vehicle motion problems for part exam

7/23/2019 Vehicle Motion Problems for Part Exam

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1 Basic Vehicle Motion Analysis: A Modern Accident Reconstruction Guide

mil

es

ho

ur

1.4

67

speed = = 52.4

4.16 Tasks that Yield Speed when a

Vehicle Decelerates from One Speed to a

Slower Speed

A. Final speed when initial speed, distance traveled

and deceleration time are known (task 40-40!

Example:

1. A vehicle is traveling at 50.0 miles per hour.

2.he vehicle decelerates covering a distance o! "50.0

!eet in 5.0 seconds.

3. #hat is the speed o! the vehicle a!ter deceleration$

%olution:

2 x distance . . . &speed = '''''''' ' initial speed

time x 1.46



mil

es

ho

ur

1.4

67

speed = = 52.4

7

B. Final speed when initial speed, distance traveledand deceleration are known (task 40-40"!

Example:


2. he vehicle decelerates at ".0 !eet per second per

second covering a distance o! 100.0 !eet.


%olution:

( )*+initial speed x 1.467,2 ' +2 x

deceleration x distance, spee ' 1.467

C. Final speed when initial speed, distance traveledand dra# factor are known (task 40-40$!

Example:

2 x "50.0 !eet5.0 seconds x

1.467

speed= hou

r

50.0 ) = 45.4 -le%

hour



" Basic Vehicle Motion Analysis: A Modern Accident Reconstruction Guide

mil

es

ho

ur

1.4

67

speed = = 52.4


2. he vehicle decelerates ith a drag !actor o! 0.15

covering a distance o! 100.0 !eet.


%olution:

speed =

1.467

speed = 50./

D. Final speed when initial speed, deceleration timeand deceleration are known (task 40-404!

Example:



!eet

second2

2 x drag !actor x"2.2

xdistance

+initial speed x1.467,

2 '

!eet

x 100.0

!eet

2 x 0.15

x"2.2

secon

dspeed= 1.4

67

(( miles &

55.0 ' ' x1.467

hour

miles

hour



mil

es

ho

ur

1.4

67

speed = = 52.4

second !or a duration o! /.0 seconds.


%olution:

deceleration x time

1.46

7

E. Final speed wheninitial speed,deceleration time and dra# factor are known(task 40-40%!

Example:


2. he vehicle decelerates ith a drag !actor o! 0.15 !or

a duration o! /.0 seconds.


speed = initialspeed '

!eetsecond2

".0

x /.0 seconds



5 Basic Vehicle Motion Analysis: A Modern Accident Reconstruction Guide

mil

es

ho

ur

1.4

67

speed = = 52.4

%olution:

!eet

drag !actor x "2.2 x time

second

F. &nitial speed when 'nal speed, distance traveledand deceleration time are known (task 40-40!

Example:

1. A vehicle is traveling at an unnon speed.

2. he vehicle decelerates covering a distance o!

"50.0 !eet in 5.0 seconds& reaching a 3nal speed o! "0.0

miles per hour.

3. #hat as the speed o! the vehicle e!ore

speed = initialspeed

0.15 x

"2.2

x /.0

seconds

mile

s

hour

mile

s

hour

speed= 55.0 =2/.7

1.4

67! eet

second1.467



mil

es

ho

ur

1.4

67

speed = = 52.4

deceleration$

%olution:

& 2 xdistance ( &

speed = '''''''' ' 3nal speed

time x 1.46



G. )&nitial speed when 'nal speed, distance traveledand deceleration are known (task 40-40)!

Example:



second covering a distance o! 100.0 !eet& reaching a

3nal speed o! 50.0 miles per hour.


deceleration$

%olution:

speed x 1.467,2 +2 x deceleration x

distance,467

H. &nitial speed when 'nal speed,distance traveled and dra# factor are known (task 40-40*!

Example:

)

+3nal speed

= ''''' '

'''''''''

miles &

50.0 ' ' x

1.467hour

!eet

2 x".0

x 100.0 !eet

miles

hour

second

speed=

=52.71.4

67



I.A vehicle is traveling at an unnon speed.


covering a distance o! 100.0 !eet& reaching a 3nal

speed o! 50.0 miles per hour.

3. #hat as the speed o! the vehicle e!oredeceleration$

%olution:

speed

=

speed = 54."

I. &nitial speed when 'nal speed, deceleration

time and deceleration are known (task 40-

40+!

!eet

+3nal speed x1.467,

2

2 x drag !actor

x "2.2

467

xdistancesecon

d

(( ( miles & .

