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Unformatted text preview: If you'd like for your Physics
professor to be in a good mood...  You might choose to vote for his daughter
Aylia Colwell to be cast in the web series
"Whatever" at:  https:/lwww.facebook.com/questions!
1 79330425473271! 0 AND at:
 http://www.whatevertheseries.com/nola.php 0 You might even choose to share this with
your friends. 0 Voting ends this weekend. Just saying... August 25 Finish off chapter 1 vector components and
some more examples. Start chapter 2. A demonstration.
First clicker questions Reminders: — Read the syllabus (available online) and sign and
return your clicker agreements.  Register your iCIickers when you can (you still get
points. even if you have not yet registered your
clicker). — Homework 1 is available on WebAssign. due one
week from tonight. 1.6 VectorAddmon and Subtraction Finishing up vectOrs... When a morons hummer!
by ~1. the magnitude olthe . vector remems me same. but
the dream of the vectons
reversed. Subtracting a vector is just
adding the negative of that
vector. The negative of a
vector is just the vector
pointing in the opposite
direction. 1.7 The Components of e Vector Finish Start i and;l are the x and y components of the vectori‘.
They are 'components' because they are
perpendicular. There is one component for each
dimension of the problem. 1.7 The Components of e Vector It IS often easier to work with the scalar components
rather than the vector components. A. and A, are the scalar
components of the vector A. We use 'unit vectors" in the x and y
directions with length 1 so that
A: A! +A,y The 'hat“ denotes unit vectors. 1.7 The Components of e Vector Vector and Vector Component Summary  Any vector can be expressed as the sum
of two perpendicular vectors, called the
components of that vector. 0 For twodimensional problems. we can
have those components be along the x
and y axes of our coordinate system. 0 This reduces the vector to a right triangle,
which we can handle mathematically with
our trigonometry rules and the
Pythagorean theorem. a. 1.7 The Components on Vector
Example A displacement vect0r has a magnitude of 175 m and pomts at
an angle of 50.0 degrees relative to the x axis. Find the x and y
components of this vector. sine  Jgfr y  rsinB  (175 mxsin 50.0" ) 134 m ~‘ c058  .1',."r
x  rcosB  (I75 mXcos 50.0“) I l2 m i" = (112m)i + (134m)? 1.8 Addmon of Vector: by Mans 01 Components 0 It 1.8 Addmon of Vector: by Mans 01 Components 0 .1 ,‘ C: Ali+Aj+Bli+Bj
(3:91, +B,)i+(A, +B:,)y C,  .4‘ + B, (1  .4, + B, Chapter 2 Kinematics in One Dimension Kinematics deals with the concepts that
are needed to describe motion. Dynamics deals with the effect that forces
have on motion. Together, kinematics and dynamics form
the branch of physics known as Mechanics. 2.1 Displacement
Displacement is a vector mth Origin dimensions of length. in, l {’0 Displacement * Ai
ﬁo_—> i
— i0 = initial position = ﬁnal position i
A}? = i — i0 = displacement 2.1 Displacement 2.1 Displacement io=—2.0mi 52:50:11}? # Ai=7.0mi Ai=i—i0 =5.0mi—(—2.0)mi=7.0mi 2.2 Speed and Velocity Average speed is the distance traveled divided by the time
required to cover the distance. ‘ Distance
Average speed  ‘—,
Elapsed time SI units for speed: meters per second (m/s) These quantities are all scalars. 2.2 Speed and Velocity Example 1 Distance Run by a Jogger How far does a jogger run in 1.5 hours (5400 s) it his
average speed is 2.22 mls? Average 5 I Distance
Elapsed time Distance  (Average speedXElapsed time)
x (2.22 m/s)(5400 s) 12000 m 2.2 Speed and Velocity Average velocity is the displacement dwided by the elapsed time. Because displacement is a vector. average velocity is
also a vector. Displacement Average velocity  .
Elapsed time x—x0 Aii v=—=—
(10 At 2.2 Speed and Velocity Example 2 The World‘s Feeteet JetEngine Cu Andy Green in the car rhmsISSC set a world record 01
341.1mls m1997. To establish such a record. the diver
nukes two am through the course. one in each direction. to nullify m effects From the date. determine the average
veloaty lor each run. 2.2 Speed and Velocity q=£=ﬂ=+3395mxs  A: 4.7405 2.2 Speed and Velocity The Instantaneous velocity indicates how fast
the car moves and the direction of motion at each
instant of time. 2.3 AMMUOII The notion of moderation emerges when a change in
velocity us combmed with the time during which the
change occurs. k}
91
>7 1'41 2.3 Acceleration Example 3 Aoceleranon and Increaan Vetooty Determne the average acceleration of the plane. v.40mls=0km‘h G=2wknuh rJOs r29s i=v— 2 ___ 260km/h—0km/h=+9.0kmm
t—tz. 293—05 5 Nottoemcmhwdwrmmitsmtuve. 2.3 Acceleration Mecca now the Games wavelet: (1an the second second :5 large'
31th (1.513009 W aumg the ﬁrst sound Th5 Is cue no a
hgner average weed curing the seams second that's no 549ch time I sad second ssoom on w; shoe 2.3 Acceleration ExmvphJAczmzbnandDeauashqa§ &.r ’1 “7—...— 2.4 Eounuom or Kinematics fer Commit Acceleration .. _. AV
v=_—_ a=—='— It is customary to dispense with the use of boldface symbols
overdrawn with arrows for the displacement. velocity. and
acceleration vectors. We will. however. continue to convey
the directions with a plus 0r minus sign. 2.4 Eounuom or Kinematics fer Commit Acceleration Five kinematic variables: 1. displacement. x 2. acceleration (constant). a
3. ﬁnal velocity (at time t). v 4. initial velocity. v0
5. elapsed time. t Our goal is to ﬁnd general expressions that connect these
variables so that we can describe motion as a function of
time. 2.4 Equauons o! Khomatlcs tor Constant Accolwatton Let the object be at the origin when the clock starts. x =0 (“=0 U 2.4 Equauons o! Khomatlcs tor Constant Accolwatton 2.4 Equauons o! Khomatlcs tor Constant Accolwatton v = v” + a! x = g(vn + v)! = “v” + v” + at )1 x  vnt + %at‘ 2.4 69000003 0! Kmamotla lb! Canton: Aocdontlon 1,0‘ _._;_.‘ :nl'll
.’ l ‘. t".
