Chapter 2:
Motion in One Dimension
Displacement and Velocity
To describe motion, we must
specify position.
For motion in one dimension, it is
convenient to use the x-axis.
From this concept, we can create a
number line.
0
0
+
-
1
Displacement
Displacement (!x)
0
!
!
Displacement of an object = the
change of position of the object
NOTE: displacement is not always
equal to the distance traveled.
0
!
!x = xf – xi
!
Where
!x = displacement
xf = final position
xi = initial position
Example
Example
If a car moves from an initial position
of 10 m to a final position of 80 m:
If a car moves from an initial position
of 80 m to a final position of 20 m:
!x =80 m – 10 m = 70 m
!x =20 m – 80 m = -60 m
Note: the negative sign indicates the
direction of displacement.
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)
1.
2.
3.
4.
Your
Your
Your
Your
displacement
displacement
displacement
displacement
is
is
is
is
0, distance traveled is 8 mi
8 mi, distance traveled is 0
0, distance traveled is 0
8 mi, distance traveled is 8 mi
VECTORS
!
A VECTOR is a physical quantity that
has both magnitude and direction
(displacement, velocity, acceleration)
2
VELOCITY – change in
displacement over time.
!
v
Average velocity (vavg ) :
"x
v
vavg =
=
"t
We can represent the
motion of a train on a
track as the position
as a function of time.
x f ! xi
t f ! ti
What’s this called???
A train leaves Denver at 8:05 a.m. and travels directly
to Colorado Springs, arriving at 9:02 a.m. The distance
between Denver and Colorado Springs is 68.5 miles.
Find the average velocity in miles/hr of the train.
v !x
!v =
!t
v 68.5 miles 60 min
!v =
!
57 min
1 hr
Time (s)
v
!v = 72 mi/hr
In this case, displacement stays
constant with time…there is no
movement.
Time (s)
For Example…
Position (m)
For Example…
Position (m)
!
Position (m)
Example
We can represent a trip on the train
track with a position vs. time graph.
Here the train has a low
constant velocity for the first
hour, and this it changes to a
higher constant velocity.
1 hr
Time (s)
3
For Example…
This time the train 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)
Position (m)
For Example…
Time (s)
Time (s)
SPEED vs. VELOCITY
!
!
!
Both describe how fast the position is
changing with respect to time.
Speed is a SCALAR quantity. It indicates
magnitude, but not direction.
Velocity is a VECTOR quantity. It
indicates both magnitude and direction.
SLOPE REVIEW
slope = m =
Velocity is not constant.
Velocity (slope) is increasing
at a constant rate.
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).
SLOPE REVIEW
rise y 2 ! y1
=
run x2 ! x1
4
SLOPE REVIEW
SLOPE REVIEW
slope =
Position (m)
SLOPE REVIEW
Rise !x
=
=
Run !t
v
v
Rise
Run
Time (s)
Position (m)
Consider this trip…
Time (s)
5
INSTANTANEOUS VELOCITY
How fast is the car going
at this instant in time?
Position (m)
Position (m)
INSTANTANEOUS VELOCITY
Time (s)
Time (s)
Describing Motion
INSTANTANEOUS VELOCITY
Position (m)
The slope of a curve at a
given point, is equal to the
slope of a tangent line at
that point.
The x(t) graph describes a 1-D
motion of a train. What must be
true about this motion?
The slope of this line
represents instantaneous
velocity at the indicated
point.
1. Speeds up all the time
2. Slows down all the time
3. Speeds up part of the time
4. Slows down part of the time
Hint: What is
the meaning of
the slope of the
x(t) graph?
5. Speeds up & then slows down
Time (s)
6. Slows down & then speeds up
Position vs. Time Graphs
x
The x(t) graph displays
motions of two trains A and
B on parallel tracks. Which
statement is true?
A
B
Motion can be described with a velocity
vs. time graph.
t
t1
1. At t1 both trains have the same velocity
2. At t1 both trains have the same speed
3. At t1 both trains have the same acceleration
4. Both trains have the same velocity sometime before t1
5. The trains never have the same velocity
6
constant velocity
n
co
st
an
c
ta
ce
le r
io
at
Velocity vs. Time 2
n
Velocity (m/s)
Velocity (m/s)
Velocity vs. Time 1
+
0
+ velocity
0 velocity
Time (s)
- velocity
-
Time (s)
The Slope of Velocity vs.
