DISPLACEMENT
AND VELOCITY
Chapter 2-1
Objectives
Describe motion in terms of frame of
reference, displacement, time and velocity.
Calculate displacement, velocity and time
interval.
Interpret position vs. time graphs.
Kinematics deals with the concepts that are
needed to describe motion.
Dynamics deals with the effect that forces
have on motion.
Together, kinematics and dynamics form the
branch of physics known as Mechanics.
Describing Motion: this slide is not in your notes!
Are these two motions different?
2.1 Displacement
 As any object moves from one position to
another, the length of the straight line drawn
from its initial position to the object’s final
position is called displacement.
 Displacement is the change in position of an
object.
 Displacement doesn’t always tell you distance
an object moved.
Displacement = final position – initial position
∆x = xf-xo
Si units – meters (m)
2.1 Displacement
Displacement is not always equal to the
distance traveled.
Displacement can be positive or negative.
If displacement is positive, the object moves to the
right.
If the displacement is negative, the object moves to
the left.
2.1 Displacement


x  final position
xo  initial position
  
x  x  xo  displaceme nt
2.1 Displacement

x o  2.0 m

x  5.0 m

x  7.0 m
  
x  x  xo  7.0 m  2.0 m  5.0 m
2.1 Displacement

x  2.0 m

x  5.0 m

x o  7.0 m
  
x  x  xo  2.0 m  7.0 m  5.0 m
2.1 Displacement

xo  2.0 m

x  5.0 m

x  7.0 m
  
x  x  xo  5.0 m   2.0 m  7.0 m
2.2 Speed and Velocity
Average speed is the distance traveled divided by the time
required to cover the distance.
Distance
Average speed 
Elapsed time
SI units for speed: meters per second (m/s)
2.2 Speed and Velocity
Average velocity is the displacement divided by the elapsed
time.
Displaceme nt
Average velocity 
Elapsed time
 

 x  x o x
v

t  to
t
SI units for velocity: meters per second (m/s)
2.2 Speed and Velocity
 Velocity gives both direction and magnitude
(amount) while speed only gives the magnitude no direction
Average velocity does not tell you the speed or velocity
of the moving object at each moment or instant.
 Average velocity can be positive or negative
depending on direction moved.
 Time can never be negative!
Average Velocity:
Position vs. Time
 The slope of a
position vs. time
graph is called the
average velocity.
7
6
x (m)
5
4
3
v AV
2
1
0
0
5
10
t (s)
15
  
x x  xo


t
t  to
‘Should get something
similar to this from
your lab data’
Average Velocity
 The definition of
average velocity
holds even if the
slope changes.

x x f  xi
v AV 

t t f  ti
8
7
6
x (m)
5
4
3
2
1
0
0
2
4
t (s)
6
 This would be the
slope of the line
connected the final
and initial points.
Average Velocity:
Position vs. Time
 Steeper slope means
a faster moving
object.
14
12
x (m)
10
8
6
4
2
0
0
5
10
t (s)
15
Speed and velocity:
Velocity and speed can be used
interchangeably in everyday language
In Physics, however, there is a distinction:
Speed is a Scalar quantity (numerical value =
magnitude only)
Velocity is Vector quantity - Describes motion with
both direction and a numerical value (magnitude and
direction)
Constant Velocity Model:
Velocity vs. Time Graph
 With constant velocity,
6
5
v (m/s)
4
position
(m)
3
Linear
(position
(m))
2
1
x  v AV t
 On a graph of velocity vs.
time, displacement can be
calculated by finding the
area under the graph.
0
0
5
t (s)
x  v AV t  5  3  15m
 If the velocity is negative,
the displacement
calculate will be negative.
Constant Velocity Model:
Velocity vs. Time Example
5
 Use the graph to
calculate the
displacement of the
object:
4
3
v (m/s)
2
1
0
0
5
-1
-2
-3
t (s)
10
From 0 s to 3 s
From 3 s to 6 s
From 6 s to 9 s
From 0 s to 9 s
More practice:- ‘GUESS’ method
A vehicle travels 2345 m in 315 s toward
the evening sun. What is its velocity?
Given: Displacement = 2345 m , time = 315 s
Unknown: Velocity = ?
Equation and model: 𝑣 = 𝑥/𝑡 (kinematics)
Separate the unknown: ----Substitute and solve: V = 2345 / 315
 7.44 m/s
 does the answer make sense?
• Yes!!
I will not accept late work
CLASS ASSIGNMENT /
HOMEWORK
Acceleration
Section 3
(REVIEW)
Displacement for constant acceleration
 The displacement
from time 0 to time t is
the area under the
velocity graph from 0
to t.
 Area = ½ b h
v
(m/s)
x  12 (t )(v)
x  12 (5)(10)
t (s)
x  25m
Instantaneous Velocity – write this under
your instantaneous acceleration paragraph
The instantaneous velocity indicates how fast
the car moves and the direction of motion at each
instant of time.


