Chapter 4 Two–Dimensional Kinematics

Section 4.1

Constant

velocity

motion

A

body

moving

with

constant

velocity

in

two

dimensions

travels

equal

distances

for

equal

time

intervals

(not

necessarily

the

same

distance

along

each

dimension,

i.e.

∆x

is

not

generally

the

same

as

∆y).

On a graph of x versus time, constant velocity still appears as a straight line where: positive

velocity

motion

has

a

positive

slope

(points

up),

zero

velocity

motion

is

a

horizontal

line,

and

negative

velocity

motion

has

a

negative

slope

(points

down).

The

slope

of

a

straight

line

on

the

graph

of

x

versus

time

is

equal

to

the

x

component

of

velocity.

Replacing the "x" in the previous paragraph with "y" would hold for the y component of

velocity as well.

Constant

acceleration

motion

A

body

moving

with

constant

vector

acceleration

travels

different

distances

for

equal

time

intervals.

In

two

dimensions,

there

are

three

possibilities:

1)

only

an

x

component

of

acceleration,

2)

only

a

y

component

of

acceleration,

or

3)

acceleration

components

along

both

x

and

y

directions.

(The

case

where

acceleration

is

zero

in

both

x

and

y

directions

is

just

the

same

as

the

constant

velocity

case

above)

Graphs

can

be

drawn

of

velocity

versus

time

for

the

two

dimensional

case,

but

separate

graphs

would

be

needed

for

v

x

and

v

y

versus

time

in

order

to

calculate

the

x

and

y

displacements

(∆x

and

∆y)

from

the

area

under

the

graph

line.

Kinematic equations in two dimensions

Motion in two dimensions (x and y) under

constant

acceleration

is

described

using

the

following

two

sets

of

equations

(Note

that

the

set

of

x

components

is

used

to

describe

one-dimensional

motion.):

x components

y components

v

x

=

v

0

x

+

a

x

t

v

y

=

v

0

y

+

a

y

t

x

–

x

0

=

1

2

(

v

0x

+

v

x

)

t

y

–

y

0

=

1

2

(

v

0y

+

v

y

)

t

x

–

x

0

=

v

0

x

t

+

1

2

a

x

t

2

y

–

y

0

=

v

0

y

t

+

1

2

a

y

t

2

v

x

2

=

v

0

x

2

+

2

a

x

(x

–

x

0

)

v

y

2

=

v

0

y

2

+

2

a

y

(y

–

y

0

)

Meaning

of

symbols

above

(for

y

components,

replace

x

below

with

y):

x – x

0

= displacement along x-axis (from origin)

a

x

=

x-component

of

acceleration

v

x

=

final

x-component

of

velocity

t

=

time

that

has

passed

(from

t

=

0)

v

0

x

= initial x-component of velocity

Displacement

and

velocity

can

also

be

represented

using

the

unit

vector

notation,

e.g.:

∆

r

→

=

v

0

x

t +

1

2

a

x

t

2

x

^

+

v

0

y

t +

1

2

a

y

t

2

y

^

v

→

=

(

)

v

0

x

+ a

x

t

x

^

+

(

)

v

0

y

+ a

y

t

y

^

Variables in equations above for specific situations:

Two-dimensional projectile motion(vertical versus horizontal):

Either

v

0

x

or

v

0y

may

be

non-

zero.

v

x

will

not

change

over

time,

but

v

y

≠

v

0

y

.

For

accelerations,

a

x

= 0 and a

y

= –g =

–9.81 m/s

2

.

Section 4.3

Case

1

(Zero

Launch

Angle):

v

0y

= 0,

v

0x

≠

0.

The

path

traced

out

by

the

object

will

be

half

a

downward

parabola

(Fig

4-5).

v

x

=

v

0x

over

the

course

of

the

path

(a

x

= 0) while v

y

=

–gt

and

becomes

increasing

larger

in

the

negative

y

direction.

Section 4.4,5

Case

2

(General

Launch

Angle,

where

∆y

=

0):

v

0y

≠

0,

v

0x

≠

0.

The

path

is

a

symmetric

parabola

opening

down

and

centered

around

the

maximum

height

(Fig

4-7,8,

11).

If

given

the initial velocity magnitude and launch angle, the time of flight, maximum height, range,

and final velocity can be calculated. In the absence of air friction, the maximum horizontal

range

occurs

when

the

launch

angle

is

45°.

The

velocity

vector

has

the

same

magnitude

going

up

through

a

value

of

y

as

it

does

coming

back

down

through

the

same

value

of

y

(Fig

4-10).

The

velocity

takes

on

its

minimum

magnitude

at

the

maximum

height

(=

v

0x

).

Case

3

(General

Launch

Angle,

where

∆y

≠

0):

v

0y

≠

0,

v

0x

≠

0.

The

path

is

no

longer

a

symmetric

parabola,

but

is

still

parabolic

in

nature

(Fig

4-20).

Problems

of

this

sort

that

ask

for

time

will

generate

a

quadratic

equation

in

time

that

must

be

solved.

Some

additional

hints

for

problem

solving

(some

of

these

are

repeated

from

Chapter

2)

:

•

Make

sure

that

you

understand

whether

the

motion

is

in

two

dimensions

or

in

one

dimension. If it is in two dimensions, you will need to consider both x and y

component

equations.

Make

sure

that

you

correctly

identify

what

each

of

the

physical

quantities (s

x

,

v

f

y

, a

x

,

etc..)

are

in

the

problem

you

are

working

–

this

includes

understanding

which

values

you

are

given

and

which

you

need

to

solve

for

in

the

problem.

•

Make

sure

that

you

understand

exactly

what

is

being

asked

for

BEFORE

you

start

looking

for

an

equation

to

apply.

For

example:

Are

initial

values

given

or

being

asked

to

solve

for?

Are

there

values

implied

but

not

stated?

(

i.e.

starting

from

rest

=

initial

velocity

of

zero,

or

object

in

free

fall

means

that

y-acceleration

is

–9.81

m/s

2

).

•

Make

sure

you

understand

whether

the

motion

is

free

fall

or

whether

the

object

experiences

some

other

acceleration

(in

either

x

or

y

direction

or

both).

•

The

component

equations

above

have

in

common

the

variable

of

time

(=

t).

Thus,

the

same

time

(t)

must

pass

for

simultaneous

x

and

y

motion.

•

If

acceleration

takes

on

different

values

over

different

parts

of

an

object's

motion,

the

problem must be divided into individual parts for each of the different accelerations.

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