50.0 ' ' x1.467

hour

!eet

x 100.0 !eet

2 x 0.15"2.2

second

miles

hour



Example:



second !or a duration o! /.0 seconds& reaching a 3nal



deceleration$

%olution:

deceleration x time1.467

&nitial speed when 'nal speed, deceleration time anddra# factor are known (task 40-40!

Example:



speed = 3nalspeed

!eetsecond2

".0

x /.0 seconds

miles

hour

miles

hour

speed=

40.0

=56.4

1.467



!or a duration o! /.0 seconds& reaching a 3nal speed o!

"0.0 miles per hour.


deceleration$

%olution:

!eet

drag !actor x "2.2

speed = 3nal speed

1.467

!eet

second2

1.46

xtimesecon

d

0.15 x"2.2

x /.0 seconds



4.17 )Tasks that Yield a Vehicle.s /onstant

Velocit

1 /onstant velocit when distance traveled and time

are known (task %0-00!

Example:

1. A vehicle is traveling at an unnon velocit8.

2. he vehicle covers a distance o! 100.0 !eet in 10.0

seconds.

3.#hat is the velocit8 o! the vehicle$

%olution:

& . distance 100.0 !eet !eet

velocit8 = 9''''''''''''''' = = 10.0

time 10.0 seconds second



4.18 Tasks that Yield 1ttained Velocit

when a Vehicle 1ccelerates from 2est

A. 1ttained velocit when distance traveled and timeare known (task %0-0!

Example:

1. A vehicle is at rest.

2. he vehicle accelerates covering a distance o! 100.0

!eet in 10.0 seconds.

3. #hat is the velocit8 o! the vehicle a!ter acceleration$

%olution:

& . 2 x distance 2 x 100.0 !eet !eet

velocit8 = '''''' = '''''''''''' = 20.0''''

time 10.0seconds second

B. 1ttained velocit when distance traveled andacceleration are known (task %0- 0"!

Example:




2. he vehicle accelerates at 7.0 !eet per second per

second through a distance o! 100.0 !eet.


%olution:

velocit8 = ;2

x acceleration x distance

< !eet !eet

velocit8 ='''''''''''2 x 7.0'''''''''''''''' x 100.0

!eet = "7.4'''''''''

; second second

C. 1ttained velocit when

distance traveled and acceleration

factor are known (task %0-0$!

Example:




2. he vehicle accelerates ith an acceleration !actor

o! 0.1 through a distance o! 100.0 !eet.


%olution:

!eet

velocit8 = 2 x acceleration !actor x "2.2 xdistance

second

f*6£t fppt

velocit8 = 12 x 0.1 x "2.2 x 100.0 !eet =

25.4

second second

D. 1ttained velocit whenacceleration and time are known (task %0-04!

Example:




2. he vehicle accelerates at 5.0 !eet per second per

second !or 10.0 seconds.


%olution:

velocit8 = acceleration x time = 5.0''''''''''''''''''''x 10.0 seconds = 50.0''''''''''

second second

E. 1ttained velocit when acceleration factor and timeare known (task %0-0%!

Example:


2. he vehicle accelerates ith an acceleration !actor o!

0.15 !or 10.0 seconds.


%olution:



velocit8 = acceleration !actor x "2.2 9)t

& x

timesecond>

fppt fpet

velocit8 = 0.15 x "2.2''''''''''''''''''''''''' x

10.0 seconds = 4/." '''''''''''''''''''''''''''

second second

4.19 Tasks that Yield &nitial Velocitwhen a Vehicle Decelerates to a Stop

A. &nitial velocit when deceleration distance and timeare known (task %0-"0!

Example:


2.he vehicle decelerates to a complete stop

covering a distance o! 100.0 !eet in 10.0 seconds.

3. #hat is the initial velocit8 o! the vehicle$



%olution:

& . 2 xdistance 2 x 100.0!eet ?.. !eet

velocit8 = ''''''' = '''''''''''' = 20.0''''''


B. &nitial velocit when deceleration distance anddeceleration are known (task %0-

"0"!

Example:


2. he vehicle decelerates to a complete stop at 7.0

!eet per second per second through a distance o! 100.0

!eet.



velocit8 = ';2 x deceleration x distance

feet feet

velocit8 = 12 x 7.0 =' x 100.0 !eet = "7.4

second second

C. &nitial velocit when deceleration distance and dra#factor are known (task %0-

"0$!

Example:


2. he vehicle decelerates to a complete stop ith a

drag !actor o! 0.1 !or a distance o!

3. #hat is initial the

velocit8 o! the vehicle$

%olution:



100.0!eet.


%olution:

!eet

velocit8 = @2 x drag !actor x "2.2 x distancesecond2

feet feet

velocit8 = 12 x 0.1 x "2.2 x 100.0 !eet = 25.4

second second

D. &nitial velocit when deceleration and decelerationtime are known (task %0-"04!

Example:


2. he vehicle decelerates to a complete stop at 5.0

!eet per second per second !or 10.0 seconds.




