Page 1
1d -Kinematics-continued Falling Bodies 1d-Kinematic Equations How to use them Homework Hints
http://nssdc.gsfc.nasa.gov/planetary/lunar/apollo_15_feather_drop.html
2)(21 tax ∆=∆
We can use 1-dim kinematics to describe falling bodies
Mechanics Lecture 1, Slide 1
Page 2
Falling BodiesExpectation for ∆t if on Earth
Real time Video
Slowed down by 0.38x
Sped up by 2.6x
ssm
mg
xt
mxsmga
tax
earth
earth
51.0/81.9
)2.1(22
3.1/81.9
)(21
2
0
2
2
==∆
=∆
===
∆=∆
%3.84%100
%0.5%100
/8.9;/622.1
/54.1)(
23.153.5983.60
3.1;)(21
22
22
2
=×−
=×−
==
=∆∆
=
=−=∆
=∆∆=∆
eatrh
earth
moon
moon
earthmoon
gga
gga
smgsmg
smtxa
ssst
mxtax
Measured acceleration
Looks like Apollo 15 was on the Moon
inftinminmtgx moon 9.5494.53)/37.39(37.137.1)(21 2 ==×==∆=∆
Mechanics Lecture 1, Slide 2
Page 3
Main Points of Unit 1
Mechanics Lecture 1, Slide 3
Page 4
Motion with Constant Acceleration
constanta(t) = a
Back to Earth!
Mechanics Lecture 1, Slide 4
Page 5
Motion with Constant AccelerationVelocity vs displacement
Let’s eliminate time…to obtain v(x) for special case of constant acceleration
Mechanics Lecture 1, Slide 5
Page 6
1ft4ft
9ft
?
16ft
At t = 0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. Suppose it moves 1 foot betweent = 0 sec and t = 1 sec.
How far does it move between t = 1 sec and t = 2 sec?
A) 1 foot B) 2 feet C) 3 feet D) 4 feet E) 6 feet
Checkpoint 2
Mechanics Lecture 1, Slide 6
Page 7
1ft4ft9ft
3
16ft
Checkpoint 2 Solution
002
21)( xtvattx ++=
200 2
1)(0;0 attxvx =⇒==
222 /2)1/()1(21)1(21)1( sftsftaftsastx ==⇒===
ftssftstx 4)2()/2(21)2( 22 =×==
ftftftstxstxx 314)1()2( =−==−==∆∴
How far does it move between t = 1 sec and t = 2 sec?
First determine acceleration a:
Now determine position at t=2s:
How far does it move between t = 1 sec and t = 2 sec?
Gather relevant formulae and facts:
ftsastx 1)1(21)1( 2 ===
Mechanics Lecture 1, Slide 7
Page 8
Smartphysics Homework
Interactive ExercisesProvides step by step guidance
Standard ExercisesProblem broken down into multiple parts
Problem with solution example“Conceptual Analysis”“Strategic Analysis”“Quantitative Analysis”
What’s going on?
How do I approach the problem? What formulae do I need? What information have I been given? Do the algebra… Plug in the numbers
Mechanics Lecture 1, Slide 8
Page 9
Homework Example – How to solve problems
Tortoise and Hare Problem…Click “See solution” link to be guided step by step to the solution…
Beware the numbers used in the solution are probably different than in your problem…the method is the same however.
Eventually you should be able to perform this type of analysis on your own…i.e in an exam!
