Physics
Notes: Motion in One Dimension
Goal: The students will gain knowledge and understanding of the fundamentals of
kinematics and apply this to motion in one dimension.
Objectives: Upon completion of this unit, the students should be able to:
1.
Describe motion in terms of frame of reference, displacement, time, and velocity.
The simplest kind of motion is the kind of motion that takes place in one direction.
A frame of reference is a coordinate system for specifying the precise location of
objects in space.
Displacement is a change in position in a certain direction, not the total distance
traveled.
Displacement
x  x f - xi
displacement = change in position = final position – initial position
The Greek letter  before the x denotes a change in the position of an object.
Displacement can be positive (moving to the right or upward), or negative (moving
to the left or downward).
Speed is a scalar quantity. It is how fast you are traveling.
Velocity is a vector quantity. It is how fast you are going and in what direction.
Acceleration is a vector quantity. It is the rate at which the velocity is changing.
Since velocity is dependent on both speed and direction, if either of those change,
your velocity changes and you have acceleration.
Deceleration is a special name given to an acceleration that is in a direction
opposite to the direction of motion of the object.
Relative velocity is expressed relative to a particular reference frame.
Sample Problem
A Mustang is being driven northbound on I-45 at 140 km/h when he passes your
car also northbound but moving at 115 km/h.
1. What is the relative velocity of the Mustang from your perspective?
2. What is your relative velocity from the perspective of the driver of the Mustang?
3. Assuming a police car is driving southbound at 115 km/h, what is the relative
velocity of the Mustang?
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The average velocity of an object during the same time interval is equal to the
displacement of the object divided by the time interval. Like displacement, velocity
indicates both speed and direction.
Average Velocity
v avg 
average velocity 
x x f - xi

t
t f - ti
change in position displaceme nt

change in time
time interval
The average velocity of an object can be positive or negative, depending on the
sign of the displacement. The time interval is always positive.
Sample Problem
If a car is traveling at a velocity of 24.5 m/s (55 mph) toward the east, what is the
displacement of the car during a 3-s interval in which the driver looks away from the
road to wave at a friend?
In everyday language, the terms speed and velocity are used interchangeably. In
physics, however, there is an important distinction between these two terms.
Velocity describes motion with both a direction and a numerical value (a
magnitude) indicting how fast something moves. However, speed has no direction,
only magnitude. An object’s average speed is equal to the distance traveled divided
by the time interval for the motion.
average speed 
distance traveled
time of travel
The velocity of an object is often not constant over a period of time. To determine
the velocity of an object at some instant, we study a small time interval near that
instant. As the intervals become smaller and smaller, the average velocity over that
interval approaches the exact velocity at that instant. This is called instantaneous
velocity.
Objects in motion are often changing velocity. The velocity is either increasing or
decreasing. An object loses speed as it slows to a stop, and gains speed as it
begins to move again.
Acceleration is the rate of change of velocity in a given time interval.
a avg 
v v f  v i

t t f  t i
average acceleration =
2
change in velocity
time required for change
The units for acceleration are m/s2.
2. Construct and interpret graphs of position versus time.
The velocity of an object can be determined if its position is known at specific times
along its path. One way to determine this is to make a graph of the motion. Time is
plotted on the horizontal axis and position is plotted on the vertical axis.
d
d
t
t
Constant Positive Velocity
Constant Position
A position-time graph for constant position is a straight line parallel to the time axis.
A negative value of a position-time graph indicates a negative displacement. If the
position-time graph is a straight (non-horizontal) line, the object is moving with
constant velocity. Position varies directly with time.
The slope of a line of a line on a position-time graph is the average velocity. If the
position-time graph is a straight line moving with a constant average velocity. A
negative slope on the position-time graph shows that the object is moving in the
opposition direction to its original motion.
Slope of a Line
slope =
Average Velocity
rise
change in vertical coordinates
=
run
change in horizontal coordinates
v ave 
d d f - d i

t
t f - ti
Sample Problem
Calculate the velocity of
the object in the graph to
the right.
Position vs. Time
Distance (meters)
100
80
60
40
20
0
0
1
2
3
Time (seconds)
3
4
5
d
d
t
t
Constant Positive Velocity
Constant Positive Acceleration
A position-time graph for constant velocity is a diagonal straight line. As time
increases, the position changes at a constant rate. The position-time graph for an
object in constant acceleration is half a parabola. Position varies as the square of
time.
To determine the instantaneous velocity on a graph of constant acceleration,
construct a straight line that is tangent to the position versus time graph at that
instant. The slope of this tangent line is equal to the value of the instantaneous
velocity at that point.
3.
Plot and interpret a velocity-time graph.
A velocity-time graph can be used to describe motion with either constant or
changing velocity. Time is the independent variable and velocity is the dependent
variable.
An object with constant position is represented by a line along the time axis, that is,
zero velocity. A velocity-time graph of constant velocity is a straight line parallel to
the time axis. A velocity-time graph for constant acceleration is a straight (nonhorizontal) line.
v
v
v
t
Zero Velocity
t
Constant Velocity
t
Constant Acceleration
The slope of the line on a velocity-time graph is the acceleration of the object. The
greater the slope, the greater the acceleration. A negative slope indicates a
negative acceleration (deceleration, or slowing down).
slope 
Δy v

