Chapter 2

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Chapter 2
Motion Along a Line
Motion Along a Line
• Position & Displacement
• Speed & Velocity
• Acceleration
• Describing motion in 1D
• Free Fall
Position & Displacement
The position (x) of an object describes its
location relative to some origin or other reference
point.
0
x1
0
The position of the red ball differs in
the two shown coordinate systems.
x2
2
1
0
The position of the ball is
1
2
x  2 cm
The + indicates the direction
to the right of the origin.
x
(cm)
2
1
0
1
2
The position of the ball is
rx  x  2 cm
The  indicates the direction
to the left of the origin.
x
(cm)
The displacement is the change in an object’s position.
It depends only on the beginning and ending positions.
x  x f  xi
All Δ quantities will have the final value 1st and the
inital value last.
Example: A ball is initially at x = +2 cm and is moved to
x = -2 cm. What is the displacement of the ball?
2
1
0
1
x  x f  xi
 2 cm  2 cm
 4 cm
2
x (cm)
Example:
At 3 PM a car is located 20 km south of its starting point.
One hour later its is 96 km farther south. After two more
hours it is 12 km south of the original starting point.
(a) What is the displacement of the car between 3 PM
and 6 PM?
Use a
coordinate
system where
north is
positive.
xi = –20 km and xf = –12 km
x  x f  xi
 12 km   20 km  8 km
Example continued
(b) What is the displacement of the car from the
starting point to the location at 4 pm?
xi = 0 km and xf = –96 km
x  x f  xi
 96 km  0 km  96 km
(c) What is the displacement of the car from 4 PM to 6
PM?
xi = –96 km and xf = –12 km
x  x f  xi
 12 km   96 km  84 km
Velocity: Rate of Change of Position
Velocity is a vector that measures how fast
and in what direction something moves.
Speed is the magnitude of the velocity. It is
a scalar.
In 1-dimension the average velocity is
vav, x
x

t
vav is the constant speed and direction
that results in the same displacement
in a given time interval.
Averagespeed 
distance traveled
timeof trip
On a graph of position versus time, the average velocity
is represented by the slope of a chord.
x (m)
x2
x1
t1
t2
Averagevelocity vav, x
t
(sec)
x2  x1

t2  t1
Instantane
ous velocity vx  lim
t 0
x
t
This is represented by the slope of a line tangent to the
curve on the graph of an object’s position versus time.
x
(m)
t (sec)
The area under a velocity versus time graph (between
the curve and the time axis) gives the displacement in a
given interval of time.
vx
(m/s)
t (sec)
Example (text problem 2.11): Speedometer readings
are obtained and graphed as a car comes to a stop along
a straight-line path. How far does the car move
between t = 0 and t = 16 seconds?
Since there is not a reversal of direction, the area
between the curve and the time axis will represent
the distance traveled.
Example continued:
The rectangular portion has an area of
Lw = (20 m/s)(4 s) = 80 m.
The triangular portion has an area of
½bh = ½(8 s) (20 m/s) = 80 m.
Thus, the total area is 160 m. This is the distance
traveled by the car.
Acceleration: Rate of Change of
Velocity
Average accelerati on  aav,x
vx

t
Instantane
ous acceleration  a x  lim
t 0
vx
t
These have
interpretation
s similar to
vav and v.
Example (text problem 2.29): The graph shows
speedometer readings as a car comes to a stop. What is
the magnitude of the acceleration at t = 7.0 s?
The slope of the graph at t = 7.0 sec is
aav 
0  20 m/s  2.5 m/s2
vx
v v
 2 1 
12  4 s
t
t2  t1
Motion Along a Line With Constant
Acceleration
For constant acceleration the kinematic equations are:
1
x  x f  xi  vix t  a x t 2
2
vx  v fx  vix  a x t
v 2fx  vix2  2a x x
x  vav, x t
Also:
vav, x 
vix  v fx
2
Visualizing Motion Along a Line with
Constant Acceleration
Motion diagrams for three carts:
Graphs of x, vx, ax
for each of the
three carts
Free Fall
A stone is dropped from the edge of a cliff; if air
resistance can be ignored, we say the stone is in free
fall. The magnitude of the acceleration of the stone is
afree fall = g = 9.80 m/s2, this acceleration is always
directed toward the Earth.
Example: You throw a ball into the air with speed 15.0
m/s; how high does the ball rise?
y
viy
Given: viy = +15.0 m/s; ay = 9.8 m/s2
x
ay
To calculate the final height, we need
to know the time of flight.
Time of flight from:
1
y  viy t  a y t 2
2
v fy  viy  ay t
Example continued:
The ball rises
until vfy = 0.
The height:
v fy  viy  a y t  0
viy
15.0 m/s
t  

 1.53 sec
2
ay
 9.8 m/s
1
y  viy t  a y t 2
2


1
2
 15.0 m/s1.53 s    9.8 m/s2 1.53 s 
2
 11.5 m
Example (text problem 2.45): A penny is dropped from
the observation deck of the Empire State Building 369 m
above the ground. With what velocity does it strike the
ground? Ignore air resistance.
Given: viy = 0 m/s, ay = 9.8
m/s2, y = 369 m
y
Unknown: vyf
x
ay
369
m
Use:
v 2fy  viy2  2a y y
 2a y y
v yf  2a y y
Example continued:


v yf  2a y y  2  9.8 m/s2  369 m  85.0 m/s
(downward)
How long does it take for the penny to strike the
ground?
Given: viy = 0 m/s, ay = 9.8 m/s2, y = 369 m
Unknown: t
1
1
2
y  viy t  a y t  a y t 2
2
2
2y
t 
 8.7 sec
ay
Summary
• Position
• Displacement Versus Distance
• Velocity Versus Speed
• Acceleration
• Instantaneous Values Versus Average Values
• The Kinematic Equations
• Graphical Representations of Motion
• Free fall

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