Mechanics Lecture 1, Slide 9
Page 10
Tortoise and Hare
Identify useful information and “translate”
Constant speed a=0
Start from rest v0=0
Identify useful equations/formulae
Solve Algebraically
“plug and chug”…
Mechanics Lecture 1, Slide 10
Page 11
Tortoise and Hare
Changing conditions Break problem down into parts (intervals)
Mechanics Lecture 1, Slide 11
Page 12
Tortoise and HareChanging conditions Break problem down into parts (intervals)
Mechanics Lecture 1, Slide 12
Page 13
Mechanics Lecture 1, Slide 13
Page 14
Mechanics Lecture 1, Slide 14
Page 15
Tortoise and Hare
Mechanics Lecture 1, Slide 15
Page 16
Homework Hints IE vs T
Changing conditions Break problem down into parts (intervals)
Mechanics Lecture 1, Slide 16
Page 17
Homework Hints
Break problem down into parts
Mechanics Lecture 1, Slide 17
Page 18
Homework Hints
Mechanics Lecture 1, Slide 18
Page 19
Homework Hints IE vs T
Mechanics Lecture 1, Slide 19
Page 20
Homework Hints IE vs T
Mechanics Lecture 1, Slide 20
Page 21
Homework Hints IE vs T
Mechanics Lecture 1, Slide 21
Page 22
Homework Hints IE vs T
Mechanics Lecture 1, Slide 22
Page 23
Homework Hints IE vs T
Mechanics Lecture 1, Slide 23
Page 24
Homework Hints IE vs T
Mechanics Lecture 1, Slide 24
Page 25
Homework Hints IE vs T
Mechanics Lecture 1, Slide 25
Page 26
Homework Hints IE vs T
Mechanics Lecture 1, Slide 26
Page 27
Homework Hints IE vs T
Mechanics Lecture 1, Slide 27
Page 28
Homework Hints IE vs T
Simplify!!!
Mechanics Lecture 1, Slide 28
Page 29
Homework Hints IE vs T
Mechanics Lecture 1, Slide 29
Page 30
Homework Hints IE vs T
Mechanics Lecture 1, Slide 30
Page 31
Homework Hints IE vs T
Mechanics Lecture 1, Slide 31
Page 32
Homework Hints IE vs T
Mechanics Lecture 1, Slide 32
Page 33
Homework Hints IE vs T
Mechanics Lecture 1, Slide 33
Page 34
Homework Hints IE vs T
Mechanics Lecture 1, Slide 34
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Homework Hints
Finally!!!
Mechanics Lecture 1, Slide 35
Page 36
Homework Hints
Mechanics Lecture 1, Slide 36
Page 37
Homework Hints
Mechanics Lecture 1, Slide 37
Page 38
Homework Hints IE vs T
Mechanics Lecture 1, Slide 38
Page 39
Maximum Height of ball thrown upward in gravitational field
smxxvv bbbbf/0)(
max===
( )
av
xx
vsmxxa
smxxvv
bbb
bbb
bbbb
f
f
f
2
/0)(2
/0)(
2
22
0
0
00
max
−+=⇒
−=−
===
What is the velocity of an object when it is at its maximum height?
avsm
avv
t
tavv
smxxvv
bbbf
bb
bbbb
f
f
f
00
0
max
/0
/0)(
−=
−=⇒
∆+=
===
Two ways to solve for maximum height…
00
2
21
bfbfb xtvatxf
++=
Use tf to solve for maximum height…
Remember a is negative
Solve for time first Solve for position directly
av
avvv
a
xxavvt bbbbbbbb
ff 0)(2
0000000
222±−
=−±−
=−−±−
=⇒
Use to solve for time:fbx
a
xxavvt
xxtvat
tvatxx
f
f
f
bbbbf
bbfbf
fbfbb
)(2
0)(21
21
000
00
00
2
2
2
−−±−=⇒
=−++⇒
+=−
Mechanics Lecture 1, Slide 39
Page 40
Quadratic Equation-which solution
( )
2
2
2222
2/
2
2
hckabh
khcahb
aacbxaxkhahxaxkhxa
−=⇒
−=⇒+=
−==
++=++−=+−
( )22
2
/)(
///21
21'
0)(21
00
000
00
gvxxhck
tgvgvavh
gaa
xxtvat
bbb
fbbb
bbfbf
f
f
−−=−=
==−−=−=
−==
=−++
t
Mechanics Lecture 1, Slide 40
Page 41
The meeting Problem in General
0000 222
112
21
1
)()(21)()(
21
)()(0
_
xtvtaxttvtta
ttxttxt
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.