 aA
Δx t
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The area under the line on a velocity-time graph is equal to the displacement of the
object from its original position to its position at time, t.
area = length x width  t  v  d
Sample Problem
Velocity vs. Time
Velocity (meters per
second)
Calculate the acceleration of
the object shown in the graph
to the right.
16
14
12
10
8
6
4
2
0
0
1
2
3
4
5
Time (seconds)
Sample Problem
What is the displacement of the
object during the first 5 seconds?
During the entire 25 s?
Velocity vs. Time
Velocity, m/s
14
12
10
8
6
4
2
0
0
5
10
15
20
25
30
Time, s
4.
Plot an acceleration-time graph and interpret its meaning.
An acceleration-time graph can be used to represent constant or changing
acceleration. Time is the independent variable and acceleration is the dependent
variable. If acceleration is constant, an acceleration - time graph is a line parallel to
the time axis.
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6
a
a
t
Constant Position
a
t
Constant Acceleration
t
Constant Velocity
The area under the curve of an acceleration-time graph represents the velocity of
the object.
area  l  w  t  a  v
Acceleration vs. Time
What is the velocity of the
object after 10 s?
Acceleration, m/s/s
Sample Problem
6
4
2
0
0
5
10
15
20
25
30
Time, s
5.
Apply kinematic equations to calculate distance, time, or velocity under conditions
of constant acceleration.
a
v f  vi
t f  ti
This equation can be solved for final velocity, where t represents the change in time.
v f  vi  at
To find displacement when velocity and time are known: Since v 
v
v f  vi
2
x
and
t
, these two equations can be combined.
x v f  vi

t
2
or
x 
To find displacement when acceleration and time are known:
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1
(v f  vi )t
2
Since x 
1
(v f  vi )t and v f  vi  at ,
2
1
x  [(vi  at )  vi ]t
2
x 
1
(2vi  at )t
2
x  v i t 
1 2
at
2
To find displacement when velocity and acceleration are known:
Since v f  vi  at and x 
1
(v f  v I )t
2
Solving the first equation for t: t 
( v f  vi )
a
Combining the two equations:
x 
(v f  vi )(v f  vi )
2a
x 
v 2f  vi2
2a
Or: v 2f  vi2  2ad
Sample Problem
A car accelerates at -2.0 m/s2. If its initial velocity is 24 m/s, to the East, what will its
velocity be 8.0 s later?
Sample Problem
During a 30.0 s interval, the velocity of a rocket increased from 200 m/s to 500 m/s.
What was the displacement of the rocket during this time interval?
Sample Problem
Find the time a car takes to travel 98 m if it starts from rest and accelerates at 4.0
m/s2.
Sample Problem
An electron is accelerated uniformly from rest to a velocity of 2.0 x 10 7 m/s. If the
electron traveled 0.10 m while it was being accelerated, what was its acceleration?
Sample Problem
A racing car reaches a speed of 42 m/s. It then begins a uniform negative
acceleration, using its parachute and braking system, and comes to rest 5.5 s later.
Find how far the car moves while stopping.
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6.
Relate the motion of a freely falling body to motion with constant acceleration.
Acceleration due to gravity is given a special symbol, g. Since acceleration is a
vector quantity, g must have both magnitude and direction. On the surface of the
earth, a freely falling object has an acceleration, g, of -9.81 m/s2 that is, directed
toward the surface of the earth.
7.
Calculate displacement, velocity, and time at various points in the motion of a freely
falling body.
Assuming no air resistance, all problems involving motion of falling objects can be
solved by using the acceleration equations with the acceleration, a, replaced by g.
v f  vi  gt
v 2f  vi2  2 gd
d  vi t 
1 2
gt
2
Freely falling objects always have the same downward acceleration. When an
object is thrown upward in the air, it has a positive velocity and a negative
acceleration. This means that the object is slowing down. The object continues to
move upward, but with a smaller and smaller speed. At the top of its path, the
object’s velocity has decreased until it is zero. Even though the velocity is zero at
this instant, the acceleration is still -9.81 m/s2.
When the object begins moving down, it has a negative velocity and its acceleration
is still negative. A negative acceleration and a negative velocity indicate that an
object is speeding up.
Sample Problem
A stone falls freely from rest for 8 seconds. What is the stone’s velocity after 8
seconds? What is the stone’s displacement during this time?
Sample Problem
Kyle is flying a helicopter that is rising at 5 m/s when he releases a bag. After 2
seconds, what is the bag’s velocity? How far has the bag fallen? How far below the
helicopter is the bag?
Freely falling objects always have the same downward acceleration. When an
object is thrown upward in the air, it has a positive velocity and a negative
acceleration. This means that the object is slowing down. The object continues to
move upward, but with a smaller and smaller speed. At the top of its path, the
object’s velocity has decreased until it is zero. Even though the velocity is zero at
this instant, the acceleration is still -9.81 m/s2.
When the object begins moving down, it has a negative velocity and its acceleration
is still negative. A negative acceleration and a negative velocity indicate that an
object is speeding up.
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REFERENCE: Holt Physics, pages 39 – 75
Motion Tutorials: http://www.physicsclassroom.com/Class/1DKin/1DKinTOC.html
Motion Tutorials: http://www.gcse.com/fm/dtg.htm
Making Scientific Graphs with Microsoft Excel:
http://www.howe.k12.ok.us/~jimaskew/excelgra.htm
http://office.microsoft.com/en-us/training/CR061831141033.aspx (Excel Reference Sheet)
Interactive Tutorial for Motion Graphs: http://jersey.uoregon.edu/vlab/block/Block.htm
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