PHY300

1. Giambattista Physics
Chapter 2
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Chapter 2: Motion Along a Line
• 2.1 Position and Displacement
• 2.2 Velocity: Rate of Change of Position
• 2.3 Acceleration: Rate of Change of Velocity
• 2.4 Visualizing Motion along a Line with Constant Acceleration
• 2.5 Kinematic Equations for Motion Along a Line with
Constant Acceleration
• 2.6 Free Fall

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2.1 Position and Displacement
The position (x) of an object describes its location relative to
some origin or other reference point.
0 x2
0 x1
The position of the ball differs in the two shown coordinate
systems.

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Position Example 1
0 x (cm)
2
1
2 1
The position of the ball is 2cm.
x  
The positive sign indicates the direction is to the right of the
origin.

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Position Example 2
0 x (cm)
2
1
2 1
The position of the ball is 2cm.
x  
The negative sign indicates the direction is to the left of the
origin.

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Displacement
The displacement is the change in an object’s position. It
depends only on the beginning and ending positions.
i
f
x x x
  
Example: A ball is initially at x = +2 cm and is moved to x = -2 cm.
What is the displacement of the ball?
0 x (cm)
2
1
2 1
2cm 2cm 4cm.
x
     

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Example: Chapter 2 Problem 4
At 3 PM a car is located 20 km south of its starting point. One
hour later it 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.
20km and 12km
i f
x
x    
12km ( 20km) 8km
i
f
x x x
 



   

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Example: Chapter 2 Problem 4 continued
(b) What is the displacement of the car from the starting point
to the location at 4 PM?
0km and 96km
i f
x x
  
96km (0km) 96km
i
f
x x x
 
 


 
(c) What is the displacement of the car from 4 PM to 6 PM?
96km and 12km
i f
x
x    
12km ( 96km) 84km
i
f
x x x

   

 


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2.2 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.

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Average vs Instantaneous Velocity and Speed
In straight-line motion (one dimension), the average velocity is
the constant speed and direction that results in the same
displacement in the given time interval:
av,x
x
v
t



The instantaneous velocity is the average velocity during a very
short time interval:
0
lim
x t
v
x
t
 



The average speed depends on the path traveled:
distance traveled
average speed
time of trip


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Motion Diagrams
A motion diagram is an image showing the location of an object
at a series of equally spaced times. It is often simplified by
depicting the object at each time as a dot.
This motion diagram shows that the speed is increasing because
the displacements are getting larger.

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Graphical Relationships: Average Velocity
On a graph of position versus time, the average velocity is
represented by the slope of a chord.
3 1
av,
3 1
Average velocity
x
x x
v
t t




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Graphical Relationships: Instantaneous Velocity
The instantaneous velocity is represented by the slope of a line
tangent to the curve on the graph of an object’s position versus
time.

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Graphical Relationships: Displacement
The (signed) area under a velocity versus time graph (between
the curve and the horizontal time axis) gives the displacement in
a given interval of time.
If the velocity is positive
(above the horizontal axis), the
area is positive.
If the velocity is negative
(below the horizontal axis), the
area is negative.

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Example: Chapter 2 Problem 10
Speedometer readings are obtained and graphed as a car skids
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 (positive) area
between the curve and the time axis will represent the distance
traveled.

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Example: Chapter 2 Problem 10 continued
The rectangular portion has an area
length width (20 m/s)(4s) 80 m.
  
The triangular portion has an area
1 1
base height (8s)(20 m/s) 80 m.
2 2
  
The total area is 160 m, the distance traveled by the car.

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2.3 Acceleration: Rate of Change of Velocity
Average acceleration:
av,
x
x
t
a
v



Instantaneous acceleration:
0
lim x
x t
v
t
a  



These have
interpretations
similar to vav
and v.

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The Direction of Acceleration
The acceleration has the same direction as the change in
velocity.
When the acceleration and velocity are in the same direction,
the object is speeding up.
When the acceleration and velocity are in opposite directions,
the object is slowing down.
Example: A car is driving east and slowing down. The
acceleration is west.

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Graphical Relationships: Acceleration
The acceleration ax is the slope on a graph of vx(t) and the
change in velocity Δvx is the area under the graph of ax(t).

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Example: Chapter 2 Problem 36
The graph shows
speedometer readings as
a car skids to a stop on a
straight roadway. What is
the magnitude of the
acceleration at t = 7.0 s?
The slope of the graph at t = 7.0 sec is
2
2 1
av
2 1
(0 20) m/s
2.5m/s
(12 4)s
x
v v v
a
t t t
  
 
 
 


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2.4 Visualizing Motion along a Line with
Constant Acceleration
Motion diagrams for three carts. Each cart is shown at 1.0 s time
intervals, and each has a (different) constant acceleration.

