The Physics of Law Enforcement - PowerPoint PPT Presentation

1 / 22
About This Presentation
Title:

The Physics of Law Enforcement

Description:

A demo drives into a car with a force of 4000 N ... Sports car r = 0.96 glass. Taxi r = 0.9 steel. m1 m2 (u1 u2)(1 r) v1 = u1 - m2 ... – PowerPoint PPT presentation

Number of Views:46
Avg rating:3.0/5.0
Slides: 23
Provided by: TWEL2
Category:

less

Transcript and Presenter's Notes

Title: The Physics of Law Enforcement


1
The Physics of Law Enforcement
New Curriculum Teaching Ideas for Secondary
School Physics Teachers
Senior Constable John H. Twelves B.Sc. Technical
Traffic Collision Investigator Western Region
Traffic
2
Collision Reconstruction
3
Sir Isaac Newton
4
Skid to Stop Formula
Speed 15.9 d x f e
d is the distance of the skid f is the drag
factor, e is the road grade
5
Build Your Own Drag Sled
Tire Cement Steel Rod
Scale
6
Vericom Accelerometer
7
Braking Distance !
S 15.9 d x f e x B
100 Braking Four Wheel Lock-up Skid or full
ABS Where f 0.75 and S 80km/hr Front
Brakes Only 70 d 48 meters Rear Brakes Only
30 d 112 meters One Front Brake 35 d 96
meters One Back Brake 15 d 225 meters
8
Multiple Surfaces Different Drag Factors
9
Combined Speed Formula
S S12 S22 S32
DRAG FACTOR f Cement 0.9 Asphalt 0.75 Wet
Asphalt 0.45 0.7 Ice 0.15
10
Demolition Derby
Non-Collinear Addition of Forces using
Trigonometry
Sine Law a b
c Sin A Sin B
Sin C Cosine Law c2 a2 b2 - 2ab
cos C


11
N
1 cm 400 N
C
FN ?
45o
30o
3000 N b
a 4000 N
45o
A
? c
E
W
B
A demo drives into a car with a force of 4000 N
N 45o W then a second demo driver hits the car
with a force of 3000 N S 30o W Net Force equals
?
S
12
Known Values a 4000 N, b 3000 N, C 75o c2
a2 b2 - 2ab cos C c2 40002 30002
- 2 x 4000 x 3000 x cos 75o c2 25000000
24000000 x 0.258819 c2 18788342 c 4334
N Using the Sine Law
a c Sin A Sin C
4000 4334 Sin A Sin 75o 4000 sin
75o 4334 sin A 67o


Net Force FN 4334 N W 7o N
13
Yaw Marks Critical Curve Speed
S 11.27 R x f e
R C2 M
8M
2
Where R radius of curve, f drag factor, e is
the superelevation across the curve and M
middle ordinate
14
Lover's Leap Formula
S 7.97d dsinØ x cosØ
hcosØ2
dhorizontal distance, Ø is in degrees, h
vertical elevation difference
15
(No Transcript)
16
Conservation of Momentum
m1u1 m2u2 m1v1 m2v2 Total momentum
Total momentum Before impact After
impact

Approximate Values of Coefficient of Restitution
Iron 0.67 Steel 0.90 Rubber 0.75 Lead 0.16
Brass 0.30 Bronze 0.52 Copper 0.22 Glass 0.96
17
HEAD-ON COLLISIONS Recoil Velocities
Ft mv mv0
m2
(u1 u2)(1 r)
v1 u1 -
m1 m2
18
Curb weight of vehicles m1 1554 kg m2 1092
kg u1 22 m/s u2 10 m/s
Coefficient of Restitution Sports car r 0.96
glass Taxi r 0.9 steel
m2
(u1 u2)(1 r)
v1 u1 -
m1 m2
1092 kg
v1 22 m/s-
(22 m/s 10 m/s)(1 .96)
1092 kg 1554 kg
V1 12 m/s
19
We need to rearrange the variables to solve for
v2
m1
(u2 u1)(1 r)
v2 u2 -
m1 m2
1554 kg
(10 m/s 22 m/s)(1 .9)
v2 10 m/s -
1554 kg 1092 kg
v2 23 m/s
20
m1u1 m2u2 m1v1 m2v2
v3 (m1 m2 ) m1
u1
In-line collision, vehicle 2 stopped, no post
impact separation
m2 v2 m1
u1 v1
In-line collision, vehicle 2 stopped, post impact
separation
21
m1u1 m2u2 m1v1 m2v2 90 degree collision
with separation, departure angle known
m1 v1 sin 0 m2
u2
v2 sin O
m2 v2 cos O m1
u1 v1 cos 0
22
Time-Distance Analysis
A person is traveling at 100 km/hr when a tree
falls across the road 90 meters ahead. Does he
hit the tree? µ 0.7
S2 254 µ
Distance in a skid to stop when speed and
coefficient of friction are known
d
100 km/hr 27.8 m/s
Reaction Time Distance 1.5 s x 27.8m/s 41.7 m
Skid to Stop Distance 56.2 m
97.9 m
Distance of Drop v1t 1/2gt2
Write a Comment
User Comments (0)
About PowerShow.com