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Physics
An Algebra Based Approach
Part I: Mechanics
Robert Goodman
Yuriy Zavorotniy
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Kinematics ........................................................................... 4
Introduction ................................................................................................................ 4
Time and Distance ..................................................................................................... 4
Constant Speed ......................................................................................................... 7
Problem Solving ........................................................................................................ 8
Instantaneous Speed............................................................................................... 12
Average Speed ......................................................................................................... 13
Position, Displacement and Velocity ..................................................................... 16
Coordinate Systems ................................................................................................ 19
Instantaneous Velocity and Acceleration .............................................................. 24
Free Fall .................................................................................................................... 26
The Kinematics Equations ...................................................................................... 29
Problem Solving with the Kinematics Equations.................................................. 41
Interpreting Motion Graphs .................................................................................... 46
Position versus time graphs for Constant Velocity ............................... 46
Velocity versus Time graphs for Constant Acceleration ...................... 56
Alternative Derivation of the First Kinematics Equation ...................................... 59
Chapter Questions................................................................................................... 62
Chapter Problems .................................................................................................... 63
Dynamics ................................. Error! Bookmark not defined.
Introduction ................................................................ Error! Bookmark not defined.
Inertial Reference Frames ......................................... Error! Bookmark not defined.
Force ........................................................................... Error! Bookmark not defined.
Newton’s First Law .................................................... Error! Bookmark not defined.
Newton’s Second Law ............................................... Error! Bookmark not defined.
Newton’s Second Law in Two Dimensions ............. Error! Bookmark not defined.
Newton’s Third Law ................................................... Error! Bookmark not defined.
Sketches and Free Body Diagrams .......................... Error! Bookmark not defined.
Some Examples of Forces ........................................ Error! Bookmark not defined.
Surface Forces ........................................................... Error! Bookmark not defined.
Normal Force .................................................... Error! Bookmark not defined.
Friction .............................................................. Error! Bookmark not defined.
Tension Force................................................... Error! Bookmark not defined.
Elastic Force – Hooke’s Law ........................... Error! Bookmark not defined.
Chapter Questions..................................................... Error! Bookmark not defined.
Chapter Problems ...................................................... Error! Bookmark not defined.
Uniform Circular Motion ......... Error! Bookmark not defined.
Introduction ................................................................ Error! Bookmark not defined.
Uniform Circular Motion ............................................ Error! Bookmark not defined.
Period and Frequency ............................................... Error! Bookmark not defined.
Multiple forces and circular motion ......................... Error! Bookmark not defined.
Chapter Problems ...................................................... Error! Bookmark not defined.
Newton’s Law of Universal GravitationError! Bookmark not
defined.
Universal Gravity ....................................................... Error! Bookmark not defined.
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Falling apples: The surface gravity of planets ........ Error! Bookmark not defined.
Gravitational Field ..................................................... Error! Bookmark not defined.
Orbital Motion ............................................................ Error! Bookmark not defined.
Launching a Satellite and Weightlessness ............. Error! Bookmark not defined.
Chapter Problems ...................................................... Error! Bookmark not defined.
Energy ...................................... Error! Bookmark not defined.
Introduction ................................................................ Error! Bookmark not defined.
Work............................................................................ Error! Bookmark not defined.
Gravitation Potential Energy .................................... Error! Bookmark not defined.
Kinetic Energy............................................................ Error! Bookmark not defined.
Elastic Potential Energy ............................................ Error! Bookmark not defined.
Power .......................................................................... Error! Bookmark not defined.
Simple Machines........................................................ Error! Bookmark not defined.
Chapter Questions..................................................... Error! Bookmark not defined.
Chapter Problems ...................................................... Error! Bookmark not defined.
Momentum ............................... Error! Bookmark not defined.
Introduction ................................................................ Error! Bookmark not defined.
Momentum of a Single Object .................................. Error! Bookmark not defined.
Momentum of a Closed System of Objects ............. Error! Bookmark not defined.
Conservation of Momentum in a closed system..... Error! Bookmark not defined.
Collisions ................................................................... Error! Bookmark not defined.
Problem Solving Strategies for Collisions .............. Error! Bookmark not defined.
Change of Momentum and Impulse ......................... Error! Bookmark not defined.
Derivation of relationship of velocities before and after an elastic collision.
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Chapter Questions..................................................... Error! Bookmark not defined.
Chapter Problems ...................................................... Error! Bookmark not defined.
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Kinematics
Introduction
Motion is fundamental to our lives and to our thinking. Moving from place to place in a
given amount of time helps define both who we are and how we see the world. Seeing
other people, objects or animals moving and being able to imagine where they came
from, where they’re going and how long it’ll take them to get there is natural to us.
All animals, not just humans, do calculations about the motion of themselves and the
world around them. Without that ability we could not survive. A basic understanding of
motion is deep in our minds and was there long before we could write or talk about
physics.
It should be no surprise then that the first, and most fundamental, questions of physics
relate to motion. Many of the first writings of physics are on this topic and date back
thousands of years. The study of motion is called kinematics. It comes from the Greek
words kinema, which means motion. Almost everything we learn in physics will involve
the motion of objects. So, kinematics must be understood well if all the other topics we
will be studying later are to make sense.
Time and Distance
Everyone knows what time and distance are until they’re asked to define them. Go
ahead; try to define what time is without using the idea of time itself in your definition.
Here are some definitions we’ve heard before. I’m sure you’ll come up with some new
ones as well.
Time is the amount of time that goes by.
Time is how long it takes for something to happen.
Time is how long I have to wait.
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The problem with these definitions is that they use the word “time” in the definition, or
imply its use. In the first definition, if you don’t know what time is, how can you use it to
define time? In the second two definitions, the phrase “how long” is just another way for
saying “the amount of time”. They don’t qualify as definitions since if you didn’t know
what time was when you started; you still don’t.
See if you can do any better in formulating a definition for distance. We think you’ll run
into the same problem.
We all believe that we know what these terms mean, but it’s impossible to define them.
We move through time and space as naturally as a fish moves through water. Time and
space are all around us, but we can’t really say what they are. They are too
fundamental to be defined. We have to take them as “givens”. You and I have a sense
that we know what they are. That sense comes directly from our minds and bodies; but
we can’t really define them beyond that.
Our sense of time and distance must have evolved in us long before we could think
about them. All animals need a basic perception about time and distance in order to
survive. The most primitive one celled animals move about through time and space and
surely they don’t have a definition of what those concepts represent. The sense of time
and distance predates our ability to think and perhaps that’s why we can’t use our minds
to define them; but we can work with them.
We can measure the flow of time with clocks and the distance between locations in
space with a ruler. Those don’t represent definitions; but they do allow us to compare
different intervals of time and space with one another. The ability to measure time and
distance represents a starting point for physics.
For instance, let’s say the amount of time it takes for me to run once around a track is 2
minutes. That means that the minute hand of my stop watch will go around two times
while I run around the track one time. That doesn’t tell me what time is, but it does tell
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me that those two processes took the same amount of it. So I can compare the time it
takes for something to happen to the time it takes for something else to happen.
Similarly, I don’t need to know what distance is in order to compare the distance
between two objects with the distance between two other objects. I can say that the
distance from the heel to the toe of my foot is the same as the distance from one end to
the other of a one foot ruler. So I can say that my foot is one foot long, even without a
definition of what length means.
Units of Time and Distance
In order that people can compare their measurements with those taken by others; an
international system of measurements was agreed upon. The Systeme International
(SI) is used by virtually all scientists in the world. In that system, the basic unit of length
is the meter and the basic unit of time is the second.
Distance is measured in meters (m).
Time is measured in seconds (s).
Length measurements are made by comparing the distance between any two locations
to the distance between the two ends of a rod that is defined to be one meter long.
Time measurements are made by measuring the time between events with the time it
takes for a second to tick off on a clock
Scientists use the meter, instead of the foot, for measuring distances because it’s
simpler. Using the SI system to measure lengths in meters, instead of the English
system of measuring lengths in feet, is no more accurate. However, it turns out to be a
lot easier. That’s because the mathematics of dealing with 12 inches in a foot and 5280
feet in a mile is just a lot more difficult than the metric system of 100 centimeters in a
meter and 1000 meters in a kilometer. When you start solving problems, you’ll be
happy not to have to deal with feet, miles and inches.
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Constant Speed
If everything in the world just stood still, we’d only need to measure time and distance
separately and we’d be done. But that’d be a pretty boring world. Many of the most
interesting things involve motion; objects moving from one location to another in a
certain amount of time. How fast they do this is the speed of the object.
Speed is not a fundamental property of the world, like distance and time, but is a human
invention. It is defined as the ratio of the distance traveled divided by the time it took to
travel that distance.
Speed ≡ Distance / Time
or
s ≡ d/t
The equals sign with three parallel lines just points out that this is a definition. We made
up the word “speed” and then defined it to mean the ratio of distance and time. There’s
no way that this could be proven right or wrong, through experiment or any other
means, since we just made it up. When we use the formula, we’ll usually just write it
with a normal equals sign, but we should remember that it’s just our definition.
Speed and distance don’t depend on the direction traveled. So if you walk two miles to
school and then return back home, the total distance you traveled is four miles. If it took
you one hour to do that, your average speed was four miles per hour. In this section
we’re assuming that your speed is constant. In later sections, we’ll talk about cases
where your speed changes.
Units of Speed
The units of speed can be derived from its formula:
s = d/t
The SI unit of distance is the meter (m) and of time is the second (s). Therefore, the
unit of speed is m/s.
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Problem Solving
When solving physics problems there is a series of steps that should be followed. In the
early problems that we’ll be doing, it’ll be possible to skip some steps and still get the
correct answer. But that won’t give you a chance to practice the methods that you’ll
need to solve more difficult problems. It’s wise to learn how to swim in the shallow end
of the pool, but if all you do is stand up there, it won’t be much help when the water gets
deep. So, please use the following approach on all the problems you solve right from
the beginning. It’ll pay off in the end.
1. Read the problem carefully and underline, or make note of, any information that
seems like it may be useful.
2. Read the problem through again, but now start writing down the information that
will be of value to you. Identify what is being asked for and what is being given.
3. If appropriate, draw a sketch.
4. Identify a formula that relates the information that you’ve been given to the
information you’ve been asked for.
5. Rearrange the formula so it’s solved for the variable you’re looking for. This
means, get that variable to be alone on the left side of the equals sign.
6. Substitute in the values you’ve been given, including units.
7. Calculate the numerical result.
8. Solve for the units on the right side of the equation and compare those to the
units that are appropriate for what you’re solving for. For instance, if you’re
solving for distance, the units should be meters not meters per second.
9. Reread the problem and make sure that your answer makes sense. It’s been
shown that successful physics students read each problem at least three times.
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Example 1: Riding your bike at a constant speed it takes you 25 seconds to travel a
distance of 1500 meters. What was your speed?
We’ve been given distance and time and we need to find speed.
s=?
