Download ÿþK i n e m a t i c s S o l u t i o n s

Name:
Unit 1: Introduction to Physics & One Dimensional Kinematics
Big Idea I: Objects and systems havc properties such as mass and chargc. Systems ma)' have
internal structure.
Essential Kuowledge I.A.!: A system is an object or a collection of
objccts. Objects are treated as having no intcrnal structure,
a. A collection of particles in which internal interactions change little
or not at all, or in which changes in these interactions
are irrelevant
thc question addresscd, can be treated as an obiect.
Hi!! Idea 3: The interactions of an obiect with other obiects
Essential Knowledge 3.A.I: An observer in a particular
reference frame can describe the motion of an object using
such quautitics as position, displacement, distance, velocity,
speed, and acceleration.
a. Displacement, velocity, and acceleration are all vector
quantities.
b. Displacement is change in position. Velocity is the
rate of change of position with time. Acceleration is the
rate of change of velocity with time. Changes in each
property are expressed by subtracting initial values from
final valucs
c. A choice of reference frame determines the direction
and the magnitude of each of these quantities.
to
can be described bv forces.
Learning Objective (3.A.I.1):
The student is able to express the motion
of an object using narrativc.
mathematical, and graphical
representations. [Science Practices 1.5,
2.1, and 2.21
Learning Objective (3.A.1.2):
The student is able to design an
experimental invcstigation of the motion
of an obiect. [Science Practice 4.21
Learning Ohjective (3.A.1.3):
Thc student is able to analyze
experimental data describing the motion
of an objcct and is able to express the
results of the analysis using narrative,
mathematical, and graphical
representations.
[Science Practice 5,11
Hi!! Idea 4: Interactions between systems C'lI1result in c1ulIl!!es in those systems.
Essential Knowledge 4.A.!: The linear
Learning Objective (4.A. 1.1):
The student is able to use representations of the centcr of
motion of a system can be described by
the displacement. velocity, and
mass of an isolated two-object system to analyze the
accelcration of its center of mass.
motion of the system qualitatively and semiquantitatively.
IScience Practices 1.2, 1.4, 2.3, and 6.41
Essential Knowledge 4.A.2: Thc
Lcaming Objective (4.A.2.!):
The student is able to make prcdictions abont the motion of
acceleration is equal to the rate of change
of velocity with time, and velocity is
a system based on the fi'Ct that acceleration is equal to thc
changc in velocity per unit time, and velocity is equal to the
equal to the rate of change of position
with time.
change in position per unit time. [Science Practice 6.4]
a. The acceleration of the center of mass
of a systcm is directly propol1ional to the
Learning Objective (4.A.2.3):
net force cxcrted on it by all objects
Thc student is able to crcate mathematical models and
interacting with the system and invcrsely
analyzc graphical relationships for acceleration, velocity,
proportional to the mass of the system.
and position of thc center of mass of a systcm and use them
to calculate properties of the motion of the center of mass
ofa svstem."rSciencc Practices 1.4 and 2.21
_
Proficicnt
Reading assignment
Read Chapter
Section 1.11.
1 and 2 then answer
Motion:
the following
questions:
A First look
Define the following terms:
motion diagramIn a motion diagram what do the following things indicate:
equally spaced images
___________
, increasing the distance between images
. ,decreasing
the distance between images
_
particle model -
Answer the "STOP TO THINK 1.1" question near the bottom of page 4 - now, stop and think about the answer to
this question, then flip to page 27 to check the correct answer!
Answer the "STOP TO THINK 1.2" question - make sure you take time to engage yourself in the ideas!
2.
Use the particle model to show a motion diagram of a car speeding up as it moves to from ieft to right across the
rectangle below.
Use 7 dots, making your diagram go all the way across the rectangle.
Section 1.2 - Position and Time: Putting Numbers on Nature
3.
Define the following terms:
position origin Displacement-
4.
Explain what is meant by the time referred to as t = 0:
S.
How is displacement calculated?
6.
STOP TO THINK 1.3 - Provide 2 examples that show why the correct answer to this question is C :
Section 1.3 - Velocity
7.
Define the following terms:
Uniform Motion Speed Velocity -
8.
How is speed different from velocity?
Reading assignment
Answer
the "STOP TO THINK 1.4" question
Section 1.4 - A Sense of Scale: Significant Figures, Scientific Notation, and Units
9.
