Chapter 2. Concepts of Motion
Kinematics – the
mathematical
description and
analysis of motion.
Chapter 2
Student Learning Objectives
•
•
•
•
•
To differentiate between concepts of position, velocity, and
acceleration.
To gain experience with graphical addition and
subtraction of vectors.
To understand why the sign of the acceleration vector does
not indicate speeding up or slowing down.
To begin the process of learning to analyze problem
statements and to translate the information into other
representations.
To make and interpret graphs of motion.
The position vector
• The position vector shows the distance (magnitude) and
direction of an object from an arbitrary origin. The SI
unit is the meter (m).
Examples of one-dimensional position vectors

x  5 .0 m

x   2 .0 m
0
For 1-D vectors parallel to a coordinate axis, the vector is
equivalent to its scalar component; therefore direction can
be specified by positive and negative signs.
In this class, negative signs attached to vector
(component) quantities mean down, left, west or south.
Negative signs do not mean going backward, slowing
down, or less than positive quantities of the same
magnitude when used with a vector quantity.
One-dimensional position vectors

x   2 .0 m
x origin
•
•
•
•
t=0s
t=1s
t=2s
t=3s
• The position vector shows where an object is, relative to the
origin, but not in which direction the object is going.
• The numbers show the object is moving to the right, but
the first two position vectors point to the left.
•The object started at the position x = -2m and moved right
The displacement vector
The displacement
vector represents
the change in
position of an
object relative to an
origin. It has a
magnitude equal to
the shortest distance
between the two
positions.
The symbol Δ
(delta) always
means final value
minus initial value.

x o  initial
position

x  final position
  
 x  x  x o  displaceme
nt
The displacement vector

 xo

x

x

x o  initial
position

x  final position
  
 x  x  x o  displaceme
nt
The displacement vector
Unlike the
position vector,
the displacement
vector always
points in the
direction of
motion.
Obtaining the
displacement
vector is an
example of
graphical vector
subtraction.

x o  initial
position

x  final position
  
 x  x  x o  displaceme
nt
Displacement Vector 1

x o  2 .0 m

x  7 .0 m

x  ?
Which way does the
displacement vector point?
Displacement Vector 2

x o  7 .0 m

x  2 .0 m

x  ?
Which way does the
displacement vector
point?
Displacement Vector 3

x o   2 .0 m

x  5 .0 m

x  ?
Which way does the
displacement vector
point?
Displacement
Three runners start at the same place. Shaun runs 4.0 km due
east and then runs 1.0 km due west. Mark runs 3.0 km due
east. Jeff runs 2.0 km due west and then runs 5.0 km due
east. Which of the following is true concerning the
displacement of each runner?
A. Mark's displacement equals Jeff's displacement, but Shaun's
displacement is different.
B. Shaun's displacement equals Mark's displacement, but Jeff's
displacement is different.
C. Shaun, Mark, and Jeff have the same displacements.
D. Shaun's displacement equals Jeff's displacement, but Mark's
displacement is different.
E. Shaun, Mark, and Jeff have different displacements.
Average velocity is the rate of change of position
during a time interval.
Since change of position (Δx) is displacement, the
average velocity can be determined by dividing
dispacement by the time interval.
Average
velocity

