Lec 03 Vehicle Motion (Transportation Engineering Dr.Lina Shbeeb)

1. Vehicle motion

2. Transportation
Engineering
Dr. Lina Shbeeb2
Definitions
• Kinematic is the study of motion irrespective of
the forces that cause it
• Kinetic is the study of motion that accounts the
forces that cause it.
• The motion of a body can be linear or curvilinear
• It can be investigated in relation to a fixed
coordinate system (absolute motion) or in
relation to a moving coordinate system (relative
motion)
Vehicle motion can be described based on kinematic and kinetic equations

3. Transportation
Engineering
Dr. Lina Shbeeb3
Equation of motion/ Rectilinear Motion
• The rectilinear position of x is measured from a
reference point and has unit of length
• The displacement is the difference in its position
between two instants.
• Velocity v is the displacement of the particle divided by
time over which the displacement occurs. It is given by
the derivative of the displacement with respect of time
• Speed is a scalar quantity and it is equal to the
magnitude of the velocity, which is a vector
dt
dx
v 

4. Transportation
Engineering
Dr. Lina Shbeeb4
Equation of motion/ Rectilinear Motion
• Acceleration a is the rate of change
of velocity with respect to time.
• It can be positive, zero or negative.
Negative acceleration or what is
common known as deceleration is
often denoted as d and its
magnitude is given in the positive
(d of 16 ft/s2 equals the same as an
acceleration of - ft/s2)
adxvdv
toleadswhich
v
dx
dv
a
dt
dx
dx
dv
a
dt
dv
a
















Equation derivation

5. Transportation
Engineering
Dr. Lina Shbeeb5
Equation of motion/ Rectilinear Motion
• The simplest case of rectilinear motion is the
case of constant acceleration where
 
oo
oo
o
t
o
v
v
xtvatx
Thus
xxavv
leadwhichadxvdv
inegratingbycedisoffunctionaasressedbecanvelocityThe
vatv
dtadv
givesttotittheoveregratingby
adtdv
tconsa
dt
dv
o









2
22
2
1
)(
2
1
,
tanexp
0limint
tan

6. Transportation
Engineering
Dr. Lina Shbeeb6
…Equation of motion/ Rectilinear Motion
• The acceleration of a vehicle from an initial speed vo is
given by the relationship
Acceleration as a function of velocity
)1()1(
)(
,
)1(
)ln(
1
tan
2
BtoBt
Bt
o
Bt
o
Bt
v
v
t
o
v
v
e
B
v
e
B
A
t
B
A
x
eBvAa
equalsaBvAainsubstituteisvif
eve
B
A
v
leadwhich
tBvA
B
dt
BvA
dv
consareBandA
BvA
dt
dv
a
o
o




















7. Dr. Lina Shbeeb
Travel Speed
12
12
tt
xx
v



Time
Distance
t2t1
x1
x2

8. Dr. Lina Shbeeb
Spot Speed
dt
dx
v 
Time
Distance
t1
x1
V

9. Dr. Lina Shbeeb
Spot Speed Measurements
t1t2t3Time
x3
x2
x1
Distance
45.0
40.0
30.0
Distance
x
(ft)
4.0
3.0
2.0
Time
t
(s)
(40-30)/(3-2)
=10.0
---
Speed 1
v
(ft/s)
---
(45-30)/(4-2)
= 7.5
---
Speed 2
v
(ft/s)
(45-40)/(3-2)
=5.0

10. Dr. Lina Shbeeb
Spot Speed Measurements
Time
(s)
Distance
(ft)
Speed
(ft)
0.0 0.0 -
0.1 2.13 21.5
0.2 4.30 21.9
0.3 6.51 22.4
0.4 8.78 22.4
0.5 10.99 21.3
0.6 13.04 -

11. Dr. Lina Shbeeb
Average Acceleration Rate
12
12
tt
vv
a



Time
Speed
t2t1
v1
v2

12. Dr. Lina Shbeeb
Spot Acceleration Rate
dt
dv
a 
Time
speed
t1
v1
a

13. Dr. Lina Shbeeb
Measuring Acceleration
Rates
Time
(s)
Distance
(ft)
Speed
(ft/s)
Acceleration
(ft/s2)
0.0 0.0 - -
0.1 2.13 21.5 -
0.2 4.30 21.9 4.5
0.3 6.51 22.4 2.5
0.4 8.78 22.4 -5.5
0.5 10.99 21.3 -
0.6 13.04 - -

14. Dr. Lina Shbeeb
Constant Acceleration Motion
consta
dt
dv

 
tv
v
adtdv 00
0vatv 
av
dx
dv

 
xv
v
adxvdv 00
a
vv
x
2
2
0
2


dtvatvdtdx )( 0
  
x t
dtvatdx0 0 0 )(
tvatx 0
2
2
1

Remark: The equation used for design is , where the
deceleration rate has a positive value.
a
vv
x
2
22
0 


15. Dr. Lina Shbeeb
Exercise
•From the following data,
calculate the acceleration
rate at the distance of 2
feet from the reference
point.
Distance
(ft)
Speed
(ft/s)
0 19.4
1 19.6
2 20.0
3 20.8
4 21.3
a=5.91ft/s2???

