Download II. Describing Motion

II. Describing Motion
Motion
Speed & Velocity
 Acceleration



Newton’s First Law of Motion
◦ An object at rest will remain at rest and an
object in motion will continue moving at a
constant velocity unless acted upon by a net
force.
motion
constant velocity
net force

Problem:
◦ Is your desk moving?

We need a reference point...
◦ nonmoving point from which motion is measured

Motion
◦ Change in position in relation to a reference point.
Reference point
Motion
Problem:
 You are a passenger in a car
stopped at a stop sign. Out of
the corner of your eye, you
notice a tree on the side of the
road begin to move forward.
 You have mistakenly set
yourself as the reference point.


You can describe the direction of an object in
motion with a reference direction.
north, south, east, west, up, or down
 Distance
 Displacement


Distance – the length traveled by an object
Practice – What is the distance?
◦ Pattie walks 4 km North, then 3 km East
◦ Joe runs 30 m East, then runs 40 meters West
◦ Aaron jogs 4 km East, 4 km North,
4 km West, and 4km South
◦ Answers: 7 km ; 70 meters ; 16 km


Distance – length
Displacement – distance AND direction of a
movement from the starting point

If an object moves in a single direction, the
displacement equals the distance + the
direction
4 km
start here 
total displacement
=
4km North

If an object moves in two opposing
directions, the displacement is the difference
between the two.
total distance = 4 + 3 = 7 km
4 km
start here 
3 km
Displacement can be positive or negative. A
negative direction can be either the opposite of
the original movement, or can follow the sign of
a typical graph [North/East vs South/West]
total displacement =
4 km North + -3 km South
= 1 km North
(of original starting point)

If an object moves in two directions, a
triangle will be formed. If the angle is 90º ,
use a2 + b2 = c2 to solve.
3 km
4 km
start here

displacem
ent

If an object moves in two directions, a
triangle will be formed. If the angle is 90º ,
use a2 + b2 = c2 to solve.
3 km
displacement =
4 km
5 km
42 + 32 = c2
c2 = (16) + (9) = 25
start here

c2 = √(25)
c = 5km

Displacement can be given in:
◦ units with direction
 example: x number of meters North
 30 meters East + 40 meters West
 displacement = 10 meters West

Displacement can be given in:
◦ positive or negative units [like on a graph]
 positive = North {up} or East {right}
 negative = South {down} or West {left}
 30 meters east + 40 meters west =
30 meters + (-40) meters = -10 meters

Practice – What is the displacement?
◦ Pattie walks 8km North, then 3 km East
 5 km displacement
◦ Joe runs 30 m East, then runs 40 meters West
 10 meters [or -10] displacement
◦ Aaron jogs 4 km East, 4 km North,
4 km West, and 4km South
 0 meters displacement
go to website!!!!
Draw each scenario and show work.
2. Amy runs 2 miles south, then turns
around and runs 3 miles north.
a. Distance ___ b. Displacement ___
3. Jermaine runs exactly 2 laps around
a 400 meter track.
4. Joe turns around 5 times.
5. Ray runs 30 feet north, 30 feet
west, and then 30 feet south.

d
Speed
◦ rate of motion
◦ distance traveled per unit time
v t
distance
speed 
time

Instantaneous Speed

Average Speed
◦ speed at a given instant
total distance
avg. speed 
total time

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!

Velocity
◦ speed in a given direction
◦ can change even when the speed is constant!
 Find
the velocity in m/s of a
swimmer who swims 110 m
toward the shore in 72 s.
v
=d
t
 v = 110m = 1.5 m/s
72 s
vf - vi

Acceleration
◦ the rate of change of velocity
◦ change in speed or direction
a
v f  vi
t
a:
vf:
vi:
t:
a t
acceleration
final velocity
initial velocity
time
Positive acceleration
“speeding up”
Negative acceleration
“slowing down”
Your neighbor skates at a speed of 4 m/s
towards home. You can skate 100 m in
20 s. Who skates faster?
GIVEN:
WORK:

d = 100 m
t = 20 s
v=?
d
v t
v=d÷t
v = (100 m) ÷ (20 s)
v = 5 m/s
You skate faster!
A roller coaster starts down a hill at 10 m/s.
Three seconds later, its speed is 32 m/s.
What is the roller coaster’s 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 = 7 m/s
Sound travels 330 m/s. If a lightning bolt
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
=
3
s
d
v t
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
Distance-Time Graph
A


