Motion in 1 Dimension
Physics is all about describing motion. For now we are going to discuss
motion in 1 dimension, which means either along the x axis or the y axis.
To describe an object’s motion, we must be able to specify its position (x)
at all times. An object’s position is the x or y coordinate of its location.
● Since we are discussing motion along the x-axis first, we will assign an
object’s initial position the label xi and its final position the label xf.
When an object changes its position, it experiences a displacement.
● This is a vector that points from an initial position xi to a final
position xf.
● The magnitude is the shortest distance between the two positions.
● Mathematically, displacement is given by…
∆x ≡ x f − xi
∆ x is positive if x f > xi
( object moves right )
Positive displacement
The 3 bar
equal sign
means
mathematical
definition.
∆ x is negative if x f < xi
( object moves left )
Negative displacement
● We will use the meter as our standard S.I. unit of measurement.
Do you remember
your
conversions?
1 meter = 100 centimeters
1 meter = 1000 millimeters
1 kilometer = 1000 meters
● FROM THIS POINT ON, YOU REALLY NEED TO KNOW THOSE
METRIC CONVERSIONS!!!
Now it’s time to realize that there is a HUGE difference between
distance traveled and displacement!!
As an example… A baseball player hitting a home run travels around the bases.
Questions…
What is his initial
position?
What is his final
position?
What distance did he
run?
What is his
displacement?
SAMPLE PROBLEM A
A marble rolls alongside a meter
stick. It begins moving at the 98cm mark and stops on the 87-cm
mark. Calculate its displacement.
∆x = xf – xi = 0.87 – 0.98 = -0.11 m
SAMPLE PROBLEM B
A “lazy susan” is spun 3 times. A container of parsley flakes
travels around a circular path having a radius of 6.0 cm. (a) What
was the distance traveled by the parsley? (b) What was the
parsley’s displacement?
a) d = 3 (2πr) = 6π(0.06) = 1.131 m
b) ∆x = 0 m
Graphing Position vs. Time (x vs. t)
Important relationships can be discovered from graphs, so let’s plot a
sample position vs. time (x vs. t) graph to analyze.
4 Steps to Graphing Position vs. Time
● Step 1: Label the x-axis with Time (units)
Label the y-axis with Position (units)
Entitle the graph Position vs. Time
● Step 2: Determine an appropriate scale for each axis. The idea is to
make the graph as large as possible!
● Step 3: Plot the data neatly and connect the points with a smooth line.
● Step 4: Calculate the slope for each segment of the graph using the
equation below. Don’t forget units as this slope will now be a
meaningful quantity!!!
rise
∆y
∆x
m=
=
or for our graph :
run
∆x
∆t
SAMPLE (x vs. t) GRAPH
Position (m)
0
200
400
600
800
1000
1200
1200
1200
1200
1200
1100
1000
900
800
700
600
500
400
Step 1:
Step 2:
Step 3:
Step 4:
Time
(s)
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
Label, label, label!
Determine an appropriate scale for each axis.
Plot the data neatly and connect the points with a smooth line.
Calculate the slope for each segment of the graph.
Slope 1 = ∆x / ∆t = 1200 m / 600 s = 2 m/s
Slope 2 = ∆x / ∆t = 0 m / 400 s = 0 m/s
Slope 3 = ∆x / ∆t = -800 m / 800 s = -1 m/s
What is significant about the slope’s units?
m/s. Indicates that the slope is telling you velocity.
Average Velocity
This slope is so important that it has been given a special name;
Average Velocity. Average velocity is the change in position with time.
● Since velocity depends on displacement, velocity is also a vector
quantity.
● meters per second (m/s) is the standard S.I. unit of measurement
● We can find
v
between any two points by drawing a straight line
between them and calculating the slope, regardless of the graph’s
shape.
Average Velocity is defined as:
∆x
vx ≡
= slope of x vs. t graph
∆t
v x is positive if ∆ x is positive
v x is negative if ∆ x is negative
or
slope of x vs . t is positive
or
slope of x vs . t is negative
PRACTICE
PRACTICE PROBLEM
PROBLEM A
A
NAME: ___________________________
A farmer collected the following data after lopping the head off one of his chickens. Set up a position
vs. time graph to help the farmer understand the motion of the bird. Label each straight-line segment
A − J.