x
v  lim
t  0 t
2.3 Acceleration
• The notion of acceleration emerges when a change in
velocity is combined with the time during which the
change occurs.
• Acceleration is the measure of how fast something
speeds up or slows down
2.3 Acceleration
DEFINITION OF AVERAGE ACCELERATION
 

 v  v o v  m 
a

 
t  to
t  s.s 
SI units for acceleration: (m/s2)
2.3 Acceleration
Example 3: Acceleration and Increasing Velocity
Determine the average acceleration of the
plane.


t  29 s
v

0
m
s
v  260 km h to  0 s
G: o

U: a
 

v
E: a   v (kinematics)
o
t  to
S:
S:
 
 v  v o 72.2 m s  0 m s
a

t  to
29 s  0 s

a  2.49 m s 2
2.3 Acceleration
2.3 Acceleration
Example 3: Acceleration and decreasing Velocity
Calculate the average acceleration


G: v o  28 m s v  13 m s to  9 s t  12 s

U: a
 

v
E: a   v o (kinematics)
t  to
S:
S:
 
 v  v o 13 m s  28 m s
a

t  to
12 s  9 s

a  5.0 m s 2
2.3 Acceleration
Graph for speeding up motion – Acceleration
describe the following motion
position (m)
Position vs. Time
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
time (s)
 Everyday examples of such motion include:
 Driving a car
 Throwing a ball
 Kids sliding at the playground
2.5
3
Changes in velocity:
Instantaneous acceleration:
Since in some situations, the acceleration of an
object can be constant , as a result, it will be the
same value at any instant of time.
The plus or minus sign before the number
indicates the two possible directions for the
acceleration vector when the motion is along a
straight line.
Interpret The Graph Below:
Distance vs. time
The graph shows
__________
Interpret The Graph Below:
Distance vs. Time (constant velocity):
The object is
__________
Interpret The Graph Below:
Distance vs. Time ( constant negative velocity)
This object is
________________
Interpret The Graph Below:
Distance vs. Time (acceleration)
The curve in the
graph shows that
_______________
Interpret The Graph Below:
Distance vs. Time (deceleration)
The curve in the
graph shows that
the objects
_______________
Interpret The Graph Below:
Distance vs. Time
In the first part of the
graph the object is
____________
In the second part of
the graph the object is
_______________
In the third part the
object is
___________
Interpret The Graph Below::
Velocity vs. Time (constant velocity)
The objects is _______
Interpret The Graph Below::
Velocity vs. Time
The objects is _______
Interpret The Graph Below:
Velocity vs. Time (constant acceleration)
The objects velocity
is ______________
this is known as
____________
The straight line
shows that the
object is moving at
a _____________
Interpret The Graph Below:
Velocity vs. Time
The objects velocity is
_________________
The straight line shows
that _______________
Interpret The Graph Below:
Velocity vs. Time
The objects velocity is
_________________
Interpret The Graph Below:
Velocity vs. Time
The objects velocity is
_________________
2.4 Equations of Kinematics for Constant Acceleration
Equations of Kinematics for Constant Acceleration
v  vo  at
x
1
2
vo  v  t
v  v  2ax
2
2
o
x  xo  vot  at
1
2
2
2.4 Equations of Kinematics for Constant Acceleration
G:
U: x  ?
E: x  vot  12 at 2
(kinematics)
S:  6.0 m s 8.0 s   1 2.0 m s 2 8.0 s 2
2
S:
 110 m
2.4 Equations of Kinematics for Constant Acceleration
Example 6 Catapulting a Jet
Find its displacement.
2.4 Equations of Kinematics for Constant Acceleration
G:
vo  0 m s
a  31m s 2
v  62 m s
U:
E:
S:
x  ??
v 2  vo2
x
2a
(kinematics)
2
2

