative? ___________________________
zero? ___________________________
8 to 10, 16 to 19 sec
20
t (sec)
Section 2-8: Graphical Analysis of Linear Motion
x (m)
5
10
20
Acceleration
–5
Positive
Negative
Negative
0
t (sec)
Velocity
Positive
Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following x vs. t graph.
x (m)
5
0
10
20
t (sec)
–5
What is the object’s average velocity between 0 and 10 seconds?
–0.8 m/s
_______________
What is the object’s acceleration between 0 and 4 seconds?
–1 m/s
_______________
Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following v vs. t graph.
v (m/s)
5
0
10
20
–5
What is the acceleration at:
–2 m/s2
t = 11 s? __________
0 m/s2
t = 15 s? __________
t (sec)
Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following v vs. t graph.
v (m/s)
5
Area = 24
0
10
20
Area = –16
–5
How far, and in which direction, did the object travel:
24 m right
from t = 4 to t = 11 s? __________
16 m left
from t = 13 to t = 18 s? __________
t (sec)
Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following v vs. t graph.
v (m/s)
5
0
10
–5
During what time intervals is the object
0 to 4, 11 to 13 seconds
speeding up? ______________________________
9 to 11, 16 to 18 seconds
slowing down? ______________________________
20
t (sec)
Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following v vs. t graph.
v (m/s)
5
0
10
–5
During what time intervals is the acceleration
0 to 4, 16 to 18 seconds
positive? ______________________________
9 to 13 seconds
negative? ____________________________
zero? ____________________________
5 to 9, 13 to 16 seconds
20
t (sec)
Section 2-8: Graphical Analysis of Linear Motion
Example: Consider the following v vs. t graph.
v (m/s)
5
Area = –12
Area = –4
0
Area = 8
Area = 20
Area10
=4
Area = –4
20
t (sec)
Net Area = 12
–5
If the object is at x = –5 m at t = 0 s, where is the object at t = 20 s?
+7 m
__________
v (m/s)
5
0
10
20
10
20
t (sec)
Slope = 1
–5
a (m/s2)
2
1
0
–1
–2
Value = 1
t (sec)
v (m/s)
5
0
10
20
10
20
t (sec)
Slope = 0
–5
a (m/s2)
2
1
0
–1
–2
Value = 0
t (sec)
v (m/s)
5
0
10
20
t (sec)
Slope = –2
–5
a (m/s2)
2
1
0
–1
–2
10
Value = –2
20
t (sec)
v (m/s)
5
0
10
–5
20
t (sec)
Slope = 0
a (m/s2)
2
1
0
–1
–2
10
20
Value = 0
t (sec)
v (m/s)
5
Slope = 2
0
10
20
10
20
t (sec)
–5
a (m/s2)
2
1
0
–1
–2
Value = 2
t (sec)
v (m/s)
5
Slope = 0
0
10
20
10
20
t (sec)
–5
a (m/s2)
2
1
0
–1
–2
Value = 0
t (sec)
v (m/s)
5
0
10
20
10
20
t (sec)
Area = 8
–5
x (m)
25
20
15
10
5
0
–5
Value = 8
Now x =3
t (sec)
v (m/s)
5
0
10
20
t (sec)
Area = 20
–5
x (m)
25
20
15
10
Value = 20
Now x = 23
5
0
–5
10
20
t (sec)
v (m/s)
5
0
10
Area = 4
20
t (sec)
–5
x (m)
25
20
15
Value = 4
Now x = 27
10
5
0
–5
10
20
t (sec)
v (m/s)
5
Area = –4
0
10
20
t (sec)
–5
x (m)
25
20
Value = –4
Now x = 23
15
10
5
0
–5
10
20
t (sec)
v (m/s)
5
Area = –12
0
10
20
t (sec)
–5
x (m)
25
20
15
10
Value = –12
Now x = 11
5
0
–5
10
20
t (sec)
v (m/s)
5
Area = –4
0
10
20
10
Value = –4
Now x = 7 20
t (sec)
–5
x (m)
25
20
15
10
5
0
–5
t (sec)
v (m/s)
5
Area = 0
0
10
20
10
Value = 0
Still x = 7 20
t (sec)
–5
x (m)
25
20
15
10
5
0
–5
t (sec)
Example: For each pair of graphs, use the given graph to make a sketch of the other
graph. Dashed lines are for reference.
x
SLOPE
t
SLOPE starts positive and SLOPE goes from zero into
decreases to zero
the negatives.
v
VALUE goes from zero into
the negatives.
VALUE
t
VALUE starts positive and
decreases to zero
Example: For each pair of graphs, use the given graph to make a sketch of the other
graph. Dashed lines are for reference.
v
SLOPE starts
negative and
goes to zero
SLOPE is zero.
SLOPE goes from
zero into
the positives.
SLOPE
t
a
VALUE starts
negative and
goes to zero
VALUE is zero.
VALUE
t
VALUE goes from
zero into
the positives.
Example: For each pair of graphs, use the given graph to make a sketch of the other
graph. Dashed lines are for reference.
a
VALUE
t
VALUE
starts at
zero and
goes positive
VALUE is a constant
positive
VALUE goes
from positive
to zero
v
SLOPE
SLOPE
starts at
zero and
goes positive
t
SLOPE is a constant
positive
SLOPE goes
from positive
to zero
Example: For each pair of graphs, use the given graph to make a sketch of the other
graph. Dashed lines are for reference.
v
VALUE
starts
negative and
goes to zero
VALUE
goes from
zero to
negative
VALUE
t
VALUE goes from VALUE goes from
zero to a
maximum positive
maximum positive
to zero
x
SLOPE
SLOPE goes from
starts
zero to a
negative and
goes to zero maximum positive
SLOPE
t
SLOPE goes from
SLOPE
maximum positive goes from
to zero
zero to
negative