021
21)2(
21
21
00000 222
11122 =−−−+++++ xtvatxtvtvatttaat meetmeetdmeetdmeetdmeet
0)21())((
00000 2112
21 =−+++−+ xxtvattvvat ddmeetd
Define your problem; Identify useful equations
Start solving…
)()(
))((
)21(
00
000
21
2112
d
d
d
dd
meet ttvttx
vvat
xxtvatt
=∆=∆
=−+
−++=
Gather terms…Does the equation look reasonable?
Solve for desired quantity…)
21(
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)(
Mechanics Lecture 1, Slide 41
Page 42
Two thrown balls problem #2
mxmxsmvsmvstsma
ttvttx
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(21()
21( 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
−=−+−=+
smsmsmttvmmmttx
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
sttvttxt
dmeetred
d
dmeet
22.1
82.0)()(
=+=
==∆=∆
=
Note that acceleration term after 2nd object is released drops out!!!
Mechanics Lecture 1, Slide 42
Page 43
I-74 problem
mihrhrmitvx
hrhrmihrmi
mimivvxx
t
mihrhrmimitvxxhrmivhrmiv
hrtmixxvvxx
t
tvxxvv
xtvxt
xtvxvvt
xttvxtvxxtt
xttvx
xtvx
meetameet
ca
acmeet
ccc
ca
ca
ca
acmeet
ccc
ca
accmeet
acccameet
ccameeta
cameet
cdelaycc
aaa
66.73)9821.0)(/75(
9821.0)/65(/75(
05.137)(
5.137)5.0()/65(105/65;/75
;5.0;105;0)(
)(
)(
)(;@
)(
00
00
00
00
00
00
00
00
0
0
'
'
'
'
===
=−−−
=−
−=
=−×−+=∆+=⇒
−==
−=∆==−
−=
∆+≡
−
−∆+=
−∆+=−
+∆+=+⇒==
++=
+=
Mechanics Lecture 1, Slide 43
Page 44
Car-ride problem
222
2
2
22
220
2
22
2
2
2
2
/539.1)385.14(
27.1592)(27.1592
)(2127.159
385.14)3.35(28.9
)8.9(213.35)8.9(
/359.3))8.9((2
)/4.15(
)/4.15(0))8.9((2
97.123)5.3(*)5.38.9()5.3)(/4.4(21)8.9(
/4.15)5.3)(/4.4()5.3(
)0(
27.159;/)8.9(
;/0)8.95.3(;/4.4)5.30(
sms
mt
ma
tamx
sa
mt
tamtxx
smtxx
sma
smvvtxxa
mtvssssmtx
smssmtv
atta
mxsmatta
smtasmta
stopy
stopyy
bstopstop
stopbstopbb
bbstopbstop
fbbbstop
bb
b
ystopy
b
bstopstopb
b
b
f
f
f
f
=×
=×
=⇒
==
=×
+=⇒
−===−
−==−
−=⇒
−=−==−
==−+==
===
=<<
=
=<<
=<<
=<<
Mechanics Lecture 1, Slide 44
Page 45
Two thrown balls problem
mmsssmsssmstx
mxstvstatx
ttt
mx
mssmssmxtvatx
ssm
sma
vvt
smxxvv
smamxsmvstmxsmv
r
rrr
delayr
b
bfbfb
bbf
bbbb
rrdelaybb
f
f
f
f
04.192.27)7.251.3)(/1.6()7.251.3)(/81.9(21)51.3(
04.19)7.2()7.2(21)(
174.26
6.0)283.2)(/4.22()283.2)(/81.9(21
21
283.2/81.9
/4.220
/0)(
/81.9;2.27;/1.6;7.2;6.0;/4.22
22
2
222
2
2
00
00
0
max
0000
=+−−+−−==
=+−+−=
−=
=
++−=++=
=−−
=−
=⇒
===
−==−====
Mechanics Lecture 1, Slide 45
Page 46
Two thrown balls problem
ssmm
smmmmt
smsms
mmssmssm
vvsa
xxsvsat
xxsvsatvvsa
