A yellow light at a traffic intersection should last long enough that a car traveling at the suggested speed can either
[li]apply the brakes and decelerate to a stop prior to reaching the front of the intersection, or [/li]
[li]maintain the same speed and pass through the intersection before the yellow light turns red[/li]

If a driver traveling at the suggested speed cannot do either of the two options, then the traffic signal (specifically the time duration of the yellow light) is considered unsafe.

If the speed of the car is v, the friction coefficient between tires of the car and the road is $\mu$.
The maximum brake force $F= -\mu N = -\mu m g $, and $F=ma $
So $a=- g\mu$. The distance required for the car to fully stopped (after brake is activated) is $s=\frac{v^2}{2a} =\frac{v^2}{2g\mu}$.
Assume the reaction for the driver is $\Delta t$, then the total distance required to stop the car after the driver find the yellow light just turn on is: $D_{min}=v \Delta t + \frac{v^2}{2g\mu}$.

The car has to be $D_{min}$ away from the intersection., for the car to be fully stopped behind the intersection. of the road.
If the distance is smaller than $D_{min}$, the time for the yellow should be enough for the car to pass the interaction. Assume the width of the intersection is $W$, and the time for the yellow is $T$.
Then it requires that $v T\ge D_{min}+W= v \Delta t + \frac{v^2}{2g\mu}+W$.
So $T\ge \Delta t +\frac{v}{2 g\mu}+\frac{W}{v}$ This is the minimum required time for car to pass the intersection.

However, if the car need to stop before the traffic light, the minimum distance is $D_{min}=v\Delta t + \frac{v^2}{2g\mu}$
For the car to stop from initial velocity v and acceleration $a=-g\mu$, it need $t_{brake}=\frac{v}{gu}$ from $v(t)=v_0+at$
So the total time required is $T'_{min}=\Delta t+\frac{v}{gu}$

It means that the time for the yellow light $T_{yellow}$ need to satisfy two equations:
$T_{yellow}\ge T_1= \Delta t+\frac{v}{2gu}+\frac{W}{v} $
$T_{yellow}\ge T_2= \Delta+\frac{v}{gu} $
So the time for yellow should be larger that the maximum of $T_1,T_2$
The condition for $T_2>T_1$ is $\frac{v}{gu}>\frac{v}{2gu}+W/v$, which imply
  $\frac{v}{2gu}>W/v$ i.e.  $ \frac{v^2}{2gu}>W$ 
The above condition is the same as stopping distance $ge$ width of intersection which is the case for normal speed limit and traffic light.
However, if  $W \ge \frac{v^2}{2gu}$, then the minimum time for traffic light is $\frac{v}{2gu}+W/v$

For v=72km/hr=20m/s, $\mu=1, \Delta t=0.8s, W=20 m$.
$T_1= 0.8+ \frac{20}{2*10*1}+\frac{20}{20} =2.8 s$
$T_2= 0.8+ \frac{20}{10*1}=2.8s$
So the minimum time required is 2.8s.
However, if the width of the interaction is less than 20m, then $1.8The minimum time required is still 2.8s

The following simulation let you play as a traffic light control manager:
You can change the width W of the interaction, the reaction time for the driver, the time for the green light and yellow light. (If you click the right most checkbox, the program will show suggested time for yellow light)

Code for the car:
green: moving at constant speed.
red: decelerate  
yellow: accelerate

*** the maximum speed and maximum acceleration for each car is randomly selected in the simulation , to make the simulation closer to the real case. I hope you can enjoy it!


/htdocs/ntnujava/ejsuser/2/users/ntnu/fkh/streetyellowlight_pkg/streetyellowlight.propertiesFull screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License
Download EJS jar file(1353.7kB):double click downloaded file to run it. (28 times by 23 users) , Download EJS source (23 times by 18 users) View EJS source

Let's apply physics principle to estimate yellow light time duration.

Suppose the reaction for the driver is RT, the speed of the car is V, the friction coefficient between tire and the road is mu, mass of the car is m, gravity is g.

Then the friction force Fr= - m*g*mu = m*a so the deceleration a=g*mu
The minimum stopping distance when driver saw the light term yellow is D[sub]min[/sub]=V*RT+ V*V/(2*g*mu)
You can adjust the deceleration a directly with slider control.
he friction mu= 1.0-1.2 for normal tire. But it is a strong brake.
Normally, we did not brake the car with maximum deceleration.  So the default value is set to a=0.5

The above analysis ignore the width of the car.
If the car want to stop before s/he  reach the front of the interaction, the minimum distance is D[sub]min[/sub].
If distance is less than Dmin, the car has to pass the interaction before the end of the yellow light.
Suppose the length of the car is d, the width of the interaction is W, and the time for yellow light is YT.
V*YT >= D[sub]min[/sub]+ W+d

For the car to pass the traffic light, the minimum time for yellow light should be
$YT_{min}= \frac{D_{min}+W+d}{V} = \frac{W+d}{V} + RT + \frac{V}{2*g*mu}$

If the yellow light is too short, then some car would not be able to pass the intersection safely.
However, if the driver do not want to brake the car so quickly (want to be more comfortably), replace 2*g*mu with 2*g*mu/k. the above simulation use k=2 to estimate the time for yellow light).

If the yellow traffic light last too long, the driver might not want to stop the car, and ,when the light turn RED, s/he would not be able to fully stopped before the interaction.

If we want the car to stop before the traffic light, the minimum time for yellow light is $RT+\frac{v}{g\mu}$
[li]For very long intersection $W\ge\frac{v^2}{2g\mu}-d$, (i.e. width of interaction + width of car larger than stopping distance for the car),
the minimum time required is $RT+ \frac{v^2}{2g\mu}+\frac{W+d}{v}$ : Reaction time + braking time+ time to pass intersection.
For short intersection where $W\le\frac{v^2}{2g\mu}-d$,
the minimum time required is $RT+ \frac{v^2}{g\mu}$: Reaction time + braking time *2
[/list] The extra time is required because we need to make decision ahead of time.

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