Kinematics - Upper Sandusky High School

doutfanaticalΜηχανική

14 Νοε 2013 (πριν από 3 χρόνια και 11 μήνες)

103 εμφανίσεις

Kinematics

BRANCH OF SCIENCE WHICH

STUDIES MOTION

Motion


Continuous change in position relative to a frame of
reference


Frame of Reference


Objects that are known to be stationary from past
experience

2 types of Motion


Constant motion


Described by speed and direction (velocity)


Change in motion


Acceleration


Speed


Instantaneous speed


Actual speed at any given moment


Average Speed


Cars speed up and slow down all the time

V =

s



t

V = average speed

s = total distance traveled

t = time it took

Quick Fact


May 12, 2006
-

Justin Gatlin ran a 100 m in 9.76 s.



V = s/t


100 m / 9.76 s = 10.2 m/s



At this speed Justin covers 10.2 m per second.


But because he starts from rest and accelerates up to
speed, his top speed is more than this


about 10%
more.

Historic 100 M Records

"A race is won not with the feet, but with the heart"

Another quick fact


Downhill skiers attain speeds of 70
-
80 mph on winding
runs inclined about 10
-
15 degrees.


A speed of 70 is 102.7 ft/s which means they cover 10.3 ft in
0.1 s.


Quicker skiers on 50 degree slopes can reach 139 mph or
204 ft/s.



At this speed a skier could cover the length of a football field
in 1.5 s.


This is faster than a skydiver falling in a spread
-
eagle position.

Example of
-

Speed


A car travels a distance of 450 km during a 10 hr period.
What is it’s average speed in m/s?

V =

s



t

V =

450km

= 45
km




10 hr hr

45 km

1 hr

x

1,000 m

1 km

= 12.5
m


s

x

1 hr

3,600 sec

v = s / t

s = vt

Constant Motion


Traveled over time

0 min 15min

30min

45 min

60 min

75 min

40 km

hr

40 km

hr

40 km

hr

40 km

hr

40 km

hr

40 km

hr

Here we have constant speed


no acceleration occurred

Graphing Constant motion


Linear relationship
-

straight line graph


Generally described by the formula y=kx


Where k is a constant

From pt A to pt B

You have 2 test grades.

80% on one and 90% on another

What is your average test grade?


You would add them and divide by two.

Change in motion


Average speed


The average of ALL the speed data


OR use distance traveled and time traveled

0 min 15min

30min

45 min

60 min

75 min

40 km

hr

60 km

hr

80 km

hr

100 km

hr

0 km

hr

65 km

hr

V =

Vo + Vf




2

v = Average speed

Vo = initial velocity

Vf = final velocity

Velocity


Represents


Speed with direction



If an airplane travels 500 km due north in 1 hr., what is its
velocity?


250 km due north in ½ hr?


125 km due north in ¼ hr?



All the same


500 km/hr

Example of
-

Velocity


d = 94 km, t = 1.5 hr, v = ? m/sec

V =

s



t

V =

94 km = 62.7 km



1.5 hr hr

62.7 km

1 hr

x

1 hr

60 min

= 17.4
m


s

x

1 min

60 sec

x

1,000 m


1 km

End for Velocity

Answer Problem Set #1

Change in motion


Acceleration


ANY change in motion


Positive acceleration


increase in speed


Negative acceleration


decrease in speed


Use positive or negative #’s to represent the direction of change

a = V
f



V
o





t

V
f

= Final Velocity

V
0

= Initial Velocity

t = Time

a =

V



t

Change in Motion


Uniform Acceleration

0 min 1 sec

2 sec

3 sec

4 sec


5 sec

44 m

s

48 m

s

52 m

s

56 m

s

60 m

s

64 m

s

a = V
f



V
o



t

Example of
-

acceleration


t = 11 sec


Initial velocity = 44 m/sec


Final velocity = 88 m/s


a = ?

a = V
f



V
o





t

a = 88
m

-

44
m


s s





11 sec


44
m


s

=



11 sec


44 m

x
1

1 sec


11 sec


= 4 m


sec
2

End for Acceleration

Answer Problem Set #2

Final velocity with uniform acceleration


Lets first manipulate the equation


V
0

+ at = V
f


a = V
f



V
o



t

t

1

t

1

at = Vf
-

Vo

+ Vo

Vo +

Example of
-

Final Velocity


A ball rolling down a hill undergoes uniform acceleration of 4m/sec
2

for 5 sec.
If the ball has an initial speed of 2 m/sec, what is it’s final speed.

V
f

= V
0

+ at


= 2
m
/
sec

+ (4
m
/
sec
2

x 5
sec
)

Final Velocity = 22
m
/
sec


Identify variables:

V
f
=



t =

V
o

=



a =

Formula

?



5 sec

2 m/s



4 m/sec
2


V
f

= V
o

+ at

End Final Velocity

Answer Problem Set #3

Distance traveled during uniform acceleration


Here we find the middle speed of the car by adding
the initial and final speed and dividing by 2

0 min 1 sec

2 sec

3 sec

4 sec


5 sec

44 m

s

48 m

s

52 m

s

56 m

s

60 m

s

64 m

s

V =
Vo + Vf


2

Distance traveled during uniform acceleration


The key is uniform acceleration.