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d = 1500 m
t = 25 s
We can directly use the equation s = d/t. That gives the relationship between the three
variables and is already solved for the variable that we’re looking for. After writing down
the formula we just have to substitute in the values with units.
s = d/t
s = 1500m/25s
s = 60 m/s
Note that not only is 1500 divided by 25 equal to 60 but also that m divided by s yields
m/s which is the correct unit for speed. By always doing the same mathematical
operations on the units as well as the numbers you should end up with the correct units
in your answer. That is a good way to check if you did the problem correctly.
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Let’s now looks at an example where the formula can’t be used directly.
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Example 2: How far will you travel if you are driving at a constant speed of 25 m/s for a
time of 360s?
We’ve been given speed and time and we need to find distance.
s = 25m/s
d=?
t = 360s
We’ll use the same formula (s=d/t) since it relates the known values (speed and time) to
the unknown value (distance). But, in this case while we need to solve for distance (d),
the formula that we have (s = d/t) is solved for speed (s). We must first use algebra to
rearrange the formula so that it is solved for d. Only at that point should values be
substituted into the formula.
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We’ll make use of three rules to rearrange the formula.
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1. If the variable that we are solving for is in the numerator and isn’t alone, then it is
mathematically connected to other numbers and/or variables. We can get it
alone by performing the inverse operation on each of the other variables or
numbers. For instance, if d is divided by t; we can get d alone by multiplying by t
(since multiplication is the opposite of division).
2. We can do anything we want to one side of an equation as long as we also do it
to the other side (except dividing by zero). So if we multiply the right side of the
equation, d/t, by t we also have to multiply the left side of the equation, s, by t.
3. We can always switch the right and left sides of an equation.
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Let’s use this approach to solve the equation, s=d/t, for d.
s = d/t
Since we are solving for d, and d is being divided by t,
we must multiply d/t by t. But we can only do that if
also multiply s by t. So multiply both sides by t.
st = (d/t)t
Cancel t on the right since it’s in the numerator and
the denominator and t/t =1
st = d
Switch the d to the left side
d = st
Substitute in values for s and t
d = (25m/s) (360s)
d = 9000m
Note that not only is 25 times 360 equal to 9000 but also that meters per second times
seconds is equal to meters since the seconds will cancel out. That gives us units of
meters, which makes sense since we are solving for a distance.
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Example 3: How much time will it take you to travel 3600 m if you are driving at a
constant speed of 20 m/s?
We’ve been given the distance and the speed, and we need to find the time.
s = 20m/s
d = 3600m
t=?
In this case we are solving for time (t) but the formula we have (s = d/t) is solved for
speed (s). We must first use algebra to rearrange the formula so that it is solved for t.
Only at that point should values be substituted into the formula.
s = d/t. We need to add one additional rule to rearrange the formula in this case.
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4. The unknown for which we are solving must be in the numerator, not the
denominator. So if we are solving the formula s = d/t for t, our first step must be
to move t to the numerator on the left instead of leaving it in the denominator on
the right. To do this, we need to multiply both sides of the equation by t, giving
us st = d. Then we can proceed just as we did above.
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s = d/t
Multiply both sides by t to cancel the t on the right and
get t on the left
st = (d/t)t
Cancel t on the right since t/t = 1
st = d
Since t is not alone, because it is multiplied by s, we
must divide both sides by s
(st)/s = d/s
Cancel s on the left side since s/s = 1
t = d/s
Substitute in values for d and s
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t = (3600m)/(20m/s)
d = 180s
The units may be a bit harder to understand in this case. We have meters divided by
meters per second. But you may recall from fractions that dividing by a fraction is the
same as multiplying by its reciprocal. (Dividing by 1/3 is the same as multiplying by 3.)
So dividing by m/s is the same as multiplying by s/m. This makes it clear that the
meters will cancel out, when we multiply s/m by m, and we are left with seconds, an
appropriate unit for time.
Instantaneous Speed
There’s an old joke about a person who’s pulled over for speeding. The police officer
tells the speeder that he was going 60 miles per hour in a forty mile per hour zone. The
speeder’s response is that he couldn’t have been going sixty miles per hour since he’d
only been driving for fifteen minutes.
The reason that argument doesn’t work is that speed is a ratio of distance and time.
There are an infinite number of ways that you can calculate a speed of ten meters per
second. Some are shown in the table below.
Distance
(m)
Time
(s)
Speed
(m/s)
1000
500
100
10
1
0.1
0.01
0.001
100
50
10
1
0.1
0.01
0.001
0.0001
10
10
10
10
10
10
10
10
You can see that at the bottom of the chart that if you travel one thousandth of a meter
in one ten thousandth of a second you are traveling at a speed of ten meters per
second. That time and distance can be made as small as you like. When the time over
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which the speed is measured is very small, the speed that is calculated is called the
instantaneous speed. This is the speed that you read on your speedometer or that a
policeman reads on his radar or laser gun.
Average Speed
While traveling along, your speed is likely to vary; to go up and down along the way.
You might even stop for a while to have lunch. Your instantaneous speed at some
moment during your trip and your average speed for the total trip are often not the
same. Your average speed is calculated by determining the total distance that you
traveled and dividing by the total time that it took you to travel that distance.
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Example 4:
You ride your bike home from school by way of your friend’s house. It takes you 7
minutes to travel the 2500m to your friend’s house. You then spend 10 minutes there.
You then travel the 3500 m to your house in 9 minutes. What was your average speed
for your total trip home?
Your average speed will be obtained by dividing the total distance traveled by the total
time it took to travel that distance. In this case, the trip consisted of three segments.
The first segment (I) is the ride to your friend’s house, the second segment (II) was the
time at your friend’s house and the third segment (III) was your ride home from your
friend’s house. In the chart below, the speed is calculated for each segment, even
though that is not necessary to get the answer that was requested, the average speed
for your total trip. All calculated figures are shown in bold type.
Segment
I
II
III
Total /
Average
Distance
(m)
2500
0
3500
Time
(s)
420
600
540
Speed
(m/s)
6.0
0.0
6.5
6000
1560
3.8
For instance the speed for the first segment is given by:
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s=?
d = 2500m
t = 7 minutes = 420 seconds
Note that we need to convert the given time in seconds in order to use SI units. Since
there are 60 seconds in a minute, this requires multiplying seven minutes by the fraction
(60 seconds / 1 minute). This leaves us with 420 seconds.
s = d/t
s = 2500m / 420s
s = 6.0 m/s
For the second segment, your speed was zero since you were within the house. But
even though you weren’t moving, time was going by. So the 10 minutes, or 600
seconds, still counts towards the total elapsed time.
The third segment is calculated in the same manner as was the first.
s=?
d = 3500m
t = 9 minutes = 540 seconds
s = d/t
s = 3500m / 540s
s = 6.5 m/s
The average speed is calculated by taking the total distance, 6000m, and dividing it by
the total time, 1560s, to get an average speed of 3.8 m/s.
While it wasn’t necessary in this case to calculate the speed for each interval, it’s
important to note that the average speed is not the average of the speeds. The average
of 6.0 m/s, 0.0 m/s and 6.5 m/s is 4.2 m/s. But this is not the correct answer. The
correct answer can only be obtained by first finding the total distance and dividing that
by the total time. In this way you get the answer of 3.8 m/s.
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Example 5:
You run a distance of 210 m at a speed of 7 m/s. You then jog a distance of 200 m in a
time of 40s. Finally, you run for 25s at a speed of 6 m/s. What was the average speed
of your total run?
Your average speed will be obtained by dividing the total distance traveled by the total
time it took to travel that distance. In this case, the trip consisted of three segments. In
the chart below, different calculations are required for each segment in order to obtain
the average speed for your total trip.
Segment
I
II
III
Total /
Average
Distance
(m)
210
200
150
Time
(s)
30
40
25
Speed
(m/s)
7.0
5.0
6.0
550
95
5.8
The time for the first segment is given by:
s = 7.0 m/s
d = 210m
t=?
s=d/t
st = d
t = d/s
t = (210m) / (7.0 m/s)
t = 30 s
We don’t really need to calculate your speed for the second segment, but we’ll do it
anyway.
s=?
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d = 200m
t = 40s
s = d/t
s = 200m/40s
s= 5.0 m/s
The distance needs to be calculated for the third segment.
s = 6.0 m/s
d=?
t = 25s
s = d/t
st = d
d = st
d = (6.0 m/s) (25s)
d = 150 m
The average speed is calculated by taking the total distance, 550m, and dividing it by
the total time, 95s, to get 5.8 m/s.
Position, Displacement and Velocity
So far our analysis has not required, or even allowed, us to know anything about the
direction of the motion under study. But in real life, direction is usually very important.
Whether you’re driving 60 miles per hour north or 60 miles per hour south makes a
great deal of difference as to where you end up.
Scalars are quantities that are defined only by their magnitude; how big they are.
Speed, time and distance are all examples of scalars. When we speak of 40 m/s, 20
minutes or 3 miles we’re not giving any information about direction.
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Vectors are quantities that are defined by both their magnitude and their direction.
So, if instead of saying that I traveled a distance of 400m, I were to say that I traveled
400m north; I am now defining a vector. The vector defined by combining a distance
with a direction is called displacement. The symbol for displacement is “Δx”. We’ll talk
more about that symbol a little later, but you can use it in the meantime. Also, in order
to keep track of what’s a scalar and what’s a vector, we’ll always show vectors in a bold
typeface.
There are important differences between how we work with scalars and vectors. That
can be most easily seen by using distance and displacement as examples. For
instance, while distances are always positive, since they have no direction associated
with them, displacement can be positive or negative. That means that if I were to take a
trip which involved going 200m north and then 200m south I get very different answers
for the total distance I traveled and my total displacement. I get my total distance by
adding 200m to 200m and getting 400m. That’s the total distance that I walked.
On the other hand, my displacement represents the sum of the two displacements. My
initial displacement north is equal and opposite to my final displacement south, so they
will cancel each other out. If I think of north as the positive direction, then that first
displacement is +200m, while my second displacement would be -200m, since it’s to the
south. The sum of +200m and -200m is zero. That’s because the direction of the
motion matters with regard to displacement while it doesn’t with regard to distance. As
a result, displacement tells you how far you are from where you started. In this case, I
am zero distance from where I started, since I end up back where I started off.
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Example 6:
You drive 1500m north and then 500m south. Determine both the total distance you
traveled and your total displacement from where you started.
The distance traveled is just the sum of the two distances, 1500m and 500m, 2000m.
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In order to determine your total displacement we need to first define our directions.
Let’s call movement to the north positive and movement to the south negative (which
direction we call positive won’t affect our answer as long as we’re consistent.). That
means that for the first part of the trip your displacement is +1500m and for the second
part of the trip your displacement is -500m. Your total displacement is the sum of those
two, +1000m. Since we decided that we’d call the north direction positive, your final
displacement is 1000m north. The last step of converting from +1000m to 1000m north
is important in that our choice of + or – was arbitrary so we need to translate back to the
original directions we were given in the problem.