Position (m)
Time (s)
0
4
8
2
2
2
4
4
6
1
2
3
3
3
4
5
6
0
1
2
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Calculate VAVG for the following time
intervals:
Describe the motion of the bird for each segment:
A:
(0 – 2 sec): ∆x/∆t = 8m/2s = 4 m/s
B:
C:
(2 – 4 sec): ∆x/∆t = -6m/2s = -3 m/s
D:
(1 – 6 sec): ∆x/∆t = -2m/5s = -0.4 m/s
E:
(4 – 9 sec): ∆x/∆t = 4m/5s = 0.8 m/s
F:
G:
(4 – 6 sec): ∆x/∆t = 0m/2s = 0 m/s
(0 – 17 sec): ∆x/∆t = 6m/17s = 0.353 m/s
H:
I:
J:
Velocity vs. Time Graphs
Instead of graphing just position vs. time, we are also able to graph
velocity (m/s)
velocity (m/s)
velocity vs. time. This graph will show you how velocity changes
with time. Therefore, the rules that applied to a position vs. time
graph do not necessarily apply to a velocity vs. time graph.
time (s)
∆x = positive
v = increasing
a = positive
velocity (m/s)
velocity (m/s)
∆x = positive
v = increasing
a = positive
time (s)
time (s)
∆x = positive
v = constant
a = zero
time (s)
∆x = positive
v = decreasing
a = negative
Instantaneous Velocity
v
Why would
not be useful if you wanted to know your
velocity the instant you noticed the flashing lights of a police car behind
you?
We need to be able to specify our velocity just as precisely as we can our
position by knowing what is happening at a specific time.
● We could not actually do this mathematically until the late 1600’s
when Calculus gave us this ability.
● We can do this in Physics by taking smaller and smaller intervals of
time on our x vs. t graph.
The limiting value of the ratio
∆x
∆t
as ∆t approaches zero is called
Instantaneous Velocity.
● Notice that this is just the limit of the average velocity as time
approaches zero.
● We can interpret V graphically by drawing a line tangent to the
curve at a specific time. We can then calculate the slope of the
tangent line.
Instantaneous Velocity is defined as:
∆x
v ≡ lim
= slope of tangent line
∆t → 0 ∆ t
From here on, we use the term velocity to designate instantaneous
velocity.
SAMPLE PROBLEM
An object moves along the x-axis as shown
in the position vs. time graph to the right.
(a) Determine the object’s displacement
during the time intervals t = 1 s to t = 3 s
and t = 4 s to t = 6 s. (b) Calculate the
average velocity during these two time
intervals. (c) Find the object’s instantaneous
velocity at t = 2.0 s and at 6.0 s.
∆x = xf – xi = 3m – 11m = -8 m
∆x = xf – xi = 5m – 2m = 3 m
v = ∆x / ∆t = -8 / 2 = -4 m/s
v = ∆x / ∆t = 3 / 2 = 1.5 m/s
v = ∆x / ∆t = -13 / 3.5 = 3.714 m/s
Instantaneous Velocity (Using Calculus)
SAMPLE PROBLEM
A person moves along the x-axis. Her position varies with time according to
the expression x = 2t2 - 4t where x is her instantaneous position in meters
while t represents the time at that position. The graph for this motion is
shown below. Find the instantaneous velocity at t = 0s, t = 1s, and t = 3s.
Step 1 = Take the derivative of the function:
x = 2t2 – 4t
v = 4t – 4
Step 2 = Plug in your value of time
v(0) = 4(0) – 4 = -4 m/s
v(1) = 4(1) – 4 = 0 m/s
v(3) = 4(3) – 4 = 8 m/s
In everyday usage, the words speed and velocity are used
interchangeably. However, there is a HUGE difference!!!
EXAMPLE
During the 2008 Summer Olympics,
Michael Phelps swam four 50-m laps
in 1.52 minutes to win the men’s 200-m
Butterfly and set a new world record.
QUESTIONS:
What distance was covered?
4 x 50 m = 200 m
What was the swimmer’s displacement?
∆x = 0m
What was the swimmer’s average velocity?
V = ∆x/∆t = 0m / 1.52 min = 0 m/s
This is not very useful, so we still need to be able to quantify
“how fast” she swam.
This is where the term Average Speed comes in. Speed is a scalar
quantity so the direction is meaningless in describing it.
● Average Speed is given by:
total distance
s=
total time
Instantaneous Speed is defined as the magnitude of the instantaneous
velocity. Your car’s speedometer measures instantaneous speed!!
You need to very careful here! Average Speed IS NOT the magnitude of
Average Velocity because average speed is dependent on distance and
average velocity depends on displacement!
Acceleration
We have already seen that velocity
can change while a particle moves.
We can quantify changes in velocity as a function of time just as we could
with x(t).
When velocity changes with time, a particle is undergoing Acceleration.
● Mathematically, acceleration is defined in the following way:
∆v
a≡
= slope of v vs. t graph
∆t
● As with velocity, when the motion is 1-Dimensional, we can simply
use +/- to indicate direction.
2
2
● meters per second (m/s ) is the standard S.I. unit of measurement.
SAMPLE PROBLEM A
Suppose a plane starts from rest (vi = 0 m/s) when ti = 0 s.