B


slope =speed
steeper slope =
faster speed
straight line =
constant speed
flat line =
no motion
Distance-Time Graph

A


B

Who started out faster?
◦ A (steeper slope)
Who had a constant speed?
◦A
Describe B from 10-20 min.
◦ B stopped moving
Find their average speeds.
◦ A = (2400m) ÷ (30min)
A = 80 m/min
◦ B = (1200m) ÷ (30min)
B = 40 m/min
Distance-Time Graph

400
Distance (m)
300
200
100

0
0
5
10
Time (s)
15
20
Acceleration is
indicated by a
curve on a
Distance-Time
graph.
Changing slope
= changing
velocity
Speed-Time Graph
slope =acceleration
 +ve = speeds up
 -ve = slows down
 straight line =
3

Speed (m/s)
2
1
constant accel.
 flat line =
no accel.
0
0
2
4
6
Time (s)
8
10
(constant velocity)
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
moving at a constant
speed
◦ 3 to 5 seconds
not moving
◦ 0 & 10 seconds
III. Defining Force
Force
Newton’s First Law
 Friction



Force
◦ a push or pull that one body exerts on
another
◦ What forces are being
exerted on the football?
Fkick
Fgrav

Balanced Forces
◦ forces acting
on an object
that are
opposite in
direction and
equal in size
◦ no change in
velocity

Net Force
◦ unbalanced forces that are not
opposite and equal
◦ velocity changes (object
accelerates)
Fnet
Ffriction
Fpull
N
N
W

Newton’s First Law of Motion
◦ An object at rest will remain at rest and an
object in motion will continue moving at a
constant velocity unless acted upon by a net
force.


Newton’s First Law of Motion
◦ “Law of Inertia”
Inertia
◦ tendency of an object to resist
any change in its motion
◦ increases as mass increases
TRUE or FALSE?
The object shown in the diagram
must be at rest since there is no net
force acting on it.
FALSE! A net force does not
cause motion. A net force
causes a change in motion,
or acceleration.
Taken from “The Physics Classroom” © Tom Henderson, 1996-2001.
You are a passenger in a car and not wearing
your seat belt.
Without increasing or decreasing its speed, the
car makes a sharp left turn, and you find
yourself colliding with the right-hand door.
Which is the correct analysis of the situation? ...
1. Before and after the collision, there is a
rightward force pushing you into the door.
2. Starting at the time of collision, the door
exerts a leftward force on you.
3. both of the above
2. neither
Starting
at the
time
4.
of the
above
of collision, the
door exerts a leftward force on you.

Friction
◦ force that opposes motion
between 2 surfaces
◦ depends on the:
 types of surfaces
 force between the
surfaces


Friction is greater...
◦ between rough
surfaces
◦ when there’s a greater
force between the
surfaces
(e.g. more weight)
Pros and Cons?
IV. Force & Acceleration

Newton’s Second Law
 Gravity
 Air Resistance
 Calculations

Newton’s Second Law of Motion
◦ The acceleration of an object is
directly proportional to the net
force acting on it and inversely
proportional to its mass.
F = ma
F
a
m
F = ma
F
m a
F: force (N)
m: mass (kg)
a: accel (m/s2)
1 N = 1 kg ·m/s2

Gravity
◦ force of attraction between any
two objects in the universe
◦ increases as...
 mass increases
 distance decreases


Who experiences more gravity the astronaut or the politician?
Which exerts more gravity the Earth or the moon?
less
distance
more
mass

Weight
◦ the force of gravity on an object
W = mg
W: weight (N)
m: mass (kg)
g: acceleration due
to gravity (m/s2)
MASS
WEIGHT
always the same
(kg)
depends on gravity
(N)

