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Mechanics Lecture 1, Slide 1
1d -Kinematics-summary
Main points of 1d-kinematics againHints on doing homeworkHomework Examples
Homework Deadline extended (Sunday 1/25 11:30PM)
1d-Kinematic Equations for constant acceleration
Mechanics Lecture 1, Slide 4
))((2)))((
2
1)(
)(
)(
020
2
002
0
00
0
xtxavtv
xtvtatx
vtatv
ata
Basic Equations to be used for 1d – kinematic problems.
Need to apply to each object separately sometimes with time offset
When acceleration changes from one constant value to another say a=0 The problem needs to be broken down into segments
Homework Hints
Start Early and allow yourself enough time Write out solutions on paper BEFORE entering answers online Ensure that stated problem is understood
Identify the useful equations Identify the information to be used
Sometimes the required information is not stated explicitly Make diagrams and rough graphs of quantities Break the problem up into parts Solve algebraically first Check that equations are reasonable Carefully perform calculations Stay calm…
Mechanics Lecture 1, Slide 5
Interactive Example V vs T
Mechanics Lecture 1, Slide 6
20
0
0
/66667.65.1//10)05.1/())0()5.1(()0(
0)0(
0)0(
smssmssstvstvtaaa
tvvv
txxx
i
i
i
?
Interactive Example V vs T
Mechanics Lecture 1, Slide 7
Region 1: 0<t<1.5 s0)0( stxxi
Region 2: 1.5s<t<3.0s2
2 /05.1/)/0(/ smssmtvaa
Region 3: 3.0s<t<5.0ssssttt if 0.20.30.5
Region 1 Region 2 Region 3
What happened in region 3?Smartphysics Conclusion
ssttt if 05.1 2
1 /667.65.1/)/10(/ smssmtvaa smssmtatavtastv /0.10)5.1)(/667.6(00)0()5.1( 2
1101
mssmtatvxstx 5.7)5.1)(/667.6(2
1)(
2
1)5.1( 222
100
smsmtstvtatv /0.10/100)5.1()( 2
mmmssmmssstvstxstx 5.220.155.7)5.1)(/0.10(5.7)5.10.3)(5.1()5.1()0.3( smstvtstv /0.10)5.1(0)0.3(
23 /100.2/)/0.10/10(0.2/)(/ smssmsmsvvtvaa if
smsmssmstvtastv /0.10/10)2)(/10()0.3()5( 22
mssmsmtstvtastxstx 0)0.2)(/10()0.2)(/10(2
1)0.3()(
2
1)0.3()0.5( 222
3
mstxmmstxstx 5.22)0.3()20()20()0.3()0.5(
smsmsmstvtastv /0.10/10)0.2)(/10()0.3()0.5( 22
)5( stxx final
I-74 problem
Mechanics Lecture 1, Slide 9
mihrhrmitvx
hrhrmihrmi
mimi
vv
xxt
meetameet
ca
acmeet
66.73)9821.0)(/75(
9821.0)/65(/75(
05.137
)(00
dcdcc
ccc
aaa
tvttxx
xtvtx
xtvtx
)(
)(
)(
0
0
0
00
)()(;@
cmeetcameeta
meetcmeetameet
xtvxtv
ttxttxtt
)(
)(
00
00
ca
acmeet
accameet
vv
xxt
xxvvt
delaycdcc tvttxx )(0
hrmivhrmiv
hrtmittxx
ca
delaydca
/65;/75
;5.0;105)(;00
Facts Equations
Where are the cars when they meet? The same place
Solve for meeting time Where was Chuck when Anna passed prospect avenue?
)(00
ca
acmeet vv
xxt
Express meeting time in terms of separation of Chuck and
Anna at t=0:
mihrhrmimitvttxx dcdcc 5.137)5.0()/65(105)(0
Solve for meeting time and location
Be careful with your facts!