K/ ld’Oms'. c:,9¢Drv,l I ‘6 x = vat + 2Lal2  (6.0 m/stm s)+ § (2.0 m/s2 18.0 s)’ =+110m 2.4 690.000: of Kmanufla lb! Constant Accounaon Example 6 Catapulting a Jet
Fund its displacement.
v0  Omxs a  +3 I mfs2
.r  ?? v  +62 mfs 2.‘ Equallqu o! Khmer!“ tar Cerium Acceleration v2  v2 (62 mis),  (0 s)2 0 x=———= V 5 20 26]m;"s‘ ) 2.‘ Equallqu o! Khmer!“ tar Cerium Acceleration Equations of Kinematk: for Constant Acceleration v = v" + at
x = av" + v)!
v2  v: + Zax _ I 2
x vut+3at 2.5 Applicant»: 0! the Equarlom o! Kinematics Reasoning Strategy
1. Make a drawing. 2. Decide which directions are to be caled positive (+) and
negative H. 3. Write down the values that are given for any of the ﬁve
kinematic variables. 4. Verify that the information contains values for at least three
0! the ﬁve kinematic variables. Select the appropriate equation. 5. When the motion is divided into segments. remember that
the ﬁnal velocity of one segment is the iniial velocity for the next. 6. Keep in mind that there may be two possible answers to a
kinematim problem. 2.5 Applications of the Equations 0! Kinematics Example 8 An Accelerating Spacecraft A spacecraft is traveling with a velocity of 93250 mls. Suddenty the retrorockets are ﬁfed. and the spacecraft begins to slow down
with an acceleration whose magnitude is 10.0 rn/s’. What is
the velocity of the spacecraft when the displacement of the cralt is +215 km. relative to the point where the retmrockets began
ﬁring? +215000m 1o.o m/s? +3250 m/s I 2.5 Applications of the Equations 0! Kinematics 2.5 Applications 01th. Equations 0! Kinematics
l
+21sooo m 40.0 rnls’ +3250 this I v  + 2ax = v = JV: + 2:.
v  4 3250mm + 2110.0m/s2 izisooo m)  12500 m/s umme Intheabsencaotairresistance. itislomdthatalbodies
at the same location above the Earth tall verticaly with
the same acceleration. If the distance of the fall is smdl
compared to the tadius at the Earth. then the acceleration remains essentially constant throughout the descent. This idealized motion is caled freefat! and the acceleration
of a freely fall'ng body is caled the axeleration due to
gravity. g=9.80m/s2 or 32.2ﬁ/s2 umme Example 10 A Fall'ng Stone A stone is dropped from the top ota tﬂ building. Alter 3.00s
of free fall. what is the displacement yet the stone? 300$ 100 MI I
\
I ,.
.. 2.6 may Falling Baum ." = V"! + #1::
= (Om/5X3.00‘1)+#4180111“2 13.00
' 44.1 m 2.6 may Falling Baum
Example 12 Haw High Does it Go? The referee tosses the coin up
with an initial speed of 5.00rw's
In the absence if air resisiance.
how high does the coin go above its point of release? 2.6 may Falling Baum "II ? 9.80 m/s’ 0 mls +5.00
mls 261%.:er 800“! "ll
? 9.80 m/s? 0 mls +5.00
rnls
‘! 1 2 2 ‘) — ‘7‘
v  V” + 20y = ;:« V  _" ' 20 a 1 v‘ vo (0m (5.00m I 7 ‘ 8 m
20 2(— 9.80m s‘ j 261%.:er 8m Conceptual Example 14 Acceleration Versus Velocity There are three parts to the motion of the coin. On the way
up. the coin has a vector velocity that is directed upward and
has decreasing magnitude. At the top of its path. the coin
momentarily has zero velocity. On the way down. the coin
has dommardpohting velocity with an ‘ncreasing magnitude. In the absence 0! air resistance. does the acceleration of the
coin. like the velocity. change from one part to another? 2.7 Graphical Analysis of Why and Acceleration +16 9
._.
N Position .1 (m)
L & 1"=25 17WAMW NWWAWM 17WAMW NWWAWM Tangent line \ on
.0
o Ax: +26 m~\ m
P
o E
)1
C
3400
m
0
CL N
.°
0 O 0 5.0 10.0 15.0 20.0 25.0
Time I (s) 17WAMW NWWAWM +36 4'
N
b Velocity v (tn/S)
I.
N ...
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 Spring '09
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