Download 5.1 Speed, velocity and acceleration

Speed, velocity and acceleration
What This Means

If you want to understand how an
object (like a car, ball, person, or
rocket) moves, you have to understand
three things about what it means "to
be moving." These three things "stick"
to any object that moves, and are
numbers that scientifically describe
just how an object's motion is working.
These three things are:
1. Position. This is precisely where an
object is located.
2. Speed. Precisely how fast an object is
moving. and..(the most difficult for
most people)...
3. Acceleration. Precisely how fast an
object's speed is changing.

LINEAR MOTION
Constant Motion-the body is moving at
constant velocity. The acceleration is zero
Uniformly Accelerated Motion- The body is
moving at constant accelaration. The
velocity is changing with change in time
•
Definitions:
– Speed
• The rate at which something moves a given distance.
• Faster speeds = greater distances
– General formula for speed:
• Speed = distance / time
• Abbreviations commonly used:
– d = distance t = time v = speed
• v = d/t
1
Speed
How can we describe how fast an object moves?
Speed is a measure of how fast something
moves.
Speed = distance travelled per unit of time
SI unit: m s
or km hr (for long distances)
1
Speed
a
Average speed
A car travels at 50 km hr,
slows down to 0 km hr,
and speeds up again to 60 km hr.
Its average speed over the whole journey
overall distance travelled
=
total time of travel
Speed formula
Distance
(meters)
Speed
Time
(m/sec) (seconds)
Constant and Avg. Speed



Constant speed is when a moving object
does not speed up or slow down, but
continues along at a constant rate over time.
Average speed is the average of many
different speeds over a long period of time.
Instantaneous Speed is how fast an object is
moving at any one instant.

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Magnitude of a
quantity tells how
large the quantity
is.
Scalar quantities
have magnitude
only.
Vector quantities
have both
magnitude and
direction.
Speed practice problems

1.
2.
3.
4.
5.
6.
7.
8.
What are the average speeds of the following
objects?
A car traveling a distance of 10m in 5 seconds.
A bicycle traveling 25 miles in 0.5hours.
A car travels 2 miles in 25 seconds.
A person walking travels 3m in 2 sec
A bicycle takes 15 seconds to travel 35m.
A boy running travels 100m in 12s.
A baby crawling takes 10 seconds to travel 3m.
A girl walking takes 20sec to walk 25m.
Answers to practice problems
1.
2.
3.
4.
5.
6.
7.
8.
9.
10m/5s = 2m/s
25mi/0.5hr = 50mi/hr
2mi/25s = .08mi/s
3m/2s = 1.5m/s
35m/15s = 2.33m/s
100m/12s =8.33m/s
3m/10s = 0.3m/s
25m/20s =1.25m/s
20mi/45min. = 0.44mi/min or 26.67mi/hr
1
Speed
b
Instantaneous speed
Speedometer tells the car’s speed
at any instant!
2
Velocity
Velocity is...
a speed in a given direction or
rate of change of displacement.
direction
velocity
magnitude
(speed)
a vector
quantity
2
Velocity
a
Speed with direction
MRT drivers concern
speed only.
speed = 90 km hr
Pilots concern velocity
(direction & speed).
speed = 300 km hr
direction = west
vf  vi
a
t
ACCELERATION
 Acceleration of an object is
the rate of change of its
velocity and is a vector
quantity. For straight-line
motion, average acceleration
is the rate of change of
speed:
change in speed
Accelerati on 
time interval
4.1 Acceleration of a car
Acceleration is the rate of
change in the speed of an
object.
4.1 Acceleration vs. Speed

Positive acceleration
and positive speed
4.1 Acceleration vs. Speed

Negative acceleration
and positive speed
4.1 Acceleration
Acceleration
(m/sec2)
a = Dv
Dt
Change in speed (m/sec)
Change in time (sec)
4.1 Calculate Acceleration

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
A student conducts an
acceleration experiment by
coasting a bicycle down a steep
hill.
The student records the speed of
the bicycle every second for five
seconds.
Calculate the acceleration of the
bicycle.
4.1 Acceleration and Speed

