ch 02 motion in one dimension

8/13/2019 Ch 02 Motion in One Dimension

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Chapter 2Motion in One Dimension

2.1 Displacement and Velocity

2.2 Acceleration

2.3 Falling Objects


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Objectives for Section 2.1

1. Describe motion in terms of

displacement, time, and velocity.

2. Calculate the displacement of an object

traveling at a known velocity for aspecific time interval.

3. Solve problems involving displacement,

time, and velocity

4. Construct and interpret graphs of

position versus time.


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5. Use an appropriate problem solving

technique when solving numerical problemsincluding the use of variables, a statement of

knowns and unknowns, the use of a labeled

diagram including an explicit display of acoordinate system, listing of appropriate

equations, use of algebra to isolate an

unknown in terms of variables, conversion ofquantities using the factor-label method,

display of numerical quantities with their

accompanying units and correct use of

significant figures.


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Motion - Reference Frame

Train moving on linear track1D

motion

But

Earth spinning on axis Earth revolving around sun

Sun moving through galaxy

EtcSimplifychoose frame of reference to

measure changes in position

For train, use train station


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Train moving on linear track1D

motion

But

Earth spinning on axis Earth revolving around sun

Sun moving through galaxy

EtcSimplifychoose frame of reference to

measure changes in position

For train, use train station


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Object is at rest if its position does notchange with respect to its reference

frame

Free to choose any frame of reference Must be consistently used once

chosen

Some choices make life easier Station as x = 0 point


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Displacement

Displacement = change in position= final positioninitial position

x = xf - xi

Value can be positive or negative

depending upon relative values ofxf

andxi

Vectorquantityhas magnitude and

direction (+ / -, north / south)


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Positive & Negative Displacements


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Velocity

Average velocity: displacement dividedby time interval

vavg= x / t

SI unitsm/svavgcan be + ordepending upon sign

of x


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Velocity

Sample problem 2A, page 44


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Practice

Average velocity and displacementPractice 2A, page 44

Problems 1-6

Chapter Review, pages 69-70

Problems 1, 6, 8-15


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Velocity vs Speed

Average velocityvavg= x/ t

Vector- has directionassociated with it

- net direction of x

Average speed

Speedavg= distance traveled / tSpeed: scalar(non-directional) quantity


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VelocityGraphical Interpretation

Average velocityvavg= x/ t

Same as slope of given segmentof

displacement vs time graph

Instantaneous velocity

Tangentto displacement vs time graph

at given point


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3 different cases of constant velocity


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VelocityGraphical InterpretationInstantaneous Velocity


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PracticeSection Review, page 47 Problem 3

Greater vavg? Greater v@ 8 min?

vbear A always +? vbear B ever -?


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Practice

Conceptual Challenge, page 45Problems 1-2

Section Review, page 47

Problems 1-6Problem 6 good challenge, especially part b!

Chapter Review, pages 69-70Problems 2-5


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Problem Solving Steps - Kinematics

1. Read problem carefully2. Write down givenquantities with

their appropriate symbols(including subscripts and units)

x = 3.5 m

viy= 4.2 m/s

t = 1.7 s

vfx= - 7.43 m/s


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3. Write down any other quantitieswhich may not be directly given,but which you can figure out

without the use of equationsa) a= g = - 9.8 m/s2 (free fall)

b) vyf= 0 (for highest part oftrajectory)

c) y= 0 (for projectile launch/land

at same elevation)


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4. Draw diagram describing theproblem

a) Establish and label coordinate

system, including an origin foreach direction (optional for some)

b) Label displacements, velocitieswith their symbols (v0, vyf, x)applies to both knownandunknownvalues


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4. Draw a diagram describing theproblem (cont.)

c) Draw labeled vectors at position

where that vector acts

d) Label any angles used


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5. Identify appropriate equation(s)and write down in symbolicform(no numbers)

6. Solve equations for unknownquantities in symbolicform

7. Substitute known quantities intoequations using both numerical

values and units


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8. Write answer so it is identified byits symbol, has the appropriatenumber of significant digits, and

has the proper units. Put a boxaround your answer.

