> JLI7 #bjbjUU .D77lffff$\8'?%?%?%&&&&&&&$"( B*x&?%M"?%?%?%&?%[ &?%?%?%?%&?%?%&?%d?%&& H2Gf?%&&'08'&*?%*&?%LECTURE NOTES 2
Deceleration rates  Highway Code:
Example:
20 mph 6 m reaction 6 m braking
20 mph = 20 x 0.44794 = 8.94 ms1
reaction time = 6/8.94 = 0.67 seconds
v2 = u2 + 2as v = 0 ms1
u = 8.94 ms1
s = 6 m
a = ?
t = ?
solving, gives a =  6.66 ms2
t = 1.34 seconds
Example:
40 mph 12 m reaction 24 metres braking
The same procedure gives a reaction time of 0.67 seconds and a = 6.66 ms2
t = 2.69 seconds
All cars which are braked very firmly on a dry and level road will decelerate at 6.66 ms2 irrespective of the speed of the car.
A deceleration rate of 6.66 ms2 is the same as 0.68g. The figure 0.68 is called the coefficient of friction (.
However, road surfaces differ. Tyres differ in constitution. Deceleration rates will change with these differences.
Typical values of ( and a are:
Dry and level road ( = 0.7 to 0.9
Wet and level road ( = 0.5 to 0.6
Ice or snow ( = 0.3 to 0.4
Black ice ( = 0.05 to 0.2
Crash into standing car a = 5g
Crash into concrete block a = 20g
Gradual stop a = 0.1g
Normal braking a = 0.2g
Out of gear < 20mph a = 0.01g
Out of gear 20 to 40 mph a = 0.02g
Out of gear >40 mph a = 0.04g
Normal acceleration, cars:
<20 mph a = 1.5 ms2
20 to 40 mph a = 1.0 ms2
>40 mph a = 0.5 ms2
Rapid acceleration, cars:
<20 mph a = 3.0 ms2
20 to 40 mph a = 1.5 ms2
>40 mph a = 1.0 ms2
Medium truck accelerations:
<20 mph a = 1.0 ms2
20 to 40 mph a = 0.5 ms2
>40 mph a = 0.3 ms2
Heavy truck accelerations:
<20 mph a = 0.5 ms2
20 to 40 mph a = 0.3 ms2
>40 mph a = 0.1 ms2
Actual deceleration rates:
These have to be determined by braking the actual car to a halt from a known speed and recording the braking distance.
(a) Measuring test skid marks
Noting the speed of car from a calibrated speedometer or from a radar gun
(b) Using a mechanical decelerometer
(c) Using an electronic decelerometer
(d) Using a chalk gun and noting the
speed from a calibrated speedometer
or a radar gun.
In our example, we had a test skid of 12.0 metres from 30 miles per hour. We need to use the equations of motion to find a and (.
Newton's equations of uniform linear motion
v = u + at
v2 = u2 + 2as
s = ut + at2
also 1mph = 0.447 ms1
a = (g
Force = ma
= m(dv/dt) = d(mv)/dt
= rate of change of
momentum
For the first test skid:
u = 30 x 0.447 = 13.4 ms1
v = 0 ms1
s = 12.0 m
a = ?
t = ?
Choose equation containing u, v, s, a
v2 = u2 + 2as
Make 'a' the subject of the equation:
a = (v2  u2)/(2s)
= (02  13.42)/(2 x 12.0)
or a = 7.48 ms2
For the second test skid:
a = 7.24 ms2
The two values for a are within 10% of each other. Choose the lower value to be on the safe side.
(a) Initial speed of the car:
v = 0 ms1
u = ?
s = 22 m
a = 7.24 ms2
t = ?
We need the equation containing v, u, s, a
v2 = u2 + 2as
02 = u2  2 x 7.24 x 22
or u = 17.8 ms1 = (17.8/0.44704) mph
= 39.9 mph
Initial speed of the car = 39.9 mph.
This calculation neglects any braking that occurred prior to the start of the tyre marks and also any speed lost in the collision.
The speed of the car was therefore something higher than 39.9 mph.
(b)
What was the impact speed of the car?
v = 0
u = ?
s = 10 m
a = 7.24 ms2
t = ?
v2 = u2 + 2as
02 = u2  2 x 7.24 x 10
u = 12.0 ms1 = 26.9 mph
The impact speed was at least 26.9 mph
(c) Did the driver react quickly enough?
