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Traffic Shockwave Physics Topic

  • Thread starter Pmand92
  • Start date
  • #1
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1. Homework Statement :

An abrupt slowdown in concentrated traffic can travel as a pulse, termed a shock wave, along the line of cars, either downstream (in the traffic direction) or upstream, or it can be stationary. figure below shows a uniformly spaced line of cars moving at speed v = 26.0 m/s toward a uniformly spaced line of slow cars moving at speed v_s = 5.20 m/s. Assume that each faster car adds length L = 15.0 m (car length plus buffer zone) to the line of slow cars when it joins the line, and assume it slows abruptly at the last instant. (a) For what separation distance d between the faster cars does the shock wave remain stationary? If the separation is twice that amount, what are the (b) speed and (c) direction (upstream or downstream) of the shock wave?


2. Homework Equations :

Incoming Flux (Fast zone): (1 car/d) * (26 m/s)
d=Unknown distance

Outgoing Flux (Slow zone): (1 car/15 m)* (5.20 m/s)

1) Δt= Δr/ΔV(avg)
2) r=V(avg)*t

3. The Attempt at a Solution :

a) I have used the conversion about the outgoing and incoming flux equations and set them equal to each other to find d, in meters. I got 75.0 meters, which was not correct in WileyPlus. Then I tried for this same part using equation 1: Δt= 15.0 m/5.20 m/s, which gave me 2.8 seconds for each car to join the "slow zone" every 2.8 seconds. Then I used the 2nd equations with the new found time: r=26.0 m/s*2.8 seconds, which gave me 72.8 m. This again was not correct in WileyPlus.

b) I tried to follow an answer that was posted on YahooAnswers about this part, except replacing their variables with mine:
"If the cars are separated by twice this distance, then 12m get added to the end of the shock zone every 4.8s (120/25). So that comes out to 2.5m/s (12/4.8). Thus the shockwave advances down the roadway at 2.5m/s (5-2.5)". After I tried this with my variables, again the answer was not correct in WileyPlus.
 

Answers and Replies

  • #2
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I also forgot to mention on Part A, I tried to follow the YahooAnswers reply using their method, with my variables (it didn't work either):

"Since the slow zone is moving at 5m/s, if the shock front is to remain stationary, then there must be 5m/s of cars being added to it. Each car will add 12m to the length of the slow zone, so a car must join the slow zone once every 2.4 seconds (12/5). Given that time separation, and the speed of 25m/s for the fast zone, the cars must be separated by 60m (25*2.4)."
 
  • #3
1,762
59
What is the time interval for the cars traveling at 26 m/s and spaced at 15 m to pass a stationary point?

For every such time interval we are adding 15 m to the shockwave and every second we are subtracting 5.2 m.

Can you develop a formula, in which spacing is the only variable, that describes how much the shock wave advances or recedes each second?
 
  • #4
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I am not sure what you are asking...there is no time interval given in the problem.
 
  • #5
1,762
59
I am not sure what you are asking...there is no time interval given in the problem.
That's true, but since the slow zone cars are moving at 5.2 m/s, we need to find a spacing between the fast zone cars that will add only 5.2 m/s to the shockwave which is receding at 5.2 m/s. If we add more space between the cars, the cars pass a stationary point less frequently but add more than 15 m with each arrival.

Let's find a formula for how many fast zone cars pass a stationary point per second, then we can calculate how much distance is added with each car. We already know the cars are traveling at 26 m/s and we're looking for a spacing that will add 5.2 m to the shockwave every second.
 
  • #6
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Where are you getting 5.2 m? In the problem, its 5.20 m/s, a velocity. The only distance given in meter is for the buffer zone of 15.0 m.
 
  • #7
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Here is the part A answer:

Incoming flux = outgoing flux
(1/(d + 15.0 m))(26.0 m/s) = (1/15.0 m)(5.20 m/s)
d = 60.0 m
 
  • #8
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Ok, so I found someone else's thread about this similar problem for part B, but I cannot seem to follow their equation fully:
"vsw = (qb – qa)/(kb – ka)

Where
vsw = propagation velocity of shock wave (m/s)
qb = flow prior to change in conditions (vehicles/s)
qa = flow after change in conditions (vehicles/s)
kb = traffic density prior to change in conditions (vehicles/m)
ka = traffic density after change in conditions (vehicles/m)"

The q's are the fluxes, but I cannot seem to follow it.

https://www.physicsforums.com/showthread.php?t=281230
 
Last edited:
  • #9
1,762
59
That's more complicated than it needs to be.

If a stream of cars is traveling 26 m/s and they're spaced 15 m apart, then the cars are passing a stationary point at the rate of one car every 15/26 per second or s/26 for any spacing. Each car adds s meters to the shockwave.

What is the formula for the amount of space s added to the shockwave each second. Note that without fast zone cars, the shockwave would move forward at 5.2 m/s so we want to find a spacing that will add exactly 5.2 m/s.
 
  • #10
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Never mind, I figured it out.
 
  • #11
13
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How I figured out Part B:

Since I found d to be 60 m, the total distance was 75.0 meter. Part B wanted to know what the speed was when the separation was twice the amount. I multiplied 75.0 m*2 to get 150 meters. I then took the double separation distance and divided by the original velocity: 150 m/26.0 m/s= 5.77 seconds. Thus, 15 m is added at the end of each shock zone every 5.77 seconds. So I divided 15.0 m/5.77 s to get a velocity of 2.6 m/s. I then subtracted 5.20 m/s-2.6 m/s= 2.6 m/s. The shockwave advances down the road at 2.6 m/s in a downstream direction.
 
  • #12
1,762
59
Congratulations
 
  • #13
let Vt be the speed of the shockwave relative to a stationary observer,taking the upstream direction to be +ve.Let Vs be the speed of the slow traffic.The time taken for the fast car to reach the slow car would be=d/(26-5.2)=d/20.8
In the meanwhile,the shockwave travels through one car of length 15m.Therefore the speed of the shockwave,relative to the traffic would be=15/(d/20.8)=312/d.
now we know that
Vt - Vs = velocity of the wave relative to the traffic=312/d
For the wave to be stationary,Vt=0
Therefore, -Vs=312/d
Vs=-5.2 (Taking upstream direction to be +ve)
We get 5.2=312/d
Therefore d=312/5.2=60m
Hence the answer is 60m :)
 

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