9. Check your work. Make sureyou answered question that wasasked! Check to see that you

answered all parts of question.


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Sample Problem2 objects2 cars drive in same direction down straight road

at constant speedone at 15 m/s & the other at35 m/s. Both cross a starting line at same time.

a) How much sooner does faster car cross a

finish line 1.5 km away?

b) How far away is (different) finish line if faster

car arrives 1.50 minutes before slower one?

v1i= 15 m/s v2i= 35 m/s

x= 1.5 km = 1.5x103m

tdiff= t1t2= ??v1i

x

v2i

S


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Sample Problem2 objectsa) How much sooner does faster car cross a

finish line 1.5 km away?v1i= 15 m/s v2i= 35 m/s

x= 1.5 km = 1.5x103m

tdiff

= t1

t2

= ??v1i

x

v2i

vavg= v= x/t t = x/v

t1=x/v1i t2=x/v2i tdiff= x(1/v1i1/v2i)tdiff= 1.5x10

3m (1/15 m/s 1/35 m/s)tdiff= 1.5x10

3m (6.7x10-2s/m 2.9x10-2s/m)

tdiff= 1.5x103m 3.8x10-2s/m = 57 s

S l P bl 2 bj


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Sample Problem2 objectsb. How far away is (different) finish line if faster

car arrives 1.50 minutes before slower one?v1i= 15 m/s v2i= 35 m/s

tdiff= 1.50 min = 90.0 s

x= ??v1i

x

v2i

tdiff= x(1/v1i1/v2i) derived in part a

x = tdiff/ (1/v1i1/v2i) = 90.0 s / 3.8x10-2s/m

x = 2400 m = 2.4 km


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2.1 Displacement and Velocity2.2 Acceleration

2.3 Falling Objects

Obj ti f S ti 2 2


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1. Describe motion in terms of changing

velocity.

2. Interpret and/or generate graphical

representations of accelerated andnonaccelerated motions.

3. Properly relate the signs of

displacement, velocity and accelerationto the type of motion that is occurring.

Obj ti f S ti 2 2


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4. Apply kinematic equations to solve one-

dimensional motion problems involvingdistance, displacement, time, and/or

acceleration using proper problem-

solving techniques (see objective 5 for

section 2.1), including one-dimensional

motion along a ramp.

A l ti


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Acceleration

Acceleration measures the rate ofchange of the velocity Stepping on the gas

Hitting the brakesaavg = v / t = (vf-vi)/(tf-ti)

Average acceleration =

change in velocity / time intervalUnits = (m/s) / s = m/s2

A l ti


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AccelerationAvg Acceleration prob. 2B, page 49


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Practice

Average AccelerationPractice 2B, page 49

Problems 1-5

Chapter Review page 71Problems 20, 30

A l ti


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AccelerationCan deduce values from velocity time

graphsee figure 2-10

+ acc?

- acc?

0 acc?

A l ti


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Acceleration

Acceleration has magnitude anddirection

Have to worry about sign of velocity,

which in turn depends upon sign ofdisplacement

For a train leaving a station in thenegative direction, acceleration is

negative as the train is speeding up

Acceleration


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Acceleration

Table 2-3 (page 51)Interpreting aspects

of motion from signs of viand avi a Motion

+ + Speeding up

- - Speeding up

+ - Slowing down

- + Slowing down- or + 0 Constant velocity

0 - or + Speeding up from rest

0 0 At rest

M ti ith C t t A l ti


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Motion with Constant Acceleration

Motion of falling ball (constanttime intervals)

Constant acceleration due to

gravity

Simple formulas exist for

constant acceleration whichallow relatively easy calculations

of parameters of motion

M ti ith C t t A l ti


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Average velocity[1] vavg= x / t