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Constant acceleration is different from constant
speed.
Motion with zero acceleration appears as a straight
horizontal line on a speed versus time graph.
zero acceleration
constant speed
4.1 Acceleration and Speed


Constant acceleration is sometimes called uniform
acceleration.
A ball rolling down a straight ramp has constant
acceleration.
constant acceleration
increasing speed
4.1 Acceleration and Speed
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
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An object can have acceleration, but no speed.
Consider a ball rolling up a ramp.
As the ball slows down, eventually its speed becomes
zero.
constant negative
acceleration
decreasing speed
4.1 Slope and Acceleration

Use slope to recognize when
there is acceleration in speed vs.
time graphs.
– Level sections (A) on the graph
show an acceleration of zero.
– The highest acceleration (B) is the
steepest slope on the graph.
– Sections that slope down (C) show
negative acceleration (slowing
down).
4.2 A Model for Accelerated
Motion
Key Question:
How do we describe and predict accelerated motion?
*Students read Section 4.2 AFTER Investigation 4.2
4.2 Solving Motion Problems
4.2 Solving Motion Problems
4.2 Calculate speed



A ball rolls at 2 m/sec onto a
ramp.
The angle of the ramp creates
an acceleration of 0.75 m/sec2.
Calculate the speed of the ball
10 seconds after it reaches the
ramp.
4.2 Calculate position
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A ball traveling at 2 m/sec rolls onto a ramp that
tilts upward.
The angle of the ramp creates an acceleration of 0.5 m/sec2.
How far up the ramp does the ball get at its
highest point?
(HINT: The ball keeps rolling upward until its
speed is zero.)
4.2 Solving Motion Problems
4.2 Calculate time

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A car at rest accelerates at 6 m/sec2.
How long does it take to travel 440 meters (about a
quarter-mile) and how fast is the car going at the end?
4.2 Calculate position



A ball starts to roll down a ramp with
zero initial speed.
After one second, the speed of the
ball is 2 m/sec.
How long does the ramp need to be
so that the ball can roll for 3 seconds
before reaching the end?
FREE
FALL
Free fall

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The constant acceleration of an
object moving only under the
force of gravity is "g".
The acceleration caused by
gravity is 10 m/s2
If there was no air, all objects
would fall at the same speed
Doesn’t depend on mass
After 1 second falling at 10 m/s
After 2 seconds 20 m/s
3 seconds 30 m/s
FREELY FALLING BODIES
* Objects move vertically due to the action of gravity alone
* Objects thrown vertically upwards from the ground are considered
to move as freely falling because it is only acted upon by gravity
once released.
Assumptions:
1) Acceleration due to gravity is constant
g= 9.8 m/s2
g = 32 ft/s2
2)
Air resistance is negligible
Galileo
1600’s
 Studied how things
fell
 Didn’t have a
good clock
 Rolled balls down
an inclined plane
 Found that the
speed increased as
it rolled down the
ramp