Would you weigh more on
Earth or Jupiter?
 Jupiter because...
greater mass
greater gravity
greater weight

Accel. due to gravity (g)
 In the absence of air
resistance, all falling objects
have the same acceleration!
 On Earth: g = 9.8 m/s2
W
g
m
elephant
g
W
m
feather
Animation from “Multimedia Physics Studios.”
http://www.animations.physics.uns
w.edu.au/jw/Newton.htm#Newton
2
Go to the astronaut’s, David Scott,
demonstration

Air Resistance
◦ a.k.a. “fluid friction” or “drag”
◦ force that air exerts on a moving
object to oppose its motion
◦ depends on:
•
•
•
•
speed
surface area
shape
density of fluid

Terminal Velocity
◦ maximum velocity reached
by a falling object
◦ reached when…
Fair
Fgrav = Fair
 no net force
 no acceleration
 constant velocity
Fgrav

Terminal Velocity
 increasing speed  increasing air
resistance until…
Fair = Fgrav
Animation from “Multimedia Physics Studios.”

Falling with air resistance
 heavier objects fall faster
because they accelerate
to higher speeds before
reaching terminal
velocity
Fgrav = Fair
 larger Fgrav
 need larger Fair
 need higher speed
Animation from “Multimedia Physics Studios.”

What force would be required to
accelerate a 40 kg mass by 4 m/s2?
GIVEN:
WORK:
F=?
m = 40 kg
a = 4 m/s2
F = ma
F
m a
F = (40 kg)(4 m/s2)
F = 160 N
F = 200 N

A 4.0 kg shotput is thrown with 30 N of
force. What is its acceleration?
GIVEN:
WORK:
m = 4.0 kg
F = 3.0 N
a=?
a=F÷m
F
m a
a = (30 N) ÷ (4.0 kg)
a = 7.5 m/s2

Mrs. J. weighs 557 N. What is her
mass?
GIVEN:
WORK:
F(W) = 557 N
m=?
a(g) = 9.8 m/s2
m=F÷a
F
m a
m = (557 N) ÷ (9.8 m/s2)
m = 56.8 kg
m = 57 kg

Is the following statement true or
false?
◦ An astronaut has less mass on the
moon since the moon exerts a
weaker gravitational force.
 False! Mass does not depend on
gravity, weight does. The astronaut has
less weight on the moon.
VI. Action and Reaction

Newton’s Third Law
 Momentum
 Conservation of
Momentum

Newton’s Third Law of Motion
◦ When one object exerts a force on a second
object, the second object exerts an equal but
opposite force on the first.

Problem:
 How can a horse
pull a cart if the cart
is pulling back on
the horse with an equal but
opposite force?
 Aren’t these “balanced forces”
resulting in no acceleration?
NO!!!

Explanation:
◦ forces are equal and opposite
but act on different objects
◦ they are not “balanced forces”
◦ the movement of the horse
depends on the forces acting
on the horse

Action-Reaction Pairs

The hammer exerts
a force on the nail
to the right.

The nail exerts an
equal but opposite
force on the
hammer to the left.

Action-Reaction Pairs
The rocket exerts a
downward force on the
exhaust gases.
 The gases exert an
equal but opposite
upward force on the
rocket.

FG
FR

Action-Reaction Pairs

Both objects accelerate.

The amount of acceleration
depends on the mass of the object.
F
a 
m

Small mass  more acceleration

Large mass  less acceleration
I. Newton’s Laws of Motion
“If I have seen far, it is because I have stood
on the shoulders of giants.”
- Sir Isaac Newton
(referring to Galileo)

Newton’s First Law of Motion
◦ An object at rest will remain at rest and an
object in motion will continue moving at a
constant velocity unless acted upon by a net
force.

Newton’s Second Law of Motion
◦ The acceleration of an object is directly
proportional to the net force acting on it and
inversely proportional to its mass.
F = ma

Newton’s Third Law of Motion
◦ When one object exerts a force on a second
object, the second object exerts an equal but
opposite force on the first.