Consider constant acceleration case

See next slide or fig 2-12, page 52 forvelocity vs time profile

[2]vavg= (vf+vi) / 2

Combining equations 1 and 2 yields

x= (vf+vi) t

Displacement = f(vi,vf,t) Constant Acceleration


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Constant Acceleration

and Average Velocity

vavg= (vf+vi) / 2


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Practice

Displacement with uniform accelerationPractice 2C, page 53

Problems 1-5

Chapter Review pages 71-72Problems 23, 25, 27,



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If aconstant, a= aavg

a =aavg= (vf-vi) / (tfti)= (vf-vi) / t

a t = (vf

-vi

)

[1]vf= vi+ a t

vf = f(vi, a, t)

Combine [1]with previous expressionx= (vf+vi) t

x= vit+ a(t)2

Dis lacement = x = f v a t Acceleration


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AccelerationV and xwith uniform acceleration

Problem 2D, page 55


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Practice

V and xwith uniform accelerationPractice 2D, page 55

Problems 1-4

Chapter Review pages 71-72Problems 21, 22, 24, 26, 28, 29, 31



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By doing some algebra (see page 56 of

text), can derive an expression whichdoes not depend on t

vf

2= vi

2+ 2ax

Caution: because solution for either vi

or vf comes from square root operation,

the signof the velocityis not knownfrom the formulamust determine by

reasoning

Acceleration


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AccelerationVfafter any displacement

Problem 2E, page 57


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Practice

Final Velocity After Any DisplacementPractice 2E, page 58

Problems 1-6

Chapter Review, page 72Problems 32-33

S f ( )


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Summary of Basic Equations (1)

Three basic definitionequations, oneeach for:

Displacement

Average VelocityAverage Acceleration

(See next slide)

Basic (Defining) Equations


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Displacement: change in position

= final positioninitial position

x = xfxi

Average velocity: displacement / timeinterval

vavg= x / t = (xf-xi)/(tf-ti)

Average acceleration: change in velocity /

time interval

aavg =v /

t = (vf- vi) / (tf- ti)

Basic (Defining) Equations



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Fourequations for uniformlyaccelerated motion expressed in terms

of fivevariables

xdisplacementttime interval

vi initial velocity

vf final velocityaacceleration

Any single equation uses only four of

these variables


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Dont need a

Dont needx

Dont need vf

Dont needt

Equations for Straight Line Motion


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Equations for Straight Line Motion

with Constant Accelerationpg 58

Practice Problems 5 6 page 59


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PracticeProblems 5,6 page 59

Time intervals during which vis constant?Time intervals during which ais constant?Time intervals during which ais zero?Is bus always moving in same direction?

Time (s)

V

elocity(m

/s)



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vavg= (vi+ vf)aavg= (vfvi)/t

x= (vi+ vf)

t = vavg

t

x= vit+ at2

vf= vi+ a

tvf

2= vi2+ 2 ax

Sample Problem


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Sample Problem

How fast is it moving at end of 5.0 seconds?

How far has it gone in first 5.0 seconds?

Average speed for first 5.0 seconds?How far does it go between 5.0 and 10.0

seconds?

Average velocity over 10.0 seconds?

Police car starts from rest and acceleratesat constant rate of 3.0 m/s2for 5.0 seconds,

then continues moving with constant speed

Diagram Givens Unknowns


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Diagram, Givens, Unknowns

Police car starts from rest and accelerates

at constant rate of 3.0 m/s2for 5.0 seconds,then continues moving with constant speed

vi1= 0 m/s

t1= 5.0 st2= 5.0 s

a1= 3.0 m/s2 (for t 5 s)

a2= 0 (for t > 5 s)

v

vf1 = ?

x1= ?vavg1 = ?

x2= ?

vavg = ?

