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

Falling BodiesExpectation for ∆t if on Earth

Real time Video

Slowed down by 0.38x

Sped up by 2.6x

mxsmga

51.0/81.9

)2.1(22

3.1/81.9

∆=∆

%3.84%100

%0.5%100

/8.9;/622.1

/54.1)(

23.153.5983.60

3.1;)(21

=×−

=∆∆

=−=∆

=∆∆=∆

earthmoon

smgsmg

Measured acceleration

Looks like Apollo 15 was on the Moon

inftinminmtgx moon 9.5494.53)/37.39(37.137.1)(21 2 ==×==∆=∆

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

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

Checkpoint 2 Solution

21)( xtvattx ++=

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===

vsmxxa

smxxvv

−+=⇒

−=−

What is the velocity of an object when it is at its maximum height?

smxxvv

−=⇒

Two ways to solve for maximum height…

bfbfb xtvatxf

Use tf to solve for maximum height…

Remember a is negative

Solve for time first Solve for position directly

xxavvt bbbbbbbb

ff 0)(2

0000000

222±−

=−±−

=−−±−

Use to solve for time:fbx

xxavvt

xxtvat

tvatxx

−−±−=⇒

=−++⇒

Quadratic Equation-which solution

hckabh

khcahb

aacbxaxkhahxaxkhxa

−=⇒

−=⇒+=

++=++−=+−

gvxxhck

tgvgvavh

xxtvat

−−=−=

==−−=−=

=−++

The meeting Problem in General

0000 222

)()(21)()(

xtvtaxttvtta

ttxttxt

launchedfirstobject

meetmeetdmeetdmeet

meetmeet

++=++++

Whenever possible solve symbolically.Then plug in the numbers!!!! You can understand what is going on better if you keep the equations organized.

00000 222

11122 =−−−+++++ xtvatxtvtvatttaat meetmeetdmeetdmeetdmeet

0)21())((

00000 2112

21 =−+++−+ xxtvattvvat ddmeetd

Define your problem; Identify useful equations

Start solving…

meet ttvttx

xxtvatt

=∆=∆

−++=

Gather terms…Does the equation look reasonable?

Solve for desired quantity…)

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

xxtvatt

1;26;/8.23;/2.1;4.0;/81.9

===−==−=

=∆=∆

−++=

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

00=+−+−=++

Solve for individual elements of equation…can check if things make sense

smsmssmvatd /124.5)/2.1()4.0)(/81.9()( 210

−=−+−=+

smsmsmttvmmmttx

/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

sttvttxt

dmeetred

82.0)()(

==∆=∆

Note that acceleration term after 2nd object is released drops out!!!

I-74 problem

mihrhrmitvx

hrhrmihrmi

mimivvxx

mihrhrmimitvxxhrmivhrmiv

hrtmixxvvxx

tvxxvv

xtvxvvt

xttvxtvxxtt

meetameet

acmeet

accmeet

acccameet

ccameeta

cameet

cdelaycc

66.73)9821.0)(/75(

9821.0)/65(/75(

05.137)(

5.137)5.0()/65(105/65;/75

;5.0;105;0)(

=−−−

=−×−+=∆+=⇒

−=∆==−

∆+≡

−∆+=

−∆+=−

+∆+=+⇒==

Car-ride problem

/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(

27.159;/)8.9(

;/0)8.95.3(;/4.4)5.30(

tamtxx

smvvtxxa

mtvssssmtx

smssmtv

mxsmatta

smtasmta

stopyy

bstopstop

stopbstopbb

bbstopbstop

fbbbstop

ystopy

bstopstopb

−===−

−==−

−=⇒

−=−==−

==−+==

Two thrown balls problem

mmsssmsssmstx

mxstvstatx

mssmssmxtvatx

smxxvv

smamxsmvstmxsmv

delayr

rrdelaybb

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

283.2/81.9

/4.220

/81.9;2.27;/1.6;7.2;6.0;/4.22

=+−−+−−==

=+−+−=

++−=++=

=−−

−==−====

smmmmt

mmssmssm

xxsvsat

xxsvsatvvsa

xtvtaxsvtvsatsaat

xtvtaxstvsta

ttxttx

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()29.7(21

0)7.2()29.7(21))7.2(

0)()(21)7.2()29.7(

21)7.2)(2(

)()(21)7.2()7.2(

=−−

−+−+−−=

−++−=⇒

=−+−+−+

=−−−+−++−

++=+−+−

smsmsm

xxavvt

xxtvat

tvatxx

mmmsmma

vvxxasolutionsealternativ

283.2/81.9/4.220

/81.9)574.25)(81.9(2)/4.22(/4.22)(2

174.26574.256.0)81.9(2

)/4.22(6.02

=−−

=±−

−−−−±−

=−−±−

=−++⇒

=+=−

−+=⇒

−=−

Two thrown balls problem #2

msssmsssmstx

msmsmm

smssmsmatvv

smmmsmsmsm

axxavv

xxtvatxtvatx

smamxsmvstmxsmv

rrdelaybb

70.24)8.1(

0.1)4.08.1)(/8.23()4.08.1)(/81.9(21)8.1(

87.29)/81.9(2)/8.23(00.1

/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(

/81.9;0.26;/2.1;4.0;0.1;/8.23

+−+−−==

=−−

−=−

=−+−=+=

−−−−±−−=⇒

−−±−=⇒

=−++⇒++=

−==−=−===

A “real world” ramp experiment

hrkmttvhrkmtv

f /4.251)(/0)0(

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

A “real world” ramp experiment

?/8.69/4.251)(

=∆==∆=

tsmhrkmttv

Could not find elapsed times…to check.

Assume constant acceleration

96.22/04.3/8.69)(

/04.31600

)/8.69()(2

==∆=

==∆=∆∆=

Verbier- Mont Fort

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

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

HyperphysicsMotion

Displacement vs timet

Velocity vs timet

Acceleration vs timet

HyperphysicsMotion

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