%olution:



velocit8 = deceleration x time = 5.0 x 10.0 seconds =50.0

E. &nitial velocit whendra# factor anddeceleration time are known (task %0-"0%!

Example:

1. A vehicle traveling at an unnon velocit8.

2. he vehicle decelerates to a complete stop ith a

drag !actor o! 0.15 !or 10.0 seconds.


%olution:

velocit8 = drag !actor x "2.2 9 0 xtime

@ second2

fppt fppt

velocit8 = 0.15 x "2.2 9x 10.0 seconds = 4/."second second

x 10.0 seconds =50.0

B' second second



4.20 Tasks that Yield Velocit when a

Vehicle 1ccelerates from One Velocit toa Faster Velocit

A. Final velocit when initial velocit, distancetraveled and acceleration time are known (task %0-$0!

Example:

1. A vehicle is traveling at 55.0 !eet per second.

2. he vehicle accelerates covering a distance o! 1/0.0

!eet in ".0 seconds.


%olution:

& . 2 x distance . . . . & .

velocit8 = '''''' ' initial velocit8

time

. . 2 x 1/0.0 !eet ) !eet ' !eet

velocit8 =''''''''''' 55.0 ''' = 65.0 '''

B. seconds second secondFinal velocit when



initial velocit, distance traveled and acceleration are

known (task %0-$0"!

Example:


2. he vehicle accelerates at ".0 !eet per second per



%olution:

velocit8 = )*+initial velocit8,1 +2 x acceleration x

distance,

1 x acceleration !actor x "2.2

second

velocit8 =

!eetsecond



C. Final velocit when initial velocit, distance

traveled and acceleration factor are known (task %0-$0$!

Example:



o! 0.15 covering a distance o! 100.0 !eet.


%olution:

!eet

seco

nd

velocit8 = velocit8 = 6".2



D. Final velocit when initial velocit, accelerationtime and acceleration are known (task %0-$04!

Example:


2. he vehicle accelerates at ".0 !eet per second persecond !or a duration o! /.0 seconds.


%olution:

velocit8 = initial velocit8 +acceleration x time,

E. Final velocit when 'nal velocit, accelerationtime and acceleration factor are known (task %0-$0%!

Example:



o! 0.15 !or a duration o! /.0 seconds.

!eetsecond2

!eet

second

!eet

second

velocit8=

".0

x /.0 seconds

=7C.0

55.0




%olution

:

feet feet feet

velocit8 = 55.0 0.15 x "2.2 ' x /.0 seconds = C".6

second@ 6 second @ secon

!eetsecond2

acceleration !actorx "2.2

xtime

velocit8 = initialvelocit8



F. d&nitial velocit when 'nal velocit, distancetraveled and acceleration time are known (task %0-$0!

Example:


2. he vehicle accelerates covering a distance o!

400.0 !eet in 6.0 seconds& reaching a 3nal velocit8 o!

75.0 !eet per second.

3. #hat as the velocit8 o! the vehicle e!ore

acceleration$

%olution:

. . 2 xdistance .

velocit8 = ''''''' ' 3nal velocit8

time

2 x 400.0 !eet !eet co !eet



velocit8 =''''''''''' 75.0 ''' = 5/." ' '

6.0 seconds secondsecond

G. &nitial velocit when 'nal velocit, distancetraveled and acceleration are known (task %0-$0)!

Example:




3nal velocit8 o! 60.0 !eet per second.


acceleration$

%olution:

velocit8 = 'B&*+3nal velocit8,2 ' +2 x acceleration x



distance,

H. &nitial velocit when 'nal velocit, distancetraveled and acceleration factor are known (task %0-$0*!

Example:

I.A vehicle is traveling at an unnon velocit8


o! 0.15 covering a distance o! 100.0 !eet& reaching a



acceleration$%olution:

velocit8 = ...

...)+3nalvelocit8,

2

!eet

2 x acceleration !actor x "2.2 r' xdistance

secon

d

!eetsecond2

!eet

second

x 100.0 !eet

=54./

velocit8 =

2 x".0

!eet

second

x 100.0 !eet

velocit8=

65.0

2 x 0.15"2.2 secon

d

!eet



velocit8 = 57.1

& &nitial velocit when 'nal velocit, acceleration timeand acceleration are known (task %0-$0+!

Example:



second !or a duration o! /.0 seconds& reaching a 3nal

velocit8 o! /0.0 !eet per second.