The important point here is the answers are different and have different uses. The
distance you traveled, 2000m, tells you something about how tired you may be because
it tells you the total distance you had to move yourself during this trip. Your
displacement, 1000m north, tells you where you are at this point in your travels relative
to where you started.
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The same difference exists between speed and velocity. The symbol for velocity is v
and the symbol for average velocity is vavg. The average velocity is determined by
dividing your total displacement by the time it took for that displacement. This is similar
to how we calculated average speed by dividing the total distance traveled by the total
time it took to travel that distance.
s ≡ d/t
while
vavg ≡ Δx/t
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Example 7:
If the travel in Example 6 was done at constant speed and required a total time of 500s,
determine the average speed and the average velocity.
s=?
d = 2000m
t = 500s
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s = d/t
s = 2000m/1200s
s = 4.0 m/s
vavg = ?
Δx = 1000m north
t = 500s
vavg = Δx/t
v = (1000m north)/500s
v = 2m/s north
Note that the numerical answers are different and that the answer for velocity includes a
direction while the answer for speed does not.
Coordinate Systems
The displacement of an object tells us how its position has changed. In order to better
understand what that means we need a way of defining position; we need a coordinate
system.
The requirements of any coordinate system are an origin and an orientation. In other
words, you need to pick a zero from which you’ll be making measurements and you
need to know the direction in which you will be measuring. The simplest type of
coordinate system is one-dimensional, in which case the coordinate system becomes a
number line, as shown below.
The origin is located at zero, negative positions are to the left of the origin and positive
positions are to the right. We can identify different locations on the number line as such
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as x0, x1 and x2. In the diagram above, x0 is located at 0, x1 is located at +5m and x2 is
located at -5m.
We can now refine our definition of displacement, the change in the position of an
object, as being the difference between an object’s final position, x, and its initial
position, xo. It now becomes clearer why the symbol for displacement is “Δx”. The
Greek letter delta, “Δ”, means “the change in” so “Δx” can be read as “delta x” or “the
change in x”. Symbolically this becomes;
Δx ≡ x - xo
____________________________________________________________________
Example 8:
An object moves from an initial position of +5m to a final position of +10m in a time of
10s. What displacement did it undergo? What was its average velocity?
x = +10m
xo = + 5m
Δx = ?
Δx = x - xo
Δx = (+10m) – (+5m)
Δx = +5m
vavg = ?
Δx = +5m
t=2s
vavg = Δx/t
v = +5m/2s
v = +2.5m/s
_______________________________________________________________
Example 9:
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An object moves from an initial position of +5m to a final position of -10m in a time of
0.25s. What displacement did it undergo? What was its average velocity?
x = -10m
xo = + 5m
Δx = ?
Δx = x - xo
Δx = (-10m) – (+5m)
Δx = -15m
vavg = ?
Δx = -15m
t = 0.25 s
vavg = Δx/t
vavg = -15m/0.25s
vavg = -60m/s
Once again, notice that the answer includes a magnitude, 15m, as well as a direction,
“-”.
________________________________________________________________
Vectors, such as displacement or velocity, can be depicted as an arrow. The length of
the arrow represents the magnitude of the vector and the direction it is pointed
represents the direction of the vector.
Vectors can be added either graphically or algebraically. (Even if you’re solving a
problem algebraically it’s helpful to also sketch the addition graphically so you can make
certain that your answer makes sense.) The way to add vectors graphically is to draw
the first vector starting at the origin of the problem. It must be drawn to scale and
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pointed in the correct direction. The second vector should be drawn in the same
manner, but starting where the first one ended. The sum of the two vectors is simply a
third vector which starts where the first vector started and ends where the last vector
ended. In other words, the solution is a third vector that connects the beginning of the
first to the end of the last vector drawn. The arrow tip of that vector should point away
from the location from which it started.
The following example is solved both graphically and algebraically.
________________________________________________________________
Example 10:
Beginning at a location that’s 400 m east of your home, you travel 500m east and then
300m west. How far are you now from your home? What displacement have you
experienced during your trip? If this travel took a total time of 20s, what was your
average velocity?
The graphical solution, shown below, starts by drawing a sufficiently large east-west
axis. If your home is located at x = 0m, then your initial position, xo, is 400m east of
that. Draw a vector that describes the first part of your trip by drawing an arrow that
begins at the location 400m to the east of your house, is 500m long and points towards
the east. That tip of that arrow should then be 900m east of your house. Then draw the
vector for the second part of your trip by drawing an arrow that begins at the tip of the
first arrow, 900m east of your house, is 300m long and is pointed to the west, towards
your house. The tip of that arrow should now lie at a location 600m to the east of your
house; that is your final location, x.
Your displacement is the difference between your final and initial positions. This is
obtained, graphically, by drawing an arrow that starts at your initial position and ends at
your final position. The length of this arrow, which can be physically measured or read
off the scale, is the magnitude of your displacement. The direction of the arrow is the
direction of your displacement. As is shown below, it can be seen that your
displacement is 200m east.
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Your average velocity is your total displacement divided by the total time it took to
undergo that displacement. In this case, we graphically determined that your
displacement is 200m east and we were told that your travel time was 20s. So,
vavg = ?
Δx = 200m east
t = 20s
vavg = Δx/t
vavg = 200m east / 20s
vavg = 10m/s east
The same problem can be solved algebraically, although an initial sketch is still a good
idea. The key step to an algebraic solution is to convert directions to be either positive
or negative. In this case, we can define east as positive and west as negative (The
choice won’t matter as long as we’re consistent throughout the problem.)
Your initial position then becomes +400m (400m east), your initial travel is +500m
(500m east) and the last leg of your trip is -300m (300m west). You can now just add
these together to get your final position, +600m. This translates into a final position of
600m east of your house.
Your displacement is just the change in your position.
x = +600m
xo = +400m
Δx = ?
Δx = x - xo
Δx = +600m – (+400m)
Δx = +200m
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Δx = 200m east
note that this last step is required since our choice of positive
or negative was arbitrary
The calculation of average velocity can be done just as it was above for the graphical
solution.
Instantaneous Velocity and Acceleration
The most boring world would be one in which the positions of all objects were
constant...nothing would move: velocity would have no meaning. Fortunately our world
is a lot more interesting than that. Objects are changing their positions all the time, so
velocity is an important concept.
But our world is even more interesting than that...objects are also changing their velocity
all the time: they are speeding up, changing direction and/or slowing down. Just as
change in position over time leads to the idea of velocity, change in velocity over time
leads to the concept of acceleration.
In the same manner that we defined instantaneous speed as the speed measured
during a very short period of time, we can now define instantaneous velocity as the
velocity measured during a very short period of time. The symbol, “v” will be used for
instantaneous velocity. In a world with acceleration, the idea of instantaneous velocity
is very important since an object’s velocity may often be changing from moment to
moment.
v ≡ Δx/t
for a very short period of time...an instant
We can now define acceleration as the change in velocity over time.
a ≡ Δv/t
or
a ≡ (v- vo)/t
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Units of Acceleration
The units of acceleration can be derived from its formula:
a ≡ Δv/t
The SI unit of velocity is meters/second (m/s) and of time is the second (s). Therefore,
the unit of acceleration is (m/s)/s or m/s/s.
This is the same as (m/s) x (1/s) since dividing by s is the same as multiplying by 1/s.
This results in m/s2 which, while not having any intuitive meaning, is a lot easier to keep
track of than m/s/s, meters per second per second, the alternative way of writing the
units for acceleration.
Since velocity is a vector, it has a magnitude and a direction. For the rest of this
chapter, we’ll be focused on accelerations that change the magnitude of an object’s
velocity. However, in later chapters a key aspect of acceleration will involve changing
the direction of an object’s velocity. These are both examples of acceleration. But let’s
first start with accelerations that change only the magnitude of an object’s velocity.
_______________________________________________________________
Example 11
An object is traveling at a velocity of 20m/s north when it experiences an acceleration
over 12s that increases its velocity to 40m/s in the same direction. What was the
magnitude and direction of the acceleration?
Let’s solve this algebraically by defining velocities towards the north as positive and
towards the south as negative. Then,
v = +40m/s
vo = +20m/s
t= 12s
a=?
a ≡ Δv/t
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a ≡ (v - vo)/t
a = (+40m/s - (+20m/s))/12s
a = (+20m/s)/12s
a = +1.7m/s2
_______________________________________________________________
Example 12
What will an object’s velocity be at the end of 8.0s if its initial velocity is +35m/s and it is
subject to an acceleration of -2.5m/s2?
v=?
vo = +35m/s
t= 8.0s
a = -2.5m/s2
a ≡ Δv/t
a ≡ (v - vo)/t
Solve for v: First, multiply both sides by t
at = v - vo
Then add vo to both sides
vo + at = v
Switch so that v is on the left side of the =
v = vo + at
Substitute in values and solve
v = +35m/s + (-2.5m/s2) (8.0s)
v = +35m/s + (-20m/s)
v = +15m/s
Free Fall
You now know enough to be able to understand one of the great debates that marked
the beginning of what we now call physics. The term “physics” was being used by the
ancient Greeks more than 2000 years ago. Their philosophy, much of it described in
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the book titled “Physics” by Aristotle, included some ideas that stood until Galileo made
some important arguments and measurements that showed the ancient Greek physics
to be of limited value.
The physics of ancient Greece included the idea that all objects were made up of a
combination of four elements (the fifth element was reserved for objects that were
beyond the earth). The four elements to be found in our world were earth, water, air
and fire. Each of these elements had their natural place. If you removed an element
from its natural place, it would, when released, immediately move back to that place;
and it would do so with its natural (constant) velocity.
Their view of the world could be thought of a set of concentric circles with each of the
elements occupying a layer. Earth occupied the center of the circle, so rocks, which are
predominantly made of earth would naturally move down, towards the center of our
world. Above earth was water, which would fill the area above the rocks, like a lake or
an ocean above the land that forms the lake or ocean bed. Above water is air, which is
seen everywhere in our world, above both earth and water. Finally, fire rises up through
the air, searching for its natural location above everything else.
All objects were considered to be a mixture of these four elements. Rocks were
predominantly earth: so if you drop a rock it falls as it tries to get back to its natural
location at the center of the earth. In so doing, it will pass through water and air: If you
drop a rock in a lake, it sinks to the bottom. Fire passes upwards to the highest
location, so if you make a fire, it always passes upwards through the air.
One conclusion that this led to is that objects which were made of a higher percentage
of earth would feel a greater drive to reach their natural location. Since earth is the
heaviest of the elements, this would mean that heavier objects would fall faster than
lighter objects. Also, they would fall with a natural constant velocity.
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That philosophy stood for more than 2000 years until Galileo, in the 1600’s made a
series of arguments, and conducted a series of experiments, that proved that neither of
those two conclusions was accurate. He showed that the natural tendency of all
unsupported objects is to fall towards the center of the earth with the same acceleration:
9.8m/s2. That number 9.8m/s2 is used so often that it as given its own symbol: “g”. In
modern terms, his conclusion can be stated as follows.