Car-ride problem
Mechanics Lecture 1, Slide 12
smsmssmvtavstv bb /4.4/0)1(/4.4)1( 20
ystopy
b
bstopstopb
b
b
atta
mx
smatta
smta
smta
f
)0(
27.159
;/)8.9(
;/0)8.95.3(
;/4.4)5.30(
2
2
2
)5.3()1.6( tvtv bb
22
/359.3))8.9((2
)/4.15(sm
txx
sma
bbstopbstop
mtvssssmtx bb 97.123)5.3(*)5.38.9()5.3)(/4.4(2
1)8.9( 22
220
2 )/4.15(0))8.9((2 smvvtxxa fbbbstop f
smssmtvb /4.15)5.3)(/4.4()5.3( 2
sm /4.15
Gather factsAnd equations
))((2)))((
2
1)(
)(
)(
020
2
002
0
00
0
xtxavtv
xtvtatx
vtatv
ata
Car-ride problem
Mechanics Lecture 1, Slide 13
stst
ssm
smsmsmt
sm
msmsmsmt
a
xtxaatv
t
xtxttvta
tt
tattvtxttx
stop
bstop
bbbstopbstopb
bbbbstop
stop
bstopbbstopb
f
f
f
38.148.9
58.4/359.3
/2.237/2.237/4.15
/359.3
3.35/68.14/4.15/4.15
)8.9(21
421
)8.9(
0)8.9())(8.9()(2
1
8.9
)(2
1))(8.9()8.9()(
2
2222
2
22
2
2
2
2
22
2
/539.1)385.14(
27.1592
)(
27.1592
)(2
127.159
sms
m
t
ma
tamx
stopy
stopyy f
Quadratic Equation-which solution
Mechanics Lecture 1, Slide 14
2
2
2222
2/
2
2
hck
abh
khc
ahb
aa
cbxaxkhahxaxkhxa
22
2
/)(
///2
1
2
1'
0)(2
1
00
000
00
gvxxhck
tgvgvavh
gaa
xxtvat
bbb
fbbb
bbfbf
f
f
t
a
acbbx
cbxax
2
4
0
2
2
Two thrown balls problem
Mechanics Lecture 1, Slide 16
mmsssmsssmstx
mxstvstatx
ttt
r
rrr
delayr
04.192.27)7.251.3)(/1.6()7.251.3)(/81.9(2
1)51.3(
04.19)7.2()7.2(2
1)(
22
2
00
2/81.9;2.27;/1.6;7.2;6.0;/4.220000
smamxsmvstmxsmv rrdelaybb Facts
smxxvv bbbb f/0)(
max
ssm
sm
a
vvt
bb
ff 283.2
/81.9
/4.2202
0
mx
mssmssmxtvatx
f
f
b
bfbfb
174.26
6.0)283.2)(/4.22()283.2)(/81.9(2
1
2
1 222
00
Two thrown balls problem
Mechanics Lecture 1, Slide 17
ssm
m
sm
mmmt
smsms
mmssmssm
vvsa
xxsvsat
meet
br
rbr
meet
638.3/01.2
3126.7
/01.2
6.2647.1675.35
/1.6/4.22)7.2)(81.9(
2.276.0)7.2)(/1.6()29.7)(/81.9(21
)7.2(
)7.2()29.7(21 222
00
000
)()( meetbmeetr ttxttx SAME HEIGHT MEET
Use relevant kinematic equations
0000)()(
2
1)7.2()7.2(
2
1 22bmeetbmeetrmeetrmeet xtvtaxstvsta
Simplify equation
0)7.2()29.7(2
1))7.2(
2
1(
0)()(2
1)7.2()29.7(
2
1)7.2)(2(
2
1
2
1
00000
00000
2
222
brrmeetbr
bmeetbmeetrrmeetrmeetmeet
xxsvsatvvsa
xtvtaxsvtvsatsaat
Solve for tmeet
Two thrown balls problem-Alternative Solution
Mechanics Lecture 1, Slide 18
ssm
sm
a
vt
sm
smsm
a
xxavvt
xxtvat
tvatxx
bf
bbbb
f
bbfbf
fbfbb
f
f
f
283.2/81.9
/4.220
/81.9
)574.25)(81.9(2)/4.22(/4.22)(2
0)(2
12
1
2
2
22
2
2
0
000
00
00
mmmsm
ma
vvxx
vvxxa
bb
bb
bbbb
f
f
ff
174.26574.256.0)81.9(2
)/4.22(6.0
2
)(2
222
22
0
0
00
Use equation relating velocity and position to solve for height
Then solve for time at maximum height
Continue solving the rest of the problem as before…
The meeting Problem in General
Mechanics Lecture 1, Slide 19
0000 222
112
21
1
)()(2
1)()(
2
1
)()(
0
_
xtvtaxttvtta
ttxttx
t
launchedfirstobject
meetmeetdmeetdmeet
meetmeet
d
Whenever possible solve symbolically.Then plug in the numbers!!!! You can understand what is going on better if you keep the equations organized.
