the physics of law enforcement new curriculum teaching ideas for secondary school physics teachers...
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![Page 1: The Physics of Law Enforcement New Curriculum Teaching Ideas for Secondary School Physics Teachers Senior Constable John H. Twelves B.Sc. Technical Traffic](https://reader036.vdocument.in/reader036/viewer/2022062321/56649ee85503460f94bf9047/html5/thumbnails/1.jpg)
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
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Sir Isaac Newton
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Speed = 15.9 d x f ±e
“d” is the distance of the skid
“f” is the drag factor, “e” is the road grade
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Tire Cement Steel Rod
Scale
Build Your Own
Drag Sled
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Vericom Accelerometer
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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
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Multiple Surfaces
Different Drag Factors
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S = S12 + S2
2 + S32 …
DRAG FACTOR f
Cement 0.9
Asphalt 0.75
Wet Asphalt 0.45 – 0.7
Ice 0.15
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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
= =
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N
S
EW
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 ?
1 cm = 400 N
B
C
A
3000 N b a 4000 N
? c
45o
45o
30o
FN = ?
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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]
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S = 11.27 R x f ±e
R = C2 + M8M 2
Where R = radius of curve, f = drag factor, e is the superelevation across the curve and M = middle ordinate
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S= 7.97d
dsinØ x cosØ ± hcosØ2
d=horizontal distance, Ø is in degrees,
h = vertical elevation difference
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m1u1 + m2u2 = m1v1 + m2v2
Total momentum Total momentum
Before impact After impact
=
Approximate Values of Coefficient of Restitution
Brass 0.30
Bronze 0.52
Copper 0.22
Glass 0.96
Iron 0.67
Steel 0.90
Rubber 0.75
Lead 0.16
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HEAD-ON COLLISIONS
Recoil Velocities
m1 + m2(u1 – u2)(1 + r)v1 = u1 -
m2
Ft = mv – mv0
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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
m1 + m2(u1 – u2)(1 + r)v1 = u1 -
m2
1092 kg + 1554 kg(22 m/s – 10 m/s)(1 + .96)v1 = 22 m/s-
1092 kg
V1 = 12 m/s
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We need to rearrange the variables to solve for v2
m1 + m2(u2 – u1)(1 + r)v2 = u2 -
m1
1554 kg + 1092 kg(10 m/s – 22 m/s)(1 + .9)v2 = 10 m/s -
1554 kg
v2 = 23 m/s
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v3 (m1 + m2 )
m1
u1 =
In-line collision, vehicle 2 stopped, no post impact separation
m1u1 + m2u2 = m1v1 + m2v2
m2 v2
m1
u1 = v1 +
In-line collision, vehicle 2 stopped, post impact separation
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m1u1 + m2u2 = m1v1 + m2v2
90 degree collision with separation, departure angle known
m1 v1 sin 0
m2
u2 = + v2 sin O
u1 = v1 cos 0 +m2 v2 cos O
m1
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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
d = S2
254 µ100 km/hr = 27.8 m/sDistance in a skid to
stop when speed and coefficient of friction are known
Reaction Time Distance 1.5 s x 27.8m/s = 41.7 m
Skid to Stop Distance = 56.2 m
97.9 mDistance of Drop = v1t + 1/2gt2