All unsupported objects fall towards the center of the earth with an acceleration
of g: 9.8m/s2.
This statement requires some explanation and some caveats.
1. Unsupported means that nothing is holding the object up. So if you release
something and nothing is stopping it from falling, then it is unsupported. In that
case, all objects will experience the same acceleration downwards. It does not
depend on how heavy the object is: all objects fall with that same acceleration.
2. Support can also come from air resistance. So a parachute provides support by
catching air that slows down the sky diver. In that case, the parachutist is not an
unsupported object: he or she is supported by air resistance. But this is
generally true to a lesser extent. So a feather or an uncrumpled piece of paper
also receives support from the air: so they don’t fall with a constant acceleration
either. Galileo’s conclusion is an idealization, it assumes that we can ignore air
resistance, which is never completely true near the earth (or airplanes and
parachutes would have a hard time of it) but will work for the problems we’ll be
doing.
3. His conclusion does not depend on the motion of the object. So baseballs
thrown toward home plate, dropped, or thrown straight up all fall with the same
acceleration towards the center of the earth. This is an area of great confusion
for students, so you’ll be reminded of it often. Whenever nothing is stopping an
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object from falling, it will accelerate downwards at 9.8m/s2, regardless of its
overall motion.
In this book, we will assume that air resistance can be ignored unless it is specifically
stated to be a factor.
_______________________________________________________________
Example 13
An object is dropped near the surface of the earth. What will its velocity be after it has
fallen for 6.0s?
v=?
vo = 0
t= 6.0s
a = g = -9.8m/s2
All unsupported objects have an acceleration of
9.8m/s2 downwards
a ≡ Δv/t
a ≡ (v - vo)/t
Solve for v: First, multiply both sides by t
at = v - vo
Then add vo to both sides
vo + at = v
Switch so that v is on the left side of the =
v = vo + at
Substitute in values and solve
v = 0 + (-9.8m/s2) (6.0s)
v = -59m/s
The Kinematics Equations
So far we have two definitions of motion that we will be using the foundation for our
study of motion: vavg ≡ Δx/t and a ≡ Δv/t. We need to add just one more equation to
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complete our foundation; and then we can start building the set of equations that we’ll
be using to solve a range of problems throughout this book. The last equation tells how
to calculate an object’s average velocity if we know its initial and final velocity. It turns
out that under the condition of constant acceleration an object’s average velocity is just
the average of its initial and final velocities. This average is computed just by adding
the two velocities, v and v0, together and dividing by 2:
vavg = (v0 + v) / 2
Or, since dividing by 2 is the same as
multiplying by ½
vavg = ½ (v0 + v)
This will be true whenever the acceleration is constant. However, that condition of
constant acceleration will hold true not only for this course, but for most all the high
school or university physics that you will take.
It’d be possible to solve all problems involving the location, velocity and acceleration of
an object just using the two definitions and the computation for average velocity shown
above. However, in physics it’s sometimes best to do some harder work up front so as
to make our work easier later on. In this case, we’ll use the above three equations to
create a set of kinematics equations that are easier to work with. We’ll first derive those
equations algebraically; then we’ll derive them using a graphical approach. Then we’ll
practice working with them.
Let’s start with by using our definition of acceleration to derive an equation that will tell
us an object’s velocity as a function of time.
a ≡ Δv / t
Substitute: Δv = v - vo
a = (v - vo) / t
Multiply both sides by t
at = v - vo
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Add vo to both sides
vo + at = v
Rearrange to solve for v
v = vo + at
This equation tells us that an object’s velocity at some later time will be the sum of two
terms: its velocity at the beginning of the problem, vo, and the product of its
acceleration, a, and the amount of time it was acceleration, t. If its acceleration is zero,
this just says that its velocity never changes. If its acceleration is not zero, this equation
tells us that the object’s velocity will change more as more time goes by and it will
change faster if the size of its acceleration is greater. Often in physics we want to know
the final position or velocity of an object after a certain amount of time. The bolded
equation above gives us a direct way of calculating velocities at later times given initial
conditions: This is a key kinematics equation.
_______________________________________________________________
Example 14
What will an object’s velocity be at the end of 15s if its initial velocity is -15m/s and it is
subject to an acceleration of +4.5m/s2?
v=?
vo = -15m/s
t= 15s
a = +4.5m/s2
v = vo + at
v = -15m/s + (+4.5m/s2) (15s)
v = -15m/s + 68m/s
v = +53m/s
____________________________________________________________
Example 15
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How long will it take an object to reach a velocity of 86m/s if its initial velocity is 14m/s
and it experiences an acceleration of 1.5m/s?
v = 86m/s
vo = 14m/s
t= ?
a = +1.5m/s2
v = vo + at
Solve for t: First subtract vo from both sides
v - vo = at
Then divide both sides by a
(v - vo) / a = t
Finally switch sides so t is on the left
t = (v - vo) / a
Now substitute in the values and solve
t = (86m/s – 14m/s) / (1.5m/s2)
t = 72 m/s / 1.5m/s2
t = 48s
Note that just as 72 divided by 14 equals 48, so too does (m/s) / (m/s2) equal seconds.
This can be seen if one remembers that dividing by a fraction is the same a multiplying
by its reciprocal: so (m/s) / (m/s2) is the same as (m/s) (s2/m). In this case, it can be
seen that the meters cancel out as does one of the seconds in the numerator, leaving
only seconds in the numerator; which is the correct unit for time.
_______________________________________________________________
Example 16
What acceleration must an object experience if it is to attain a velocity of 40m/s to the
north in a time of 18s if it’s starts out with a velocity of 24m/s towards the south?
For this problem, let’s define north as positive and south as negative. Then,
v = +40m/s
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vo = -24m/s
t= 18s
a=?
v = vo + at
Solve for t: First subtract vo from both sides
v - vo = at
Then divide both sides by t
(v - vo) / t = a
Finally switch sides so a is on the left
a = (v - vo) / t
Note that this is just our original definition for
acceleration. We could have just used that definition,
but it’s easy enough to recover from the equation we’ll
be using...either way works.
Now substitute in the values and solve
a = (40m/s – (-24m/s)) / (18s)
Note that in substituting -24m/s for vo we put it into its
own parentheses. That’s so we don’t lose the
negative sign...a common mistake. Now we can see
that -(-24m/s) equals +24m/s
a = 64 m/s / 18s
a = 3.6 m/s2
Since the answer is positive, the acceleration must be
towards the north based on our original decision that
north was positive
a = 3.6 m/s2 towards the north
Note that just as 64 divided by 18 equals 3.6, so too does (m/s) / s equal m/s2. This can
be seen if one remembers that dividing by a fraction is the same a multiplying by its
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reciprocal: so (m/s) / s is the same as (m/s) x (1/s).
____________________________________________________________________
We now have a useful equation for determining how an object’s velocity will vary with
time, given its initial velocity and its acceleration. We need to derive a similar
expression that will tell us where an object is located as a function of time given its initial
position and velocity and its acceleration.
We need to combine three of our equations together to do that
v = vo + at
The equation we just derived from the definition of
acceleration
vavg ≡ (x - x0) / t
The definition of average velocity
vavg = ½ (v + v0)
The equation for average velocity in the case of constant
acceleration
Since we have two equations for average velocity, vavg, they must be equal to each
other.
vavg = vavg
We can then substitute in the two different equations for vavg
from above: one on the left side of the equals sign and the
other on the right
[(x - x0) / t] = [½ (v + v0)]
Let’s solve this for x: first multiply both sides by t to get it out
of the denominator on the left
x - x0 = ½ (v + v0)t
Then add x0 to both sides to get x by itself
x = x0 +½ (v + v0)t
Distribute t into the parentheses on the right
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x = x0 + ½ vt + ½v0t
Now substitute in our new equation for v: v = vo + at
x = x0 + ½ (vo + at) t + ½v0t
Distribute ½t into the parentheses
x = x0 + ½ vot + ½at2 + ½v0t
Combine the two ½ vot terms
x = x0 + vot + ½at2
This is another of the key kinematics equations. It allows us to determine where an
object will be as time goes by based on a set of initial conditions. In this case, there are
three terms: x0 tells us where the object started; vot tells us how fast it was moving
initially and how long it’s been traveling; and ½at2, tells us how much its acceleration
has affected the distance it has traveled. The reason that t in the last term is squared is
that not only does it’s velocity change more as time goes by, it also has had more time
for that change in velocity to affect how far it’s gone.
____________________________________________________________
Example 17
A car is at rest when it experiences an acceleration of 2.0m/s2 towards the north for
5.0s. How far will it travel during the time it accelerates?
For this problem, let’s define north as positive and south as negative. Also, since we’re
not told where the car starts, let’s just define its initial position as the origin for this
problem, zero. In that case, the distance it travels will just be its position, x, at the end
of the problem. Then,
x0 = 0
x=?
vo = 0
t= 5.0s
a = 2.0m/s2
x = x0 + vot + ½at2
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The equation is already solved for x so we just have
to substitute in numbers. However, a good first step
is to cross out the terms that will clearly be zero, in
this case the first and second terms. (Since vo = 0,
anything times vo will also be zero.)
x = ½at2
That vastly simplifies the equation and avoids some
algebra mistakes. Now we can substitute numbers in
for the variables.
x = ½(2.0m/s2)( 5.0s) 2
Make sure to square 5.0s before multiplying it by
anything else
x = (1.0m/s2)(25s 2)
x = 25m
The car will travel 25m to the north during the time that it accelerates
_______________________________________________________________
Example 18
An object accelerates from rest. How long will it take for it to travel 40m if its
acceleration is 4m/s2?
Since we’re not told where the object starts, let’s just define its initial position as the
origin for this problem, zero. In that case, the distance it travels, 40m, will just be its
position, x, at the end of the problem. Also, since it’s initial at rest that means that its
initial velocity is zero. Then,
x0 = 0
x = 40m
vo = 0
t= ?
a = 2.0m/s2
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x = x0 + vot + ½at2
Let’s cross out the terms that will clearly be zero, in
this case the first and second terms. (Since vo = 0,
anything times vo will also be zero.)
x = ½at2
Now let’s solve this for t: Multiply both sides by 2 and
divide both sides by a
2x/a = t2
Now take the square root of both sides, to get t
instead of t2, and switch t to the left
t = [2x/a]½
Now we can substitute in the values
t = [2(40m)/( 2.0m/s2)]½
t = [80m)/( 2.0m/s2)]½
t = [40s2]½
t = 6.3s
The first equation that we derived allows us to determine the velocity of an object as a
function of time if we know its acceleration. The second equation allows us to
determine the position of an object as a function of time if we know its initial position and
velocity and its acceleration. Sometimes we use both equations to solve one problem.
_______________________________________________________________
Example 19
A plane must reach a speed of 36m/s in order to take off and its maximum acceleration
is 3.0m/s. How long a runway does it require?
Solving this problem requires us to use both our kinematics equation. First, let’s figure
out how much time it must accelerate to reach takeoff velocity. Then, let’s figure out
how far it will travel in that time.
