“Drivers’ Ed.” Physics
Interesting examples from the Alabama Driver Manual
Albert A. GapudUniversity of South Alabama
AAPT-Alabama Meeting
Apr 5 2014
Issues:
1. Using a seat belt
2. Reducing speed when turning
3. Blind spots
4. Stopping distance
5. Following distance
Seat belt• Page 31: Alabama safety belt and child restraint laws
Newton’s Law of Inertia!
Can hurt you…– Consider: If your car hits a wall at 40 mph, you will go
through your windshield at 40 mph.– You keep moving even if the car does not
… unless you make yourself a part of the car.– Also explains why:
• We don’t put things on the dashboard
Seat belt• Page 31: Alabama safety belt and child restraint laws
Newton’s Law of Inertia!
Can help you…!– How seat belt locks up!– Clutch lever snaps when belt is jerked– Weighted pendulum tilts when accelerated
Reducing speed when turning• Page 22: Reduce speed before making turns.• Centripetal force is required!
– Fc = mv2/R. Reducing v : Required Fc reduced more rapidly.– Usually provided by: static friction.
• Static friction: required for rolling• Coefficient, μs :
– rubber on dry concrete : about 0.8– rubber on wet concrete : about 0.3, less than half!
• Maximum friction = μs N = μs x weight = μs x (1600x 10 N)
• = mv2/R. So max
Example: 90 degree intersection turn: curb radius:
local streets: R = 15 ft = 4.6 m : max v : dry = 6.1 m/s = 14 mph, wet = 3.7 m/s = 8
mph
arterial streets: R = 35 ft = 10.7 m: max v : dry = 9.3 m/s = 21 mph , wet = 5.7 m/s =
13 mph
sm
Rv 16000
Mirrors and blind spots• Page 65: Check “blind spot” before changing lanes.• Geometric optics: law of reflection, ray trace
Mirrors and blind spots• Page 65: Check “blind spot” before changing lanes.• Geometric optics: law of reflection, ray trace
Blind area on either side due to:
Limited field of view of mirrors
(m.1 and m.2 angled inward)
Mirrors and blind spots• Page 65: Check “blind spot” before changing lanes.• Geometric optics: law of reflection, ray trace
Blind area on either side due to:
Limited view of mirrors
(m.1 and m.2 angled outward)
Stopping / following distance• Page 38: eye to brain to foot to wheel to road.• Stopping distance: static target (e.g., debris)
– Student exercise? Find reaction time that was used.• tr = 0.74 s
– Student exercise? Find the acceleration (assume constant).• a = -19.4 ft/s2
– Issue with significant digits??
Stopping / following distance• Page 37: avoiding rear end collisions
• Comparison: One c.l. per 10 mph vs. 2-sec rules:One c.l. per 10 mph equivalent to: “1.36-sec rule”
2-sec rule equivalent to: “1.47 c.l. per 10 mph”
Stopping / following distance• Page 37: avoiding rear end collisions
• How well do these help avoid rear end collisions?Graphed x(t). x1 = x2 means collision.
Other assumptions: same acceleration (deceleration) for the two cars, one shifted by the reaction time.
Stopping / following distance• Rules for following distance: comparison
– How do these help, in a typical situation?
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.00
2
4
6
8
10
12
14
16
18
20
22black = your carred = car in front
tr = 0.74 s
1cl/10mph = 1.36 sec
d
ista
nce
, in
ca
r le
ng
ths
time (s)
initial = 55 mphreaction time = 0.74 s
acceleration = 19.6 ft/s2
1 car length = 20 ft
2sec = 1.47 cl/10mph
Stopping / following distance• Rules for following distance: comparison
– How do these help, in a typical situation?
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.00
2
4
6
8
10
12
14
16
18
20
22black = your carred = car in front
tr = 0.74 s
+ 2.5 cl
1cl/10mph = 1.36 sec
d
ista
nce
, in
ca
r le
ng
ths
time (s)
initial = 55 mphreaction time = 0.74 s
acceleration = 19.6 ft/s2
1 car length = 20 ft
2sec = 1.47 cl/10mph
+ 5 cl
Separation at stop = d – votr (ft)
or D – (vo/20)tr (cl)
Stopping / following distance• Rules for following distance: comparison
– How do these help, in a typical situation?
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.00
2
4
6
8
10
12
14
16
18
20
22black = your carred = car in front
0.54 cl/10mph
tr = 0.74 s
+ 0 cl+ 1 cl+ 2.5 cl
1cl/10mph = 1.36 sec
1sec
d
ista
nce
, in
ca
r le
ng
ths
time (s)
initial = 55 mphreaction time = 0.74 s
acceleration = 19.6 ft/s2
1 car length = 20 ft
2sec = 1.47 cl/10mph
+ 5 cl
Separation at stop = d – votr (ft)
or D – (vo/20)tr (cl)
Further Reading: “Car physics”The Isaac Newton School of Driving:
Physics & Your Carby Barry Parker: Professor Emeritus of Physics,
Idaho State University and son of a car mechanic
Mainly a good source of conceptual examples from all fields of physics– Driving, Engines, Electrical system, Brakes,
Suspension & Transmission, Aerodynamics, Collisions
– Racing
– Traffic and Chaos
– Future technologies
(hybrid, fuel cell)
Further Reading: “Car physics”Fast Car Physicsby Chuck Edmondson: Professor of Physics, US
Naval Academy, racing enthusiast
More quantitative, detailed treatment, with ultimate goal of fast-car racing: Possible source of student projects?– Finding shift points, shifting properly
– Drag force, gear ratios
– Rounding curves efficiently
– Tires and load transfer
– Steering and suspension
– Future/green technologies
Final notes• Many other possible topics• Goal: Show physics in our daily life• Driver’s ed still fresh w/ HS students• “Car physics”: entry point for HS students?
Spare slides
Stopping / following distance• Rules for following distance: comparison
– How do these help, in a typical situation?
Separation at stop = d – votr (ft)
or D – (vo/20)tr (cl)
Following rule up to t = 0
0 2 4 6 8 10 12 14 16 180
10
20
30
40
50
60
70
80
900.54 cl/10mph 1cl/10mph = 1.36 sec
2sec = 1.47 cl/10mph
black = your carred = car in front
spe
ed
, in
ft/s
distance, in car lengths
initial = 55 mphreaction time = 0.74 s
acceleration = 19.4 ft/s2
1 car length = 20 ft
Here, condition is:
x2 (v=0) > x1 (v=0)