Download II. Describing Motion

Motion
PSc.1.1 OBJECTIVE: Understand
motion in terms of speed, velocity,
acceleration, and momentum.
Objectives
 PSc.1.1.1
 Interpret all motion as relative
to a selected reference point.
 Identify distance and
displacement as a scalar-vector
pair.
Objectives
 PSc.1.1.1,
cont.
 Describe motion qualitatively
and quantitatively in terms of
• an object’s change of
position,
• distance traveled,
• and displacement.
Reference Point

For all motion problems we need a
reference point...
 a non-moving point from which
motion is measured.
Reference Point

A reference point is a point or object
that is used to measure what the
distance and direction to another
object is.

In the picture above, the reference
point is the yard line.
Reference Point

In the picture above your house is used
as a reference point to determine where
your friend’s house is.
Reference Point
Motion
 An
object is in motion when the
distance between the object and
the reference point is changing.
Reference point
Motion
Motion
 The
person is in
motion when the
fence is used as a
reference point
because the
distance between
the person and the
fence is changing.
Motion
 The
person is NOT
in motion when the
ground is used as a
reference point
because the
distance between
the person and the
ground is not
changing.
Distance

Distance is the space traveled
between the reference point and the
object or ending point.

The distance between the reference
point and the blue football player is
3 m (meters).
Distance
• Problem #1:
Suppose a runner jogs to the 50-m
North mark and then turns around
and runs back to the 20-m mark
South. Determine her distance.
The runner travels 50 m in
the original direction
(north) plus 30 m in the
opposite direction (south),
so the total distance she
ran is 80 m.
Position
Position is an object’s location
compared to a reference point.
 It includes both a distance and
direction.

Position

If City Hall is the reference point, the
position of the library is 500 m East.
Displacement

Displacement is a change of
position in a certain direction, not
the total distance traveled.
Displacement

The displacement is the shorter
directed distance from start to stop
(green arrow).
start
stop
Displacement
If two displacement vectors are going
in the same direction add to find the
total displacement.
 If two displacement vectors are going
in opposite directions subtract to find
the total displacement.

Displacement

Problem #2:
A man walks 54.5 meters east and
then and an additional 30.0 meters
east. Calculate his displacement
relative to where he started.
54.5 m, E
30.0 m, E
84.5 m, E
Displacement
Problem #3:
A man walks 54.5 meters east and
then 30.0 meters west. Calculate his
displacement relative to where he
started.
54.5 m, E
30.0 m, W
24.5 m, E
Distance vs Displacement

Problem #4: Suppose a bus starts
from terminal A, travels 1500 m to
reach terminal B and then returns to
terminal A. What is
 a) the distance traveled and
3000 m
 b) the displacement traveled?
0m
Vectors and Scalars


A scalar quantity
is any quantity
that has a
magnitude, but
NO direction
associated with
it.
Scalar
Example
Speed
Magnitude
Distance
10 m
Age
15 years
Heat
1000
calories
20 m/s
Magnitude – A numerical value with
units.
Vectors and Scalars

Distance is a scalar quantity.

The distance
between the
two books is
15 units.
Vectors and Scalars

The distance points A and B
is 6 units.
Vectors and Scalars
A vector is represented by an arrow.
 A vector gives 2 pieces of
information:
 magnitude and
 direction.

The magnitude
is 5 m/s.
 The direction is
42°.

5 m/s
42°
Vectors and Scalars
The length of the arrow represents
the magnitude (how far, how fast,
how strong, etc, depending on the
type of vector).
 The arrow points in the direction of
the force, motion, displacement,
etc. It is often specified by an
angle.

Vectors and Scalars

Displacement is a vector quantity.

The
displacement
between
points A and
B found by
using the
Pythagorean
Theorem.
Vectors and Scalars
The x-displacement is 5 units.
 The y-displacement is 1 unit.

(5)2 + (1)2 = (AB)2
25 + 1 = (AB)2
√ 26 = AB
5.1 units = AB
Vectors and Scalars

The direction is northeast.
Vectors and Scalars

Distance and displacement are
referred to as a scalar-vector pair.
Vectors and Scalars
Problem #5:
John runs 3 km north, then turns and
walks 4 km south.
a) Calculate the total distance covered.

