Chapter 2. Kinematics in One Dimension - GSU P&Aphysics.gsu.edu/dhamala/Physics2211/Chapter2.pdf ·...

In this chapter we study kinematics of motion in one dimension—motion along a straight line. Runners, drag racers, and skiers are just a few examples of motion in

Chapter 2. Kinematics in One Dimension

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

few examples of motion in one dimension.

Chapter Goal: To learn how to solve problems about motion in a straight line.

Topics:

• Uniform Motion

• Instantaneous Velocity

• Finding Position from Velocity

Chapter 2. Kinematics in One DimensionChapter 2. Kinematics in One Dimension


• Finding Position from Velocity

• Motion with Constant Acceleration

• Free Fall

• Motion on an Inclined Plane

• Instantaneous Acceleration

1 s 2 s 3 s 4 s

Motion along a straight line

Can be illustrated by position-

versus-time graph:

x Continuous

(smooth)

curve


3

xOrigin

(x=0)10

cm

20

cm40

cm

70

cm

t(s)1 2 3 4

Position graph

x1

sec

2

sec

3

sec

4

sec

xv

∆∆∆∆====

x∆∆∆∆

t∆∆∆∆


4

t(s)1 2 3 4

vrrrr

Velocity is the

same – zero

acceleration

0avg

va

t

∆∆∆∆= == == == =

∆∆∆∆

rrrrrrrr

Straight line – uniform

motion – the same velocity

Velocity is the slope

xv

t

∆∆∆∆====

∆∆∆∆

Position graph: uniform motion

x

0x

vt

∆∆∆∆= == == == =

∆∆∆∆

Motions in opposite directions 0v >>>>


5

t(s)

0v <<<<

Uniform motion: motion with constant velocity

x0

0

tx xx

vt t t

−−−−∆∆∆∆= == == == =

∆ −∆ −∆ −∆ −

0x

0 0( )

tx x v t t= + −= + −= + −= + −


6

t0t

0 0t

v

t0t

Position graph


7

unrealistic realistic


8

Instantaneous velocity

xv

t

∆∆∆∆====

∆∆∆∆

x∆∆∆∆

t∆∆∆∆

Very small t∆∆∆∆

Instantaneous

velocity:

x


9

Instantaneous velocity - slope

If dependence is given,

then

( )x t

( )x dx tv

t dt

∆∆∆∆= == == == =

∆∆∆∆- derivative

Example:4( ) 5x t t====

3( )20

dx tv t

dt= == == == =

( ) 5sin( )x t t====( )

5cos( )dx t

v tdt

= == == == =

Finding position from velocity

( )dx tv

dt==== ( ) ( )x t v t dt==== ∫∫∫∫


10

2

1

2 1 2 1( )

t

t

x x x v t dt−−−−

= − == − == − == − = ∫∫∫∫

v

t1t

1( )x

2( )x

2t

The displacement is the area, displacement is t


11

The displacement is the area, displacement is

positivev

1t

2t t

Displacement is

negative

1( )x

2( )x

2

1

2 1 2 1( ) 0

t

t


= − = >= − = >= − = >= − = >∫∫∫∫

2

1

2 1 2 1( ) 0

t

t


= − = <= − = <= − = <= − = <∫∫∫∫

( ) 0v t <<<<

( ) 0v t >>>>

Example:

Find the net displacement

A B

C D

1Bt

∫∫∫∫


12

1( ) 2 2 2

2

B

A

t

A B B A

t

x x x v t dt m−−−−

= − = = − = −= − = = − = −= − = = − = −= − = = − = −∫∫∫∫ ��

1( ) 2 2 2

2

C

B

t

B C C B

t


= − = = == − = = == − = = == − = = =∫∫∫∫ ��

area

( ) 2 (10 4) 12D

C

t

C D D C

t


= − = = − == − = = − == − = = − == − = = − =∫∫∫∫ ��

2 2 12 12

A D D A D C C B B A C D B C A Bx x x x x x x x x x x x

m

− − − −− − − −− − − −− − − −= − = − + − + − = + += − = − + − + − = + += − = − + − + − = + += − = − + − + − = + +

= − + + == − + + == − + + == − + + =

Velocity is the

slope dsv

dt====

x

t

A

B

C

0A

v <<<< 0C

v >>>>


13

t

A B Cv v v< << << << <

v dv∆∆∆∆

Velocity is the

slope dsv

dt====

Acceleration is the slope


14

v dva

t dt

∆∆∆∆= == == == =

∆∆∆∆

Motion with constant acceleration

Motion with constant velocity:

0

0

tx xx

vt t t

−−−−∆∆∆∆= == == == =

∆ −∆ −∆ −∆ −0 0

( )t

x x v t t= + −= + −= + −= + −

Motion with constant

acceleration:0t

v vva

−−−−∆∆∆∆= == == == = ( )v v a t t= + −= + −= + −= + −


15

t

t

dxv

dt====

0 0

2

0 0 0 0 0 0 0 0

1( ( )) ( ) ( )