02
1
2
1)2(
2
1
2
100000 22
2111
22 xtvatxtvtvatttaat meetmeetdmeetdmeetdmeet
0)2
1())((
00000 2112
21 xxtvattvvat ddmeetd
Define your problem; Identify useful equations
Start solving…
)(
)(
))((
)21(
00
000
21
2112
d
d
d
dd
meet ttv
ttx
vvat
xxtvatt
Gather terms…Does the equation look reasonable?
Solve for desired quantity…
)2
1(
00 112 xtvat dd
Position of object 1 when object 2 is launched
)(01vatd
Speed of object 1 when object 2 is launched
This time is w.r.t object 2. To obtain time w.r.t object 1 need to add delay time
dmeetmeet ttt 1)(
Two thrown balls problem #2
Mechanics Lecture 1, Slide 21
mxmxsmvsmvstsma
ttv
ttx
vvat
xxtvatt
d
d
d
d
dd
meet
1;26;/8.23;/2.1;4.0;/81.9
)(
)(
))((
)21(
0000
00
000
2121
21
2112
Position of object 1 at moment object 2 is launched
Velocity of object 1 at moment object 2 is launched
Use your equation in symbolic form and gather the values to be used….
mmssmssmxtvat dd 735.2426)4.0)(/2.1()4.0)(/81.9(2
1()
2
1( 22
112
00
Solve for individual elements of equation…can check if things make sense
smsmssmvatd /124.5)/2.1()4.0)(/81.9()( 210
smsmsmttv
mmmttx
d
d
/94.28/8.23/124.5)(
735.231735.24)(
Solve for displacement and relative velocity at time when object 2 launches
Divide to obtain time when objects meet
sttt
sttv
ttxt
dmeetred
d
dmeet
22.1
82.0)(
)(
Note that acceleration term after 2nd object is released drops out!!!
Two thrown balls problem #2
Mechanics Lecture 1, Slide 22
mstx
msssmsssmstx
stt
msm
smm
a
vvxx
vvxxa
smssmsmatvv
ssm
smsmt
sm
mmsmsmsmt
a
xxavvt
xxtvatxtvatx
mtx
smamxsmvstmxsmv
rb
rb
rb
bb
bb
bbbb
frr
f
f
fr
rrdelaybb
f
f
ff
f
70.24)8.1(
0.1)4.08.1)(/8.23()4.08.1)(/81.9(2
1)8.1(
4.0
87.29)/81.9(2
)/8.23(00.1
2
)(2
/586.22)18.2)(/81.9(/2.1
18.2/81.9
/617.22/2.1/81.9
)00.26)(/81.9(2)/2.1()/2.1(
)(2
02
1
2
1
0.0)(
/81.9;0.26;/2.1;4.0;0.1;/8.23
22
2
222
22
2
2
2
22
0200
002
002
2
0
0
00
0
0000
Displacement,Velocity & Acceleration
Mechanics Lecture 1, Slide 24
Need to become comfortable with displacement, velocity and acceleration and how they are related!!!
Hyperphysics-Motion
Mechanics Lecture 1, Slide 27
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html
Hyperphysics-Motion
Mechanics Lecture 1, Slide 28
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html
HyperphysicsMotion
Mechanics Lecture 1, Slide 29
Displacement vs timet
Velocity vs timet
Acceleration vs timet
Enter Question Text
A. TrueB. False
Mechanics Lecture 1, Slide 31
6%
94%
I experienced some degree of “frustration” with the Homework Problems