falling bodies 1d-kinematic equations how to use them...

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

http://nssdc.gsfc.nasa.gov/planetary/lunar/apollo_15_feather_drop.html

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 ==×==∆=∆


http://www.physics.utah.edu/%7Espringer/phys1500/Lectures/PHYS1500-2015/01.b/Ham-feath1.0x.mp4

http://www.physics.utah.edu/%7Espringer/phys1500/Lectures/PHYS1500-2015/01.b/Ham-feath0.38.mp4

http://www.physics.utah.edu/%7Espringer/phys1500/Lectures/PHYS1500-2015/01.b/Ham-feath2.6x.mp4

Main Points of Unit 1


Motion with Constant Acceleration

constanta(t) = a

Back to Earth!


Motion with Constant AccelerationVelocity vs displacement

Let’s eliminate time…to obtain v(x) for special case of constant acceleration


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


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( =−==−==∆∴


First determine acceleration a:

Now determine position at t=2s:


Gather relevant formulae and facts:

ftsastx 1)1(21)1( 2 ===


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


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!


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”…


Tortoise and Hare

Changing conditions Break problem down into parts (intervals)


Tortoise and HareChanging conditions Break problem down into parts (intervals)


Tortoise and Hare


Homework Hints IE vs T

Changing conditions Break problem down into parts (intervals)


Homework Hints

Break problem down into parts


Homework Hints



Simplify!!!


Homework Hints

Finally!!!


Homework Hints


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

−−±−=⇒

=−++⇒

+=−


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


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)(


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!!!


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

'

'

'

'

===

=−−−

=−

−=

=−×−+=∆+=⇒

−==

−=∆==−

−=

∆+≡

−

−∆+=

−∆+=−

+∆+=+⇒==

++=

+=


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

=×

=×

=⇒

==

=×

+=⇒

−===−

−==−

−=⇒

−=−==−

==−+==

===

=<<

=

=<<

=<<

=<<


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

=+−−+−−==

=+−+−=

−=

=

++−=++=

=−−

=−

=⇒

===

−==−====



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

=−

=−−

=

++−

−+−+−−=

−+

−++−=⇒

=−+−+−+

=−−−+−++−

++=+−+−

===



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

=−−

=±−

=⇒

−−−−±−

=−−±−

=⇒

=−++⇒

+=−

=+=−

−+=

−+=⇒

−=−


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

==

+−+−−==

−=

=−−

+=−

+=

−=−

=−+−=+=

=−

±=

−

−−−−±−−=⇒

−−±−=⇒

=−++⇒++=

=

−==−=−===


A “real world” ramp experiment

hrkmttvhrkmtv

f /4.251)(/0)0(

====

Kilometre lance http://www.kl-france.com/


http://www.kl-france.com/

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


http://www.youtube.com/watch?v=wl_lAeLU6X0&feature=player_embedded%23t=71

http://www.youtube.com/watch?v=wl_lAeLU6X0&feature=player_embedded%23t=71

Schaum’s Outline


Displacement,Velocity & Acceleration

Need to become comfortable with displacement, velocity and acceleration and how they are related!!!


Constant Acceleration


Hyperphysics-Motion

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html



Hyperphysics-Motion




HyperphysicsMotion

Displacement vs timet

Velocity vs timet

Acceleration vs timet


HyperphysicsMotion


falling bodies 1d-kinematic equations how to use them...

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