x1 x2

Part 1 vf1


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Part 1 vf1v

x1x2

vi1= 0 m/s

t1= 5.0 s

a1= 3.0 m/s2 (for t 5 s)

vf1= vi1+ a1t1

vf1= 0 + 3.0 m/s25.0 svf1= 15 m/s

Part 2 - x1


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Part 2 x1v

x1x2

vi1= 0 m/s

t1

= 5.0 s

a1= 3.0 m/s2 (for t 5 s)

x1= vi1t+ a1t12

x1= 0 + 3.0 m/s2(5.0 s)2x1= 1.5 m/s

225 s2

x1= 38 m

Part 3 - v 1


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Part 3 vavg1v

x1x2

vi1= 0 m/s

t1

= 5.0 s

a1= 3.0 m/s2 (for t 5 s)

x1= vavg1t1

vavg1= x1/ t1 [have x1 from part 2]vavg1= 38 m / 5.0 s

vavg1= 7.6 m/s


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Part 5 v


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Part 5 vavgv

x1x2

x1= 38 m [ from solution to part 2]

x2= 75 m [ from solution to part 4]

t1= 5.0 s t2= 5.0 sxtot= vavgttotvavg= xtot/ ttot

xtot= x1 + x2= 38 m + 75 m = 113 m

ttot= t1 + t2= 5.0 s + 5.0 s = 10.0 s

vavg = xtot / ttot = 113 m/10.0 s = 11.3 m/s Rolling Ball Down CPO Straight Ramp


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Rolling Ball Down CPO Straight Ramp

g

v0= 0 (ball released from rest), = 57.0

x1= 20.0 cm, x

2= 60.0 cm

v1= ?, v2= ?, t12= ?

Convert calculated velocities into photogate

times using steel ball diameter = 18.9 mm

cos() =gx/ggx= gcos() = ax= 5.34 m/s

2

???

x1= 20.0 cm, x2= 60.0 cm


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v12

= v02

+ 2 axx1

v22= v0

2+ 2 axx2Gate time = d/ v

v2= v1+ axt12

Convert calculated velocities into photogate

times using steel ball diameter = 18.9 mm

v12= 2 5.34 m/s20.200 m = 2.14 m2/s2

v1= 1.46 m/sgate1 = 1.89x10

-2m / 1.46 m/s = 12.9 ms

1 2

v1= ?, v2= ?, t12= ?

gate1 = 12.9 ms

v22= 2 5.34 m/s20.600 m = 6.41 m2/s2

v2= 2.53 m/sgate2 = 1.89x10

-2m / 2.53 m/s = 7.47 ms

gate2 = 7.47 ms

t12= (v2v1) / ax

t12= (2.53 m/s1.46 m/s) / 5.34 m/s2= 0.200 s

t12= 0.200 s

gate1 = 15.6 ms gate2 = 9.0 ms t12= 0.247 s

Measuredtimes different than calculatedtimesWhy?Assume measured gate1is correct. Calculate

new ax& use it to calculate gate2and t12.

ax(exptl) = v12/(2x1) = (1.21 m/s)2/(0.400 m)ax(exptl) = 3.66 m/s

2

Practice


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Practice

Section Review, page 59Problems 1-4

Chapter 2 Motion in One Dimension


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2.1 Displacement and Velocity

2.2 Acceleration

2.3 Falling Objects

Objectives for Section 2 3


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1. Relate the motion of a freely falling body

to motion with constant acceleration.

2. Solve problems involving displacement,

velocity, and time for a freely falling

object or objects using proper problem-

solving techniques (see objective 5 for

section 2.1).

3. Compare the motion of different objects

in free fall.

