Transcript II. Describing Motion - Manchester Local School District

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.

Speed
◦ rate of motion
◦ distance traveled per unit time
distance
speed 
time
d
v t

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

Unbalanced forces
◦ unbalanced forces that are not
opposite and equal
◦ velocity changes (object accelerates)
Fnet
Ffriction
Fpull
N
N
W
 Net
force
the combination of all the
forces acting on an object

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
Since these two forces are of equal magnitude and in
opposite directions, they balance each other. The book
is said to be at equilibrium. There is no unbalanced
force acting upon the book and thus the book maintains
its state of motion.

The force of gravity pulling downward and
the force of the table pushing upwards on the
book are of equal magnitude and opposite
directions. These two forces balance each
other. Yet there is no force present to balance
the force of friction. As the book moves to
the right, friction acts to the left to slow the
book down.


Free-body diagrams are diagrams used to
show the relative magnitude (strength) and
direction of all forces acting upon an object
in a given situation.
The size of the arrow in a free-body diagram
is shows the magnitude of the force. The
direction of the arrow reveals the direction in
which the force acts.

It is customary in a free-body
diagram to represent the
object by a box or a small
circle and to draw the force
arrow from the center of the
box or circle in an outward
direction.
F = force
Ffrict = friction
Fgrav = gravity
Fnorm = normal force
Fapp = applied force
Object lies motionless on a surface.
A rightward force is applied to a book
in order to move it across a desk at
constant velocity. Consider frictional
forces. Neglect air resistance.
Object slows due to friction (rough
surface).
An object is suspended from the ceiling.
A new force is tension.


A flying squirrel is gliding (no wing flaps)
from a tree to the ground at constant
velocity. Consider air resistance.
A new force is air resistance. It pushes an
object up.


http://www.mrwaynesclass.com/freebodies/r
eading/index01.html
Free body diagrams with movement
A
stationary object remains
stationary if the sum of the forces
acting upon it - resultant force is zero.
 A moving object with a zero
resultant force keeps moving at
the same speed and in the same
direction.
 If
the resultant force acting on an
object is not zero, a stationary
object begins to accelerate in the
same direction as the force. A
moving object speeds up, slows
down or changes direction.
 Add
forces going in the same
direction.
 Subtract
forces going in opposite
directions.
 2N
2N

Resultant Forces
http://physicsnet.co.uk/gcse-physics/the-effects-offorces-resultant-force-and-motion/
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.

https://www.youtube.com/watch?v=VutAx3R
DFbI
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
4. neither of the above
2. Starting at the time 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?

Static Friction: friction between
surfaces that are stationary
Kinetic Friction: friction between
moving surfaces
 The force necessary to make a
stationary object start moving is
usually greater than the force needed
to keep it moving.




Sliding friction : when objects slide
past each other
Rolling friction: If a round object rolls
over a flat surface
Fluid friction: an object moving
through a fluid such as air (car moving
hits air molecules)
 How
can a car minimize its fluid
friction?
smooth surface – wax car surface
shape of the car -streamlining


http://www.mrwaynesclass.com/freebodies/r
eading/index01.html
Free body diagrams with movement
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.