To answer this question we need to find the time from the "point of perception"
to the impact point. That time is given by (reaction time + time from start of braking to impact).
What was the time to impact?
V = 12.0 ms1
u = 17.8 ms1
s = 12 m
a = 7.24 ms2
t = ?
Choose either V = u + at or s = ut + at2
v = u + at
12.0 = 17.8  7.24 x t
or t = 0.80 seconds
Time from start of braking to impact is 0.80 s
What was the reaction time?
This is a grey area in accident investigation!
The Highway Code assumes a reaction time for an alert driver in ideal conditions as being 0.68 seconds. This value is highly artificial and much shorter than the vast proportion of real reaction times.
TRRL Laboratory Report 1004, 1981
Age 15%ile 50%ile 85%ile
<25 1.76 s 1.88 s 2.00 s
2534 1.54 s 1.62 s 1.70 s
3544 1.45 s 1.52 s 1.59 s
4554 1.33 s 1.40 s 1.47 s
5564 1.67 s 1.86 s 2.05 s
65+ 1.69 s 1.96 s 2.15 s
Choose 1.33 seconds to be on the safe side.
At 17.8 ms1, the car will cover (17.8 x 1.33) = 23.7 metres in a reaction time of 1.33 s.
Thus, the time from the point of perception to the the collision point is (1.33 + 0.8) = 2.13 seconds.
The distance between the point of perception and the impact point is
(12 + 23.7) = 35.7 metres.
(d)
What would have happened from 30 mph?
Assume that the driver would start to react at exactly the same point of perception
(35.7 m from impact).
At 30 mph, the car would travel
(30 x 0.44704) x 1.33 = 13.4 x 1.33 =
17.8 metres in the driver's reaction time.
He would then start braking and he would be (35.7  17.8) = 17.9 metres from the impact point.
The braking distance from 30 mph is 12.4 m
Thus, the car would come to a halt
(17.9  12.4) = 5.5 metres before reaching the child and so the accident would not have occurred.
For how long was the child in view?
Distance from gateway to collision point = 1.9 + 2.1 + 1.6 = 5.6 metres
What was his walking speed?
J.J.Eubanks measured the walking speeds of children. For a sample of 72 children aged 7 years, he found
15%ile = 1.25 ms1
50%ile = 1.52 ms1
85%ile = 1.92 ms1
On that basis, it would take between 2.9 seconds and 4.5 seconds for the child to appear and reach the impact point.
This range of times is greater than the 2.13 seconds quoted above and so the driver may have been slow to react but he does appear to have reacted marginally before the child stepped into the road.
The time fallacy:
Time to stop from 30 mph = 1.33 + 1.85
= 3.18 seconds
Time for child to cross the road to the collision point lies between 1.9 s and 3.0 s.
Therefore, the accident was inevitable because the driver could not have stopped in time from 30 mph.
Speed at which the accident is just avoided
Distance available in which to stop = 35.7 m
Let the initial velocity of the car be denoted by U ms1.
Distance travelled in reaction time = 1.33U
Braking distance is given by U2/(2a)
Therefore, U2/(2a) + 1.33U = 35.7
Substituting a = 7.24 ms2 gives
0.0691U2 + 1.33U  35.7 = 0
and so U = 15.1 ms1 or  34.3 ms1
Only the positive value is practical and so the fastest speed at which there would have been no collision is 15.1 ms1 or 33.8 mph.
10m 12m 0.8s 23.7m 1.33s RT POP
39.9 mph  RTA
30 mph
5.6m 12.4m 17.8m 1.33s
33.8 mph
15.6m 20.1m 1.33s RT
Diagrammatic Summary
Note: If the speed was above 33.8 mph, the collision might still not have occurred because it would have taken the car longer to reach the impact point and in that extra time, the child would have moved beyond the path of the car.
Note: It is not always to the benefit of the car driver to assume that he has a rapid reaction time. The longer the reaction time, the more likely it is that the accident would have been avoided from some lower speed.
Note: Children under 10 years (Highway Code, Rule 7) cannot be held liable.
6!#24zCUVc}78# ) E G c e
9
;
W
Y
:
;
?