62 m s   0 m s 
x

2 31 m s 2
S: x  62 m

I will not accept late work
CLASS ASSIGNMENT /
HOMEWORK
Chapter 2
Section 6:
Falling Objects
Free Fall acceleration:
An object thrown or dropped in the
presence of Earth’s gravity experiences a
constant acceleration directed toward the
center of the earth.
This acceleration is called the free- fall
acceleration, or the acceleration due to
gravity.
Free fall acceleration is the same for all objects,
regardless of their mass.
Free fall acceleration:
The value for free fall acceleration used in
this book is:
 g = 9.81 m/s2.
In this book, the direction of the free fall
acceleration is considered to be negative
because the object accelerates toward
Positive
earth.
Negative
2.6 Freely Falling Bodies
g  9.80 m s
2
Free fall acceleration:
Since gravity on earth is constant at
9.81m/s2,
If 2 objects of different masses are dropped
from the same height, they will both hit the
ground at the same time considering no air
resistance. No matter what their masses are.
If there is air resistance, some objects may
be slowed down
The more surface area an object has, the
more air resistance it will experience.
ex. A wadded piece of paper will fall faster than a flat
piece of paper which will experience more air
resistance due to its larger surface area
An apple and a feather falling at the
same rate absent air resistance
(vacuum)
Free fall acceleration:
Describe the
acceleration of the
object as it falls off
the cliff – notice
the change in
velocity after each
second of fall
Free fall acceleration:
All objects, when thrown up will continue
to move upward for some time, stop
momentarily at the peak, and then change
direction and begin to fall.
Velocity at the
top of the arc is 0 m/s
What goes up must come down!!
In an upward displacement of an
object, as soon as the ball is
released with an initial positive
velocity, it has an acceleration of
-9.81m/s2, causing the object to
slow down until it stops
momentarily. It then changes
direction moving downward with
an increasing velocity.
Distance vs. Time and Velocity vs. time
graph of an object thrown up in the
presence of gravity
2.6 Freely Falling Bodies
Example 10 A Falling Stone
A stone is dropped from the top of a tall building. After 3.00s
of free fall, what is the displacement, y, of the stone?
Example 10: A falling stone
2.6 Freely Falling Bodies
G: 𝑎 =
𝑚
−9.81 2 ,
𝑠
U: 𝒚 = ?
2
1
E: y  vot  2 at
S:
𝑣𝑜 = 0 𝑚/𝑠 , 𝑡 = 3.00 𝑠
(kinematics)


y  0 m s 3.00 s   12  9.80 m s 2 3.00 s 
S: y  44.1 m
2
2.6 Freely Falling Bodies
Example 12 How High Does it Go?
The referee tosses the coin up
with an initial speed of 5.00m/s.
In the absence of air resistance,
how high does the coin go above
its point of release?
Example 12: How high does it go?
2.6 Freely Falling Bodies
G: 𝑎 =
𝑚
−9.81 2 ,
𝑠
𝑣𝑜 = 5.00 𝑚/𝑠 , v= 0 𝑚/𝑠
U: 𝒚 = ?
v 2  vo2
y
2a
E: v  v  2ay
2
S:
S:
2
o
2
2

0 m s   5.00 m s 
y

2  9.80 m s 2
y  1.28 m

(kinematics)
2.6 Freely Falling Bodies
Conceptual Example 15 Taking Advantage of Symmetry
Does the pellet in part b strike the ground beneath the cliff
with a smaller, greater, or the same speed as the pellet
in part a?
2.7 Graphical Analysis of Velocity and Acceleration
x  8 m
Slope 

 4 m s
t
2s
2.7 Graphical Analysis of Velocity and Acceleration
2.7 Graphical Analysis of Velocity and Acceleration
2.7 Graphical Analysis of Velocity and Acceleration
v  12 m s
Slope 

 6 m s 2
t
2s
I will not accept late work
CLASS ASSIGNMENT /
HOMEWORK