Why do scientist use significant
10. What is precision?
figures (what does it tell them about a measurement)?
Why is it important
in taking data?
Answer the "STOP TO THINK 1.5" question on Page 16
Section 2.1- Describing Motion:
11. A motion diagram is a basic picture of motion, why is a position versus time graph not a picture of motion?
12. What does a steep slope on a position versus time graph mean about the objects motion?
13. How can a velocity vs. time graph be created from a position versus time graph?
14. How can displacement
be determined
from a velocity vs. time graph?
Answer the "STOP TO THINK 2.1" question on Page 33
Section 2.2 - Uniform Motion:
15. From a position vs. time graph how can you tell if the objects motion is uniform?
16. Why is the area under a velocity graph displacement?
Answer the "STOP TO THINK 2.2" question on Page 35
Section 2.3 - Instantaneous Velcoity:
17. Define instantaneous
18. How can instantaneous
19. When is instantaneous
velocity.
velocity be determined
from a graph of position versus time?
velocity the same as the average velocity of an object? When is it different?
20. Using conceptual example 2.4 on page 37, write a story about the motion another object (other than a
hockey player that exhibits the motion shown on the graph
Reading assignment
Answer the "STOP TO THINK 2.3" question on Page 38
Section 2.4 - Acceleration:
21. Define acceleration.
22. Using a motion diagram, how can determine
23. How can acceleration
be determined
24. What does the sign of acceleration
information
about the acceleration
of an object?
from a graph of position versus time?
mean in terms of whether
25. Can an object have negative acceleration
an object is speeding up or slowing down?
and be speeding up?
Answer the "STOP TO THINK 2.S" question on Page 41
Section 2.5 - Motion with Constant Acceleration:
26. What are the three constant acceleration
equations?
27. What is the significance of the Y, at' in the position equation?
Where does this come from?
Answer the "STOP TO THINK 2.6" question on Page 45
Section 2.6 - Solving One-Dimensional Motion Problems:
28. Describe the steps in problem solving strategy.
Read through example problems to see how the book using
the strategy.
Section 2.7 - Free Fall:
29. Define free fall
30. Explain why in the absence of air resistance, two objects of significantly
different
mass will hit the ground at
the same time
31. What is the value for the acceleration
due to gravity on the earth's surface?
the book says are worthy to note about free fall.
List some important
points that
Graphing Motion Homework Problems (answers on last page)
A. Description:
x
Two cars travel on the parallel lanes of
Car I
a two-lane road. The cars' motions are represented by
the position versus time graph shown in the figure.
Answer the questions using the times from the graph
indicated by letters.
1. At which of the times do the two cars pass each
other? How do you know?
t) ..,S~
p\o.( t. eM- 'l>~
~ rw
C.ar
2. Are the two cars traveling in the same direction
when they pass each other? Explain!
tJb, Cctr 2. 'M~ t--)
S\op( et (g.r
_1
1 \-ID.~
3. At which of the lettered times, if any, does car 111
momentarily
stop? How do you know?
:n. d~sn
'+, L~r '1
hoJ
vtloti~(-tv\''( e.V\\1rt
(OY\Sr-~-t.
4. At which of the lettered times, if any, does car 112momentarily
c..1W.("~ \')
~\Ot>-e lLt C
t'O
-\1(V'4 ')
stop? How do you know?
,,=O~)
lS\op!. i••~~r1) "'0
5. At which of the lettered times are the cars moving with nearly identical velocity? How do you know?
A- ~
o..n. (>o.xo-\\<,\
tIN')
s\
'il0 ~
B. You are running in a race (I'm
impressed!) and we have decided to
graph your movement.
e.o-r 1.~ 1. u.-a.
:l~S 6j
Displacement
1800
1600
running
L
What
time
'/-=2.
in those
first 90 seconds.
is your average
period
(a-90s)?
YSO-Q..
'to-o
velocity
"2
?C>
0 _10
_
Are your answers
different?
'"'"
4.
.;"'"
- '"
1000
~
800
o
,Ill:;'
H'
•• 0"
'1";":
~~',1'~t:.
)! Po-:
'I~
,~
"~
o..r
RI>"
!LiT
~,!,
'
"
i::'"::;
,\.,
;;:",:
:':;"
,'t! ;.::.,1-:. ~.. ;y,, ;~
'~'-
~-,
.