3. #hat is the velocit8 o! the vehicle a!teracceleration$

%olution:

velocit8 = 3nal velocit8 ' +acceleration x time

,

!eet

second

!eet

second

!eet

second

!eet

second2

=56.0

velocit8=

".0

x /.0 seconds

/0.0



&nitial velocit when 'nal velocit, accelerationtime and acceleration factor are known (task %0-$0!

Example:



o! 0.15 !or a duration o! /.0 seconds& reaching a 3nal

velocit8 o! C0.0 !eet per second.


acceleration$

%olution:

acceleration !actor x "2.2 9)eet

2 x time

second4.21 Tasks that Yield Velocit when a

Vehicle Decelerates from One Velocit to aSlower Velocit

A. Final velocit when initial velocit, distancetraveled and deceleration time are known (task %0-

velocit8 = 3nalvelocit8

C0.0!eeD = 51.4

!cetvelocit8= secon

d

secon

d

feet

0.15 x "2.2'''r' x /.0

secondssecond



40!

Example:


2. he vehicle decelerates covering a distance o!

200.0 !eet in 5.0 seconds.

3. #hat is the velocit8 o! the vehicle a!ter

deceleration$

%olution:

& . 2 x distance . . . . & .

velocit8 = ''''''' ' initial velocit8

time

& . 2x 200.0 !eet (( ( !eet .. ( !eet

velocit8 =''''''''''' 50.0 ''' = "0.0

seconds second secon



d

5.0

B. Final velocit when initial velocit, distancetraveled and deceleration are known (task %0-40"!

Example:




3.#hat is the velocit8 o! the vehicle a!ter

deceleration$

%olution:

velocit8 = ')*+initial velocit8,2 ' +2 x deceleration x

distance,

C. Final velocit when initial velocit, distancetraveled and dra# factor are known (task %0-40$!

Example:

!eet

second

!eetsec

ond2

!eet

second

velocit8=

55.0

2 x".0

x 100.0 !eet

=4C.2





covering a distance o! 100.0 !eet.


deceleration$

%olution

:

velocit8 = 45.

D. 4Final velocit when initial velocit, decelerationtime and deceleration are known (task %0-404!

Example:



!eet

... *+initialvelocit8,

2 '2 x drag !actor x"2.2

xdistancesecon

d>

!eet

second

!eetvelocit8=

55.0

2 x 0.15"2.2

x 100.0 !eetsecon

d

!eet

second



second !or a duration o! /.0 seconds.


deceleration$

%olution:

velocit8 = initial velocit8 ' +deceleration x time,

E. Final velocit when initial velocit, decelerationtime and dra# factor are known (task %0-40%!

Example:



!or a duration o! /.0 seconds.


deceleration$

%olution

:

!eet

second

!eet

!eet

second

velocit8=

55.0

".0

x /.0 seconds

="1.0secon

d

!eet

drag !actor x"2.2

xtimesecon

d

velocit8 = initialvelocit8



F. &nitial velocit when 'nal velocit, distancetraveled and deceleration time are known (task %0-40!

Example:


he vehicle decelerates covering a distance o! 200.0

!eet in 5.0 seconds& reaching a 3nal velocit8 o! "0.0

!eet per second

.

2.

& . 2 xdistance ( & & .

velocit8 = ''''''' ' rinal velocit8

time

!eet

second

!eetsecond2

!eet

second

velocit8=

55.0

x /.0 seconds

0.15 x"2.2

=16.4'

3. #hat as the velocit8 o! thevehicle e!ore deceleration$ %olution:



& . 2 x 200.0 !eet ( !eet !eet

velocit8 ''''''''''''''''''''''' "0.0 ''''' = 50.0'''''''

5.0 seconds secondsecond

G. Final velocit when initial velocit, distancetraveled and deceleration are known (task %0-40)!

Example:






deceleration$

%olution:



velocit8 = )*+3nal velocit8,2 +2 x deceleration x

distance,

H. &nitial velocit when 'nal velocit, distancetraveled and dra# factor are known (task %0-40*!

Example:

I.A vehicle is traveling at an unnon velocit8.


covering a distance o! 100.0 !eet& reaching a 3nal

velocit8 o! 45.0 !eet per second.


deceleration$

%olution

!eet

second

!eetsecond2

!eet

second

x 100.0 !eet

=55.7

velocit8=

50.0

2 x".0



:!eet

2 x drag !actor x"2.2second

&

)

?

! e

e

t

+3nalvelocit8,

2

xdistance

velocit8 =



v

e

l

o

c

i

t

8

=

5

4

.

7

second

& &nitial velocitwhen 'nal velocit,deceleration time



and deceleration are known (task %0-40+!

Example:

1. A vehicle is

traveling at anunnon velocit8.