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x0 = 0
x=?
vo = 0
v = 36m/s
t= ?
a = 3.0m/s2
v = vo + at
Solve for t
v - vo = at
t = (v - vo) / a
t = ((36m/s) – 0) / 3.0m/s2
t = 12s
Now we can add that new piece of information to what we knew before:
x0 = 0
x=?
vo = 0
v = 120m/s
t = 12s
a = 3.0m/s2
We can now use the second equation to solve for the location of the plane when it takes
off. That will be the minimum required length of the runway.
x = x0 + vot + ½at2
Eliminating the zero terms
x = ½at2
x = ½(3.0m/s2)(12s)2
x = (1.5m/s2)(144s2)
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x = 216m
_______________________________________________________________
The final kinematics equation that we’ll develop combines the first two so that we can
determine the velocity of an object as a function of its position, rather than as a function
of time. That would allow us to have solved Example 18 in just one step. Essentially
we derive this equation by doing exactly what we did in Example 18, but without
inserting in any numbers, we leave everything as variables. The result is a solution that
we can use in the future to save a lot of work.
Let’s first solve our velocity versus time equation for time.
v = vo + at
v - vo = at
at = v - vo
t = (v - vo) / a
Then we’ll use that expression for time in the equation that tells us an object’s position
as a function of time. That will eliminate time from that equation.
x = x0 + vot + ½at2
Now we’ll substitute in [(v - vo) / a] wherever we see a
“t”. The brackets just help us see what we’ve done.
Review this next equation carefully to see that
wherever there used to be “t” there’s now [(v - vo) / a].
x = x0 + vo[(v - vo) / a] + ½a[(v - vo) / a] 2
Let’s subtract x0 from both sides, distribute the vo into
the second bracket and square the contents of the
third bracket.
x - x0 = (vov - vo2) / a] + ½a(v - vo) 2 / a 2
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We can now cancel one of the a’s in the last term and
use the fact that
(v - vo) 2 = v2 - 2v vo + vo2
x - x0 = (vov - vo2) / a] + ½(v2 - 2v vo + vo2) / a
Since a is in the denominator of both terms on the
right we can simplify a bit by multiply all the terms by
a
a(x - x0) = vov - vo2 + ½(v2 - 2v vo + vo2)
Now let’s distribute the ½ into the last parentheses on
the right
a(x - x0) = vov - vo2 + ½v2 - v vo + ½vo2
By combining the two vvo terms, they cancel out. At
the same time we can combine - vo2 and ½vo2 to get ½vo2
a(x - x0) = ½v2 - ½vo2
Now multiply both sides by 2 to cancel out the ½’s
2a(x - x0) = v2 - vo2
Switching the terms from left to right completes this
derivation
v2 - vo2 = 2a(x - x0)
This is sometimes written by substituting d for (x - x0)
since that is just the distance that the object has
traveled and it’s easier to read
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v2 - vo2 = 2ad
This equation lets us determine how the velocity of an object will change as its position
changes. Having done all this work now, will save us work later. In Example 19, let’s
take a look at how we would use this equation to solve the same problem as was posed
in Example 18
________________________________________________________________
Example 20
As was the case in Example 18, a plane must reach a speed of 36m/s in order to take
off and its maximum acceleration is 3.0m/s2. How long a runway does it require?
x0 = 0
x=?
vo = 0
v = 36m/s
a = 3.0m/s2
v2 - vo2 = 2a(x - x0)
Simplify by eliminating the zero terms, vo & x0
v2 = 2ax
Solve for x by dividing by 2a and switching sides
x = v2 / 2a
x = (36m/s) 2 / [2(3.0m/s2)]
x = 216m
______________________________________________________________
Problem Solving with the Kinematics Equations
The Kinematics Equations
x = x0 + vot + ½at2
v = vo + at
v2 - vo2 = 2a(x - x0)
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All kinematics problems can be solved by using either one or two of the above
equations. It is never necessary to use all three equations to answer one question. The
biggest question students have in working with these equations is which one to use?
First, you need to relax and understand that you can’t get the wrong answer by using
the wrong equation; you just won’t get an answer at all. You’ll just find that you’re
missing the information that you need to solve the problem using that equation. At that
point, you should realize that you need to use a different equation, or read the problem
again to see if you’re missing a piece of information that you need but have overlooked.
For instance, if a problem indicates or implies that an object was at rest at the beginning
of the problem that means that its initial velocity was zero. Sometimes this is obvious...
sometimes it isn’t. For instance, if I “drop” something, the implication is that it had a
zero initial velocity, but that isn’t explicitly stated; it’s implied. Physics will help you learn
to read very carefully to understand what the author meant when they wrote the
problem or described the situation.
So, the first step in solving the problem is to read it very carefully.
The second step is to read it again. This time writing down the information you’ve been
given in terms of the variables with which you’ll be working. For instance, translating
“dropped” into “vo = 0”. One of the pieces of information that you’ll be given is what
you’re supposed to be looking for: what’s the question. If the author asks “What is its
final velocity?” that is translated as “v =?” and becomes another of the facts to add to
your list of facts that you’ll use to solve the problem.
The next step is to determine which of the kinematics equations relates your collection
of facts to one another. Each equation represents a relationship between a different set
of facts: picking the correct equation is just a matter of determining which equation
relates to this specific set of facts. If you pick the wrong one, no harm will be done
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(except some wasted time) since you’ll find you just don’t have the right facts to use that
equation.
If the problem that you’re working with doesn’t have time as one of its facts, then you’ll
be using the third of the equations listed above: v2 - vo2 = 2a(x - x0); it’s the only one that
doesn’t include time as a factor. If time is included, you’ll be using one of the first two.
In that case, you just need to determine which of those first two equations to use. If the
problem deals with the position of the object as a function of time, then you’d use the
first equation: x = x0 + vot + ½at2. If it is dealing with the velocity of the object as time
goes by, you’ll use the second equation: v = vo + at. It’s really as simple as that.
_______________________________________________________________
Example 21
In this example, we’re just going to decide which equation(s) will be needed to solve
each problem.
1. A ball is subject to an acceleration of -9.8 m/s2. How long after it is dropped does
it take to reach a velocity of -24m/s?
2. A ball is released from rest and subject to an acceleration of -9.8 m/s2. How far
will it travel before it reaches a velocity of 24m/s?
3. A dropped ball is subject to an acceleration of -9.8 m/s2. How far will it travel in
the first 5.0s?
4. You throw an object upwards from the ground with a velocity of 20m/s and it is
subject to a downward acceleration of 9.8 m/s2. How high does it go?
5. You throw an object upwards from the ground with a velocity of 20m/s and it is
subject to a downward acceleration of -9.8 m/s2. How much later does it
momentarily coming to a stop?
6. You throw an object upwards from the ground with a velocity of 20m/s and it is
subject to an downward acceleration of 9.8 m/s2. How high is it after 2.0s?
Take a second to write down the facts in each problem and then determine which
equation you would use. Then compare your results to those shown below.
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1. “an acceleration of -9.8 m/s2”
means a =-9.8 m/s2
“How long after”
means t = ?
“it is dropped”
means vo = 0
“to reach a velocity of -24m/s”
means v = -24m/s
Since time, t, is a factor, we need to only choose between the first two equations.
Since velocity, v, is a factor, it must be the second equation:
v = vo + at
2. “released from rest”
means vo = 0
“an acceleration of -9.8 m/s2”
means a =-9.8 m/s2
“How far will it travel”
means x0=0 and x=?
“reaches a velocity of 24m/s”
means v = -24m/s
Since time, t, is not a factor, we need to use the third equation
v2 - vo2 = 2a(x - x0)
3. “A dropped ball”
means vo = 0
“an acceleration of -9.8 m/s2”
means a =-9.8 m/s2
“How far will it travel”
means x0=0 and x=?
“in the first 5.0s”
means t = 5.0s
Since time is a factor, we need to only choose between the first two equations.
Since position, x, is a factor, it must be the first equation:
x = x0 + vot + ½at2
4. “upwards from the ground with
a velocity of 20m/s”
means vo = + 20m/s and x0 = 0
“a downward acceleration of
9.8 m/s2”
means a = -9.8 m/s2
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“How high does it go?”
means x = ? and v = 0 (the second fact, v = 0,
is less obvious but is implied by the fact that
when it reaches it’s greatest height it must
momentarily stop...or it would go higher
Since time, t, is not a factor, we need to use the third equation
v2 - vo2 = 2a(x - x0)
5. “upwards from the ground with
a velocity of 20m/s”
means vo = + 20m/s and x0 = 0
“a downward acceleration of
9.8 m/s2”
means a = -9.8 m/s2
“How much later”
means t = ?
“momentarily coming to a stop”
means v = 0
Since time, t, is a factor, we need to only choose between the first two equations.
Since velocity, v, is a factor, it must be the second equation:
v = vo + at
6. “upwards from the ground with
a velocity of 20m/s”
means vo = + 20m/s and x0 = 0
“a downward acceleration of
9.8 m/s2”
means a = -9.8 m/s2
“How high is it”
means x = ?
“after 2.0s”
means t = 2.0s
Since time is a factor, we need to only choose between the first two equations.
Since position, x, is a factor, it must be the first equation:
x = x0 + vot + ½at2
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Interpreting Motion Graphs
There are two types of motion graphs that we’ll be considering, “Position versus Time”
and “Velocity versus Time”. In both cases the horizontal axis, the x- axis, is used to
record the time. In a Position versus Time graphs, the vertical axis, the y-axis, is used
to record the position of the object. In a Velocity versus Time graph, the vertical axis,
the y-axis, is used to record the object’s velocity. In this section, we’re going to learn
how to make and interpret these graphs and see their relationship to each other.
Position versus time graphs for Constant Velocity
If while you were moving you were to record your position each second, it’d be easy to
make a position versus time graph. Let’s take the case that you’re walking away from
your house with a constant velocity of +1m/s (note that since velocity is a vector defining
it requires a direction, “+”, and a size, “1m/s”). If you define your house as “zero” and
define your starting time as zero then the first five seconds of your walk would give you
the following data.
Time (t)
seconds
Position (x)
meters
0
1
2
3
4
5
0
1
2
3
4
5
To create a position versus time graph all you need to do is graph those points and then
connect them with a straight line.
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Position versus Time
Position (meters)
5
4
3
2
1
0
0
1
2
3
4
5
Time (seconds)
This graph allows you to directly read your position at any given time. In fact, it allows
you to determine your position for times at which you made no measurement, that’s the
meaning of the line connecting the points. For instance, your position at 1.5 seconds
can be seen to be at x = 1.5m. Now this assumes that you were traveling at a constant
velocity, but that’s the assumption that was made in creating this graph.
You can also indirectly read your velocity from this chart. The velocity of an object
will be the slope of the line on its Position versus Time graph.