16. Transportation
Engineering
Dr. Lina Shbeeb16
Constant Acceleration Motion
consta
dt
dv

 
tv
v
adtdv 00
0vatv 
av
dx
dv

 
xv
v
adxvdv 00
a
vv
x
2
2
0
2


dtvatvdtdx )( 0
  
x t
dtvatdx0 0 0 )(
tvatx 0
2
2
1

Remark: The equation used for design is , where the
deceleration rate has a positive value.
a
vv
x
2
22
0 


17. Transportation
Engineering
Dr. Lina Shbeeb17
Braking Distance
a
g
w

w
sinw
u
coswf
Db
G
1.0
Distance to stop vehicle

18. Transportation
Engineering
Dr. Lina Shbeeb18
Braking on Grades
 sincos WWfa
g
W






a
vv
x
2
22
0 

x
Db
 cos
2
cos
22
0
a
vv
xDb


bD
vva
2
cos
)( 22
0




cos
sincos
2
cos
)(
1 22
0 





f
D
vv
g b


cos
sin
2
1
)(
1 22
0 





f
D
vv
g b
G 


tan
cos
sin
)(2
22
0
Gfg
vv
Db




19. Transportation
Engineering
Dr. Lina Shbeeb19
Braking distance
• Braking Distance (Db)
• Db = distance from brakes enact to final speed
• Db = f(velocity, grade, friction)
• Db = (V0
2 – V2)/[30(f +/- G)]
• or
• Db = (V0
2 – V2)/[254(f +/- G)] metric
– Db = braking distance (feet or meters)
– V0 = initial velocity (mph or kph)
– V = final velocity (mph or kph)
– f = coefficient of friction
– G = Grade (decimal)
30 or 254 = conversion coefficient

20. Transportation
Engineering
Dr. Lina Shbeeb20
Braking Distance
Db = braking distance
u = initial velocity when brakes are
applied
a = vehicle acceleration
g = acceleration of gravity (32.2 ft/sec2)
G = grade (decimal), level roads G=zero
• AASHTO represents friction as a/g which is a function
of the roadway, tires, etc
• Can use when deceleration is known (usually not) or
use previous equation with friction
Db = _____u2_____
30({a/g} ± G)

21. Transportation
Engineering
Dr. Lina Shbeeb21
Vehicle Braking Distance
• Factors
• Braking System
• Tire Condition
• Roadway Surface
• Initial Speed
• Grade
• Braking Distance Equation
• db = (V2 - U2) / 30( f + g )

22. Transportation
Engineering
Dr. Lina Shbeeb22
Coefficient of friction
Pavement condition Maximum Slide
Good, dry 1.00 0.80
Good, wet 0.90 0.60
Poor, dry 0.80 0.55
Poor, wet 0.60 0.30
Packed snow and
Ice
0.25 0.10

23. Transportation
Engineering
Dr. Lina Shbeeb23
Motion on Circular Curves
dt
dv
at 
R
v
an
2


24. Transportation
Engineering
Dr. Lina Shbeeb24
 coscossin ns amWfW 
 cos
cos)(cossin
2
WR
v
g
W
WfW s 
e 


tan
cos
sin
gR
v
fe s
2

Motion on
Circular
Curves

25. Transportation
Engineering
Dr. Lina Shbeeb25
Minimum Radius of a Circular Curve
• where u = vehicle velocity (mph)
• e = tan  (rate of superelevation)
• fs = coefficient of side friction (depends on design speed)
• Example
– design speed = 65 mph
– rate of superelevation = 0.05
– coefficient of side friction = 0.11
• Solution
– minimum radius
– R = (65)2/[15(0.05+0.11)] = 1760 ft
)(15
2
sfe
u
R



26. Transportation
Engineering
Dr. Lina Shbeeb26
Relative Motion
• It is common to examine the motion of one
object in relation to another, for example the
motion of vehicles on a highway may be studies
from the point of view of the driver of a moving
vehicle.
• The simplest case of relative motion involves the
motion of one object B relative to a coordinate
system (x, y, z) that is translating but not rotating
with respect to a fixed coordinate system (X, Y,
Z)

27. Transportation
Engineering
Dr. Lina Shbeeb27
Relative Motion
• The relationship between the position vectors of the two objects in relation to the fixed
system, RA and RB and the position vector rB/A with respect to the moving object A is
Y
Z
y
X
x
z
RA
RB
RA/B
ABAB
ABAB
ABAB
aaa
and
vvv
givestimetorespectwithatingDifferenti
rrr
/
/
/


