The Physics of Law Enforcement New Curriculum Teaching Ideas for Secondary School Physics Teachers...

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