The definition of the slope of a line, m, is m ≡ Δy / Δx. This means that the slope of a
line is determined by how much the vertical value, the y-value, of the line changes for a
given change in its horizontal value, its x-value. If it doesn’t change at all, then the line
is horizontal, it has no tilt. If it has a positive value it is sloping upwards, since that
means that the y-coordinate gets larger as you move to the right along the x-axis. A
negative slope means that the line is tilted downwards, since its y-value decreases as
you move to the right.
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If you make a graph of the position versus time for a moving object, with position shown
on the y-axis and time on the x-axis, then the slope of that line is given by m ≡ Δy / Δx,
but in this case the y-values are the position, x, and the x-values are time, t.
This can be confusing since the “x” in the definition of slope is different than the
“x” used in kinematics equations. In the definition of slope, x means the
horizontal axis. In the discussion of motion, x means the position of the object.
When we graph the position of an object versus time, we always put the position
on the vertical, y-axis, and time, t, on the x-axis. This can lead to confusion since
the y-coordinate on the position versus time graph gives the position, which is x
in kinematics equations.
So the slope of a line on that graph becomes:
m ≡ Δy / Δx
but since the y-values represent the position, x, and the x-values
give the time, t, this becomes
m ≡ Δx / Δt
but the definition for velocity is the same, v ≡ Δx / Δt, therefore
m=v
the slope of a line show on the Position versus Time graph give us
its velocity
For the graph shown above, the slope of the line is:
m ≡ Δy / Δx
using the first and last point (any pair of points will work)
m = (5m - 0m) / 5s – 0s)
m = 5m/5s
m = 1 m/s
v = 1 m/s
_________________________________________________________________
Example 22
Determine the location and velocity of the object in the below graph at when t = 2.5s.
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Position versus Time
Position (meters)
14
12
10
8
6
4
2
0
0
1
2
3
4
5
Time (seconds)
Solution: The location of the object can be read directly off the graph by noting that
when the time is equal to 2.5s then the position is equal to 7 meters. The velocity is
constant during the entire trip (hence the straight line) and so it can be found by
determining the slope of that line between any two points. Usually we pick points that
are easy to read and are as far apart as possible. In this case, let’s use the points (0,0)
and (4,12).
m ≡ Δy / Δx
m = (12m - 0m) / 4s – 0s)
m = 12m/4s
m = 3 m/s
v = 3 m/s
Now it’s possible that the velocity of an object will change during the time it is being
observed. For instance, if you were to walk away from your house at a velocity of 1m/s
for 6 seconds, stop for 3 seconds and then run back to your house in 3 seconds, the
Position versus Time graph for your trip would look like this.
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Position versus Time
Position (meters)
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10 11 12
Time (seconds)
You can read this graph to determine your position at any time during the trip. You can
also see from this that during your trip you had three different velocities.
Your initial velocity is given by the slope of the line during the first 6 seconds.
m ≡ Δy / Δx
m = (6m - 0m) / 6s – 0s)
m = 6m/6s
m = 1 m/s
v = 1 m/s
During the time that you’re standing still, your velocity must be zero. The slope of the
line must be zero if your velocity is zero, so this is the flat portion of the curve between 6
and 9 seconds. Just for completeness, you get the same results analytically.
m ≡ Δy / Δx
m = (6m - 6m) / 9s – 6s)
m = 0m/6s
m = 0 m/s
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v = 0 m/s
Finally, during your return trip the slope of the line is negative, meaning that you have a
negative velocity.
m ≡ Δy / Δx
m = (0m - 6m) / 12s – 9s)
m = (-6m)/3s
m = -2 m/s
v = -2 m/s
Velocity versus Time graphs for Constant Velocity
Any motion that can be recorded using a Position versus Time graph can also be
recorded using a Velocity versus Time graph. Your choice of graph will have different
benefits, but it’s important that you be able to see how they relate to one another.
Let’s take the first Position versus Time graph that we did above and recast it as a
Velocity versus Time graph. In this case, the vertical axis records velocity; the
horizontal axis continues to indicate the time. In the first graph, you maintained a
constant velocity of 1 m/s for 6 seconds so this becomes:
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Velocity versus Time
Velocity (m/s)
3
2
1
0
0
1
2
3
4
5
6
Time (s)
In this case, the object’s velocity can be read directly off the graph, but its displacement
and the distance it has traveled cannot. However, we can determine the object’s
displacement, how far it is from where it started, and the distance it has traveled by
measuring the area under the curve (in this case the horizontal line at v = 1 m/s). (If the
velocity is always positive, the distance traveled and displacement will be the same.)
A rectangle has two pairs of opposing sides. In this case, one side will be the horizontal
line indicating the velocity that the object is traveling and the side opposing that is the
horizontal axis. The second pair of sides is a vertical line drawn straight up from the
time we start measuring and the side opposing that is a vertical line drawn straight up
from the time we stop measuring.
So for an object moving at constant velocity we can define a rectangular shape whose
height is its velocity and whose length is the time interval that we’re studying. The area
of a rectangle is given by its height multiplied by its length, so the area of this rectangle
is it’s velocity, v, multiplied by the elapsed time, t.
Area = (height)(length)
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A = (velocity)(time)
A = vt
But early in the chapter we determined that Δx, so
A = Δx
The displacement of an object is determined by the area between its velocity
graph and the horizontal axis.
In the current example the horizontal sides are the horizontal line at v = 1 m/s and the
horizontal axis, and the vertical sides are formed by the vertical axis and t = 6s. The
dimensions of that rectangle are 1m/s tall by 6s long. So the object’s displacement is
simply the product of those two, or 6m. This is the same result you would expect, an
object moving at a constant velocity of 1 m/s for 6s would be displaced 6m. It has also
traveled a distance of 6m since no negative motion was involved
If the displacement is only positive, it is equal to the distance that the object has
traveled. While we derived this for constant velocity, it will always be true that the
displacement will be equal to the area under the Velocity versus Time curve and that, if
the velocity is always positive, that the distance traveled will equal the displacement of
the object.
____________________________________________________________________
Example 23
Determine the displacement of the following object, and the distance it has traveled,
during its first 3 seconds of travel.
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Velocity versus Time
Velocity (m/s)
3
2
1
0
0
1
2
3
4
5
6
Time (s)
Solution: The area under the curve of its Velocity versus Time graph is its displacement.
Since we are only considering the first 3 seconds of its travel, the height is 2 m/s and
the length is 3 seconds, so its displacement is 6 meters. This is also equal to the
distance it has traveled.
In the case that an object has both a negative and positive velocity, the distance it has
traveled and its displacement will not be the same. This is because distance does not
depend on direction while displacement does.
____________________________________________________________________
Example 24
Determine the distance traveled and the displacement of the following object during its
entire trip. It travels at a velocity of +2m/s for the first six seconds and then a velocity of
-3m/s during the last four seconds.
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Velocity (m/s)
Velocity versus Time
5
4
3
2
1
0
-1 0
-2
-3
-4
-5
1
2
3
4
5
6
7
8
9
10
Time (s)
The object’s displacement during the first six seconds is given by:
Δx = Area
Δx = (+2m/s)(6s)
Δx = 12m
During the last four seconds of its trip its displacement is
Δx = Area
Δx = (-3m/s)(6s)
Δx = -18m
So its total displacement is the sum of those two contributions:
Δx = 12m + (-18m)
Δx = -6m.
On the other hand, the total distance it traveled is given by the sum of the two areas,
treating both as positive numbers since distance traveled can never be negative. So
the distance traveled is the sum of 12m and 18m, or 30m.
d = 12m + 18m
d = 30m
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If we take motion to the right as positive and motion to the left as negative, the physical
interpretation of this is that the object traveled 12 m to the right, momentarily stopped
and then traveled 18m to the left. It moved a distance of 30m, but ends up 6 meters to
the left of where it started.
Velocity versus Time graphs for Constant Acceleration
So far we have only considered motion at constant velocity. However, the same
principles apply to motion at constant acceleration. If an object’s velocity is changing
with time, it’s Velocity versus Time graph will have a slope. That slope will give its
acceleration.
The slope of a line on the Velocity versus Time graph is:
m ≡ Δy / Δx
but since the y-values represent the velocity, v, and the x-values
give the time, t, this becomes
m ≡ Δv / Δt
but the definition for acceleration is the same, a ≡ Δv / Δt, therefore
m=a
The slope of a line on the Velocity versus Time graph for an object give us its
acceleration.
Let’s calculate the acceleration of the following object.
Velocity versus Time
Velocity (m/s)
6
5
4
3
2
1
0
0
1
2
3
4
Time (s)
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6
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For the graph shown above, the slope of the line is:
m ≡ Δy / Δx
Now use the first and last point (any pair of points will
work)
m = (6m/s - 0m/s) / 6s – 0s)
m = (6m/s)/6s
m = 1 m/s2
a = 1 m/s2
It is still the case that the area under the Velocity versus Time graph will give us the
displacement and the distance traveled. However, the shape is not longer a rectangle,
it’s a triangle. The area of a triangle is given by the formula: Area = ½ (base) (height).
We can use this to determine the displacement and distance traveled; in this case they
will be the same since all motion is positive. The base is given by the elapsed time and
the height is the greatest velocity attained, since that’s the highest point for the triangle.
Let’s determine the object’s displacement during its six seconds of travel.
Δx = Area
Δx = ½ base x height
Δx = ½ vt
where v is the velocity at time t
Δx = ½ (+6m/s)(6s)
Δx = 18m
since all motion is at positive velocity, this is also the
distance traveled
d = 18m
In the following example we need to divide the motion into two triangles, since one will
have a positive area, below the horizontal axis, and one will have a negative area. That
will give us different answers for displacement and distance.
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Velocity (m/s)
Velocity versus Time
6
5
4
3
2
1
0
-1
0
-2
-3
-4
-5
-6
1
2
3
4
5
6
7
8
9
10
Time (s)
_______________________________________________________________
Example 25
Determine the displacement and distance traveled by the object whose motion is
described in the above graph.
Solution: Since the motion included both negative and positive velocity, we need to
separate out those two parts.
For the motion with positive velocity we need to determine the area of the triangle
formed above the horizontal axis:
Δx = Area
Δx = ½ base x height
Δx = ½ vt
where v is the maximum positive velocity and t is the total
time that the object moved with a positive velocity
Δx = ½ (+6m/s)(8s)
Δx = 24m
This is the displacement due to positive velocity
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For the motion with negative velocity we need to determine the area of the triangle
formed below the horizontal axis:
Δx = Area
Δx = ½ base x height
Δx = ½ vt
where v is the maximum negative velocity and t is the total
time that the object moved with a negative velocity
Δx = ½ (-6m/s)(2s)
Δx = -6m
This is the displacement due to positive velocity
In this case the displacement and distance traveled will be different. The time spent
traveling with a negative velocity will reduce the displacement, but will add to the total
distance the object moved during its trip.
Δx = +24m + (-6m)
Δx = +18m
To find the distance traveled we just treat the areas of both triangles as positive since
distance is never negative.
d = 24m + 6m
since distance is always positive
d = 30m
So after traveling a total distance of 30m, the object is 18m to the right of where it
started.