"~
:~r
'e~!;--:
o
f:~
m.
ii
'".
,,~.
;p.~ ~~:''',
60
.i"
,,!i.'
120
~.f;.~
180
1'7, -
1'":.>1
240
,.
h
v<,1t> cAhj
and now you're
out of breath!
./ .. / A
1z:
?~ '"
/
'/1.
:'1 ..... ;
I
'/
..
-
, >-, ~'., ,--
300
Time(s)
period?
]f
lc),
5
this time
,!",,;
"If, :'IF !~: • h :, ••••
'';=,.. 7~i~
...
',~ 1!;ii
:;;~
.. , i'.;~ ,.
~~
. rj~.'!:'
~ ~ 3!., "
;:"'1
600 Jr.1"
"";.(.~' :-"--w
.,'
:i::ij1~P,t~
.. ~i .. ~
~~;.;1-1
400
mj i:j~~;r,; .. , :.::
.•.. :;1~
:~~
o Z
0... (OV\& \-Wl t-
during
;;;;
:'~~; ,. ...
I':,ri
':'r..l
:'!~,'
~
--or,
~'".':i
';.,
::l~r, 'il'j;.
Are you accelerating?
What is your distance
~
,"Ii; il.
::.~,I]::i:
~"l
j:'!'
200
at 60
5.
Nt>.
90s to 1505 Yikes! You ran too fast at the start
6.
E
,~
to 1 and 2 the same or
Why?
f'l\ov \ Y\ ~
~
1200
.!!?
velocity
_
1400
'"
a.
oS
. V ~ -\#0- 0
3.
this
=tC;~ j
What is your instantaneous
seconds?
during
E
:=
c:
vs Time
{!;I
:I;; :,!' ,. , .".,
I\'
., . -, i;~~ '"i' .,.~', "
.~
.
// /'
;;,~:~yii1i7
., , f" "n
~r~
'tn<+
.
.
..
"....:.
,
,
? -,
;:~:'Wi ~~''I, ,",' ;;;;1 ':,~' ;,," ," '/ .~
-;v::
~:.~~ffi~
(:~
~~'
Zero to 90s: Let's look at how you are
~,S(1...l"Y'II..
360
420
480
54'
Graphing Motion Homework Problems (answers on last page)
What is your velocity during this time period?
7.
O~~
What happened?
8.
o\-<rp~,
150s toQ:
By looking at the graph are you moving lfaster
9.
or __
l';)\-~
10. What is your velocity during this time peri~d?
V
-=
2405 to 300s
\0100- 4&St> ::
~()O
- ~
'2..110 - 1St)
OJt>
-
'J-=
~C5c- \0$'0
~~ - '2.&.tt:>
:.
-_'2.00
L'
S\~~
J
l# ~
@i1
-:..o~
What is your velocity in this period?
11.
slower than you did in the first 90 seconds?
~
-:
CD C
12. What does Uils mean (which way are you going.)
bo..tf
~()..f"ds
Ceo
Of"
Or'ijh'''(Mt>,jf\~
o\\(t~M.),
~~Jt
\'\
3005 to 360s
13. How would you describe your motion during this time interval?
V\0t- Movi~lStor~).
'1>\~ '"D.
3605 to 5105: You know that you have only one chance to still win the race... run as fast as you can!
flO. 3 ~ J
'l_
@
-= =,=,0 ~ 11.~~
14. What is your average velocity during this time period?
V~
15.
\~OO-\000,
SLfo -
-=
'b'1o
(Y)
a. What is your instantaneous velocity at 420 s? b. What is your instantaneous velocity at 480 s?
000
lqc!)O
.\f -= 1
(*1.\1.0) . 42e - .Bo
Cio
'-J ::.
11,,00-
(o..\"""4~'.> S\O
-
(000
~,\D
_ \000
I2-D
16. What do the differences in 14 & 15 represent? (what does a curved line on a position vs time graph mean?)
\H\o(,i~ is if\cr.faSw4' We
(}x..(
sp-uc;l;J
lAf'
Now Construct a Velocity-Time (v-t) Graph for your motion
You need to remember that the rules for dot graphs do not apply to v-t graphs. A common mistake is to assume
that all three types of graphs work the exact same way. The graphs can be related to each other, but that
doesn't mean you look at them the same way.
