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 = S S S 3 2 … DRAG FACTOR f Cement0.9 Asphalt0.75 Wet Asphalt0.45 – 0.7 Ice0.15

Non-Collinear Addition of Forces using Trigonometry Sine Law a b c Sin A Sin B Sin C Cosine Law c 2 = a 2 + b 2 - 2ab cos C ==

N S EW A demo drives into a car with a force of 4000 N [N 45 o W] then a second demo driver hits the car with a force of 3000 N [S 30 o W] Net Force equals ? 1 cm = 400 N B C A 3000 N b a 4000 N ? c 45 o 30 o F N = ?

Known Values a = 4000 N, b = 3000 N, C = 75 o c 2 = a 2 + b 2 - 2ab cos C c 2 = x 4000 x 3000 x cos 75 o c 2 = – x c 2 = c = 4334 N Using the Sine Law = = a c Sin A Sin C Sin A Sin 75 o 4000 sin 75 o = 4334 sin A = 67 o Net Force F N = 4334 N [W 7 o N]

S = R x f ±e R = C 2 + M 8M 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 ± hcosØ 2 d=horizontal distance, Ø is in degrees, h = vertical elevation difference

m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 Total momentum Before impact After impact = Approximate Values of Coefficient of Restitution Brass0.30 Bronze0.52 Copper0.22 Glass0.96 Iron0.67 Steel0.90 Rubber0.75 Lead0.16

HEAD-ON COLLISIONS Recoil Velocities m 1 + m 2 (u 1 – u 2 )(1 + r)v 1 = u 1 - m2m2 Ft = mv – mv 0

Curb weight of vehicles m 1 = 1554 kgm 2 = 1092 kg u 1 = 22 m/su 2 = 10 m/s Coefficient of Restitution Sports carr = 0.96 glass Taxir = 0.9 steel m 1 + m 2 (u 1 – u 2 )(1 + r)v 1 = u 1 - m2m kg kg (22 m/s – 10 m/s)(1 +.96) v 1 = 22 m/s kg V 1 = 12 m/s

We need to rearrange the variables to solve for v 2 m 1 + m 2 (u 2 – u 1 )(1 + r)v 2 = u 2 - m1m kg kg (10 m/s – 22 m/s)(1 +.9) v 2 = 10 m/s kg v 2 = 23 m/s

v 3 (m 1 + m 2 ) m 1 u 1 = In-line collision, vehicle 2 stopped, no post impact separation m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 m 2 v 2 m 1 u 1 = v 1 + In-line collision, vehicle 2 stopped, post impact separation

m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 90 degree collision with separation, departure angle known m 1 v 1 sin 0 m 2 u 2 = + v 2 sin O u 1 = v 1 cos 0 + m 2 v 2 cos O m 1

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 = S µ 100 km/hr = 27.8 m/s Distance 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 m Distance of Drop = v 1 t + 1/2gt 2