Let’s first work with a few formula manipulations.

V =
Vo + Vf


2

Remember


V is average velocity

V =
s


t

s

= V

t

V =
Vo + Vf


2

s

=
Vo + Vf


t 2

t

1

t

1


s =
Vo + Vf

t


2

Example of Distance traveled
-

Uniform acceleration


What distance is traveled by a train as it is accelerated
uniformly from 22 m/sec to 44 m/sec in a 20 sec period?


s =
Vo + Vf

t


2

22 m/sec + 44 m/sec

x 20 sec


2

s = 660 m

Identify Variables:

Vf =

Vo =

t =

Formula =

44 m/sec

22 m/sec

20 sec


End Distance Traveled

Answer Problem Set #4

What if?


You are given: acceleration and time and are asked
to solve for distance.


Can we solve for distance with the formulas we have so far?



You are given: acceleration and final velocity and are
asked to solve for distance


Can we solve for distance with the formulas we have so far?

NOT YET

But you have the means to

What needs to be done then?


We need to combine formulas to solve for the
desired variable.



If we look at the following numbers.

5 + 3

= 4


2

4 =

8 x 2


4

THEN

5 + 3

=
8 x 2


2 4


s = V
o

+ V
f

t


2

Uniform Acceleration


Starting & Stopping


If you are given acceleration, time, and starting from rest.

s =
at
2


2

If initial velocity = 0

Then V
f

= V
o

+ at Vf =
0
m
/
s

+ at



Becomes V
f

= at


s =
0
m
/
s
+V
f

t


2


s = V
f


t


2


s =
V
f

t


2

V
f
=


s = at


t


2

Can be rewritten


s = ½ at
2

Example of
-

Starting from rest


A car starting from rest is accelerated at a constant rate of
6.2 m/s
2
. What distance does the car travel during the first
7 sec of acceleration?

s =
1
/
2
at
2

S =
(6.2
m
/s
2
)(7 sec)
2


2

S = 151.9 m

Identify Variables:

a =

t =

s =

Formula =

6.2
m
/
sec
2

7 sec

?

Solve for final velocity when given a and s

V
f

=V
o

+at and s =
V
f

+ V
o

t


2

Solving for t


V
f


= t


a

If Vo is zero


V
f

= at and s =
V
f

t


2

Now we can

combine

V
f

=
2s

a V
f

Solved for V
f

V
f

= 2as

When initial is 0

V
f
2

= 2as



t =
2s


V
f

When final is 0


V
o

= 2as

How did we get that.

V
f

=V
o

+at and s =
V
f

+ V
o

t


2

Solving for t


V
f


= t


a

If Vo is zero


V
f

= at and s =
V
f

t


2

Now we can

combine

V
f

=
2s

a V
f

Solved for V
f

V
f

= 2as

When initial is 0

V
f

1

V
f

1

a

1

a

1

V
f
2

= 2as



2


1

1 Vf

2


1

1 V
f

1

a

1

a

t =
2s


V
f

When final is 0


V
o

= 2as

Example of
-

deceleration


An airplane flying at 63 m/s lands on a runway and travels
1,000 m before stopping. At what rate is the plane
decelerated?

When final is 0


V
o
2

= 2as

a = V
o
2



2s

a = (63 m/s)
2



2 x 1,000 m

a = 2.0
m
/
sec
2

Identify variables:

Vo =

a =

s =

Formula =

63 m/s

?

1,000 m

1

2s

1

2s

Graphing Uniform Acceleration


Quadratic relationship


A parabola


y varies with the x
2


y = kx
2





A hyperbola


Indirect or inverse


relationship


As x increases, y decreases


Y = k/x

End Starting / Stopping at Zero

Answer Problem Set #5 & 6

Acceleration due to gravity


Neglecting any friction due to air


All free falling objects gain speed toward the Earth at the same
rate


Acceleration due to gravity = 9.8 m/s
2

V
f

= V
o

+ gt and V
f
2

= 2gs

Simply Put


As an object fall


Every second it falls the object’s speed increases by 9.8 m/s

Example of


Acceleration due to gravity


What is the speed (in m/s) of a brick that drops from a high
scaffold after 4 sec of free fall? How far does the brick fall
during the first 4 sec? (brick begins at rest)

V
f

= V
o

+ gt and V
f
2

= 2gs

V
f

= 0
m
/
sec

+ 9.8
m
/
sec
2

x 4sec


= 39
m
/
sec

s =
1
/
2
gt
2



s = 9.8 m/sec
2

(4sec)
2


2

How far will she fall


How far will a freely falling object that is released
from rest, fall in 2 sec?, 10 sec., 20 sec.?

S = ½ gt
2

S =
10 m/s
2
(2sec)
2


2 = 20 m

S =
10 m/s
2
(10sec)
2


2 = 500 m

Example of


Deceleration of gravity


Calculate the time it takes an object, starting from rest, to
reach a speed of 98 m/s when it falls freely. Using V
f
= gt

t = 98
m
/
s


9.8
m
/
s
2

t = V
f


g

t = 10 sec

Since it takes 10 sec to bring the ball

to a halt it will take another 10 sec

to bring it back to the Earth for
a total of

20 seconds.

End Acceleration due to Gravity

& End of Kinematics unit

Answer Problem Set #7