7 km
b) Calculate the total displacement.
1 km south
Vectors and Scalars
Problem #6:
During a ride in a hot air balloon, a
group of people are carried 50 km
[North], 65 km [West] and then 75 km
[South].
a) Calculate the total distance covered.
190 km
b) Calculate the total displacement.
70 km southwest

Objectives
 PSc.1.1.2
 Compare speed and velocity as a
scalar-vector pair.
 Velocity is a relationship between
displacement and time.
 Apply concepts of average speed and
average velocity to solve conceptual
and quantitative problems.
 Explain acceleration as a relationship
between velocity and time.
Motion
• Distance and time are important
when considering motion. In order
to win a race, you must cover the
distance in the shortest amount of
time.
• How would
you describe
the motion of
the runners in
the race?
Speed

Speed is the measurement of the
change in distance for a given period
of time.

The car traveled 40 m in 6 s.
Speed
Instantaneous speed is speed at a
given instant.
 The speed shown on a speedometer
is the instantaneous speed.

Speed
When something is speeding up or
slowing down, its instantaneous
speed is changing.
 Constant speed means the speed
stays the same.
 If an object is moving with constant
speed, the instantaneous speed
doesn't change.

Speed

Average speed is total distance
divided by total time.
total distance
avg. speed 
total time
Speed
distance
speed 
time
d
v t
Velocity

Velocity is the measurement of the
change in displacement for a given
period of time.
Velocity

Velocity is speed in a given direction
and can change even when the
speed is constant!
Speed and Velocity
Speed and velocity are a scalarvector pair.
 Speed is a scalar quantity that refers
to "how fast an object is moving."
 Velocity is a vector quantity that
refers to "the rate at which an object
changes its position."

Speed and Velocity
Since a vector quantity has a
direction associated with it, velocity
values can be negative if they are
directed west or south.
 Speed cannot be negative.

Speed and Velocity

To calculate speed/velocity you need
the distance/displacement traveled
and the time the object traveled.
d
v
t
Speed/Velocity Problem #7

A car traveled 40 m in 6 s. Determine
the speed.
d
v
t
40m
v
6s

The speed of the car is 6.7 m/s.
Problem
Distance
Displacement
1
135 miles
85 miles
2
6m
0m
3a
7 km
5 km NE
3b
5 blocks
3.6 blocks SE
3c
800 m
0m
3d
5 blocks
3.6 blocks NW
3e
300 m
0m
4
2750 km
750 km
5
180 km
170 km N
6
390 m
30 m E
7a
30 m
0m
7b
45 m
45 m E
8
9 km
6.7 km NE
9
17 km
11.7 km SW
10
12 m
8.9 m NE
11
12 km
2 km N
Speed/Velocity Problem #8
Your neighbor skates at a speed of
4 m/s. You can skate 100 m in 20 s.
Who skates faster?
GIVEN:
WORK:

Δd = 100 m
Δt = 20 s
v=?
d
v t
v = Δd ÷ Δt
v = (100 m) ÷ (20 s)
v = 5 m/s
You skate faster!
Speed/Velocity Problem #9
A baseball pitcher throws a fastball at 42 m/s. If
the batter is 18 m from the pitcher, how much
time does it take for the ball to reach the batter?
GIVEN:
WORK:
Δd = 18 m
Δt = ?
v = 42 m/s
Δt = Δd ÷ v
Δt = (18 m) ÷ (42 m/s)
d
v t
Δt = 0.43 s
Speed/Velocity Problem #10
A toy car moves at a constant velocity of
1.6 m/s. If it continues, how far will the car travel
in 3.0 s?
GIVEN:
WORK:
Δd = ?
Δt = 3.0
v = 1.6 m/s
Δd = v • Δt
Δd = (1.6 m/s) • (3.0 s)
d
v t
Δd = 4.8 m
Speed/Velocity Problem #11
You travel 35 km in 0.4 h, followed by
53 km in 0.6 h. What is your average
speed?
GIVEN:
WORK:

Δd = (35 + 53) km v = Δd ÷ Δt
Δt = (0.4 + 0.6) h
v = (88 km) ÷ (1.0 h)
v=?
v
=
88
km/h
d
v t
Speed and Velocity

Problem:
 A storm is 10 km away and is
moving at a speed of 60 km/h.
Should you be worried?
 It depends
on the
storm’s
direction!
Speed and Velocity
Problem
Answer
1
62.2 mi/hr
2
4.76 mi/hr
3
4400 mi
4
238.2 mi
5
800 mi
6
381.5 mi
7
68.7 mi/hr
8
4.23 hr
Acceleration
Acceleration is the rate of change of
velocity (change in speed or
direction).
 Acceleration occurs whenever there
is a change in speed, direction or
both.

