# Traffic shock wave problem

• agnimusayoti

#### agnimusayoti

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 2-25 shows a uniformly spaced line of cars moving at speed v = 25.0 m/s toward a uniformly spaced line of slow cars moving at speed vs = 5.00 m/s. Assume that each faster car adds length L = 12.0 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?
Relevant Equations
average velocity = displacement / time taken
This was a Halliday Resnick Problem on the linear motion. I could have see the solution but I was not satisfied because I have not get the idea behind the solution. Could you please help me to explain what is exactly the problem is? Or, how to model the shock wave? So that I can make the equation of the fast car position, slow car position, and shock wave position at any instance time, t?

• Delta2

If you want the textbook solution explained, you will need to post it.

Suppose car A has just joined the slow queue at position x along the road. How far behind is B, the next car? How much faster is B moving than A? How long will it take to join the slow queue? Where will it be then?

Not really. I do not only want the solution, but also the understanding of the idea of how to solve the problem..

Well, does B will join the slow queue if
$$x_{B} - x_{A} = L$$
Since ##x_{A} = (L+d) + v_{s}(t)## and ##x_{B} = vt##
then ##t(v-v_{s})=d##

But, how long, or what is the time? Besides, i don't really see where the shock wave occur in that model.

• Delta2
Not really. I do not only want the solution, but also the understanding of the idea of how to solve the problem..

Well, does B will join the slow queue if
$$x_{B} - x_{A} = L$$
Since ##x_{A} = (L+d) + v_{s}(t)## and ##x_{B} = vt##
then ##t(v-v_{s})=d##

But, how long, or what is the time? Besides, i don't really see where the shock wave occur in that model.
I see from your second equation that your are taking d to mean the separation additional to L, which is fair enough.
Your notation is a bit confusing. Are you defining ##x_{A}, x_{B}## as functions of t or as initial positions?
If as functions of t then
##x_{A}(0) - x_{B}(0) = L+d##
##x_{A}(t) = x_{A}(0) + v_st##
and until B joins the queue:
##x_{B}(t) = x_{B}(0) + v_ft##
But somehow you got the right equation, ##t=(v_f-v_s)/d##.

At this time, where is the start of the queue? What is its displacement?

Yes, I take (0,0) at B at t = 0. So at instance time, t, I get the A dan B position . Or,
$$x_{A}(0) = L + d$$
$$x_{B}(0) = 0$$
Therefore,
$$x_{A}(t) - x_{B}(t) = x_{A}(0) - x_{B}(0)+ v_{s}t - v(t)$$
$$x_{A}(t) - x_{B}(t) = L+d+(v_{s} - v)t$$
Then,
The condition that B starts geting in the slow car queue is
$$x_{A}(t) - x_{B}(t) = L$$
Or it means that the rear part of A and the rear part of B separated by L.
Therefore, I got
$$d =(v_{s} - v)t$$

What do you mean by your question? is this equation enough to show the traffic shock wave? because I don't see it.

is this equation enough to show the traffic shock wave?
No, you are right, it isn't, which is why I asked those further questions.
At t=0, the tail of the queue is at ##x_A(0)##.
When B is distance L behind A, where is the tail of the queue?
How far is that from ##x_A(0)##?

What is the definition of the shock wave we are using here? Haven't seen one in classical mechanics book unless it is classical fluid mechanics book.

What is the definition of the shock wave we are using here?
It is the movement of the boundary between the two states.

• Delta2
No, you are right, it isn't, which is why I asked those further questions.
At t=0, the tail of the queue is at ##x_A(0)##.
When B is distance L behind A, where is the tail of the queue?
How far is that from ##x_A(0)##?
Well, I think when B is distance L behind A, so the tail of the queue is ##x_{B}(t)## or when ##x_{B}(t) = x_{A}(0)##? Is this correct?

Well, I think when B is distance L behind A, so the tail of the queue is ##x_{B}(t)##
Yes.
or when ##x_{B}(t) = x_{A}(0)##? Is this correct?
No, that's the point. The tail was at ##x_{A}(0)##, now it is at ##x_{B}(t)##. How far has it moved? Use your equations in post #5.

