CHAPTER 2 Motion In One Dimension

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CHAPTER 2
Motion In One Dimension
Mechanics     Motion in General

Kinematics                 Dynamics
How Things Move                                 Why Things Move

Translational Motion       Rotational Motion
Change of Position                         Motion Around an Axis

1 Dimensional Motion 2 Dimensional Motion       3 Dimensional Motion
Projectiles (x & y)          Not Studied
Horizontal (x)

Vertical (y)
Kinematics
“Frame of Reference”
•   What you compare your movement against
•   Examples:
1. Earth
2. Sun
3. Vehicle you are in
4. Second vehicle
5. Coordinate System (of your choice)
• You usually get to choose your Frame of
Reference. Make it a convenient one.
Vectors and Scalars

Vector:   • A physical quantity that requires the
specification of both a magnitude and a
direction
• A starting point need not be specified.

Examples:
Wind “From the NE at 20 miles per hour”
River “Flows South at 2 meters per second”
Scalars:    A physical quantity that has a
magnitude but no direction

Examples:
Distance     “I drove my car 25 miles

Energy       “2500 Joules”
Horizontal Motion
Displacement:        A change in position in a given
direction
Vector          x = x - x0

x = change in position
x0 = Initial position
x = Final position (“Position of Interest”)

NOTE: The text uses a slightly different notation.
xf = Final Position
Distance:   An amount of movement without any
consideration of direction
Scalar
Average Velocity:       A change in displacement
divided by a change in time

Vector       v = x = x - xo
t    t - to
v = Average velocity
to = Initial Time Corresponding to xo, the Initial
Position
t = Final Time Corresponding to x, the Final Position
Speed:      A distance traveled divided by the time involved
Scalar
NOTE: Time is a scalar quantity

Vector Quantity = Vector Quantity
Scalar Quantity
Example Problem Solving Strategy
Problem:
A car initially at rest at the starting line traveled the 250.
meter drag strip in 4.1 seconds. What was his average
velocity?
Extract the Data
xo = 0. meters         to = 0.0 seconds
x = 250. meters t = 4.1 seconds
Choose the Appropriate Formula
v = x = x - xo
t      t - to
Plug in the Data (with Units)
v = 250.m – 0.0m
4.1s – 0.0s
Calculate the Answer (with Units & Proper S/F)
m
v = 61 m/s or 61 s             “meters per second”
Acceleration

Average Acceleration = a = v
t
“Average rate of change of velocity with
respect to time”

NOTE: In this class, acceleration = average acceleration

Acceleration can be Positive.
Acceleration can be Negative.
Acceleration can be Zero.
Acceleration
a = v = m/s
t    s
=m
s “meters per second per second”
s
= m/s2 “meters per second squared”
Example:
A car initially traveling at 35 m/s began to slow
down and came to a stop 10.3 seconds later. Find
the acceleration of the car.
vo = 35 m/s                            a = v = -35m/s
v = 0.0 m/s                                t     10.3s
t = 10.3 s
v = 0.0m/s – 35m/s = -35m/s             a = -3.4 m/s2
Acceleration & Velocity

Case 1
Velocity and Acceleration are in the same
direction  object’s speed increases
Case 2
Velocity and Acceleration are in opposite
directions  object’s speed decreases

Case 1 Examples           Case 2 Examples
Ball Dropped           Ball Thrown Upward
Rocket Launching       Person Diving into Water
(while in the water)
Graphical Interpretations
x
B
Position (m)

graph of time &
position data
A

t
to    Time (s)    t
The slope of the segment connecting points A and B is the
Average Velocity for the period of time from to to t.
The object’s velocity may not have been constant (as in
this example)
The object’s “Instantaneous Velocity” can be found at any
time by calculating the slope of the tangent to the position
time curve.
v = lim x
 t  0 t
Interpreting Position Time Graphs
P (x)
10.0 m

t
3.0 sec
Moving Forward
Stopped
Moving Backward
Total Displacement?   Zero meters forward
Total Distance?       20. Meters
Average Velocity?     0.0 m/s forward
Average Speed?        6.7 m/s
Interpreting Position Time Graphs