@
V
W
5EHH*OJQJ jm5CJ,OJQJ jm5CJ$OJQJ5CJ H*OJQJ5CJ OJQJ5CJ$OJQJ5CJ(OJQJH56?fg$;Dop[\BSSd\*$]S^S#BCcd@^} & H f 7SMd\*$]7^S`M7Sd\*$]7^SSd\*$]S^`
<
Z
v
ghBgwx'
7d\*$]7^`7SMd\*$]7^S`M7d\*$]7^7Sd\*$]7^S'
(
6
G
X
Y
r

"#IJZ7d\*$]7^`7@d\*$]7^@`7Sd\*$]7^SW
X
o
q
y
z
>HLMNQRSWY
Zxz{5CJ H*OJQJ56CJ OJQJ5H*OJQJ5CJ$H*OJQJ56CJ$OJQJ5CJ(OJQJ5CJ$OJQJ jm5CJ$OJQJ5EHH*OJQJ5CJ OJQJ5OJQJ=Z[YZx 2D7Sd\*$]7^SEk4_*,FH}!#<AZ~]_rtkmǿ5CJ,OJQJ5CJ(OJQJ5>*CJ OJQJ5CJ$OJQJ5>*CJ$OJQJ56CJ OJQJ5CJ H*OJQJ5CJ OJQJHDEkl349_`hp{
:;7d\*$]7`7Sd\*$]7^S;
8IR;<*C]y7Sd\*$]7^Syrs!<=AghG7Sd\*$]7^SYZ~Mau7@Md\*$]7^@`M7@d\*$]7^@`7Sd\*$]7^Ss5op
(ST7Sd\&d*$P]7^S7Sd\*$]7^SACPR189<
Q T !!.!@!y!{!!!!}""X#^###ļļļ5CJ$OJQJ5CJ,OJQJ5CJ(OJQJ5CJOJQJjUmHnHu5OJQJ5CJ H*OJQJ5CJ OJQJ,19 Q !.!y!!}!!!"}"W#X###$7Sd\*$]7^Sa$7Sd\*$]7^Sd\*$,&P 1. A!"n#$%7
i<@<NormalCJOJQJ_HmH sH tH <A@<Default Paragraph FontD56?fg$;Dop[\BCcd@^}&Hf<ZvghBgwx' ( 6 G X Y r 
"
#
I
J
Z
[
YZx 2DEkl3
4
9
_
`
h
p
{
:;
8IR;<*C]yrs!<=AghGYZ~Maus5op
(ST19Q.y}}WX00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000W
# B '
ZD;y#!#8@ (
DB
?"DB
?"VB
Co?" VB
Co?"
DB
?"DB
?"VB
@
C?"VB
C?"VB
C?"JB
@
#Ԕ?"JB
#Ԕ?"VB
Co?"
VB
Co?"DB
?"DB
?"VB
C?"VB
C?"VB
C?"JB
@
#Ԕ?"JB
#Ԕ?"VB
Co?"VB
Co?"DB
?"VB
C?"VB
C?"B
S ?12345679:; QRSyz
t tt+t+tttwGtwGttZZtwtwt
btZXtbZbXtbtZZtbZZtZbZt
t
t
t
KKtK
KtG.R.GreatrixMC:\WINDOWS\Application Data\Microsoft\Word\AutoRecovery save of Document2.asdG.R.GreatrixC:\My Documents\AAC\lec2.docG.R.GreatrixC:\My Documents\AAC\lec2.doc@yy$yy@@UnknownG:Times New Roman5Symbol3&:Arial3"Swiss"1hmYxY
7!Z0
2LECTURE NOTES 2G.R.GreatrixG.R.Greatrix
Oh+'0
<H
T`hpxLECTURE NOTES 2ECT
G.R.Greatrix 2.R..R.Normala
G.R.Greatrix 22R.Microsoft Word 9.0@@30G@:2G
՜.+,0hp
R7
LECTURE NOTES 2Title
!"$%&'()*+,./012345678:;<=>?@BCDEFGHKRoot Entry F H2GM1Table#*WordDocument.DSummaryInformation(9DocumentSummaryInformation8ACompObjjObjectPool H2G H2G
FMicrosoft Word Document
MSWordDocWord.Document.89q