2. he vehicle

decelerates at ".0

!eet per second per

second !or a

duration o! /.0

seconds& reaching a

3nal velocit8 o!

"0.0 !eet per

second.

3. #hat as the

velocit8 o! the vehicle



e!ore deceleration$

%olution:

velocit8 = 3nal

velocit8

+deceleration x time, &nitial velocitwhen 'nal velocit,deceleration timeand dra# factor are known (task %0-40!

Example:

1. A vehicle is

traveling at an

unnon velocit8.

2. he vehicle

decelerates ith a

drag !actor o! 0.15

!or a duration o!

!eet

secon

d

!eet

velocit8=

"0.0

".0

x /.0 seconds

second



/.0 seconds& reach'

ing a 3nal velocit8

o! 20.0 !eet per

second.

3. #hat as thevelocit8 o! the vehicle

e!ore deceleration$

%olution

:!eet

drag !actor x "2.2 r'x time

second

4.22 Tasks that Yield 1ccelerationwhen a Vehicle1ccelerates from

2est A. 1ccelerationwhen accelerationdistance and time are known (task 0-0!

velocit8 = 3nalvelocit8

!eet

second

!eetsecond2

velocit8=

x /.0 secon

20.0

0.15 x"2.2



Example:

1. A vehicle is at

rest.

2. he vehicle

accelerates covering a

distance o! "0.0 !eet

in ".0 seconds.

3. #hat is the

acceleration o! thevehicle$

%olution:

& 2 xdistance

2 x"0.0!eet

( !eet

acceleration =''''''''''''''''''



5

'''''''''''''''''' ='''''''''''''''''' = 6.7''''''''''''''''''r

ti

me

+".0

seconds,

second



B. 1ccelerationwhen attained speedand time are known(task 0-0"!

Example:

1. A vehicle is at

rest.

2. he vehicle

accelerates to a speed

o! "0.0 miles per

hour.

3. he duration o!

the acceleration is 5.0

seconds.

4. #hat is the

acceleration o! the

vehicle$



%olution:

miles

acceleration 3

%%66i)6 3 $00

4)

//

(@ee)

time 5.0

seconds

second

C. 1ccelerationwhen attained speedand distance are known (task 0-0$!

Example:

A vehicle is at rest

.



1.

2. he vehicle

accelerates to a

speed o! "5.0 miles

per hour through a

distance o! 150.0

!eet.

3. #hat is the

acceleration o! the

vehicle$

%olution:

D. 1ccelerationwhen attained velocit and time are known (task 0-04!

Example:

1. A vehicle is at

2 x 150.0!eet

acceleration =

+speed x1.467,

2 2

x distance

.. ( miles & .

"5.0 ' ' 1.467

hour



rest.

2. he vehicle

accelerates to a

velocit8 o! "0.0 !eet

per second.

3. he duration o!

the acceleration is 5.0

seconds.

4. #hat is theacceleration o! the

vehicle$

%olution:

"0.0'F

)acceleration3

seco5d 3 6



time 5.0 seconds

second

E. 1cceleration

when attained velocit and distanceare known (task 0-0%!

Example:



1. A vehicle is at

rest.

2. he vehicle

accelerates to a

velocit8 o! "5.0 !eetper second in a

distance o! 150.0 !eet.

3. #hat is the

acceleration o! the

vehicle$

%olution

:!eet

second

"5.0

velocit82

2 xdistance

acceleration =

=42 x 150.0

!eetsd



4.23 Tasks that Yield 1cceleration when a Vehicle 1ccelerates from One Speed or Velocit to a Faster Speed or Velocit

A. 1cceleration when initial speed, 'nal speed andtime are known (task 0-"0!

Example:

1. A vehicle is traveling at a speed o! 40.0 miles per

hour.

2. he vehicle accelerates to a speed o! 50.0 miles per

hour in 2.0 seconds.

3. #hat is the acceleration o! the vehicle$

%olution:

+3nal speed x 1.467, ' +initial speed x1.467,

acceleration = '''''''''''''''''''''''''''''''''time

B. 1cceleration when initial speed, 'nal speed anddistance are known (task 0-"0"!

!eetsecond2

=7."

acceleration =

2.0 seconds



Example:

1. A vehicle is traveling at a speed o! 40.0 miles perhour.

2. he vehicle accelerates to a speed o! 50.0 miles perhour in a distance o! 100.0 !eet.


%olution:

+3nal speed x 1.467,2 ' +initial speed x

1.467,2

acceleration =E'''''''''''''''''''''

H ''''''''''''

2 x distanc

e

C. 1cceleration when initial velocit, 'nal velocit andtime are known (task 0-$0!

Example:

1. A vehicle is traveling at a velocit8 o! 40.0 !eet per

second.