Acceleration

Acceleration occurs whenever
there is a change in speed,
direction, or both.
Acceleration
 Positive
acceleration
 “speeding up”
 Negative
acceleration
 “slowing down”
Signs for Velocity and
Acceleration
Acceleration
vf - vi
a t
a
v f  vi
t
a:
vf:
vi:
t:
acceleration
final velocity
initial velocity
time
Acceleration Problem #12
A roller coaster starts down a hill at
10 m/s. 3.0 seconds later, its speed is
32 m/s. What is its acceleration?
GIVEN:
WORK:

vi = 10 m/s
t=3s
vf = 32 m/s
vf - vi
a=?
a t
a = (vf - vi) ÷ t
a = (32m/s - 10m/s) ÷ (3s)
a = 22 m/s ÷ 3 s
a = 7.3 m/s2
Acceleration Problem #13
Sound travels 330 m/s. If lightning
strikes the ground 1 km away from you,
how long will it take for you to hear it?
GIVEN:
WORK:

v = 330 m/s
t=d÷v
d = 1km = 1000m
t = (1000 m) ÷ (330 m/s)
t=?
t
=
3.03
s
d
v t
Acceleration Problem #14
How long will it take a car traveling
30 m/s to come to a stop if its
acceleration is - 3 m/s2?
GIVEN:
WORK:

t=?
vi = 30 m/s
vf = 0 m/s
a = -3 m/s2
t = (vf - vi) ÷ a
t = (0m/s-30m/s)÷(-3m/s2)
vf - vi
a t
t = -30 m/s ÷ -3m/s2
t = 10 s
Speed, Velocity and Accel.
Problem Answer
1
75.9 km/hr
2
-2 m/s2
3
30 km
4
5 m/s2
5
-3.33 m/s2
6
24 m/s
7
0.1 s
8
8.6 m/s
Objectives
 PSc.1.1.2,
cont.
 Using graphical analysis, solve for
displacement, time, and average
velocity. Analyze conceptual trends in
the displacement vs. time graphs
such as constant velocity and
acceleration.
Objectives
 PSc.1.1.2,
cont.
 Using graphical analysis, solve for
velocity, time, and average
acceleration. Analyze conceptual
trends in the velocity vs. time graphs
such as constant velocity and
acceleration.
Graphing
•
•
The independent variable is the
variable that the experimenter
manipulates or changes, on purpose.
The dependent variable is a variable
that changes depending on some
other factors.
Graphing
Put the
dependent
variable on the
‘y-axis’
and
the independent variable on the ‘x-axis.’
Slope of a graph

The slope of a
graph is equal to
the ratio of rise
to run.
Graphing Motion
• The motion of an object over a
period of time can be shown on a
distance-time graph.
• Time is plotted along the horizontal
axis of the graph and the distance
traveled is plotted along the vertical
axis of the graph.
Graphing Motion
• Each axis must have a scale that
covers the range of number to be
plotted.
• Once the scales for each axis are in
place, the data points can be
plotted.
• After plotting the data points, draw
a line connecting the points.
Distance-Time Graph
Distance-Time Graph
A
B

slope = speed

steeper slope =
faster speed

straight line =
constant speed

flat line =
no motion
(at rest)
Distance-Time Graph
Distance-Time Graph
Curve =
acceleration
Distance-Time Graph
Distance-Time Graph
400

Acceleration is
indicated by a
curve on a
distance-time
graph.

Changing slope =
changing velocity
Distance (m)
300
200
100
0
0
5
10
Time (s)
15
20
Distance-Time Graph
What would these look
like on a distance-time
graph?
1. stopped
2. slow
3. fast
4. accelerating
Distance-Time Graph
Distance-Time Graph
A
B
A or B: Who
started out faster?
 A (steeper
slope)
 A or B: Who had a
constant speed?
 A (continuous
straight line)

Distance-Time Graph
Distance-Time Graph
A
B

Describe B’s motion
from 10-20 min.
 B stopped moving

Find their average
speeds.
 A = (2400 m) ÷ (30 min)
 A = 80 m/min
 B = (1200 m) ÷ (30 min)
 B = 40 m/min
Speed-Time Graphs
Speed-Time Graph
3

