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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
Sir Isaac Newton
Speed = 15.9 d x f ±e
“d” is the distance of the skid
“f” is the drag factor, “e” is the road grade
Tire Cement Steel Rod
Scale
Build Your Own
Drag Sled
Vericom Accelerometer
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
Multiple Surfaces
Different Drag Factors
S = S12 + S2
2 + S32 …
DRAG FACTOR f
Cement 0.9
Asphalt 0.75
Wet Asphalt 0.45 – 0.7
Ice 0.15
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
= =
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 = ?
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]
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
S= 7.97d
dsinØ x cosØ ± hcosØ2
d=horizontal distance, Ø is in degrees,
h = vertical elevation difference
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
HEAD-ON COLLISIONS
Recoil Velocities
m1 + m2(u1 – u2)(1 + r)v1 = u1 -
m2
Ft = mv – mv0
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
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
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
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
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