Alternative Derivation of the First Kinematics Equation
We used a lot of algebra to derive the following equation earlier in this chapter:
x = x0 + vot + ½at2
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However, this same equation was also derived graphically in the 1400’s by Oresme
using the approaches that we just developed: recognizing that the displacement of an
object is given by the area under the Velocity versus Time curve.
For those who prefer a great picture to some complicated algebra, it’s worth seeing how
he did it. We’ll also use a similar approach to derive some equations in some later
chapters.
All we have to do is leave variables in our graphs instead of numbers. For instance let’s
start out with an object moving at a constant velocity v0 for a time t. That means that
the graph of its velocity versus time will form a rectangle whose height is v0 and whose
length is t. The area of that rectangle, will give its displacement.
Displacement due to initial velocity
Δx = Area
Δx = height x length
Δx = v0t
Due to initial velocity alone
Now let’s add onto that the displacement due to a constant acceleration. If the
acceleration is positive, that means it gets a bigger velocity as time goes by and you get
the graph shown below (assuming for simplicity that v0 = 0.). We showed earlier that
the area under this curve is equal to an object’s displacement. In this case, its greatest
velocity will be the height of the triangle and the base of the triangle will be the time of
its acceleration.
Displacement due to acceleration
Δx = Area
Δx = ½ base x height
Δx = ½ vt
But recall that that v = v0 + at. In this case, v0 is zero, so v =
at. Substituting that in for v we get
Δx = ½ (at)t
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Δx = ½ at2
Due to acceleration alone
Total Displacement
An object that has an initial velocity and experiences acceleration has a displacement
due to both of these terms. So its total displacement will be
Δx = v0t + ½ at2
But recall that Δx = x - x0, so
x - x0 = v0t + ½ at2
Solving for x yields our kinematics equation
x = x0 + v0t + ½ at2
The position of an object at any time will be given by three terms: where it started, x 0,
how far it moved due to its initial velocity, v0t, and how far it moved due to its
acceleration, ½ at2. Adding those terms together gets you the same result we got using
algebra earlier in the chapter: x = x0 + vot + ½at2.
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Chapter Questions
1. When you drive a car and take a quick look on the speedometer, what do you
see velocity or speed?
2. A ball is thrown vertically up and is caught at the starting point; compare the
traveled distance and displacement of the ball.
3. Can an object with a constant acceleration reverse it direction of travel? Explain.
4. Can an object have a varying velocity when its speed is constant? Explain.
5. Is it possible for an object to have average velocity equal to its instantaneous
velocity?
6. What quantity describes how quickly you change your speed?
7. Can an object have a southward velocity and a northward acceleration?
8. Is it possible for an object to have average speed equal to the magnitude of its
average velocity?
9. If the speedometer of a car reads constant speed 50 mi/h, does it mean that the
car has constant velocity?
10. What does the area under the curve on a velocity-versus-time graph represent?
11. What does the slope of the curve on a displacement-versus-time graph
represent?
12. What does the slope of the curve on a velocity-versus-time-time graph
represent?
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Chapter Problems
Motion at Constant Speed
1. A rabbit can cover a distance of 80 m in 5 s. What is the speed of the rabbit?
2. During the first 50 s a truck traveled at constant speed of 25 m/s. Find the
distance that it traveled.
3. An elevator ascends at a constant speed of 4 m/s, how much time is required for
the elevator in order to travel 120 m upwards?
4. A racing car can complete a 900 m long lap in 15 s. What is the speed of the
car?
5. It is known that a shark can travel at a speed of 15 m/s. How far can a shark go
in 10 s?
6. How long does it take an airplane to fly 1500 miles if it maintains a speed of 600
miles per hour?
7. A person can run 280 m in 68 s. At what speed are they running?
8. How far does a runner travel if they maintain a speed of 8.5 m/s for 240 s?
9. A sportsman can develop a maximum speed of 12 m/s when he is swimming in a
pool. Calculate the time interval that is required to travel a distance of 25 m.
10. A train, of length 200m, with a speed of 30 m/s crosses a bridge that is 850 m
long. How much time does it take for the train to completely cross the bridge
(from the moment it reaches the bridge until it has completely crossed it)?
11. A train with a length of 150 m crosses a 390 m long bridge in 180 s. Calculate the
speed of the train.
12. A battalion of soldiers that extends over a distance of 250 m crosses a 125m
long bridge. If they march at a speed of 2.5 m/s, how much time will elapse from
when the first soldier reaches the bridge to when the last soldier steps off of it?
13. A car travels at a constant speed of 20 m/s. Convert the speed from m/s into
km/h.
14. A train travels at constant speed of 54 km/h. Convert the speed from km/h into
m/s.
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Non-Uniform Motion and Average Speed
15. A boat travels 40 miles in two hours, speeds up to travel the next 80 miles in three
hours then slows down to travel the last 40 miles in three hours. What is the average
speed of the boat for the entire trip?
16. A car travels for an hour at a speed of 20 km/hr, the next two hours at a speed of 65
km/hr and the final hour at a speed of 85 km/hr. What is the average speed of the
car for the entire trip?
17. A bicyclist travels the first 800 m of a trip with a speed of 10 m/s, the next 500 m with
an average speed of 5 m/s and the final 1200 m at a speed of 13 m/s. Find the
average speed of the bicyclist for this trip.
18. A bicyclist travels the first 700 m of a trip at an average speed of 8 m/s, travels the
next 600 m in 90 s and the last 50s at a speed of 21 m/s. Find the average speed of
the bicyclist for this trip.
19. A car travels 100 km in the first two hours, stops for half an hour and then travels
200 km in the next four hours. Find the average speed of the car.
20. An airplane travels at a speed of 300 km/hr for 2 hours, speeds up to 400 km/hr for
the next hour and then travels the final four hours at a speed of 500 km/hr. What is
the average speed of the plane for this trip?
21. An airplane travels 2800 km at a speed of 700 km/h, decreases its speed to 500
km/h for the next 1500 km and travels the last 1000 km at a speed of 400 km/hr.
Find the average speed of the plane for the trip.
22. An airplane travels 2100 km at a speed of 500 km/h, encounters a head wind that
decreases its speed to 400 km/h for the next three hours and then travels the last
400 km to complete the trip at an average speed, for the entire trip of 440 km/hr.
What was the speed of the plane for the last part of the trip?
23. A runner runs the first 400m of a race in 80s, the second 400m in 70s and the final
800m in 130s. What was her average speed for this run?
24. A train travels from Boston to New York. It travels at a speed of 180 km/hr for two
hours, speeds up to 200 km/hr for the next four hours and then slows down to 120
km/hr for the next six hours. What is the average speed of the train for this trip?
25. A train travels 120 km at a speed of 60 km/h, makes a stop for 0.5 hr, and then
travels the next180 km at a speed of 90 km/h. What is the average speed of the train
for this trip?
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26. A train travels a total distance of 600km in eight hours. It travels the first 120 km at a
speed of 60 km/h, travels the next 180 km in two hours and then completes the trip
at an unknown speed. What was the average speed of the train for the last leg of
the trip? What is the average speed of the train for the entire trip?
Position, Displacement and Velocity
27. A polar bear goes 4 miles north and then 7 miles south. Draw a sketch that
shows its first displacement; its second displacement; and its total displacement
for the entire trip. Find the total distance it traveled as well as its total
displacement. Compare those two results.
28. An object moves from the initial position x1 = -3 m to the position x2 = 4 m and
then to the position x3 = 8 m (see diagram above). Show both displacement
vectors and the vector representing the object’s total displacement. Find the total
traveled distance and the total displacement of the object. Compare those two
results.
29. An object moves from the initial position x1 = -2 m to the position x2 = 7 m and
then to the position x3 = 3 m (see diagram above). Draw the two displacement
vectors and the vector representing the total displacement. Find the total
distance traveled and the total displacement. Compare those two results.
30. An object moves from the position +16 m to the position +47 m in 12 s. What is
its total displacement? What is its average velocity?
31. An object moves from the position +34 m to the position -15 m in 15 s. What is its
total displacement? What is its average velocity?
32. A balloon drifts 140 m toward the west in 45 s; then the wind suddenly changes
and the balloon flies 90 m toward the east in the next 25 s.
a. What distance did it travel during the first 45 s?
b. What distance did it travel during the next 25 s?
c. What total distance did it travel?
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d. What was its average speed during the first 45 s?
e. What was its average speed during the next 25 s?
f. What was its average speed for the entire trip?
g. What was its displacement during the first 45 s?
h. What was its displacement during the next 25 s?
i. What was its total displacement?
j. What was its average velocity during the first 45 s?
k. What was its average velocity during the next 25 s?
l. What was its average velocity for the entire trip?
33. An object starts at a point +25 m goes 40 m toward +X direction in 5 s then
suddenly changes its direction to the opposite and covers 50 m in 10 s.
a. Where is the object after the first 5 s?
b. Where is the object after the next 10 s?
c. What total distance did it travel?
d. What was its average speed during the first 5 s?
e. What was its average speed during the next 10 s?
f. What was its average speed for the entire trip?
g. What was its displacement during the first 5 s?
h. What was its displacement during the next 10 s?
i. What was its total displacement?
j. What was its average velocity during the first 5 s?
k. What was its average velocity during the next 10 s?
l. What was its average velocity for the entire trip?
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Problem Solving with the Kinematics Equations
Using the Equation: v = v0 + at
34. If an object accelerates from rest, with a constant acceleration of 5.4 m/s 2, what
will its velocity be after 28 s?
35. An object is traveling at a constant velocity of 15 m/s when it experiences a
constant acceleration of 3.5 m/s2 for a time of 11 s. What will its velocity be after
that acceleration?
36. An object is moving at a velocity of 23 m/s. It accelerates to a velocity of 85 m/s
over a time of 8.3 s. What acceleration did it experience?
37. An object initially at rest experiences an acceleration of 9.8 m/s2. How much time
will it take it to achieve a velocity of 58 m/s?
38. An object accelerates to a velocity of 34 m/s over a time of 13 s. The
acceleration it experienced was 15 m/s2. What was its initial velocity?
39. A car is driving at a velocity of 24 m/s. If its brakes can supply an acceleration of
-5.0 m/s2, how much time will be required to bring the car to a stop?
40. A car’s speedometer reads 20 m/s after accelerating, from a standing start, for
25s. What was the magnitude of its acceleration?
41. An object increases its velocity from 22 m/s to 36 m/s in 5 s. What is the
acceleration of the object?
42. A bicyclist is traveling at a speed of 5 m/s when it suddenly accelerates, at a
constant rate of 0.6 m/s2, for a time of 10s. What is the speed of the bicycle at
the end of that 10s?
43. A train departs from its station at a constant acceleration of 5 m/s 2. What is the
speed of the train at the end of 20s?
44. What is the velocity of a dropped object after it has fallen for 3.0s?
45. What is the velocity of a dropped object after it has fallen for 12s?
46. A ball is thrown straight up with a velocity of 16m/s; what will be its velocity 2.0s
after being released?