It is the movement of the boundary between the two states.
Hmm, let me see if I understand something in the intuitive level, since you avoid to give some equations.
In order for the boundary to move in the same direction as the traffic, the velocity at the downstream (down to the direction of traffic) must be greater than the velocity of the upstream and in order for it to move opposite to the direction of traffic the velocity at the down stream must be smaller than the velocity of the upstream like in this problem statement. Is my intuition right?

Is my intuition right?
No. If the downstream velocity were greater than the upstream no queue would exist.

• Delta2
Yes.

No, that's the point. The tail was at ##x_{A}(0)##, now it is at ##x_{B}(t)##. How far has it moved? Use your equations in post #5.
So the tail of the queue moved forward with distance L, with the velocity vs?

So the tail of the queue moved forward with distance L, with the velocity vs?
Please post your working. Use your equations in post #5 to write ##x_{A}(0)## and ##x_{B}(t)## in terms of ##d, L, v, v_s##.

At the time B enter the queue, that is ##t = t_{q}## then
$$t_{q}=\frac{d}{(v-v_{s})}$$
Therefore,
$$x_{A}(0) = L+d$$
$$x_{B}(t_{q}) = v(\frac{d}{(v-v_{s})})$$

Is it correct?

At the time B enter the queue, that is ##t = t_{q}## then
$$t_{q}=\frac{d}{(v-v_{s})}$$
Therefore,
$$x_{A}(0) = L+d$$
$$x_{B}(t_{q}) = v(\frac{d}{(v-v_{s})})$$

Is it correct?
Yes. So what is the change in position of the tail of the queue?

The change in position of the tail of the queue then ##x_{B}(t_{q}) - x_{A}(0)##? So, we have:
$$x_{B}(t_{q}) - x_{A}(0) = d(\frac{v}{v - v_s} - 1) - L$$?

The change in position of the tail of the queue then ##x_{B}(t_{q}) - x_{A}(0)##? So, we have:
$$x_{B}(t_{q}) - x_{A}(0) = d(\frac{v}{v - v_s} - 1) - L$$?
yes, but is that the shift in the direction of the traffic or the other way?
Also, you can simplify the term in parentheses.
Can you now answer questions a, b, c?

• agnimusayoti
Now I understand. So, shock wave basically is the rate of the tail shifting in the low speed queue?
Therefore, we have:
$$v_w \frac{d}{v-v_s} = d\frac{v_s}{v-v_s} - L$$
But, because the value of ##\frac{v_s}{v-v_s}## equal to 0,25 then ##0,25d## must be bigger than L if we want to produce downstream shockwave. Either way, if we have 0,25d = L then the shock wave is stationary?

Hence,
(a) d = 4L
(b) ##v_{w} = v_{s} - \frac{L}{d}(v-v_s) = 2,5 m/s##
(c) downstream

But, how do you know the shock wave is the shifting of the tail? I mean, how can you get that idea?

It is the movement of the boundary between the two states.
Did the idea come from this definition?

• Delta2
how do you know the shock wave is the shifting of the tail?
A wave in any medium consists of the linear movement of a structural relationship of parts of the medium relative to the medium itself. If you take two snapshots at different times it looks as though the medium has moved, but it is only the form that has.
In a shock wave, the medium is in a different state (in this case density) on the two sides of the wave front.

I don't really understand what do you mean by linear movement of a structural relationship of parts of the medium relative to the medium. But, yes I know that the 2 snapshots is different in the form. The difference is caused by the movement of the vibration or the disturbance of the medium...

In this shock wave case, what is the medium, btw?

Anw, if we have this understanding, could we solve the problem with another way?

By structural relationship I mean some abstraction from a spatial distribution of entities.
In a body of water, the abstraction may be the height of the surface. It is an abstraction because we do not care about exactly which molecule is where; we only care about the overall appearance of the form. If, over time, that shape moves (other than by simply the bodily movement of the water mass as though a solid) we call it a wave.
Other examples… a "Mexican" wave, where people in a crowd raise and lower their arms in a coordinated manner; in an epidemic, the geographic distribution of cases; in the present case, the density distribution of cars along the road, the cars being the medium.