Speed (m/s)
2
1
0
0
2
4
6
Time (s)
8
10
slope = acceleration
 + slope = speeds
up
 negative slope =
slows down
Speed-Time Graphs
Speed-Time Graph

straight line =
constant accel.
2

flat line = no accel.
(constant velocity)
Speed (m/s)
3
1
0
0
2
4
6
Time (s)
8
10
Speed-Time Graphs
Speed-Time Graph
Specify the time
period when the
object was...
 slowing down
 5 to 10 seconds
 speeding up
 0 to 3 seconds
3
Speed (m/s)
2
1
0
0
2
4
6
Time (s)
8
10
Graphing Motion
Speed-Time Graph
Specify the time
period when the
object was...
3
2
moving at a
constant speed
 3 to 5 seconds
 not moving
 0 & 10 seconds
Speed (m/s)

1
0
0
2
4
6
Time (s)
8
10
Objectives
 PSc.1.1.2,
cont.
 Infer how momentum is a relationship
between mass and velocity of an
object.
 Explain change in momentum in
terms of the magnitude of the applied
force and the time interval that the
force is applied to the object.
Momentum
Momentum refers to inertia in
motion.
 Momentum is a measure of how
difficult it is to stop an object or a
measure of “how much motion” an
object has.

Momentum

Momentum
 quantity of motion
p = mv
p
m v
p:
m:
v:
momentum (kg ·m/s)
mass (kg)
velocity (m/s)
Momentum Problem #15
Find the momentum of a bumper car
if it has a total mass of 280 kg and a
velocity of 3.2 m/s.
GIVEN:
WORK:
p=?
p = mv
m = 280 kg
p = (280 kg)(3.2 m/s)
v = 3.2 m/s
p = 896 kg·m/s
p

m v
Momentum Problem #16

The momentum of a second bumper car
is 675 kg·m/s. What is its velocity if its
total mass is 300 kg?
GIVEN:
WORK:
p = 675 kg·m/s
m = 300 kg
v=?
v=p÷m
p
m v
v = (675 kg·m/s)÷(300 kg)
v = 2.25 m/s
Change in Momentum
•
•
Sometimes the momentum of
an object changes.
Change in momentum can be
called impulse.
Change in Momentum
Force applied on everyday
objects results in a change
in velocity.
Change in Momentum
• When playing with a paddleball, the
less time the ball is in contact with
the paddle, the more force applied to
the ball.
Change in Momentum
• If time is extended, less force will be
applied to the object.
Change in Momentum

The purpose of seat belts, air bags,
and padded dashboards is to
extend the time during which you
come to rest during a crash. These
safety devices help reduce the
forces exerted on you.
Change in Momentum

A 1000 kg car moving at 30 m/s
(p = 30,000 kg m/s) can be stopped
by 30,000 N of force acting for
1.0 s (a crash) OR
Change in Momentum

by 3000 N of force acting for
10.0 s (normal stop).
Change in Momentum

The same change in momentum
can be accomplished by a small
force acting for a long time or by
a large force acting for a short
time.
F·t = m·v
where v = vf - vi
Change in Momentum
Use the diagrams below to
determine the magnitude of
changes in velocity of the same
ball.
∆v = 15 m/s
∆v = 58 m/s
Change in Momentum
Problem #17
Calvin throws a 0.450 kg snowball at
Susie. Calculate the initial speed of
the snowball if its change in
momentum is 3.40 kg.m/s.
(7.56 m/s)
Change in Momentum
Problem #18
A 0.40 kg soccer ball approaches
Joe horizontally with a velocity of
18 m/s north. He strikes the ball
and causes it to move in the
opposite direction with a velocity
of 25 m/s. Calculate the magnitude
of the change in momentum of the
ball.
(17 kg.m/s)
Change in Momentum
Problem #19

Calvin strikes a 0.0450 gram golf
ball with a club. The force applied
to the ball is 1900. N. If the club is
in contact with the ball for 0.00500 s,
what is the change in velocity of the
golf ball?
(211 m/s)
Change in Momentum
Problem #20
• If a pitcher pitches the ball at a
speed of 40.2 m/s, and Calvin sends
it back with a speed of 49.1 m/s,
determine the time the ball is in
contact with the bat. Calvin applies a
force of 12,000 N to the
0.142 kg baseball.
(0.00106 s)
Conservation of Momentum

Law of Conservation of Momentum
 The total momentum in a group
of objects doesn’t change unless
outside forces act on the objects.
pbefore = pafter