Time Graphs
Acceleration
Velocity (m/s)
Acceleration – rate of change in
velocity over time.
Rise = ! v
v
"v v f ! vi
v
aavg =
=
"t t f ! ti
Run = ! t
Time (s)
Slope =
!v
v
= acceleration = a
!t
v
where aavg is average acceleration
A train travels at 5 miles per hour for 1
hour. What is its displacement after 1 hour?
v !x
v=
!t
v
!x = v " !t
!x = (5 mi/hr)(1 hr)
!x = 5 mi
Displacement can also be determined by finding the
area under the curve of a velocity vs. time graph.
Using a graph
Velocity (m/s)
Consider a trip…
5 mi/hr
1 hour
Time (s)
7
Velocity (m/s)
Find the “area under the curve” (AUC).
Graphical analysis summary
Displacement vs. Time graph:
Slope = velocity
5 mi/hr
Velocity vs. Time graph:
1 hour
Time (s)
Area = l x w = 1 hour x 5 mi/hr = 5 miles.
Slope = acceleration
AUC = displacement
Graphing Examples
Draw a “v vs. t” graph for:
!
The answer
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 clock is reset to 0 seconds and the velocity-time
data for both cars is collected and plotted.
The red car accelerates from rest at 4 m/s/s for three
seconds and then maintains a constant speed. The blue
car maintains a constant speed of 10 m/s for the entire
12 seconds.
8
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.
Velocity with Constant Acceleration
velocity
v
vo
Rise = !v
= v f - vo
vo
vf
velocity
velocity
vf
time
t
time
Find the AUC…
vi
Run = !t = (t – 0) = t
t
time
t
time
Break the area into two parts…
"v v # v
slope = = f o = a
"t
t
v f =vo +at
Area = (l x w) + (!b x h)
!x = vit + ! t(vf-vi)
(p. 35)
!
velocity
vf
Find the AUC…
vi
We now know…
1) vf = vo + at
time
t
2) !x = ! (vo + vf)t
By substitution of equation #1 into #2
AUC = !x = vot + ! t(vf-vo)
!x = vot + ! vft – ! vot
!x = ! vot + ! vft
!x = ! (vo+vf)t
and
distribution
simplification
p.(35)
!x = ! [vo + (at +vo)]t
!x = ! (2vo + at)t
!x = vot + ! at2
substitution
combining terms
(p. 36)
9
Solve Eq #1 for t & substitute into Eq #2…
1) vf = vo + at
t=
vf " vo
a
2) !x = ! (vo + vf)t
(v o + v f ) (v f $ v o )
#
2
a
2
2
vf $ vo
"x =
2a
2
2
v f = v o + 2a"x
(p. 36)
"x =
!
5 Parameters of Motion
velocity m
= 2
time
s
1. a = acceleration
2. !x = displacement
m
m
s
m
s
3. vf = final velocity
4. vi = initial velocity
5. t = time
sec
!
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:
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
Example…
required for the plane to come to rest, and (b)
the minimum size of the runway.
vf = vo + a!t
a) vi = +100m/s
vf= 0 m/s
a = -5.0 m/s2
!x = !(vo + vf)!t
b) Solve for !x
!x = vo!t +
!a(!t)2
vi = 20 m/s
a = -1
m/sec2
t = 6 sec
!x = ?
!x = vit + ! at2
0m/s = 100m/s + (-5m/s2)t
t = 20s
vf2 = vi2 + 2a!x
0m/s = (100m/s)2 + 2(- 5m/s2) !x
vf2 = vo2 + 2a!x
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?
vf = vi + at
!x = 1000 m
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 = ?
vf 2 = vi2 + 2a!x
!x = (20m/s)(6s) + ! (-1m/s2)(6s)2
vf 2 = 0 + 2(5m/s2)(30.5m)
!x = 120m – 18m = 102 m
vf = 305 m 2 /s 2
!x = 100 m
(sig. figs!)
vf = 17.5 m/s
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