Note: A straight sloped line on a v-t graph means acceleration.
The slope of the line is ~qual to the
acceleration; a positive slope is a positive acceleration, and a negative slope is a negative acceleration.
There is one other trick you need to know about v-t graphs.
If you multiply velocity by time, what do you get? Displacement, since d
•
= vt
!
So, if I have a v-t graph and J calculate the area under the line (which means I'm calculating velocity X
time). I will know how far the object has gone.
Graphing Motion Homework Problems (answers on last page)
17. Graph your velocity vs. time on the grid below. label your axis with proper units and use an appropriate
scale for both time and velo~ity.
:
/
rT
/
..
-
,
if
I
I
1-
s
-f--C -
- 1-- .-
-- -
- - I-
-
I
-
--
I
J
I
/
,
I
0
60
120
180 2\10
3
360 420 4~O 540
I
-r
1-'---
1-
-1=
-t-
f-
-s
I
--
:
I
,
18. In the table below properly identify the type of velocity you exhibited during the race
Time interval
Canstant ar Changing
+, -, zero
Average Velacity (m/s)
Ave. Acceleratian (m/s')
Zero to 90s
eMS\-o.nJr
-r
90s ta 1505
(J>\I\~to.nt
t~rb
150 ta 2405
toy\"oTvV'~
-t
240 ta 3005
c..t)Y'\S>~ ~
300 ta 3605
Ct> n.'t.\o..v\k"
re~t>
360 ta 5105
c...,",,-~i~
-t
""5
0
0
+<1.(,
0
0
-~.?;
0
0
0
S:~
..•
+ .0\115
AV
()..---
6\-
n"\
~1.
Graphing Motion - Velocity vs. Time graph (answers on last page)
SHOW All WORK IN SPACESBelOW
Modified from J. Kova/cin 9/17/01
C. The following graph describes the velocity
of an automobile as a function of time
30
~J
f-\(J!
'il
1. What was the velocity of this c
t = 3S SeCOndS?~
-::
2. During which time interval/intervals
car at rest? How do you know?
eM \ S ~
~
~
6
was the
~
Pt~
r.t!.\-
\It \ 0 c.i hj i~ ~~
()J.;\
4. What was the displacement
t>.rto..
s
o rY\
(Mt
T
~
0
III "OJ
At> -=
"'
\00;-1'5t>
blC
35
-\
50
-t?,
60
0
90
-.1$
0
,I
I
I
I
,
\
I
"
U III
I
.>
I
~
~
~
~
W ~ W
T!MEfscc)
~
(\0 Y-'n)
",.
~
_
100 II01W
-
-l( \0 i- L.D] = it~.o \)1 ~\\Ob~\
vt\oci ~
is l:,-trl».
of this car between t = 25 and t = 40 seconds?
-t
(-'2.1..<;)
of this car between t
= \"-1\ '2..5
~cl4l
+ Lt'50 +
1<100+11'2...5
yY\
1
= 0 and t = 110 seconds?
t iY\(.\v..O.~
10r
OfI.()J
-l'
I
4 - ').
=1+\I5Dm \
= 0 and t = 110 seconds? (is it different
(~-t
<: -)
from 117 and why?)
+~,.~
=~
of the car and graph
scale the gra~:
()\~fe at
-'2.
I
of this car between t = 10 and t = 25 seconds?
the data making sure to appropriately
Time
Acceleration
(s)
(m/s')
S
20
~
I
10
e.~ 'i,..Q5~
9. Calculate the acceleration
I
I
~
I
\'So •. l.t'SO + 11;00- \\'2..; - 1>"1,1;
~.
I
,/1
-10
8. What was the total distance traveled by this car between t
A\l
,
-
V
of this car between t = 0 and t = 10 seconds?
7. What was the total displacement
+\00
10
~
I
L'oy.n') ': ~ ( V~,C\?)=
A'('flo..~ctw
.
't+J(
'),0'
"eo
YV\.ovi""3
6. What was the displacement
At.-=: ~
II
'ptLCUA.-I. ~
\).W\6-~ -\oWO.~N~:
5. What was the displacement
20
I
•••• ...,
-30
3. During which interval/intervals
was the car
moving in reverse? How do you know?
2" ~ q 1.\;
1-'
-20
\Q~ ~'2.SS{'C-tro).
