Motion
In-Class Assignment
1) Break into groups of 2 to 3 people.
2) Using your domino, execute a simple motion.
3) On a sheet of paper, write as accurately as
possible a description of the motion. Write your
group members’ names on the paper and turn it in
when you are complete.
This should take no longer than 10 minutes.
Would the Motions We Observe Everyday
Appear Differently if Observed from a
Different Vantage Point?
Motion is Relative
…To the position of the observer.
I collided with a stationary truck coming the
other way.
A pedestrian hit me and went under my car.
The guy was all over the road. I had to
swerve a number of times before I hit him.
The telephone pole was approaching fast. I
was attempting to swerve out of its path
when it struck my front end.
Just For Fun…
"In an attempt to kill a fly, I drove into a telephone pole."
"Coming home I drove into the wrong house and collided
with a tree I don't have."
"I thought my window was down, but I found it was up
when I put my head through it."
"To avoid hitting the bumper of the car in front I struck a
pedestrian."
"I was sure the old fellow would never make it to the other
side of the road when I struck him."
"The pedestrian had no idea which way to run as I ran over
him."
"The pedestrian ran for the pavement, but I got him."
What terms are necessary to
accurately describe the
motion of ANY object?
Position
Position (x): location, can be specified in one, two or
three dimensions, described in terms of a distance from
an origin (reference point), given in units of length.
Displacement
displacement: a change in position. Often
denoted by s or d or x, displacement has
units of length, and is a vector quantity.
Δx  x final  x initial
xf
xi
Δx  5 cm  2 cm  3 cm
Distance vs. Displacement
A
B
Displacement is a Vector Quantity
Vector quantities require both a magnitude AND
a direction to fully describe them.
25 meters North
Magnitude
Direction
Distance is a
scalar
quantity.
Displacement Can Be Negative
0
5
10
15
Which just indicates displacement in the negative
direction.
Position vs. Time Graph
What’s the total displacement?
Position (m)
10
(a) 10 meters
(b) 0 meters
(c) 5 meters
(d) 15 meters
5
0
10
5
Time (s)
15
Velocity
Average velocity (vavg) – the total displacement
divided by the time interval during which the
displacement occurred.
displaceme nt
Δx
v avg 

Δt
time
Instantaneous velocity – the velocity at a
particular instant.
Speed is a scalar quantity. Velocity is a vector quantity.
Speed is the magnitude of velocity…
Average Velocity
Vavg = 5 miles/0.2 hours = 25 miles/hour
Position vs. Time Graph
Position (m)
10
m 
rise
run
5

Δx
Δt
x
0
t
10
5
Time (s)
15
The Slope of the Position vs.
Time Graph is the Velocity
Positive slopes indicate positive velocities,
negative slopes, negative velocities.
x (m)
t (s)
The Slope of the Position vs.
Time Graph is the Velocity
Constant slopes indicate constant velocities.
x (m)
t
t
x
t
x
x
t (s)
The Slope of the Position vs.
Time Graph is the Velocity
Increasing slopes indicate increasing velocities,
decreasing slopes indicate decreasing
velocities.
x (m)
t (s)
Concept Check…
Get Your Cards…
The graph represents the motion of two
cars. Which statement is correct?
(a) The blue car is slower than the red car.
(b) The cars have the same speed at t = 12 sec.
(c) The red car is speeding up.
(d) All roads lead to Rome.
x(m) 20
18
16
14
12
10
8
6
4
2
0 1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16
t (s)
(a) The blue car is slower than the red car.
The slope of the red car is steeper/greater than that of
the blue car, therefore, the velocity is likewise greater.
x(m) 20
18
16
14
12
10
8
6
4
2
0 1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16
t (s)
This graph shows
(a) speeding up in the positive direction.
(b) slowing down in the negative direction.
(c) speeding up in the negative direction.
(d) slowing down in the positive direction.
x(m) 20
18
16
14
12
10
8
6
4
2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
t (s)
(c) speeding up in the negative direction.
The slope begins at nearly zero, and becomes
increasingly more negative.
x(m) 20
18
16
14
12
10
8
6
4
2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
t (s)
This graph shows
(a) speeding up in the positive direction.
(b) slowing down in the negative direction.
(c) speeding up in the negative direction.
(d) slowing down in the positive direction.
x(m) 20
18
16
14
12
10
8
6
4
2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
t (s)
(b) slowing down in the negative direction.
The graph begins with a fairly steep negative slope
which becomes less and less steep, nearly zero, at
the end of 15 seconds.
x(m) 20
18
16
14
12
10
8
6
4
2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
t (s)
Average Velocity from a Graph
v avg 
Δx
Δt

displaceme
nt
time
Position (m)
10
5
0
x
t
10
5
Time (s)
15
Which
really is the
slope of the
line
between
the two
positions.
Instantaneous Velocity
Position (m)
The instantaneous velocity for any time t may be pulled
from a position vs. time graph as the slope of the graph at
that time.
10
t
x
5
x
0
t
5
10
15
Time (s)
The Velocity vs. Time Graph…
Your new best friend…
The Velocity vs. Time Graph
Velocity (m/s)
10
5
0
10
5
Time (s)
15
Acceleration
Average acceleration (aavg) is the change in
velocity divided by the time interval during
which the change occurred.
a avg 
Δv
Δt

vf - vi
tf  ti

change in velocity
time
Instantaneous acceleration is the
acceleration at a particular instant.
Acceleration is a vector quantity.
Units for Acceleration
The units for acceleration have the dimensions
of length over time squared:
L 
2
T 
Very often, we’ll report accelerations in meters
per second squared (m/s2), but any other
combination of units that have the same
dimensions are acceptable.
What Does an Acceleration of 10
m/s2 Mean?
It means that the velocity
of an object is changing
at a rate of 10 m/s every
second.
The Velocity vs. Time Graph
Velocity (m/s)
10
m 
rise
run