2. he vehicle accelerates to a velocit8 o! 50.0 !eet per

second in 2.0 seconds.

I40.0 )

hour

mile

shour

x

1.467

x

1.467

50.

0

!eet

second2

=C.7acceleration = 2 x 100.0 !eet




%olution:

+3nal velocit8, ' +initial velocit8,time

D. 1cceleration when initial velocit, 'nal velocit anddistance are known (0-$0"!

Example:


second.

2. he vehicle accelerates to a velocit8 o! 50.0 !eet per

second in a distance o! 100.0 !eet.


%olution:

+3nal velocit8,2 ' +initial velocit8,

2

acceleration = ''''''''''''''''''''' 9

acceleration = !eet

second

!eet

second

50.

0

40.

0

!ee

tacceleration =

=5.02.0

secondssecond



2 x distanc

e!eet

second

50.0

40.0

!eetsecond2

acceleration

=4.52 x 100.0

!eet

!eet

second



4.24 Tasks that Yield 1cceleration Factor

when a Vehicle 1ccelerates from 2est A. 1cceleration factor when acceleration distance andtime are known (task )0-0!

Example:


2. he vehicle accelerates covering a distance o! "0.0

!eet in ".0 seconds.

3. #hat is the acceleration !actor o! the vehicle$

%olution:

2 x

distance acceleration

!actor = ''''''''''''''''''

2 ((( !eet

time x "2.2"

second

acceleration !actor = 2 x "0.0 !eet999( )



2 ) !eet

+".0 seconds, x "2.2

second

B. 1cceleration factor when attained speed and timeare known (task )0-0"!

Example:


2. he vehicle accelerates to a speed o! "0.0 miles per

hour.

3. he duration o! the acceleration is 5.0 seconds.


%olution:

miles & t*r (

sneed x 1 467 "0.09 x 1.467

acceleration !actor = ' ' '''''''''''''''')9'9 = '''''''''''''''''''''''''''9'9 =0."time x "2.29997 5.0 seconds x "2.29

second> second

C. 1cceleration factor when attained speed anddistance are known (task )0-0$!



Example:


2. he vehicle accelerates to a speed o! "5.0 miles per

hour in a distance o! 150.0 !eet.




%olution:. . .+speed x1.467,

acceleration !actor = ''''''''''

? ? !eet

2 x distance x "2.2second2

second

D. 1cceleration factor whenattained velocit and time are known (task )0-04!

Example:


2. he vehicle accelerates to a

velocit8 o! "0.0 !eet per second.

3. he duration o! the

acceleration is 5.0 seconds.

acceleration!actor =

=0."

eet

. ( ( miles & &

"5.0 ' ' x1.467

hour

2 x 150.0 !eet x"2.2



4. #hat is the acceleration !actor

o! the vehicle$

%olution:

30.

0!eet

acceleration !actor = ''''''''''''''''''''''''velocit8 9(9 ( '''''''''''''''''''''''''second9(9(9 ( J 2

time x "2.2 966 ?5.0

secondsx "2.2

66

second2

second)

E. 1cceleration factor whenattained velocit and distance are known (task )0-0%!

Example:



velocit8 o! "5.0 !eet per second in



a distance o! 150.0 !eet.


o! the vehicle$

%olution:

ve

locit82

acceleration !actor =

? ...

!eet

2 x

distance

x "2.2

secon

d


= 0.1

2 x 150.0 !eetx "2.2

!eetsecond2



4.25 Tasks that Yield1cceleration Factor when a Vehicle 1ccelerates fromOne Speed or Velocit to a Faster Speed or Velocit

A. 1cceleration factor wheninitial speed, 'nal speed andtime are known (task )0- "0!

Example:

1. A vehicle is traveling at a



speed o! 50.0 miles per hour in

2.0 seconds.


o! the vehicle$

%olution:

+3nal speed x1.467, '+initial speedx 1.467,

acceleration !actor = ''''''''H

''''''''''''''''''''''''''''''''''7



9''''''''''''''''''

!eet

time x"2.2''''''''r'

second

B. 1cceleration factor wheninitial speed, 'nal speed anddistance are known (task )0-"0"!

Example:




speed o! 50.0 miles per hour

through a distance o! 100.0 !eet.


o! the vehicle$

= 0

acceleration!actor = +2.0 seconds, x

"2.2

fe

etsecond

2



%olution

:+3nal speed x 1.467,

2 ' +initial

speed x 1.467,

acceleration !actor =

! ee

t

2 x distance x"2.2

C. 1cceleration factor wheninitial velocit, 'nal velocit andtime are known (task )0-$0!

Example:




velocit8 o! 50.0 !eet per second in

2.0 seconds.