'c7e\-V"-t.(,n
I
I
'I
b-? P ::-lID
v~.l)
I
I
3
I
~
:!\
,....~(
~. 0
t
I
'"
'1.
~~
II)
20
30
4IJ
~
60 ~
IlO 90
lJME(=J,
100 110 120
1-0 Problem Solving
"0-= o~
Levell problems:
1. A car is at rest on a horizontal surface. The accelerator is applied and the car accelerates at 3.00 m/s':
a. What;iiibe
the speed of this car after 6.50 seconds?
,,-=
(1::::y~...
;-~--l
=~~~-.
1" (j,) lID.'!»
b. What will be the average speed of this car during thes
\:;-0.(,.5$
VA.'"'::
\I-:;?
c.
t,=Vin s
q.i'it
e"S
-=
(tj;:: .,.1S)
q.1S ~
:>.
i.
-t
g",~'2..
-tC~)("',s)'Z..:- tl3.LtM.
A ball is dropped from the top of a building to the ground below. It takes 2.87 seconds for this ball to
reach the ground;
a. What will be the velocity of this ball as it reaches the ground?
V,::
=
~o=0 ~
3
(,."3.'"\-0
0 seconds?
How far will this car move d ring these 6.50 seconds?
)(~
2.
=-
-',ttl
tf-b
)if
'/..-::'Jf -to ~
13 -:
"= t>
\10 l' q.. ~
v. ~ tj-i:
U
:ti~.
t~.
\~J
~t.
=
~
r;~
~q.~)(1.."i')
b. What is the average velocity of the I:lal as It ralls
V,."( ""-2 S. ~ c.
0
-:
Of"
0
he ground?
l-/O.l/
V4)/(=
How tall is the building?
h= iC3t2.
=
-0 :~
1..~1
~
1
.J. (1.~){'2.~1)'2.
:> \40 .•.•
1\"\
m)
'l.
l.
_
.....
. __---1
3. A ball is thrown upward so that it just barely reac es the top of a telephone pole and then falls back to
6~-::;.2 the ground. The time from the release of the ball until its return to the ground is measured to be 5.20
~ -: ". ~~
-0
'"
2.l,., 5
':0
•
I
XD+ Vo\:
'::' tI ~
'Z.
-to 2.~,.
0 .• i. ('I.V)
'7.. 1#)
:. r3~.\
~
~.g'~...
V2.oe
t 21A./)'I-
\/0'2.
5.
e
-:.
(J'Ot .• 2("l.~(6)t.)
V'L::. I'iLlI.J
.
oJ
2~'11.2.
-::)
f-:.
38. m/s.
\V=
~W3C;.1.
0
f\'"\
4. A ball is thrown downward from the top of a building 122 meters tall with an initial speed of
W\ What will be the velocity of this ball as it reaches the ground?
.~'IC~-::\'2.2.ltliVeI2 problems:
_
)l
'Z.
'6
\.10:
t"'ll""W' -:: ~
~
V,;?
'/.. -
seconds. What is the height of the telephone pole?
".:. 0
]
v\."7 .
~----
Suppose that a car is moving with a speed of 18.5 m/s when the brakes are applied so as to slow the car
down at a rate of -2.85 m/s';
to
a. What will be the speed of this car 3.55 seconds after the brakes are a . d?
Vo- \~.S 5
4\.:-Z.?~~
b.
=
V-= Vo-t'Clt
~~
Ig.S t(-2..~SJ(3.S5) ': ~.J.ttY'o/S
far will this car move during this 3.55 second period?
b~ -=
)(0
l--+1.. ~
+VO-\:-t
0 4{\'l.,)('3~'5)+ iG'l.'~I:;)(3.~S)
c.bSJolt\I !DOg will it take for this car to stoP'-"
V =Vo + 11\..lc
d.
H~
-:!)
':l.
W~
0 '= \<l.S~ (-'2..75)~
V':: 0
u=:
\
- 1:l.L..1l'Y' -
':!J
1;=
-I~.S :-2..••.
S't:
_
-1'1'.!O -:~.5'sJ
-'2..,••••
will this car move from the time that the brakes are applied until the car comes to a stop?
-
6 y.. -:::. ~
\
t
V"
tT
01
~ 'l.-
"Z.
0.-
':a..
-:::C!~.Gx.c..S)of i(-2.'iS)(~.5)
-::: \"2.0.'2.')