Galileo
Acceleration= change in
velocity
time
t=0
t = 1 second
t = 2 seconds
t = 3 seconds
2-5. Free Fall
The acceleration of
gravity (g) for objects
in free fall at the
earth's surface is 9.8
m/s2.
Galileo found that all
things fall at the same
rate.
1) Objects thrown vertically upward will hit the
ground at the same speed that it left.
2) It will take exactly the same amount of time
going up as it does coming down.
3) At its peak height, its velocity is zero.
2-5. Free Fall
The rate of falling
increases by 9.8 m/s
every second.
Height = ½ gt2
For example:
½ (9.8 )12 = 4.9 m
½(9.8)22 = 19.6 m
½ (9.8)32 = 44.1 m
½ (9.8)42 = 78.4 m
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2-5. Free Fall
A ball thrown
horizontally
will fall at the
same rate as
a ball
dropped
directly.
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2-5. Free Fall
A ball thrown into the air
will slow down, stop, and
then begin to fall with the
acceleration due to
gravity. When it passes
the thrower, it will be
traveling at the same rate
at which it was thrown.
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2-5. Free Fall
An object thrown upward at an
angle to the ground follows a
curved path called a parabola.
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Introduction to Free Fall
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A free-falling object is an object which is falling
under the sole influence of gravity.
That is to say that any object which is moving and
being acted upon only be the force of gravity is said
to be "in a state of free fall."
This definition of free fall leads to two important
characteristics about a free-falling object:
– Free-falling objects do not encounter air
resistance.
– All free-falling objects (on Earth) accelerate
downwards at a rate of 9.8 m/s/s or 32ft/s/s
Freely Falling Body
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The acceleration of freely falling body is so
important that physicist called it
acceleration due to gravity.
Denoted by letter g which is equivalent to
32ft/sec2 or 9.8m/sec2.
Meaning in the 1st second, a falling body
accelerates from a stationary position to a
velocity of 9.8m/sec2, after 2 seconds, the
velocity is doubled to 19.6m/sec2 after 3
seconds it triples to 29.4m/sec2
Freely Falling Body
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Since accelerating objects are constantly
changing their velocity, you can say that
the distance traveled divided by the time
taken to travel that distance is not a
constant value.
A falling object for instance usually
accelerates as it falls.
The fact that the distance which the
object travels every interval of time is
increasing is a sure sign that the ball is
speeding up as it falls downward
A simple rule to bear in mind
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is that all objects (regardless of
their mass) experience the same
acceleration when in a state of
free fall.
When the only force is gravity,
the acceleration is the same value
for all objects.
On Earth, this acceleration value
is 9.8 m/s/s that it is given a
special name - the acceleration of
gravity - and a special symbol - g.
Newton

Newton's first law the law of interaction
states that for every
action there's an
equal and opposite
reaction.
Newton

Newton's second law the law of acceleration
states that the
acceleration of an object
is directly related to the
net force and inversely
related to its mass.
Newton’s Law of Acceleration

Fnet = m * a
• A=F/m


Acceleration depends upon two
factors:
force
and
mass.
The 10-kg elephant obviously has
more mass (or inertia). This
increased mass has an inverse
effect upon the elephant's
acceleration.
And thus, the direct effect of
greater force on the 10-kg
elephant is offset by the inverse
effect of the greater mass of the
10-kg elephant; and so each
object accelerates at the same rate
- approximately 10 m/s/s.
One Newton is defined as the amount of force
required to give a 1-kg mass an acceleration of 1
m/s/s.
Complete the table
Net Force
(N)
Mass
(kg)
Acceleration
(m/s/s)
1.
10
2
5 m/s/s
2.
20
2
10 m/s/s
3.
20
4
5 m/s/s
4.
10
2
5
5.
10
1
10
Example calculations for velocity

Calculate the
velocity of a freefalling object after
six, and eight
seconds
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vf = g t

Solution
At t = 6 s
vf = (10 m/s2) (6 s)
= 60 m/s
At t = 8 s
vf = (10 m/s2) (8 s)
= 80 m/s
Velocity of Freely Falling Body
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If you were to observe the
motion of a free-falling object
you would notice that the object
averages a velocity of
5 m/s in the first second,
15 m/s in the second second,
25 m/s in the third second,
35 m/s in the fourth second, etc.
Our free-falling object would be
accelerating at a constant rate.
Distance
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The distance which a free-falling
object has fallen from a position of
rest is also dependent upon the time
of fall.
This distance can be computed by
use of a formula; the distance fallen
after a time of t seconds is given by
the formula.
d = 0.5 g t2
where g is the acceleration of
gravity
Example calculations for distance
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
Calculate the distance
fallen by a free-falling
object after one, two
and five seconds
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d = 0.5gt2
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Example Calculations:
At t = 1 s
d = (0.5) (10 m/s2) (1 s)2
=5m
At t = 2 s
d = (0.5) (10 m/s2) (2 s)2
= 20 m
At t = 5 s
d = (0.5) (10 m/s2) (5 s)2
= 125 m
Distance