=0."2 x 100.0 !eet x "2.2

!eet second2

second




o! the vehicle$

%olution:

+3nal velocit8, '+initial velocit8,

!eet

time x"2.2

second

seco

ndD. 1cceleration factor wheninitial velocit, 'nal velocitand distance are known (task)0-$0"!

Example:




velocit8 o! 50.0 !eet per second in


!eet

second

!eet

second

50.0

40.0

= 0.2

!eet

2.0 seconds x"2.2




a distance o! 100.0 !eet.


o! the vehicle$

%olution:

+3nal

velocit8,2 '+initialvelocit8,

2

acceleration !actor = '''''''''''''''''''''''''''''''7979'9

. ?

?!eet

2 xdistancex "2.2'''''''''''

second!eet

second

!eet

second

50.0

40.0acceleration

!actor = 9

= 0.1

!eetsecond2

2 x 100.0 !eet x"2.2



4.26 Tasks that Yield Deceleration when a Vehicle Decelerates to a Stop

A. Deceleration when deceleration distance and time

are known (task *0-0!

Example:

1. A vehicle is traveling at a constant speed.

2. he vehicle decelerates to a complete stop

covering a distance o! "0.0 !eet in ".0 seconds.

3. #hat is the deceleration o! the vehicle$

%olution:

2 x distance 2 x "0.0 !eet & ( !eet

deceleration = ' ''' = '''''''''' = 6.7''''

time +".0seconds, second

B. Deceleration when initial speed and decelerationtime are known (task *0-0"!

Example:



1. A vehicle is traveling at a speed o! "0.0 miles per

hour.

2. he vehicle decelerates to a complete stop.

3. he duration o! the deceleration is 5.0 seconds.


%olution:

miles

. speed x 1.467 > hour

o o ! eet

deceleration = 9'' = '''' ' ''''''''''''''''''''' =

/./ ''''''''''''''''''''


Deceleration when initial speed and distance are known

(task *0-0$!

x1.467

"5.0

!eet

= 8.8

2 x 150.0!eet

second

Example:

1. A vehicle is traveling at a speed o! "5.0 milesper hour.

2. he vehicle decelerates to a complete stopthrough a distance o! 150.0 !eet.


%olution:

miles

^ II

+speed x 1.467,2

deceleration = 2 x

distance

hour



C. Deceleration when initial velocit and timeare known (task *0-04!

Example:

1. A vehicle is traveling at a velocit8 o! "0.0

!eet per second.

2. he vehicle decelerates to a complete stop.

3. he duration o! the deceleration is 5.0

seconds.


%olution:

"0.0

deceleration = ''''''second = 6 0 (@eet

time 5.0 seconds second

D. Deceleration when initial velocit anddistance are known (task *0-0%!

Example:

1. A vehicle is traveling at a velocit8 o! "5.0



!eet per second.

2. he vehicle decelerates to a complete stop

through a distance o! 150.0 !eet.


%olution:

4.27 Tasks that Yield Decelerationwhen a Vehicle Decelerates from oneSpeed or Velocit to a Slower Speed or Velocit

1 Deceleration when initial speed, 'nal speedand time are known (task *0-"0!

Example:

1. A vehicle is traveling at a speed o! 50.0 miles

per hour.

2. he vehicle decelerates to a speed o! 40.0

miles per hour in 2.0 seconds.


!eetsecond

"5.0 !eetsecond2

deceleration =

=4.12 x 150.0

!eet

velocit8

2 2 x

distance



%olution:

A((&((((((( +initial speed x 1.467, ' +3nalspeed x 1.467,

deceleration 9 ''''''''''''''''''''''''''''''

tim

e

C.Deceleration when initial velocit, 'nal

velocit and time are known (task *0-$0!

) ? miles &

40.0 ' ' x1.467

hour

? ) miles & .

50.0 ' ' x1.467

hour

!eet=7."

deceleration = 2.0 seconds second2

6 Deceleration when initial speed, 'nal speed and distance

are known (task *0-"0"!

Example:

1. A vehicle is traveling at a speed o! 50.0 miles per hour.

2. he vehicle decelerates to a speed o! 40.0 miles per

hour through a distance o! 100.0 !eet.


%olution:

+initial speed x 1.467,2 ' +3nal

speed x 1.467,2 2 x distance

deceleration

2 x 100.0 !eet second

2!eet

=C.7

deceleration =



Example:

1. A vehicle is traveling at a velocit8 o! 50.0

!eet per second.

2. he vehicle decelerates to a velocit8 o! 40.0

!eet per second in 2.0 seconds.


%olution:

+initial velocit8, ' +3nal velocit8,

tim

e

deceleration =

!eet

second

!eet

second

50.0

40.0

!eet

deceleration

=5.02.0

secondssecond



D. Deceleration when initial velocit, 'nal velocit

and distance are known (task *0- $0"!

