Chapter 2 New Kinema Tics

8/13/2019 Chapter 2 New Kinema Tics

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2.1 Linear Motion

Displacement

Velocity Average velocityis defined as the change in displacement over the time, t

Instantaneous velocityis called the limit of x/t as approaches zero or thederivative for x with respect to time, t

Direction of vis a tangent to the path at that point in the direction of motion

Magnitude of vis thespeed

r2

r1

rx

ry

r = rxi r

yj

rr2 r

1

t

r

v

dtd

ttrr

limv

tangen


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Acceleration Average accelerationdefine as the ratio of the change in the velocity !"!2#!1

to the time interval t"t2#t1

$ Instantaneous accelerationlimit of !/t as approaches zero or the derivative

for x with respect to time, t

$ Magnitude change of vonly # 1D motion

$ Direction change of vonly # circular motion

Uniform motion

2.2 Motion with %onstant a& constant acceleration means that the acceleration does not depend on time'

(ntegrating this e)uation, the velocity of the o*+ect can *e o*tained

radient " v " uniform

r

t

t

v&rea " vt " r

v

t

a "

t

v

a

dt

d

tt

vv

lima


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where v is the velocity of the o*+ect at time t " .

-rom the velocity, the position of the o*+ect asfunction of time can *e calculated'

where x0 is the position of the object at time t = 0

Graph of velocity vs time with a constant

Displacement " graph area

graph gradient " acceleration

velocity at any later time t,

Displacement s, at any later time t,

&nd

vo

t

vt

v

=+= vttvvS o 2

1

t

vva o

=

atvv o +=

2

2

1attvs

o +=

asvv o 222

+=


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2.0 -ree falling motion

hat you understand a*out freely falling *odies -ree -all 3)uation

4he 3)uation of Motion for the 1#D motion of an o*+ect in free fall near the 3arth5s

surface are o*tained from e)uations

(f we use the coordinate system shown,then,when su*stituting into e)uationa " g " #6.7 ms#2

8ince g points down

4his is a nice example of constant acceleration gravity'

(n this case, acceleration is caused *y the force of gravity'

9sually pic: y#axis ;upward< &cceleration of gravity is ;down


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-ind the time the stone reaches its original height -ind the velocity of the stone when it reaches its original height -ind the velocity and position of the stone at t"?.s

1.

2.

0.

@.

/.26

.?7.6.2

sm

tavv yyiyf

=

+=

+=

?.2A.?7.62

1.?.2.?

2

1

2

2

m

tatvyy yyiif

+=++=

++=

5-Velocity

5-Position

st

ttavv yyif

@.27.6.2

.7.6.2

==

=+== +

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@.27.62

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2

1

2

2

m

tatvyy yyiif

=+=

++=

++=

st 7.@[email protected] ==

/[email protected] smtavv yyiyf =+=+=


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=ro*lem 2.1.2

$ 4he pilot of a hovering helicopter drops a lead *ric: from a height of 1 m.Bow long does it ta:e to reach the ground and how fast is it moving when it gets

there neglect air resistance

1000 m


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8olution'

-irst choose coordinate system. Crigin andy#direction.

ext write down position e)uation'

Eealize that v0y= 0.

1000 m

y = 0

y

2y gt

2

1tvyy +=

2

2

1gtyy =


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9/13

=ro+ectile Motion

& 2#dim motion of an o*+ect under the gravitational acceleration with theassumptions

-ree fall acceleration, #g, is constant over the range of the motion

&ir resistance and other effects are negligi*le

$ & motion under constant accelerationFFFF8uperposition of two motions

Borizontal motion with constant velocity and !ertical motion under constant

acceleration8how that a pro+ectile motion is a para*olaFFF

2

22

2

cos2tan

cos21

cossin

cos

f

ii

if

ii

f

ii

fiif

ii

f

xv

gx

vxg

vxvy

v

xt

=

=

=

3xample @.1

vx= u

x+ a

xt

vy= u

y+ a

yt

X = xo+ v

xt + (1/2)a

xt2

Y = yo + vyt + (1/2)ayt2

sm

gyvy

/1@

2

=

=

( )

2

2

2

1sin

2

1

cos

sin,cos

gttv

tgtvy

tvtvx

vvvv

jgjaiaa

ii

yif

iixif

iiyiixi

yx

=

+=

==

==

=+=


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& *all is thrown with an initial velocity v"2i>@jm/s. 3stimate the time of flight and

the distance the *all is from the original position when landed.

hich component determines the flight time and the distance

Borizontal Eange and Max Beight

Gased on what we have learned previously, one can analyze a pro+ectile motion in

more detail Maximum height an o*+ect can reach Maximum range

Flight time is determined by y

component, because the ballstops moving when it is on theground after the flight.

Distance is determined by x

component in 2-dim, because the

ball is at y=0position when itcompleted its flight.

( )

( )

s77

or

7

2

1@ 2

==

=

+=

gtt

gtt

tgtyf

(

tvx xif

1H72 ==

=


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vvii

h

!t the ma"imum height the ob#ects vertical motion stopsto turn around$$

%ince noacceleration in ", it

still flies atvy&'

g

vt

gtvtavv

iA

Aiyyiyf

sin

sin

=

==+=( )

=

=

+==

g

v

g

vg

g

vv

tgtvhy

ii

iiiiii

yif

2

sin

sin

2

1sinsin

2

1

22

2

2

=

==

g

v

g

vv

tvR

ii

iiii

Axi

2sin

sincos2

2

2


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Maximum Eange and Beight

hat are the conditions that give maximum height and range in a pro+ectilemotion

3xample @.2

( & stone was thrown upward from the top of a *uilding at an angle of 0o tohorizontal with initial speed of 2.m/s. (f the height of the *uilding is @?.m,how long is it *efore the stone hits the ground

hat is the speed of the stone +ust *efore it hits the ground

=

g

vh ii

2

sin22

)his formula tells us that thema"imum hieght can be achieved

when i&*'o$$$

=

g

vR ii

2sin2

)his formula tells us that thema"imum hieght can be

achieved when 2i&*'o, i.e.,

i&+5o$$$

=

g

vR ii

2sin2

( )

st

stst

t

tttgt

gttvy

smvv

smvv

yif

iiyi

ixi

22.@

[email protected]

7.62

[email protected]

.6.27.6.6.2

2

1.@?

/.10sin.2sin

/0.1A0cos.2cos

2

22

2

===

=

==

==

===

===


13/13

( ) smvvv

smgtvgtvv

smvvv

yfxf

iiyiyf

ixixf

/[email protected]

/@[email protected]

/0.1A0cos.2cos

2222

=+=+=

====

====

Chapter 2 New Kinema Tics

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