P (y)
5.0 m

t
6.0s

-5.0 m

Moving up slowly
Stopped                    Total Displacement?   -2.0 m
Moving up quickly          Total Distance?       18.0 m
Moving down more quickly   Average Velocity?     -.33 m/s
Moving up quickly          Average Speed?        3.0 m/s
Interpreting Velocity Time Graphs
v

t
• Starting from rest and velocity increasing at a constant rate
• Constant velocity
• Slowing down at a constant rate while still moving forward
p

t
One Dimensional Motion & Constant Acceleration
Velocity
v

vo

to=0             t        time
a = v = v - vo
t     t          NOTE: t = t (usually)
at = v - vo
v = vo + at
v = vo + v
for constant acceleration
2
v = x
t
for constant velocity
x = vct            zero acceleration
Velocity
v
Area = v – vo
( 2    )
t
vo
Area = vot
to=0                 t      time
x = vot + ½ (v – vo) t
x = vot + ½ (at) t
x = vot + ½ at2
v = v + vo
x = vo + v
[2   t]                 2
x = v t
t = v - vo
v2 = vo2 + 2ax                  a
x = vot + ½ at2
Kinematics Equations
Horizontal              Vertical
x = vct           y = vct (rare)
 x = vot + ½at2       y = vot + ½gt2
x = vo + v
[    2
] t
y = vo + v
[   2
] t

v = vo + at            v = vo + gt
v2 = vo2 + 2ax        v2 = vo2 + 2gy
“g” = acceleration due to gravity = -9.8 m/s2
~ -10 m/s2
=
Last year’s class suggest you memorize
equations and begin by memorizing these!!
Vertical Motion

Free Fall:   • Air resistance is negligible
• Gravity is the only force acting on the
object
• “Free falling” objects may be going up or
down
• “Free falling” objects always accelerate
downward with an acceleration of -9.8 m/s2
• g = -9.8 m/s2
• The same kinematics equations that hold
vertical motion
Substitute g for a
Substitute y for x
Vertical Motion Examples
1. An object is thrown up in the air and returns to
the ground 4.6 sec later. What was its velocity
just prior to hitting the ground?

Extract Data   Notes: • Motion in x or in y direction
vo = ?                always involves essentially the
same 5 variables.
v =?
g = -9.80              • Every word problem gives you
3 of the 5 variables. Look for key
t =4.60 sec            words and phrases.
y = 0.00m                “at rest”
“comes to stop”
“free falling object”
Etc.
• Choose appropriate equations
to solve for unknowns.
Appropriate Equations
y = vot + ½gt2             Solve for vo
vo = 22.5 m/s
v = vo + gt              v = -22.5 m/s
Hints:     • Notice that one variable had to be solved
for before the other (frequently)
• Lots of choice on equations to solve for
second variable
• Avoid solving for time first if it involves
Notice: The object hit the ground with the
same “speed” that it left the ground
Magnitude of velocity the same
Sign is different
Question: How does the time traveling up compare
to time traveling down?
2. How high did the ball go in the previous example?

vo = ?
v = 0.00 m/s
g = -9.80 m/s2
t =2.30 sec
y = ?
Appropriate Equations
Solve for vo
v = vo + gt
vo = 22.5 m/s
y =    vo + v       t   y = 25.9 m
[    2       ]
3. A ball is thrown upward with an initial velocity
of 15.0 m/s. How long will it stay in the air and
what will be its velocity at impact?
vo = 15.0 m/s
v= ?
g = -9.80 m/s2
t= ?
y = 0.00m
Appropriate Equations
y = vot + ½gt2      Poor choice. Why?
v2 = vo2 + 2gy      Solve for v
v = [vo2 + 2gy]1/2
Don’t forget the
v = ± 15 m/s
negative sign!!
v = -15 m/s
v = vo + gt          Solve for t
t = 3.06 seconds
Question: What value do you calculate for “t”
if you use v = +15 m/s?

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