+
-l,pO.'2.
---..
\b)C. -:.1ot)l'V\
\
1. .,7Vo of:
V::
0
1,Q.bY.
+ 1.[-1.. , ••) A)(.
;,r&. ~'.f
l'll.1.~ = b)(.
c,),
-:
lo~
1-D Problem Solving
- l&.f4tf
_)(
-''l.tt
-
'l'f- -=
-'2.•.•0
.f
1?l# 1
+'2.Yc.-=f3"\~.lI"Y'\J
1~.111
1-0 Problem Solving
Suppose that instead of throwing this ball upward it is thrown downward with a speed of 38.0 m/s;
g. How long will it take for the ball to reach the ound?
t:: 4.IS
'1.-=)!.""dt-~~~t'L
O~-z.4.""~f1t1".I.~V)t"Z.
h. What will be the speed oTthe ball as it reaches the ground?
\I( "1.=
"f>.•..to ~o..6'f.
-= ~~
-tU1.i)(Zqo) :: ~l-~"'~
J
9. A car is initially traveling with a constant ve oClty of 12 m/s as it moves down the highway. The car travels
_ \'J,.~a distance of 175 m at this velocity.
~How long will it take the carto travel th~75 m while moving at a constant 12 m/s?
0..~O~
ble
"0' .,
'#.~\"~
'It,:
~:--tl) ").
~o .•..vo"", ~
111'-=O+ll'2)t ..•.~D1:
!J!::-t
1"2.
:tB.(,~J
('O:6ro..-.t-ve\otilv.
'J
--
b. After moving at lL m/s for the time found in part a, the driver pushes her foot down farther on the gas
~edal, accelerating at 1.5 mis' for 10 seconds.
i. How fast was the car moving after 10 seco
a elerating?
.•.V; 1.'1(). ~
" -:: VD +
•.•
~~\'l
o.."'c
,,::
'; l'2.t(I.~)l\o)
2.1
t!)
~
ii. How far dia the'i:ar move during this
)(:: ~D+'JOt+:!z...~'2.
acceleration?
1"
r)C- l'\S
':l.
l -
-:=o+I\'2.\t\o\+~(\oS)lIO)
fY'\.
"
c. Just as the driver r~aches the spela found in part c. above, she notices an accident has occurred farther
down the road and she applies the brakes, coming to a compl~p
after traveling an additional 50
meters.
i. What was the acceler~ion of the car while stopping?
Ol.
X \lole -+ ~~ 'to
'.J1--; "'f;)'1.-., tp..l:.)L
/
J,00
-= }-z.1
-=
;0
~h..,)(-t)+=io..-t'Z
tl'\O''t-~
1> ~ 2"1'1..
ii. How I~ng diOit take fOr the car to stop?
~O
=
'2.1 (~)
+:l (-1"'1')~-z.
Of"
V
e Vo
-t
'ii
I'" "'-'.'1-'1 l!':loz.]
~
2,.0..(150)
i-Q..t
'= 2.' •.t-,,7.."I)~
I
2.1
.: 2; ;" t
=
d. Make accurate graphs of position vs. time. and velocity vs. time on the axes below shliwlhg the motion of
this car during this entire problem.
V (mls)
X(m)
500
30
400
24
300
18
200
12
100
6
t (s)
t (5)
o
5
10
15
20
25
30
o
5
10
15
20
25
30
I
"3.1$.
l-D Problem Solving
_ \G.~
~,
St.
OJ
-\:: ~ ~~
10. Santa's elves are building an experimental rocket powered sled. The rocket sled is mounted on wheels
and placed on a set of straight railroad tracks. At t=o the rbcket is ignited, causing a constant acceleration of
15m/s2 until t= 8s. From t= 8s to t= 20s the rocket travels at a constant velocity. At t= 20s the engine is shut
down and the brakes are applied.
a. What is the velocity of the sled at t=8s?
\
.'I ':.Vo ; tt.-t
':. 0 -to \st!)
':: 1'2.0 '!l
~
b. How far does the sled travel durin
x= Vot .• to\t~
= ~ (15) (I)....
seconds?
-= {lifO rn \
fA":" 0
c. How far does the sled travel from t=8s to t=20s?
~-z.o-'t
\:. 0;.
\'l-
lC -; Vo ~
V, t
=
-t
io..t'2.
""~
X -::G'2.0')l \'2.)