Free Fall


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Free Fall

Time (s)Veloc

ity(m/s)

Constant (-acc) even when ball moving up

Shown by constantslope

Velocity is zero at peak height Feather and Apple in Free Fall


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ea e a d pp e ee a

(Vacuum)

In absence of air

resistance, all

objects regardlessof size or mass fall

with same

acceleration (that ofgravity)

Free Fall


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Free Fall

Acceleration due to gravity denoted

with symbol gValue at earths surface = 9.81 m/s2

Books convention: upis positivedirection

Thus, a= g= -9.81 m/s2

All equations developed for previoussection apply to free fall problems but

replace

x with

y Free Fall - Sample Problem


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Free Fall Sample ProblemBaseball thrown straight upward at 40.0 m/s

1. How high is it at 2.0 s?

2. How fast is it moving at 2.0 s?

3. When does it reach its maximumheight?

4. How high does it go?

5. How long is it in air (to return to start)?

6. How fast is it going when it reaches its

starting point?

Diagram, Givens, Unknowns


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ag a , G e s, U o s

vi= 40.0 m/sa=9.81 m/s2

Baseball thrown straight upward at 40.0 m/s

v

y

t

ymax

y (2.0 s) = ?v (2.0 s) = ?

tmax = ?

ymax = ?ttot = ?

vf = ?

tmax

ttot

Part 1 - y (at 2.0 s)


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y ( )

vi y

vi= 40.0 m/s

a=9.81 m/s2

t = 2.0 s

y= vit+ at2

y= 40.0 m/s

2.0 s 9.81 m/s

2

(2.0 s)

2

y= 80.0 m2.0x101m

y= 6.0x101m

Part 2 - v (at 2.0 s)v


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vi y

vi= 40.0 m/s

a=9.81 m/s2

t = 2.0 s

vf= vi+ at

vf= 40.0 m/s9.81 m/s

2

2.0 s

vf= 40.0 m/s2.0x101m/s

vf= 2.0x101m/s

( )vf

Part 3 - tmaxv


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vi= 40.0 m/s

a=9.81 m/s2

vf= vi+ atmaxvf now refers to vat highest point = 0 m/s

0= vi+ atmaxtmax=vi/ a

tmax=40.0 m/s / (9.81 m/s2)

tmax= 4.08 s

max

y

tmax

vi ymax

vf

Part 4 - ymaxv


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vi= 40.0 m/s

a=9.81 m/s2

tmax= 4.08 s [from part 3]

ymax= vitmax+ at2

max

ymax= 40.0 m/s 4.08 s

9.81 m/s2(4.08 s)2ymax= 163 m81.7 m

ymax= 81 m

ymax

vi ymax

vf

Alternate solution

vf2= vi

2+ 2 aymax

ymax= - vi2/(2a)

Part 5 - ttot


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vi= 40.0 m/s

a=9.81 m/s2

ytot= 0 when ball lands

ytot= vittot+ at2tot= 0

ttot(vi+ attot)= 0 [root when ( )=0]

ttot=2 vi/ a=2 40.0 m/s / (9.81 m/s2)

ttot= 8.15 s = 2

tmax

tot

y

ttot

vi ytotTime to go up

always = time to

come down (will

prove this)

Part 6 - vf


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vi= 40.0 m/s

a=9.81 m/s2

vf2= vi

2+ 2 aytot

ytot= 0 when ball landsvf

2= vi2

vf= vi2 [know must be opposite, so use]

vf= vi

f

vi ytot

vf

Comes down at

same speed as

was sent up!

Proof of t(up)= t(down)


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y(u) =y(d) (distance traveled same)

Trip up: vf(u) = 0 Trip down: vi(d) = 0

y= (vi+ vf)t t =2y/(vi+ vf)t(u)= 2y(u)/vi(u) t(down)= 2y(u)/vf(d)

From previous slide (part 6): vf(d) =vi(u)

t(u) = t(down)

( p) ( )

ttot

vi yup

y

ttup

Divide flight of

ball into 2 parts:

u = trip up

d = trip down tdown

2 Object Problem with 2 Versions


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j

v01= 0.0 m/s

v02=41.0 m/s

a=9.81 m/s2

t1= t2= t

Dog drops ball, boy throws it straight down

at 41.0 m/s. Both balls hit simultaneously.

Version 1

h2 - h1= 5.90x102mVersion 2

h2/h1= r= 1.580

h1, h2= ?These problems aredesigned to

challenge your

mathematical skills!