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Graphs of x, vx, ax for the Three Carts
The slope of the
x(t) graph at any
time is the
velocity vx at that
time. The slope of
the vx(t) graph at
any time is the
acceleration ax at
that time.

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2.5 Kinematic Equations for Motion Along a Line
with Constant Acceleration
A few essential relationships (if ax is constant during the entire
time interval):
f i
x x
x x
v v v a t
     av, f i
1
( )
2
x x x
v v v
 
av,
f i f i
1
( )
2
x x x
x x x v t v v t

      

24. ©McGraw-Hill Education
Kinematic Equations continued
Other useful relationships for constant acceleration:
2
i
1
( )
2 x
x
x v t a t
    
2 2
f i 2 x
x x
v v a x
  

25. ©McGraw-Hill Education
Example: Chapter 2 Problem 44
The St. Charles streetcar in New Orleans starts from rest and has
a constant acceleration of 1.20 m/s2 for 12.0 s.
(a) Draw a graph of vx versus t.
0
2
4
6
8
10
12
14
16
0 2 4 6 8 10 12 14
v
(m/sec)
t (sec)

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Example: Chapter 2 Problem 44 continued
(b) How far has the train traveled at the end of the 12.0
seconds?
2
i
2 2
1
( )
2
0 0.5(1.2 m/s )(12s) 86.4 m
x x
x v t a t
    
  
(c) What is the speed of the train at the end of the 12.0 s?
This can be read directly from the graph, vx = 14.4 m/s.

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Example: Chapter 2 Problem 51
A train is traveling along a straight, level track at 26.8 m/s.
Suddenly the engineer sees a truck stalled on the tracks 184 m
ahead. If the maximum possible braking acceleration has
magnitude 1.52 m/s2, can the train be stopped in time?
2
i f
1.52 m/s , 26.8m/s, 0
x x x
a v v
   
Strategy: using the given acceleration, find the distance traveled
by the train before it comes to rest.
2 2
f i
2 2
2
0 (26.8m/s) 2( 1.52 m/s )
236 m.
x x x
v v a x
x
x
  
    
 
The train cannot be stopped in time.

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2.6 Free Fall
If no forces act on an object other than the gravitational force,
we say the object is in free fall.
For example, a stone dropped from the edge of a cliff—if air
resistance can be ignored, the stone is in free fall.
Or a ball thrown upward—if air resistance is ignored, the ball is
in free fall.
An object in free fall has constant downward acceleration,
denoted by the symbol g. Unless otherwise specified, assume it
has the magnitude
2
free fall 9.80 m/s
a g
 

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Free fall example
You throw a ball into the air with speed 15.0 m/s; how high does
the ball rise?
Given:
2
i 15m/s, 9.8m/s
y y
v a
   
x
y
viy
ay
One method: first find the time of flight.
The ball rises until vfy = 0:
f i
i
2
0,
15.0 m/s
1.53s.
9.8m/s
y y y
y
y
v v a t
v
t
a
   
     


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Free fall example continued
The height: 2
i
2 2
1
2
1
(15.0 m/s)(1.53s) ( 9.8m/s )(1.53s)
2
11.5m.
y y
y v t a t
   




Alternate method (doesn’t need time):
2 2
f i
2 2
2
0 (15.0 m/s) 2( 9.8m/s )
11.5m.
y y y
v v a y
y
y
  
    
 

31. ©McGraw-Hill Education
Example: Chapter 2 Problem 57
A penny is dropped from the observation deck of the Empire
State Building (369 m above the ground). With what velocity
would it strike the ground if air resistance were negligible?
369 m
x
y
ay
Given:
2
i 0 m/s, 9.8m/s 369 m
,
y y
v a y
     
Find vfy:
2 2
f i
2
f
2
2( 9.8m/s )(-369 m)
85.0 m/s (downward)
y y y
y
v v a y
v
  
  

32. ©McGraw-Hill Education
Example: Chapter 2 Problem 57 continued
How long does it take for the penny to strike the ground?
2 2
i
1 1
2 2
2
8.7s.
y y y
y
y v t a t a t
y
t
a
      

  

33. ©McGraw-Hill Education
Graphs of x, vx, ax for the Three Carts Appendix
For each cart, the graph of ax(t) is a constant (zero, positive, or
negative). The graphs of vx(t) are straight lines with zero,
positive, or negative slope. The graph of position x(t) is a straight
line with positive slope for the cart with zero acceleration. For
the other two carts, the graphs of x(t) are parabolas with either
increasing slope (concave up) for positive acceleration or
decreasing slope (concave down) for negative acceleration.