Expert: Daniel Mazur - 2/10/2010

Question
QUESTION: Hi this problem is one I couldn't understand very well so I couldn't do big attempts at solving
since I have it on pdf and I don't want to mistake in typing and that there is a picture in it I made the problem into a picture
http://img15.imageshack.us/img15/9461/53086498.png

L=12
v(fast)=25
v(slow)=5
d=?

I couldn't even understand what's wanted in the problem nor all the data given
PS: this problem is from Halliday Resnick Fundamentals of physics 8th edition chapter 2 question 12

ANSWER: Hi Hamad,

I am familiar with an earlier edition of that book. The problem touches a surprising, yet practical emergence of wave mechanics (of sorts) in a system of solid bodies - cars. First must come the experience - you may or may not have a personal experience with it, depending on the kind of traffic in your hometown or region. The authors of the problem are Westerners (Americans, if I am any judge) and there it is pretty much everybody's experience to frequently spend long periods in a car on an expressway during rush hours. The expressway is full, so the cars move really slowly around exits, then between exits they move faster...

The moment one car has to slow down for some reason, every car behind (unless they change lanes) must slow. In this problem they use a very much simplified model situation. There is only one lane (we take an ordinary road instead of a highway) and every car experiences only two distinct states of motion: A) going fast, B) going slowly. Let's pick any ONE car for the moment. The model says that the slowing down is abrupt and in the last moment, so we can consider it as an instantaneous change. This change happens the moment our car has only as much distance to the bumper of the car in the front, as is the spacing between every two cars in the slow-moving line.

So far I have only re-stated the model's setup car by car. Instead of cars they could have picked for example red blood cells slowing down in a capillary vein or numbered balls inside a tube coming out of a lottery machine. They chose cars, because it is most people's everyday experience in that part of the world.

Some time ago somebody noticed that we can look at the motion of the cars NOT from the point of view of the cars (or their drivers) but "from above, far away". From this point of view the individual cars become merely part of a flowing Traffic - and the traffic can resembles a "liquid". Already our saying that traffic "flows" gives a hint of "liquid" properties, doesn't it. Those, who decided to investigate this phenomenon of liquid traffic, soon observed that there are "waves" moving through the traffic independently of the motion of any PARTICULAR car. Say we mark with a (mobile) flag the point in the traffic, where the cars suddenly slow down form the high speed state to low speed. The flag would be mobile with respect to the cars: It would hang on the tail of the slow-moving line of cars. But as soon as a new car comes from behind and slows down (ergo joins the slow line) the flag moves behind that new car in order to maintain its function as a dividing marker between the fast and slow part of the traffic. Can you visualize it? Make a moving model, if you need to, anything to help you visualize it.

Still before we jump at the problem, we must make an observation, which is: depending on the speeds of the cars in front of and behind the flag and depending on the separations between the cars (again in front of and behind the flag) the flag may move WITH RESPECT TO THE STATIONARY GROUND in all different ways. It can move forward, it can stay in place (oscillating about one spot) or it can move against the flow. How? Well the two extremes are easy to observe.

Extreme 1) The slow cars actually have a vary small speed, or they stand still. Then all incoming cars just increase the length of the slow (stopped) line and the flag, which marks the end of the slow line, will move backwards.
Extreme 2) The slow cars move still at some decent speed (say half of the fast cars) and the rate of incoming fast cars is really low. This means that by the time another fast car joins the slow line the end of the line moves forward by much more than the length L (there is one car per length L in the slow line). This means that the flag on the whole moves forward with only an occasional jump by L back. The distance traveled between each successive jumps backward is much longer than L, so the overall motion is forward.
Considering the two extremes coming together one can imagine the stationary case as such, when the rate of incoming cars (flag jumps by L backwards) exactly matches the speed with which the slow cars' line moves forward. The flag will be sliding forward at the speed of the slow cars and then after it travels the length L forward, a new car joins the slow line ant the flag jumps distance L backward and thereby ends up exactly, where it'd begun.