The solutions

illustrated are notnecessarily the only

ones possible.

Two Object ProblemVersion 1


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jvy01= 0.0 m/s

vy02

=41.0 m/s

a=9.81 m/s2

h2- h1= h= 5.90x102m

t1=

t2=

t

h1, h2= ?

y1= -h1= vy01t+ at2 = at2

y2= -h2= vy02t+ at2

h2- h1= -vy02t- at2 + at2h2- h1= -vy02t = h

t = -h/ vy02= -5.90x102m/ -41.0 m/s

t = 14.4 s



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jvy01= 0.0 m/s

vy02=41.0 m/sa=9.81 m/s2

h2- h1= h= 5.90x102m

t1= t2= t

h1, h2= ?

t = 14.4 s

h1= - at2 = 0.59.81 m/s2 (14.4 s)2

h1= 1020 m

h2= h1+ 5.90x102m= 1610 m



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jvy01= 0.0 m/s

vy02=41.0 m/sa=9.81 m/s2

h2/h1= r= 1.580

t1= t2= t

h1, h2= ?

y1= -h1= vy01t+ at2

t = (-2h1/ a)

y2= -h2= vy02t+ at2

h2= -vy02 (-2h1/a) + h1r h1= -vy02 (-2h1/a) + h1

(r 1)h1 = -vy02 (-2h1/a) Square both sides



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jvy01= 0.0 m/s

vy02=41.0 m/sa=9.81 m/s2

h2/h1= r= 1.580

t1= t2= t

h1, h2= ?

(r1)2h12+ 2 vy02

2h1/a = 0

h1[(r1)2h1 + 2vy02

2/a] = 0 Find roots

h1 = -2vy022

/[a(r1)2

]h1 = -2(-41.0 m/s)

2/ [-9.81 m/s2 (0.580)2]

h1 = 1018 m = 1020 m

h2 = h1 r = 1610 m

St ti f 2 0 b fl ll b ll hitFalling object problem, p. 63, direct solution


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Starting from 2.0 m above floor, volleyball hit

straight up, vi= 6.0 m/s.

How long in air before it hits floor?Coord. system origin at starting point

y=2.0 m a= -g = -9.81 m/s2 t=?

y= vit+ at2

at2+ vit- y= 0 use quadratic formula

A x2

+ B x + C= 0 x = -B[B

2

4AC]/2At= -vi[vi

24(a/2)(-y)]/a =

-vi[vi2+2ay]/a=(-6.0 m/s 8.7m/s )/-9.81 m/s2

t = -14.7 m/s / -9.81 m/s2 = 1.50 s

St ti f 2 0 b fl ll b ll hitFalling object problem, p. 63, book solution


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Starting from 2.0 m above floor, volleyball hit

straight up, vi= 6.0 m/s.

How long in air before it hits floor?Coord. system origin at starting point

y=2.0 m a= g = -9.81 m/s2 t=?

Use 2 equations to avoid use of quadratic formula

vf2= vi

2+ 2 ay vf= vi+ at

Once vfknown, 2d equation will give t

vf= (vi2 + 2ay)

= (36 m2/s2+29.81 m/s22.0 m) = 75 m2/s2

vf= 8.7 m/s (going down) t=(vf vi)/a = 1.50 s

Falling Objects


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g jFalling object prob. 2F, page 63

Key points: Dont need to know how high it goes

onlyneed y

Need to pick a coordinate system Need to stick with up as positive

If want to avoid quadratic formula,

need to use two different equationsthe first to get vf, the second to get t

Forced to decide the signof vfon your

own from h sics of roblem

Practice


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Falling Object

Practice 2F, page 64Problems 1- 6

Section Review, page 65Problems 1-6

Chapter Review, pages 72-73

Conceptual Problems 34(a-e), 35,36, 37(a-c)

Problems 38 42

ch 02 motion in one dimension

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