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Given these average velocity values
during each consecutive 1-second time
interval, the object falls:
– 5 meters in the first second,
– 15 meters in the second second (for a
total distance of 20 meters),
– 25 meters in the third second (for a
total distance of 45 meters),
– 35 meters in the fourth second (for a
total distance of 80 meters).
d=0.5(10g/s2 )(1s)2
v=0.5(10m/s2 )1s
Time
Interval
Average Velocity
During Time
Interval
v=0.5gt
d= 20 -5
Total Distance
Distance
Traveled from 0 s
Traveled During
to End of Time
Time Interval
Interval
d=dt -di
d=0.5gt2
0-1s
5 m/s
5m
5m
1-2s
15 m/s
15 m
20 m
2-3s
25 m/s
25 m
45 m
3-4s
35 m/s
35 m
80 m
The table illustrates that a free-falling object which is
accelerating at a constant rate will cover different
distances in each consecutive second.
Further analysis of the first and last columns of the
table above reveal that there is a square relationship
between the total distance traveled and the time of
travel for an object starting from rest and moving
with a constant acceleration.
For objects with a constant acceleration, the distance
of travel is directly proportional to the square of the
time of travel.
Equation for motion of an
object in free fall:
d
v
t
v  vo
g
t
v  vo  gt
v f  gt
d  v0t  0.5 gt
v  v  2 gd
2
2
0
 v0  v 
d 
t

 2 
d  0.5 gt
2
2
Another formula
S= -Vo
2g
S=Vot +1/2 gt2
Sample Problem
A
coin was dropped from the top
of the LTA building with a height
of 727 ft. If there is no air
resistance, how fast (ft/s) will the
coin be moving when it hits the
ground?
 215.68 ft/s
Solution A
d  0.5 gt
t  45.44
2
d  0.5(32 ft / s )t
2
t  6.74
2
727 ft  0.5(32 ft / s )t
2
(727 ft )
2
t 
2
16 ft / s
2
v  at
v  (32 ft / s )(6.74s )
2
v  215.68 ft / s
Problem 1

A marble is dropped from a bridge
and strikes the water in 5 seconds.
Calculate the speed with which it
strikes and the height of the
bridge.

(Vf = 49 m/s, d = 122.5 m)
Solution 1
v f  gt
v f  (9.8m / s )(5s )
2
v f  49m / s
d  0.5 gt
2
d  0.5(9.8m / s )(5s )
2
d  122.5m
2
P2


A feather is dropped on the moon
from a height of 1.40 meters. The
acceleration of gravity on the
moon is 1.67 m/s2. Determine the
time for the feather to fall to the
surface of the moon.
T = 1.29 sec
S2
Given :
v i  0m/s
d  1.40m
a  1.67m/s
t?
2
P3
 The
observation deck of the World
Trade Center is 420 m above the
street. Determine the time required
for a penny to free fall from the
deck to the street below.
 T = 9.26 sec
S3
Given
vi  0m / s
d  0.5gt
2
 420m  0.5(9.8m / s )t
2
 420m  (4.9m / s )t
d  420m
 420m
2

t
2
g  9.8m / s  4.9m / s 2
2
2
t ?
85.7 s  t
2
t  85.7 s  9.26s
2
2
2
P4
 With
what speed in miles/hr
(1
m/s = 2.23 mi/hr) must an object
be thrown to reach a height of 91.5
m (equivalent to one football
field)? Assume negligible air
resistance.
 V = 94.4 mi/hr
v  v  2 gd
2
f
P4
2
i
0m / s 
2
Given :
 v  2(9.8m / s )(91.5m)
2
i
2
g  9.8m / s 0  v  1793m / s
v f  om / s vi2  1793m 2 s 2
d  91.5m
vi  1793m / s  42.3m / s
vi  ?
2
t ?
2
i
2
2
(42.3m / s)( 2.23mi / hr )
vi 
1m / s
vi  94.4mi / hr
P5
A
10kg block being held at rest above the
ground is released. The block begins to
fall under only the effect of gravity. At
the instant that the block is 2.0 meters
above the ground, the speed of the block
is 2.5m/sec. The block was initially
released at a height of how many meters.
 D = 2.3 m
v  v  2 g (d  d 0 )
2
S5
2
0
v0  0
d 0  initialhei ght v  2 g (d  d 0 )
2
v0  0
d  2m
v  2.5m / s
m  10kg
g  9.8m / s
2
v
 d  d0
2g
2
0.5v
d0 
d
g
2
0.5(2.5m / s )
d0 
 2m
9.8m / s / s
d  2.3m
Assignment