Displaceme
Elapsed

v
 
x  xo
t  to

nt
time

x
t
SI unit for velocity: meters per second (m/s)
2.2 Speed and Velocity

v
 
x  xo
t  to


x
t
The direction of the velocity vector is:
•the same as that of the displacement vector
(note displacement vector divided by a
scalar)
•always in the direction of motion
•for one-dimensional motion, we use positive
and negative to determine direction, since
we are really using components.
Graph of a Straight Line
• The vertical axis value comes first
when naming graphs
• The standard format is:
vertical axis = m(horizontal axis) + b,
where m is the slope and b is the yintercept.
• The normal variable for horizontal
position is x and the quantity on the
horizontal axis is time (t) so the
equation for this line is:
x = mt + b
Position as a function of time
Graph of a Straight Line
m = “rise over run”, in this
case Δx/Δt, where:
Δx = x2-x1 and Δt = t2-t1 for
any two points on the line.
Note that the slope has
units, in this case:
Δx/Δt = 8m/2s = +4 m/s.
In this case the slope has
units of velocity and
represents the velocity of
the object being graphed.
Position as a function of time
Graphical Analysis of Contstant Velocity
If an object moved 4 m
every second, a graph of
its motion a graph of its
motion would be a
straight line. What is the
equation of this line,
using the appropriate
variables?
A.
B.
C.
D.
y = 4t
y = 4x
x = 4t
t = 4x
Position as a function of time
Graphical Analysis of Constant Velocity
The equation of this line is:
x = 4t + 0
The general format for
uniform motion is:
x = vΔt + x0 where
v= constant velocity, m/s
x0= initial position (position at
the start of the time interval,
usually t=0)
One could rewrite the
equation:

v
 
x  xo
t  to


x
t
The slope of a linear position
graph is the velocity.
Position as a function of time
The slope of the position
graph tells us the rate of
change of position during the
time interval
Equation of a line
Find the equation of a line
if you know the x,t
coordinates of 2 points:
• use Δx/Δt to find m
• substitute the values of
one of the points into the
equation to find b:
x1 = mt1 + b
b = x1 – mt1
or
b = x2 – mt2
Position as a function of time
At the end of 10 seconds, the object is 5 meters away
from the origin. What is the equation of the line?
A.
B.
C.
D.
x = 0.5t
x = 2t +2
x = 0.5 t + 2
x = 0.3 t + 2
Instantaneous velocity
When the object is speeding
up or slowing down, the
position graph will be
curved (blue). By drawing a
straight line from beginning
to some known end (green),
we can determine the
average velocity:

  x 50 m
v

t
20 s
But that doesn’t tell us how
fast the object was going at
any particular time.
Instantaneous velocity
The slope of the tangent
line drawn to the curve at
a given time (t = 20 s) tells
us the instantaneous
velocity at t = 20 s.
Pick any 2nd point on the
tangent line to find the
slope (46,20), (72,25).

v inst 

x
t

26 m
5s
 5 .2 m / s
2.3 Acceleration
If an object speeds up OR slows down during some time interval,
it has an acceleration.
Acceleration is a vector that represents the rate of change of
velocity for a given time interval.
Often the term “deceleration” is used for objects slowing down.
However, it is absolutely correct to say that an object that is
slowing down is accelerating.
2.3 Acceleration

a 
 
v  vo
t  to


v
t
• SI unit for acceleration: meters per second per second
(m/s/s or m/s2).
• The acceleration vector points in the same direction as
the Δv vector.
•The a vector does NOT ALWAYS point in the same
direction as either v0 or v
Direction of Acceleration- it’s probably not
what you think
• Acceleration is a vector quantity.
Recall our previous assertion
regarding the direction of
vectors:
• “Negative signs attached to
vector quantities mean down,
left, west, or south. They do not
mean going backward, slowing
down, or less than positive
quantities….
• So what is “negative
acceleration” if not slowing
down?
Direction of Acceleration – Here’s the rule
• If the object is speeding up, the
velocity vectors and the
acceleration vectors point in
the same direction, whether
that be positive or negative.
• If the object is slowing down,
the velocity vectors and
acceleration vectors point in
opposite directions.
• The direction of the velocity
vector is always the direction
of motion.
• The direction of a is that of ∆v.
In both cases shown
above, acceleration is
positive (i.e. right). In
case b, velocity is
negative.
A skydiver is falling down (y
component negative) at a speed
of 48 m/s. She opens her chute.
11 seconds later, she has slowed
down to 26 m/s. Her acceleration
is:
A. 2.0 m/s/s
B. -2.0 m/s/s
C. 22 m/s/s
D. -22 m/s/s
Equation of a line
What is the equation of
this velocity vs time
graph? What quantity
is represented by the
slope?
Velocity as a function of time
The slope of a linear velocity vs time graph is the acceleration
Slope 
v
t

 12 m s
2s
 6 m s
2
The slope of a linear velocity vs time graph is the acceleration
v = 6m/s2 (Δt) + 5m/s