Δv
Δt
5
v
0
t
10
5
Time (s)
15
The Slope of the Velocity vs.
Time Graph is the Acceleration
v (m/s)
Positive slopes indicate positive accelerations,
negative slopes, negative accelerations.
0
t (s)
The Slope of the Velocity vs.
Time Graph is the Acceleration
v (m/s)
Constant slopes indicate constant acceleration.
t
t
v
t
v
v
t (s)
The Slope of the Velocity vs.
Time Graph is the Acceleration
Increasing slopes indicate increasing
accelerations, decreasing slopes indicate
decreasing velocities.
v (m)
t (s)
Something to Consider…
Velocity (m/s)
Let’s say that you are traveling (graph below)
southbound with a constant velocity of 10 m/s. If you
travel for 4 seconds, how far have you traveled? How
far have you traveled in 6 seconds?
10
40 m = 10 m/s x 4 s
0
2
4
6
Time (s)
Something Else to Consider…
Velocity (m/s)
Let’s say that you are start from rest and travel (graph
below) southbound with a constant acceleration of 1
m/s2. If you travel for 4 seconds, how far have you
traveled? How far have you traveled in 6 seconds?
10
vavg= (vf – vi)/2= 2 m/s
2 m/s x 4 s = 8 m
8 m = ½ bh = ½ (4s)(4 m/s)
0
4
2
Time (s)
6
What a Revelation!!!!!
The area between the graph and the
horizontal axis is the displacement
that occurred during that time
interval.
A Second Look
Velocity (m/s)
10
A1 = ½ (5 m/s)(5 s) = 12.5 m
A2 = =(5 m/s)(5 s) = 25 m
A3 = ½ (5 m/s)(5 s) = 12.5 m
5
A2
A1
0
A3
10
5
Time (s)
15
x(m)
t(s)
0
0
12.5
5
37.5
10
50
15
Position (m)
You Can Draw an X vs. T
Graph from a V vs. T Graph
(50, 15)
50
(10, 37.5)
25
(5, 12.5)
0
5
10
x(m)
t(s)
0
0
12.5
5
37.5
10
50
15
15 Time (s)
The graph represents the motion of two
cars. Which statement is correct?
v(m/s)
(a) The blue car is always slower than the red car.
(b) The cars have the same speed at t = 12 sec.
(c) The cars have the same position at r = 12 sec.
(d) The blue car has greater acceleration.
20
18
16
14
12
10
8
6
4
2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
t (s)
A rock is dropped from a hovering
helicopter. If it’s initial velocity is 0 m/s,
and the acceleration due to gravity is 10
m/s2, what is the velocity of the rock after
it has fallen for 5 seconds?
(a) 0 m/s
(b) 5 m/s
(c)50 m/s
(d)20 m/s
v=?
(c) 50 m/s
Every second the rock falls, its
velocity increases by 10 m/s.
t=0
t=1s
v=0
v = 10 m/s
t=2s
v = 20 m/s
t=3s
v = 30 m/s
t=4s
v = 40 m/s
t=5s
v = 50 m/s
(b) The cars have the same speed at t = 12 sec.
v(m/s)
At t = 12 sec, the cars have the same speed.
20
18
16
14
12
10
8
6
4
2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
t (s)
From 0 – 6 sec, the graph shows
v(m/s)
(a) speeding up in the positive direction.
(b) slowing down in the negative direction.
(c) speeding up in the negative direction.
(d) slowing down in the positive direction.
12
10
8
6
4
2
0
-2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
-4
-6
t (s)
(d) slowing down in the positive direction.
v(m/s)
The object’s motion begins with a velocity of 10 m/s
and decreases at a constant rate to 0 m/s at t = 6 sec.
12
10
8
6
4
2
0
-2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
-4
-6
t (s)
From 0 – 6 sec,
v(m/s)
(a) The velocity and acceleration are positive.
(b) The velocity and acceleration are negative.
(c) The velocity is positive, and acceleration negative.
(d) The velocity is negative, and acceleration positive.
12
10
8
6
4
2
0
-2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
-4
-6
t (s)
(c) The velocity is positive, and
acceleration negative.
v(m/s)
The slope (acceleration) is negative, and the velocity is
positive.
12
10
8
6
4
2
0
-2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
-4
-6
t (s)
Another Case of Acceleration
Since acceleration is defined as a change in velocity
over time, and velocity is a vector quantity, a change
in direction constitutes acceleration also.
This is called centripetal acceleration, and will
be discussed later
Observe the animation of the three cars below.
Which car or cars (red, green, and/or blue) are
undergoing an acceleration? Study each car
individually in order to determine the answer.