47. A ball is thrown straight down with a velocity of 12m/s; what will be its velocity
2.0s after being released?
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48. A ball is thrown straight up from the ground with an unknown velocity. It returns
to the ground after 4.0s. With what velocity did it leave the ground?
49. A ball is thrown straight up from the ground with an unknown velocity. It reaches
its highest point after 3.5s. With what velocity did it leave the ground?
Using the Equation: x = x0 + vot + ½at2
50. An object, moving with a constant velocity, travels 274m in 23s. What is its
velocity?
51. An object accelerates from rest with a constant acceleration of 2m/s2. How far
will it have moved after 9s?
52. An object is traveling with a constant velocity of 5m/s. How far will it have gone
after 7s?
53. An object, moving with constant acceleration and zero initial velocity, travels 48
m in 5.2s. What is the magnitude of its acceleration?
54. An object is moving with an initial velocity of 23 m/s. It is then subject to a
constant acceleration of 3.5 m/s2 for 12 s. How far will it have traveled during the
time of its acceleration?
55. An object is moving with a constant velocity of 278 m/s. How long will it take it to
travel 7500 m?
56. An object, initially at rest, is subject to an acceleration of 34 m/s 2. How long will it
take for that object to travel 3400 m?
57. You are 12 miles north of your base camp when you begin walking north at a
speed of 2 miles per hour. What is your location, relative to your base camp,
after walking for 5 hours?
58. An object travels at a constant velocity of 15 m/s for 5.0 seconds. How far does
it move during that time?
59. An object moves 250 m in 17 s. What is its average velocity?
60. How long will it take a person walking at 2.1 m/s to travel 13 m?
61. You are at a rest stop 250 miles north of New York City. You then travel north at
a constant velocity of 65 miles per hour for 2.0 hours. Describe your location
relative to New York City.
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62. An object is at rest when it undergoes a constant acceleration of 13 m/s 2 for 5.0
seconds. How far will it have traveled during this time?
63. An object is shot upwards, from the ground, with an initial velocity of 120 m/s.
How high will it be after 4.0s?
64. An object is shot upwards, from the ground, with an initial velocity of 40 m/s.
How high will it be after 2.0s?
65. An object is dropped from the top of a building and strikes the ground 2.0s later.
How tall is the building?
66. An object is dropped from a 42m tall building. How long does it take to reach the
ground?
Using the Equation: v2 - vo2 = 2a(x - x0)
67. An object accelerates from rest with a constant acceleration of 7.5 m/s 2. How
fast will it be traveling after it goes 21 m?
68. An object experiences an acceleration of 9.8 m/s2 over a distance of 210 m.
After that acceleration it has a velocity of 380 m/s. What was its velocity before
being accelerated?
69. An object accelerates from rest to 24 m/s over a distance of 56 m. What
acceleration did it experience?
70. An object experiences an acceleration of 6.8 m/s2. As a result, it accelerates
from rest to 24 m/s. How much distance did it travel during that acceleration?
71. A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5
m/s2 over a distance of 340 m. How fast is it going after that acceleration?
72. A car slams on its brakes creating an acceleration of -3.2 m/s2. It comes to rest
after traveling a distance of 210 m. What was its velocity before it began to
accelerate?
73. An object accelerates from rest to 85 m/s over a distance of 36 m. What
acceleration did it experience?
74. An object experiences an acceleration of -6.8 m/s2. As a result, it accelerates
from 54 m/s to a complete stop. How much distance did it travel during that
acceleration?
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75. An object is dropped from a 32m tall building. How fast will it be going when it
strikes the ground?
76. An object is dropped from a building and strikes the ground with a speed of
31m/s. How tall is the building?
77. A hopper jumps straight up to a height of 1.3m. With what velocity did it leave
the floor?
78. A hopper jumps straight up to a height of 0.45m. With what velocity will it return
to the table?
Using all three Kinematics Equations
79. An object accelerates from rest, with a constant acceleration of 8.4 m/s 2, what
will its velocity be after 11 s?
80. An object is traveling with a constant velocity of 3.0 m/s. How far will it have
gone after 4.0 s?
81. An object is moving with an initial velocity of 19 m/s. It is then subject to a
constant acceleration of 2.5 m/s2 for 15 s. How far will it have traveled during the
time of its acceleration?
82. An object accelerates from rest to a velocity of 34 m/s over a distance of 3.5 m.
What was its acceleration?
83. A car which is traveling at a velocity of 1.6 m/s undergoes an acceleration of 9.2
m/s2 over a distance of 540 m. How fast is it going after that acceleration?
84. An object is traveling at a constant velocity of 250 m/s when it experiences a
constant acceleration of 1.5 m/s2 for a time of 38 s. What will its velocity be after
that acceleration?
85. An object accelerates from rest with a constant acceleration of 7.5 m/s 2. How
fast will it be traveling after it goes 87 m?
86. An object is moving at a velocity of 8.8 m/s. It accelerates to a velocity of 85 m/s
over a time of 6.3 s. What acceleration did it experience?
87. An object accelerates from rest, with a constant acceleration of 7.4 m/s 2, what
will its velocity be after 5.4 s?
88. An object accelerates from rest to a velocity of 44 m/s over a distance of 35 m.
What was its acceleration?
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89. An object is traveling with a constant velocity of 2.0 m/s. How far will it have
gone after 38 s?
90. An object is moving with an initial velocity of 5.5 m/s. It is then subject to a
constant acceleration of 2.5 m/s2 for 11 s. How far will it have traveled during the
time of its acceleration?
91. An object accelerates from rest with a constant acceleration of 7.5 m/s 2. How
fast will it be traveling after it goes 12 m?
92. An object is traveling at a constant velocity of 11 m/s when it experiences a
constant acceleration of 1.5 m/s2 for a time of 14 s. What will its velocity be after
that acceleration?
93. An object is moving at a velocity of 5.8 m/s. It accelerates to a velocity of 25 m/s
over a time of 3.3 s. What acceleration did it experience?
94. A car which is traveling at a velocity of 9.6 m/s undergoes an acceleration of 4.2
m/s2 over a distance of 450 m. How fast is it going after that acceleration?
Free Fall
95. A stone is dropped from the roof of a building. It took 5s for the stone to reach the
ground. What is the height of the building?
96. An object falls from a height of 490 m. How much time does it take for the object
to reach the ground?
97. A ball is thrown vertically down from the edge of a cliff with a speed of 8 m/s, how
high is the cliff, if it took 6 s for the ball to reach the ground?
98. What is the landing velocity of an object that is dropped from a height of 49 m?
99. What is the landing velocity of an object that is thrown vertically down with a
velocity of 5 m/s from a height of 25m?
100.
A water drop falls down from the roof of a house during 3 s, how high is
the house?
101.
An object is dropped from a height of 100 m, how long is it in the air?
102.
A pump can throw a stream of water up to 29.6 m. What is the initial
speed of the stream when it leaves a hose nozzle?
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103.
A stone is thrown vertically up with a speed of 15 m/s. What is the traveled
distance during first 4 s?
104.
An arrow is projected by a bow vertically up with a velocity of 20 m/s, and
reaches a target in 3 s. What is the velocity of the arrow just before it hits the
target? How high is the target located?
105.
An object is thrown vertically up with a velocity of 35 m/s. What was the
maximum height it reached?
106. A boy throws a ball vertically up and catches it after 3 s. What height did the
ball reach?
107. A marble is projected vertically up by a spring gun, and reaches the maximum
height of 9.8 m. What is the initial speed of the marble? How long was the
marble in the air?
108. An arrow is shot vertically up by a bow, and after 8 s returns to the ground
level. What is the initial velocity of the arrow? How high did it go?
Graph Problems
109. Starting at the position, x0 = 4, you travel at a velocity of +2m/s for 6s.
a. Determine your position at the times of 0s; 2s; 5s; and 6s.
b. Draw the Position versus Time for your travel during this time.
c. Draw the Velocity versus Time graph for your trip.
110. Starting at the position, x0 = -15, you travel at a velocity of +5m/s for 3s.
a. Determine your position at the times of 0s; 1s; 2s; and 3s.
b. Draw the Position versus Time for your travel during this time.
c. Draw the Velocity versus Time graph for your trip.
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111. The Position versus Time graph, below, describes the motion of three different
cars moving along the x-axis.
a. Describe, in words, the velocity of each of the cars. Make sure you
discuss each car’s speed and direction.
b. Calculate the velocity of each of the cars.
c. Draw, on one set of axes, the Velocity versus Time graph for each of the
three cars.
Position versus Time
15
Velocity (m/s)
10
5
0
-5
0
1
2
3
-10
-15
Time (s)
Series1
Series2
Series3
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112. The Position versus Time graph, below, describes the motion of three different
cars moving along the x-axis.
a. Describe, in words, the velocity of each of the cars. Make sure you
discuss each car’s speed and direction.
b. Calculate the velocity of each of the cars.
c. Draw, on one set of axes, the Velocity versus Time graph for each of the
three cars.
Velocity (m/s)
Position versus Time
16
14
12
10
8
6
4
2
0
-2
-4 0
-6
-8
-10
-12
1
2
3
4
5
Time (s)
Series1
Series2
Series3
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113. The Velocity versus Time graph, below, describes the motion of three different
cars moving along the x-axis.
a. Describe, in words, the velocity of each of the cars. Make sure you
discuss each car’s speed and direction.
b. Calculate the displacement of each car during its trip.
c. Calculate the distance traveled by each car during its trip.
d. Compare your answers to b and c.
Velocity (m/s)
Velocity versus Time
6
5
4
3
2
1
0
-1
0
-2
-3
-4
-5
-6
1
2
3
4
5
Time (s)
Series1
Series2
Series3
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114. The Velocity versus Time graph, below, describes the motion of three different
cars moving along the x-axis.
a. Describe, in words, the velocity of each of the cars. Make sure you
discuss each car’s speed and direction.
b. Calculate the displacement of each car during its trip.
c. Calculate the distance traveled by each car during its trip.
d. Compare your answers to b and c.
Velocity (m/s)
Velocity versus Time
6
5
4
3
2
1
0
-1
0
-2
-3
-4
-5
-6
1
2
3
4
5
Time (s)
Series1
Series2
Series3
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115. A graph of the velocity as a function of time for two cars is presented below.
a.
b.
c.
d.
e.
What is the initial velocity of each car?
What is the acceleration of each car?
What is the traveled distance of the car I at the end of 25 s?
What is the traveled distance of the car II at the end of 20 s
Sketch graphs, on one set of axes, of each car’s acceleration as a function
of time.
116. A graph of the velocity as a function of time for two objects A and B is
presented below.
a.
b.
c.
d.
e.
What is the initial velocity of each object?
What is the acceleration of each object?
What is the traveled distance of the object A at the end of 6 s?
What is the traveled distance of the object B at the end of 10 s?
Sketch clear graphs of the object’s acceleration as a function of time.
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