1. Miguel drops a pile of roof shingles
from the top of a roof located 8.52 meters
above the ground. Determine the time
required for the shingles to reach the
ground.
2. Brandy throws his mother's crystal vase
vertically upwards with an initial velocity
of 26.2 m/s. Determine the height to which
the vase will rise above its initial height.
Assignment



3. A kangaroo is capable of jumping to a
height of 2.62 m. Determine the take-off
speed of the kangaroo.
4. A stone is dropped into a deep well and is
heard to hit the water 3.41 s after being
dropped. Determine the depth of the well.
5. Ronald McDonald is riding an Air Balloon
on his way to Subic. If Ronald free-falls for
2.6 second, what will be his final velocity and
how far will he fall?
2-7. First Law of Motion
The first law of
motion states: If
no net force
acts on it, an
object at rest
remains at rest
and an object in
motion remains
in motion at a
constant
velocity.
An object at rest will remain at rest unless
acted on by an unbalanced force. An
object in motion continues in motion with
the same speed and in the same direction
unless acted upon by an unbalanced
force. This law is often called
"the law of inertia".
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

This means that there is a
natural tendency of
objects to keep on doing
what they're doing. All
objects resist changes in
their state of motion. In
the absence of an
unbalanced force, an
object in motion will
maintain this state of
motion.
This law is the same reason
why you should always wear
your seatbelt.
FIRST LAW OF MOTION
An object will remain at rest or in uniform
motion in a straight line unless acted upon
by an external, unbalanced force.
 FORCE IS A PULL OR PUSH
Tug of war

2-9. Second Law of Motion
Newton's second law of
motion states: The net
force on an object
equals the product of
the mass and the
acceleration of the
object. The direction of
the force is the same as
that of the acceleration.
F = Ma
Second Law of Motion
Acceleration is produced when a force acts
on a mass. The greater the mass (of the
object being accelerated) the greater the
amount of force needed (to accelerate the
object).
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2-9. Second Law of Motion
A force is any
influence that
can cause an
object to be
accelerated.
The pound (lb)
is the unit of
force in the
British system of
measurement:
1 lb = 4.45 N (1
N = 0.225 lb)
1 newton  1 N  1 (kg)(m/s 2 )
Newton’s Second law of Motion
The greater the mass of an object, the greater
its inertia (mass is measure of inertia)the
greater its inertia is vice versa
Ex. Two different people on swings initially at
rest (an adult and a child)

This is an example of how Newton's Second Law works:
Mike's car, which weighs 1,000 kg, is out of
gas. Mike is trying to push the car to a gas
station, and he makes the car go 0.05 m/s/s.
Using Newton's Second Law, you can
compute how much force Mike is applying to
the car.

Everyone unconsciously
knows the Second Law.
Everyone knows that
heavier objects require
more force to move the
same distance as lighter
objects.
2-8. Mass
Inertia is the apparent
resistance an object offers to
any change in its state of
rest or motion.
2-10. Mass and Weight
• Weight
Definition: The force with which an
object is attracted by the earth’s
gravitational pull
• Example: A person weighing 160 lbs is being
pulled towards the earth with a force of 160 lbs
(712 N).
– Near the earth’s surface, weight and
mass are essentially the same
Weight  (mass)(acc eleration of gravity)
w  mg
2-11. Third Law of Motion
The third law of
motion states:
When one object
exerts a force on
a second object,
the second object
exerts an equal
force in the
opposite direction
on the first object.

When you kick the wall in your room, you
will probably end up hurting your foot.
Newton's Third Law of Motion can explain
why: when one object applies a force on a
second object, the second object applies a
force on the first that has an equal
magnitude but opposite direction. In other
words, when you kick the wall, the wall
kicks you back with equal force. As a result
you will get hurt. These forces are called
action-reaction forces
2-11. Third Law of Motion
Examples of the 3rd Law
2-12. Circular Motion
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2-12. Circular Motion
833 N is needed to make this turn.
If he goes too fast, which wheels are likely to
come off the ground first?
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2-13. Newton's Law of Gravity
Gm1m2
Gravitatio nal force  F 
2
R
G = 6.67 x 10-11
N•m/kg2