2

t t

t t

t t

x x v dt x v a t t dt x v t t a t t= + = + + − = + − + −= + = + + − = + − + −= + = + + − = + − + −= + = + + − = + − + −∫ ∫∫ ∫∫ ∫∫ ∫

0

0

tv vv

at t t

−−−−∆∆∆∆= == == == =

∆ −∆ −∆ −∆ −0 0

( )t

v v a t t= + −= + −= + −= + −

?t

x ====

Motion with constant velocity or constant acceleration

2

0 0 0 0

1( ) ( )

2t

x x v t t a t t= + − + −= + − + −= + − + −= + − + −

Motion with constant velocity: 0 0( )

tx x v t t= + −= + −= + −= + −


acceleration:0 0

( )t

v v a t t= + −= + −= + −= + −

Useful relation:


16

2Useful relation:

0

0( ) t

v vt t

a

−−−−− =− =− =− = 2

0 0

0 0

( )1

2

t t

t

v v v vx x v

a a

− −− −− −− −= + += + += + += + +

then

2 2

0 02 ( )

t tv v a x x− = −− = −− = −− = −

Motion with constant velocity or constant acceleration

Motion with constant velocity: 0tx x v t= += += += +

Motion with constant v v a t= += += += +

00t ====


17

2

0 0

1

2t

x x v t a t= + += + += + += + +


acceleration:0t

v v a t= += += += +

2 2

0 02 ( )

t tv v a x x− = −− = −− = −− = −

Free fall motion - motion with constant acceleration

29.8

ma g

s= − = −= − = −= − = −= − = −

constantg ====rrrr

y

0tv v g t= −= −= −= −


18

2

0 0

1

2t

y y v t g t= + −= + −= + −= + −

2 2

0 02 ( )

t tv v g y y− = − −− = − −− = − −− = − −

Free fall motion - motion with constant acceleration

29.8

mg

s====

02 /v m s====

====

0final

v ====

2

0 0

1

2t

y y v t g t= + −= + −= + −= + −

0tv v g t= −= −= −= −

2 2

0 02 ( )

t tv v g y y− = − −− = − −− = − −− = − −


19

00y ====

22 4.9t

y t t= −= −= −= −

2 9.8t

v t= −= −= −= −

2 22 2 9.8t t

v y− = − × ×− = − × ×− = − × ×− = − × ×

At final point: 0

finalv ====

22 4.9f f f

y t t= −= −= −= −

0 2 9.8f

t= −= −= −= −

22 2 9.8f

y− = − × ×− = − × ×− = − × ×− = − × ×

EXAMPLE 2.14 Friday night football

QUESTION:






Motion on an Inclined Plane


EXAMPLE 2.17 Skiing down an incline

QUESTION:


Chapter 2. Summary SlidesChapter 2. Summary Slides


Chapter 2. Summary SlidesChapter 2. Summary Slides

General Principles


General Principles


Important Concepts


Important Concepts


Applications


Applications


Chapter 2. QuestionsChapter 2. Questions


Chapter 2. QuestionsChapter 2. Questions

Which position-versus-time graph represents the motion shown in the motion diagram?


Which position-versus-time graph represents the motion shown in the motion diagram?


Which velocity-versus-time graph goes with the position-versus-time graph on the left?


Which velocity-versus-time graph goes with the position-versus-time graph on the left?


Which position-versus-time graph goes with the velocity-versus-time graph at the top? The particle’s position at ti = 0 s is xi = –10 m.


Which position-versus-time graph goes with the velocity-versus-time graph at the top? The particle’s position at ti = 0 s is xi = –10 m.


Which velocity-versus-time graph or graphs goes with this acceleration-versus-time graph? The particle is initially moving to the right and eventually to the left.


Which velocity-versus-time graph or graphs goes with this acceleration-versus-time graph? The particle is initially moving to the right and eventually to the left.


The ball rolls up the ramp, then back down. Which is the correct acceleration graph?


The ball rolls up the ramp, then back down. Which is the correct acceleration graph?


Rank in order, from largest to smallest, the accelerations a

A– a

Cat

points A – C.

A) aA

> aB

> aC

B) a > a > a


B) aA

> aC

> aB

C) aB

> aA

> aC

D) aC

> aA

> aB

E) aC

> aB

> aA

A) aA

> aB

> aC

B) a > a > a

Rank in order, from largest to smallest, the accelerations a

A– a

Cat

points A – C.


B) aA

> aC

> aB

C) aB

> aA

> aC

D) aC

> aA

> aB

E) aC > aB > aA

Chapter 2. Kinematics in One Dimension - GSU P&Aphysics.gsu.edu/dhamala/Physics2211/Chapter2.pdf ·...

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