CAV\.\~+
1/
-=0"1.\0'"" ]
-t C>
.
.
d. After the brakgs are applied at t=20s the sled continues to move forward for 680 m. Calculate the
acceleration while the brakes are applied.
_ -1'-1,#0 D
X 0:: 'lot"," 'io....'c 1.
(f\JI..ll.. '"
'11.-;. "D7-.•.?-tA.b y.
0'('"
MJ. ')
0
fA. - i((,?D
= Q1.0)~ i' '2. 0..( CIVt»
Q.
-::-10.V?
e. Calculate how long it takes the sled to come to a stop after the brakes are a
V,. = V" -t O.:t
t- -;.-=-10.\1
''1b
of"
-=
C ~ \ '2..0 "" (-Ib.~)-\:.
f. Graph the motion of the sled from t=Os until it comes to a stop.
140 m/s
?500rll ,. -
,120m/s
2000m
100 m/s
1~OOrn- ..
1000m
80 n\ls
, 60 m's
.
40
m's
•
20 m's
.--t-~t--t-t--,t-+--i,-4
v
4
8
..
-t,-+-+,.
12 lG 20 24 .78 32 30 40 ,.,
I
-I-f--t-I-t---i-l
4__ 8 _'2
_
Ie
20
'4
28
0'
..
32
3Ii 40
44
Answers to Graphing Problems and 1-0 kinematics Problems
A. (two cars) 1) D (when positions are the same), 2) No (car 2 is going in the reverse direction),
3) Never (constant slope
means constant velocity), 4) C (zero slope means zero velocity), 5) A (slopes are identical)
~t"cf>fHtl
B. (you running a race) 1) 5 mis, 2) 5 mis, 3) slope is constant so velocity is constant, 4) No,S) 0, 6) 0, 7) Stopped 8) ~
9) 6.7 mis, 10) -3.3 mis, 11) backwards (maybe your hat blew off and you ran to get it), 12) at rest, 13) 5.3 mis, 14) a) -4.2
mis, b) - ~m/s,
15) acceleration, 16)
17) 0-90: canst., +, 5 mis, a mis'
a
90-150: const, 0, 0,
150-240: const., +, 6.7 mis, a
240 - 300: canst., -, -3.3 mis,
mis'
a mis'
300-360: canst, 0,0,0
mis, 0.07 mis'
360-510: changing, +, 5.3
C. 1) -10m/s 2)10-25, 45, 110- velocity is zero 3) 25-45s velocity is negative 4) 100m 5) 0, 6) -112.5m 7) 1150m 8) 1450m
9) -2, 0, -1.0, 3, 0, -0.75 mis'
, ,
,
, ,, ,
, , ,
I
,
,
,
J
,,
, ,
I
,
10
,
,
,
,, ,
,
, ,
,
I
,
~
,
I
,
,
,
I
I
,
,
,
,
, ,,
,
,
I
,
,
, ,
,
,
,
,
,
, , ,
I
, ,, ,
,
,, ,
,,,
, ,,
,, ,
,
,
, ,,,, ,, ,
, ,
,
I
I
I
I
I
w ~ ~ ~ ~ m ~
I
100 IW
I~
l1ME (KG)
1-D Kinematic
1 a) 19.5
Problems:
m/s b) 9.75 m/s c) 63.4 m
2 a) -28
m/s b)-14.1 m/s c) 40.4 m
3) 33.1 m
4) -62 m/s
5 a) 8.38 m/s b) 47.7 m c) 6.5 s d) 60 m
6 a)
a b) -52
m/s c) 5.3s d) 138 m e) 14.3 m/s f) 33.2 m/s
7) same at -59 m/s
8 a)
a b) 3.9 s c) 11.7 s d) 7 & 0.7
9) a) 14.6 s bi) 27
10) a) 120
s e) -77 m/s f) 314m g) 4.1s h) -78.4
m/s
m/s bil) 195 m ci) -7.29 m/s2 cii) 3.7 s d)
."".~ f'
\.
11)-,--.•
mis, b) 480 m, c) 1440 m, d) -10.6 m/s2 e) 11.33 s f)
"'~J '. -
..
i
I''''''
It " •• _.
'."'.
"
z; rr.t
"/
\
/
I
..
/
to)
',-,
"I
-,D.C.
7~
,..-
.
l' ~~ JOI