Traffic Shock Wave - Speed & Direction of Wave

[Greychu]

An abrupt slowdown in concentrated traffic can be 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 uniform spaced line of car moving at speed v = 25.0 m/s toward a uniformly spaced line of slow cars moving at speed v_s = 5 m/s. Assume that each faster car adds length L = 12.0 m (car plus buffer zone) to the line of slow car 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?

For clearer detail, take a look on the attach files.

My works:

(a)

15 = \frac{d+nL}{\Delta t} ... (1)
(average speed of the faster car before and after it join the line.

5 = \frac{nL}{\Delta t} ... (2)

Suppose that at time = \Delta t, the faster car (closest to the slower car line) joins the slower car line, and it's speed decrease to 5 m/s.

(2) - (1) yields 10 = \frac{d}{\Delta t}

The problem is I can't find a value of \Delta t, making the part (a) cannot be solved...

I need to know how to do part (a) so that I can do part (b) and (c) as well.


http://www.amasci.com/amateur/traffic/traffic1.html for more detail about traffic shock wave.

 

[Integral]

If you can post a pdf of your word doc you'll find more people willing to open it.

 

[Dick]

This is really a flux problem. If d is the distance between the incoming cars then the flux of cars coming in is (1car/d)*(25m/sec). The units of flux are cars/sec. The outgoing flux is (1car/12m)*(5m/sec). If the shock front is stationary, then incoming flux must equal outgoing flux. This is the principle of 'conservation of cars'.

 

[Greychu]

Hey, Dick. If I followed according what u said, then the answer for part (a) is 60 m. So the speed of shock wave is actually the speed of the slower car right? As I do part (b), if I use the same way as part (a), I get answer of 2.5 m/s, which the shock wave will remain stationary...? I currently thinking about it.

 

[Greychu]

Integral, u know how to convert a Words document into pdf document?

 

[Integral]

Integral, u know how to convert a Words document into pdf document?

You have a private message

 

[Greychu]

Why the incoming flux is not (1car/(d+12))*(25m/sec)? Imagine that faster car B join the line of slower car X, it travel d+12 m if the slower car X travel 12 m, and another faster car C replace the place of faster car B that all events happened at time of t.

stationary shock wave, so speed of shock wave = 0
Incoming flux = outgoing flux
(1/(d + 12))(25) = (1/12)(5)
d = 48 m

I think the flux u mean is qb and qa as in formula below, right?

vsw = (qb – qa)/(kb – ka)

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

Note the magnitude and direction of the shock wave.

(+) Shock wave is traveling in same direction as traffic stream.
(-) Shock wave is traveling upstream or against the traffic stream.

So can I use the above formula to solve part (b) and (c)?

 

[Dick]

Yes, the fluxes are the q's. And that equation looks correct. If d is the spacing of the cars before the shock wave, why would the flux be (1car/(d+12))*(25m/sec)? In front of the shock wave the cars don't necessarily even know there is a slowdown coming up. 12 is the spacing AFTER the slowdown.