1
Chapter 2: Vehicle Motion: Kinematics & Kinetics
Introduction
* Why we care about vehicle motion?
Kinematics:
1. Need to know
s, v, a
to allow calculation of,
e.g.

determining stopping sight distance

brake distance

passing distance
etc.
Kinetics:
2. Need to know
forces
so to design

guard rail

attenuation devices

safe design of vehicles

curves of road
etc.
2
* Eq
uation
s. of Motion
2.2.1
Rectilinear motion
o
If position (x) is a function of time,
dt
dx
v
x
d
a
v
d
v
dt
dv
a
o
If
a
is constant
t
a
v
v
o
a
2
v
v
x
x
2
o
2
o
2
2
1
t
a
t
v
x
x
o
o
3
Example 2.1
Constant Acceleration
G: Vehicle appro
aches an intersection @ a speed = 30 mph
= 44 f/s
o
At t = 0, it started to decelerate @ a = 16 f/s
2
o
Vehicle was 55 ft away from stop line
R: 1) Did it stop legally?
2) Plot a
–
t, v
–
t & x
–
t graphs
1)
a
2
v
v
x
x
2
o
2
o
)
16
(
2
44
0
0
x
2
x = 60.5 ft > 55
Therefore, it did not stop legally….
2)
a
–
t
graph
a is constant
v
–
t
v = v
o
+ at
v = 44 + (

16) t
= 44
–
16 t
x
–
t
Other
x
–
x
o
= v
o
t
+
2
1
at
2
x = 44t +
2
1
(
–
16) t
2
x = 44t
–
8 t
2
or
dt
dx
v
= 44t
–
8 t
2
t
o
2
t
8
t
44
dt
)
t
16
44
(
dt
v
dx
x
55
x
a

16
t
v
44
t
t
x
4
Questions & Notes on Example:
1.
Why we have (. . . . ) pa
rt in graphs?
look @ v
–
t, final v is o. Driver will not reverse.
2.
Once you plot x
–
t, you can get the other graphs. Why?
dt
dx
v
we did this
dt
dv
a
3.
In reality, it is often that
a
is not
constant. But if assumed as in this case

it is reasonably OK.
5
Example: Variable acceleration:
G: a = (18 t + 2)m/s
2
for a vehicle.
vehicle @ rest at t = 0
R: 1) What are
x at t = 3
v at t = 3
1)
2
t
18
dt
dv
a
t
o
2
t
2
t
9
dt
a
v
s
/
m
87

v
3
t
3
o
3
2
3

t
t
3
dt
v
x
= 90
Recommended HW
Problem 1, p.94
(see the HW sheets @ the end of Notes of this chapter)
6
2.2.2 Braking Distance

Important in roadway design

Kinematic equa
tions in example 2.1 deals with a =
–
16 f/s
2
“a” there represents
“net effect” of forces according to Newton’s 2
nd
Law.

More practical, we consider the following:
A car is
decelerating
on an up

hill with incline of
º. Surface

tire friction is f.
“car is
sliding
up
when braking. So
F is down”.
+
F
y
= ma
y
= 0
N
–
w cos
= 0
N = w cos
+
F
x
= ma
x
–
w sin

F = ma
x
–
w sin
–
f w cos
=
a
g
w
a
=
–
g sin
–
f g cos
=
–
g cos
(tan
+ f)
7
In
a
2
v
v
x
x
2
o
2
o
)
f
(tan
cos
g
2
v
v
x
x
2
o
2
o
for small angles, cos
1
~
Also, tan
is G = grade in decimal = slope.
e.g.
075
.
40
3
not 7.5%
Notes:
0. For braking , v
2
= 0
1. For downhill, x
–
x
o
=
)
G
f
(
g
2
v
v
2
2
o
2. f is a variable. It varies with:

speed

condition of surface : wet , dry

the surface : concrete , asphalt
3.
Wet f is taken for design
–
ask why?
4. On asphalt:
f
wet
= 0.3
f
dry
= 0.6
)
G
f
(
g
2
v
v
x
x
2
2
o
o
8
5. x
–
x
o
=
)
G
f
(
g
2
v
v
2
2
o
; also
x
–
x
o
=
a
2
v
v
2
o
2
limit of comfortable a (8 to 10 f/s
2
)
Recommended HW
Probl
ems 2, 5, 6
a = g (f ± G)
9
2.2.3 Curvilinear Motion
Important in designing curved roads.
v is tangent
a is to the concave side
a
can be resolved into a
n
& a
t
a
n
points to center at the
instant
a
t
is tangent to path.
dt
dv
a
t
&
2
n
v
a
where
= radius of curve @ the instant
For a vehicle
FBD
KD
OR
F
t
=
dt
dv
F
n
=
2
v
m
F
z
= 0
v
a
a
a
n
a
t
Σ F
n
Σ F
t
ma
n
ma
t
10
Notes:
o
If v is constant, then a
t
= 0
o
In a circle,
= R = constant
Consider vehicle on a flat surface turning to left
tendency to overturn
Provide “
banking
”
superelevation
+
F
y
=
ma
y
:
N
–
W cos
= ma
n
sin
= 0
1
+
F
x
= ma
x
:
W sin
+ F = ma
n
cos
2
f = f N
3
1 & 3 in 2
W sin
+ f (W cos
) = ma
n
cos
=
cos
R
v
g
w
2
Dividing by W cos
gives
R
F
N
W
R
F
N
W
β
β
浡
n
11
tan
+ f =
R
g
v
2
tan
= e = superelevation rate
e + f =
R
g
v
2
Notes
:
1
This equation provides combination of the parameters (e, f, v & R) so to
avoid
slipping
.
2
f is the
side friction coefficient developed. It is different from f in SSD.
f is about 0.16
3
Tipping is about to occur when N is @ the outer wheel.
Dimension of vehicle plays a role.
o
Suburban overturns often.
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