Example:


second.

2. he vehicle decelerates to a velocit8 o! 40.0 !eet per



second in a distance o! 100.0 !eet.


%olution:

+initial velocit8,2 ' +3nal velocit8,

2 2 x

distancedeceleration =

!eet

second

!eet

second

50.0

40.0 !eetdeceleratio

n ==4.52 x 100.0

!eetsecond



Glossary

1cceleration

he mathematical opposite o! deceleration. Acceleration

is the rate o! change o! an oectKs speed or velocit8

ith respect to time ith speed or velocit8 ecoming

greater over time. <n this text& acceleration ill e

presented ith units o! !eet per second per second.

1cceleration Factor

A percentage or !raction o! the earthKs gravit8 constant

+"2.2 !eet per second per second, used to descrie the

magnitude o! an acceleration or deceleration.

Acceleration !actor is descried in g num4ers such that a

!ractional g'value is eJual to the eJuivalent !raction o!

the earthKs gravit8 constant. Lor example& 0.5 g is

exactl8 eJual to "2.2 !eet per second per second

multiplied 8 0.5& or 16.1 !eet per second per second.

+%ee Mrag Lactor,.



6rakin# Distance

he distance reJuired !or a vehicle to come to a

complete stop !rom the point at hich the actual raing

process egins.

/oe7cient of Friction (/OF!

he ratio o! lateral !orce reJuired to move one oectacross another oect to the eight o! the oect eingmoved. Lor example& i! it reJuires a !orce o! 5.0 poundsto push a 10.0' pound ox across a tale top& then thecoeNcient o! !riction eteen the ox and the tale topis the result o! dividing the lateral !orce +5.0 pounds, 8the eight o! the ox +10.0 pounds,. his 8ields acoeNcient o! !riction o! 0.5. <n realit8& the coeNcient o! !riction eteen to sur!aces has to values. One value

is that hich is evident prior to movement& hich iscalled the static coeNcient o! !riction +%OL,. he othervalue is that hich is evident during movement& hich iscalled the d8namic coeNcient o! !riction +MOL,. <ngeneral& the static coeNcient o! !riction is greater thanthe d8namic coeNcient o! !riction. <n this text& the termcoeNcient o! !riction +OL, ill re!er to the d8namiccoeNcient o! !riction.

/onstant

A mathematical Juantit8 or value that does not change.

/onversion Factor



A constant 8 hich a value is multiplied or divided to

8ield an eJual value ith a diPerent unit o! measure.

Deceleration

he mathematical opposite o! acceleration. Meceleration is

the rate o! change o! an oectKs speed or velocit8 ithrespect to time ith speed or velocit8 ecoming less over

time. <n this text& deceleration ill e presented ith units

o! !eet per second per second.

Densit

he ratio o! an oectKs eight to its volume. Lor

example& i! a loc o! ice is exactl8 one !oot in height&

idth and length its volume is exactl8 one cuic !oot. One

cuic !oot o! ice eighs aout 62.0 pounds. here!ore& the

densit8 o! ice is approximatel8 62.0 pounds per cuic

!oot. Alternativel8& a cuic !oot o! steel eighs aout

4C0.0 pounds and& there!ore& has a densit8 o! 4C0.0



pounds per cuic !oot. Qecause o! these values& it is clear

that the densit8 o! steel is almost eight times that o! ice.

Dra# Factor

A percentage or !raction o! the earthKs gravit8 constant

+"2.2 !eet per second per second, used to descrie themagnitude o! an acceleration or deceleration. Mrag !actor

is descried in g num4ers such that a !ractional g'value is

eJual to the eJuivalent !raction o! the earthKs gravit8

constant. Lor example& 0.5 g is exactl8 eJual to "2.2 !eet

per second per second multiplied 8 0.5& or 16.1 !eet per

second per second. +%ee Acceleration Lactor,.

Friction

he resistance to motion that arises hen one oect ismoved across another. Lriction is that !orce hich actseteen an oect and the sur!ace across hich the oect

is eing moved to resist the sliding motion eteen theto oects.



8ravit /onstant (#!

he acceleration o! oects resulting !rom the ePect o!

earthKs gravit8 upon the oect. Rear the earthKs sur!ace&

this acceleration is approximatel8 "2.2 !eet per second per

second. <n this text& as in most others& the areviation !or

earthKs gravit8 constant ill e g.

9erception:2eaction Distance

he distance through hich a vehicle travels hile thedriver is processing in!ormation that ill give rise to a

decision regarding a response to a given stimulus or set o! stimuli.!eet

vehicle motion problems for part exam

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