The plane accelerates down the runway and at tf = 29 s
attains a velocity of vf = + 260 km/h, where the plus sign
indicates the velocity points to the right. Determine the
acceleration of the plane.
260 km x 1000 m x 1 hr = 72.222 m/s
hr
1 km
3600s
a = (Vf – Vi) / t = (72.222 – 0) / 29 = 2.490 m/s2
SAMPLE PROBLEM B
A drag racer crosses the finish line, and the driver deploys a
parachute and applies the brakes to slow down. The driver
begins slowing down when t = 9.0 s and the car’s velocity is
v = 28 m/s. When t = 12.0 s, the velocity has been reduced
to v = 13 m/s. What is the acceleration of the dragster?
a = (Vf – Vi) / t = (13 – 28) / (12 - 9) = -5 m/s2
SAMPLE PROBLEM C
A glider’s velocity on a tilted air track increases from - 17.0 cm/s
at the time t = 0.33 s to - 55.0 cm/s at a time of t = 2.33 s. What
is the acceleration of the glider?
-17 cm/s = -0.17 m/s, -55 cm/s = -0.55 m/s
a = (Vf – Vi) / t = (-0.55 – (-0.17)) / 2 = 0.19 m/s2
1-D Motion w/ Constant Acceleration
The following kinematic equations only work for situations
involving constant acceleration!!!
v f = vi + at
v f = v i + 2 a∆x
2
2
1
∆x = (vi + v f )t
2
1 2
∆x = vi t + at
2
You need to have these memorized by
next class!
Physics
Kinematics SAMPLE PROBLEMS
Name:_______________________
Date: __________ Period: ______
Directions: Solve the following problems in the space provided. Use the tables to
organize your data. If a variable is not given, record a −− in the block. If a variable is
unknown, record a ? in the block. Convert all measurements into units of meters,
seconds and meters per second. Be sure to show all work!
SAMPLE PROBLEM A
The doors open for a back−to−school sale and a crazed teen sprints
from her mom’s car. If the girl reaches a velocity of +4.6 m/s in
1.9 s, determine how far she ran to the store.
DATA TABLE
∆x
vi
vf
a
t
?
0
4.6
1.9
∆x = ½ (Vi + Vf)t
= 0.5 (0 + 4.6)1.9
= 4.37 m
SAMPLE PROBLEM B
Starting from rest, a cow reaches a velocity of +2.5 m/s in a time of 5.0 s.
A Calculate the cow’s average acceleration.
B By how much does the cow’s speed change each second?
C Calculate the bovine’s displacement during this time interval.
∆x
?
DATA TABLE
vi
vf
a
0
2.5
0.5
t
5
A. a = (Vf – Vi) / t = (2.5 – 0) / 5 = 0.5 m/s2
B. Increases + 0.5 m/s for every second of movement
C. ∆x = Vit + ½ at2 = 0.5(0.5)52 = 6.25 m
SAMPLE PROBLEM C
A man’s 2010 Porsche 911 reaches 60 mi/h with a constant
acceleration of + 8 m/s2. Assuming the car starts from rest,
how long does it take the car to reach this speed?
(Hint: 1 mile = 1609 m)
DATA TABLE
∆x
vi
vf
a
t
0
26.817 8
?
60 mi x 1609 m x 1 hr = 26.817 m/s
hr
1 mi
3600s
t = (Vf – Vi) / a = (26.817 – 0) / 8 = 3.352 s
SAMPLE PROBLEM D
Billy Bob is going to fetch some dinner. He catches sight of
a squirrel that he fancies and “takes off a runnin’.” Billy
Bob starts from rest, accelerates at the rate of + 1.5 m/s2 and
reaches a velocity of + 4.7 m/s before catching the critter.
Calculate Billy’s displacement for the hunt.
∆x
?
DATA TABLE
vi
vf
a
0
4.7
1.5
t
-
Vf2 = Vi2 + 2a∆x
∆x = (Vf2 – Vi2) / 2a = (4.72 – 02) / (2(1.5)) = 7.363 m
SAMPLE PROBLEM E
A motorcycle accelerates from 90 mi/h to 65 mi/h in 2.3 s.
Find the motorcycle’s acceleration and displacement
during this time interval.
DATA TABLE
∆x
vi
vf
a
? 40.225 29.051 ?
t
2.3
90 mi x 1609 m x 1 hr = 40.225 m/s
hr
1 mi
3600s
65 mi x 1609 m x 1 hr = 29.051 m/s
hr
1 mi
3600s
a = (Vf – Vi) / t = (29.051 – 40.225) / 2.3 = -4.858 m/s2
∆x = Vit + ½ at2 = 40.225(2.3) + 0.5(-4.858)2.32 = 79.669 m
SAMPLE PROBLEM F
A car, initially traveling at 70.0 km/h, accelerates
at the constant rate of -1.50 m/s2. How far will the
car travel in 10.0 s?
