Time duration for the yellow traffic light

A yellow light at a traffic intersection should last long enough that a car traveling at the suggested speed can either

apply the brakes and decelerate to a stop prior to reaching the front of the intersection, or
maintain the same speed and pass through the intersection before the yellow light turns red

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:
$T_{yellow}\ge T_1= \Delta t+\frac{v}{2gu}+\frac{W}{v}$
and
$T_{yellow}\ge T_2= \Delta+\frac{v}{gu}$
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 $1.8<T_1<2.8$ s.
The 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!

-*-

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)

Full 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

Please feel free to post your ideas about how to use the simulation for better teaching and learning.
Post questions to be asked to help students to think, to explore.
Upload worksheets as attached files to share with more users.

Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!

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_min=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_min.
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_min+ 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 $RT+\frac{v}{g\mu}$
Summary:

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

The extra time is required because we need to make decision ahead of time.

You can check out Tale Of The 3-Second Yellow Light, Traffic Light Logic, THE YELLOW LIGHT for more story.

Related Topics
	Subject	Started by	Replies	Views	Last post
	Traffic Light Simulation (Java) Request for physics Simulations	null404	9	40416	April 22, 2008, 08:59:13 am by Fu-Kwun Hwang
	traffic light whole system by using matlab Request for physics Simulations	bennycyg	14	26806	August 01, 2019, 08:46:52 am by sunnysunny
	Time duration for the yellow traffic light kinematics	ahmedelshfie	0	4387	June 01, 2010, 07:21:53 pm by ahmedelshfie
	solar power traffic light replace the signals on wires kinematics	louse	1	3029	October 31, 2019, 11:08:36 am by Hanfonal
	JAVA: Counters, Traffic Light Selection, Hilbert Curve, Morse Code Request for physics Simulations	Loving	2	1107	October 12, 2019, 01:53:00 pm by Alexjohnson

	Author	Topic: Time duration for the yellow traffic light (Read 47496 times)
0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).