xtvtaxsvtvsatsaat
xtvtaxstvsta
ttxttx
meet
br
rbr
meet
brrmeetbr
bmeetbmeetrrmeetrmeetmeet
bmeetbmeetrmeetrmeet
meetbmeetr
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
0)7.2()29.7(21))7.2(
21(
0)()(21)7.2()29.7(
21)7.2)(2(
21
21
)()(21)7.2()7.2(
21
)()(
222
2
222
22
00
000
00000
00000
0000
=−
=−−
=
++−
−+−+−−=
−+
−++−=⇒
=−+−+−+
=−−−+−++−
++=+−+−
===
Mechanics Lecture 1, Slide 46
Page 47
Two thrown balls problem
ssmsm
av
t
smsmsm
a
xxavvt
xxtvat
tvatxx
mmmsmma
vvxx
vvxxasolutionsealternativ
bf
bbbbf
bbfbf
fbfbb
bbbb
bbbb
f
f
f
f
f
ff
283.2/81.9/4.220
/81.9)574.25)(81.9(2)/4.22(/4.22)(2
0)(21
21
174.26574.256.0)81.9(2
)/4.22(6.02
)(2_
2
2
22
2
2
222
22
0
000
00
00
0
0
00
=−−
=±−
=⇒
−−−−±−
=−−±−
=⇒
=−++⇒
+=−
=+=−
−+=
−+=⇒
−=−
Mechanics Lecture 1, Slide 47
Page 48
Two thrown balls problem #2
mstx
msssmsssmstx
stt
msmsmm
avv
xx
vvxxa
smssmsmatvv
ssm
smsmt
smmmsmsmsm
t
axxavv
t
xxtvatxtvatx
mtx
smamxsmvstmxsmv
rb
rb
rb
bbbb
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(21)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
021
21
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
==
+−+−−==
−=
=−−
+=−
+=
−=−
=−+−=+=
=−
±=
−
−−−−±−−=⇒
−−±−=⇒
=−++⇒++=
=
−==−=−===
Mechanics Lecture 1, Slide 48
Page 49
A “real world” ramp experiment
hrkmttvhrkmtv
f /4.251)(/0)0(
====
Kilometre lance http://www.kl-france.com/
Mechanics Lecture 1, Slide 49
Page 50
A “real world” ramp experiment
?/8.69/4.251)(
/0)0(
=∆==∆=
==
tsmhrkmttv
smtv
Could not find elapsed times…to check.
Assume constant acceleration
ssmsm
attvt
smm
smttx
ttva
96.22/04.3/8.69)(
/04.31600
)/8.69()(2
)(
2
222
==∆=
=∆
==∆=∆∆=
=
Verbier- Mont Fort
Mechanics Lecture 1, Slide 50
Page 51
Schaum’s Outline
Mechanics Lecture 1, Slide 51
Page 52
Displacement,Velocity & Acceleration
Need to become comfortable with displacement, velocity and acceleration and how they are related!!!
Mechanics Lecture 1, Slide 52
Page 53
Constant Acceleration
Mechanics Lecture 1, Slide 53
Page 54
Mechanics Lecture 1, Slide 54
Page 55
Hyperphysics-Motion
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html
Mechanics Lecture 1, Slide 55
Page 56
Hyperphysics-Motion
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html
Mechanics Lecture 1, Slide 56
Page 57
HyperphysicsMotion
Displacement vs timet
Velocity vs timet
Acceleration vs timet
Mechanics Lecture 1, Slide 57
Page 58
HyperphysicsMotion
Mechanics Lecture 1, Slide 58