DATA TABLE
∆x
vi
vf
a
? 19.444
-1.5
t
10
70 km x 1000 m x 1 hr = 19.444 m/s
hr
1 km
3600s
∆x = Vit + ½ at2 = 19.444(10) + 0.5(-1.5)102 = 119.44 m
CHALLENGE PROBLEM
Watch Out for the Speed Limit!
A car traveling at a constant speed of 45.0 m/s passes a trooper
hidden behind a billboard. One second after the speeding car
passes the billboard, the trooper sets out from the billboard to
catch it, accelerating at a constant rate of 3.00 m/s2. How long
does it take her to overtake the car?
∆xcar = 45 + 45t
∆xtrooper = Vit + ½ at2 = 0 + ½(3t2) = 1.5t2
1.5t2 = 45t + 45
1.5t2 - 45t – 45 = 0
USE QUADRATIC FORMULA
t = (-b ± √(b2 – 4ac)) / 2a = (-(-45) ± √(452 – 4(1.5 * -45))) / 2(1.5)
t = 30.968 s, -0.969 s
Freely Falling Objects
All objects dropped near the earth’s surface fall with the same
constant acceleration in the absence of air resistance.
● Aristotle (384-322 B.C.) believed that heavier objects fell
faster than lighter ones.
● This fact was not fully accepted until the early 1600s. Galileo
(1564-1642) conducted experiments with balls on inclined
planes and drew conclusions about freely falling objects since a
freely falling ball is equivalent to a ball moving down a vertical
incline.
? What kinematic equation could be used to find the
acceleration in this experimental procedure?
1 2
∆x = vi t + at
2
● On August 2nd, 1971 David Scott released a hammer and a
feather on the surface of the moon. What was discovered?
By definition, a freely falling object is any object moving freely
under the influence of gravity alone, regardless of its initial
motion. Objects thrown upward or downward and those
released from rest are all falling freely once they are released.
Any freely falling object experiences an acceleration directed
DOWNWARD, regardless of its initial motion.
The value of g does change with increasing altitude, as we will
see when we study Newton’s Law of Universal Gravitation.
At the earth’s surface (and for our discussions in this course) the
acceleration due to gravity is approximately - 9.80 m/s2.
Our kinematic equations now become:
v y f = v yi − gt
1 2
∆y = v yi t − gt
2
2
2
v y f = v yi − 2 g ∆ y
Name: ______________________________ Date: _______ Period: ______
Mr. Talboo - Physics
SAMPLE PROBLEM A
A stone is thrown downward from the top of a tall building with an initial speed of
5.00 m/s. What is the displacement of the stone after 3.00 s of free-fall? What is the
final velocity of the stone when it strikes the ground?
∆y
?
DATA TABLE
vyi
vyf
g
5
?
9.8
t
3
∆y = Vit – ½ gt2
= (-5)(3) - 1/2 (9.8)(32)
= -59.1 m
Vyf = Vyi – gt
= 5 – (9.8)(3)
= -24.4 m/s
SAMPLE PROBLEM B
You and a friend are going to toss a coin to determine who asks Mr. Talboo for help on
a kinematics problem. You toss the coin up with an initial velocity of 6.00 m/s. How
high does the coin go above the point of release? What is the total time the coin is in
the air before returning to its release point?
∆y
?
DATA TABLE
vyi
vyf
g
6
0
9.8
∆y = (Vf2 – Vi2) / -2g
= (02 – 62) / -2(9.8)
= 1.837 m
t = (Vf – Vi) / -g
t = (-6 – 6) / -9.8
t = 1.224 s
t
?
SAMPLE PROBLEM C
A stone thrown from the top of a building is given an initial velocity of 20.0 m/s
straight upward. The building is 50.0 m high, and the stone just misses the edge of the
roof on its way down. Determine (a) the time at which the stone reaches its maximum
height, (b) the maximum height, (c) the total time that it is in the air and (d) the velocity
of the stone when it hits the ground.
∆y
DATA TABLE
vyi
vyf
g
t
t = (Vf – Vi) / -g
t = (0 – 20) / -9.8
t = 2.041 s
∆y = (Vf2 – Vi2) / -2g
= (02 – 202) / -2(9.8)
= 20.408 m + 50 m = 70.408 m
∆y = Vit – ½ gt2
= (0) – ½ gt2
t = √((2∆y)/-g)
= √((2 * -70.408 )/-9.8)
= 3.791 s
+ 2.041 s
t = 5.832 s
Vyf = Vyi – gt
= 20 – (9.8)(5.832)
= -37.154 m/s