a) So, the first question should be solved straightforwardly now. Let T be the period of the rate with which new cars join the line. Then the slow line grows backwards once every T and this is related to the speed v(fast) and separation d of the fast cars as T=d/v(fast). For a stationary wave this T must be also the time, in which the slow line cars travel the distance L, so T = L/v(slow). Elimination of the T gives d = L*v(fast)/v(slow) = 60 m.
c) It takes only some common sense to observe that if the separation d increases (d1=2*L*v(fast)/v(slow)), the tail of the slow line will experience a growth less frequently than necessary to compensate for the forward motion of the slow line and the wave (the flag we have been using) will, on average, move FORWARD with the cars!
b) The speed of the wave is the difference between the forward speed of the slow line (v(slow)). Let T1 be the new, longer period at which the line's tail grows by L (one car). Then the average speed of the tail's growth, v(tail) = L/T1 = L/(d1/v(fast)) = L*v(fast)/(2*L*v(fast)/v(slow)) = v(slow)/2. And this is the answer: The wave will move forward at a speed v(slow)/2 = 2.5 m/s.

Here you are. I hope this is intelligible, I tried to explain it really step by step. Model solution, you might say. I hope I didn't make any embarrassing error in calculations, but it shouldn't have.

Cheers,
Daniel




---------- FOLLOW-UP ----------

QUESTION: Could you please explain more on this part?
"b) The speed of the wave is the difference between the forward speed of the slow line (v(slow)). Let T1 be the new, longer period at which the line's tail grows by L (one car). Then the average speed of the tail's growth, v(tail) = L/T1 = L/(d1/v(fast)) = L*v(fast)/(2*L*v(fast)/v(slow)) = v(slow)/2. And this is the answer: The wave will move forward at a speed v(slow)/2 = 2.5 m/s. "

ANSWER: Hi, yes,

the problem for b) and c) is modified in terms of the separation d between the fast cars. The slow line stays intact. If T is the time, in which a car in the slow line covers the distance L (look up definition of L), then the motion of the slow line is described by v(slow)=L/T just like before (in par a)).

The fast on-coming cars are joining the slow line one by one. One needs to get used to the idea, that the slow line grows 'jerkily', not continuously, but the growth can still be described by some 'average speed' of the line growth, a v(line growth). The key is to realize: The velocity of the end of the slow line that includes new cars joining it - this is what I labeled v(tail) previously - is the difference between the speed v(slow) and the average speed v(line growth).

In the following I have just realized I made a mistake - it doesn't pay to solve problems in the middle of a night! :-) Let me do it properly. We know that the distance covered by a fast car is s=v(fast)*t, where t is just time. In order to catch up with the end of the slow line a fast car needs to cover the distance s=d+v(slow)*t - this is where I made an error previously, because I neglected the second term. [NOTE: In this response I will use the numerical result of part a) as the input here, so that d=120m in this follow-up. You can do the full analytic substitutions if you like, it should be easy.] We join the two equations by substituting for s and we express the time t, t=d/(v(fast)-v(slow)). This is intuitive in the sense that if v(slow)>=v(fast), then the "slow" line will not grow. Importantly, the time t is no longer just some time, but it's the time between the successive events of new (fast) cars joining the slow line. The inverse this time 1/t we call the "line growth rate". Considering that each growth event adds the length L to the line then v(line growth)=L/t. After substituting for t the formula is
v(line growth) = [v(fast)-v(slow)]*L/d, which is sensible only for v(fast)>=v(slow).

Finally we obtain v(tail) as
v(tail) = v(slow)-v(line growth) = v(slow) - [v(fast)-v(slow)]*L/d
Numerically v(tail) = 5m/s - 20m/s*12m/120m = 3m/s.

So the answer to part b) is v(tail)=3m/s. As this is a positive number, the tail average motion is forward (just as with my wrong solution to b) ) and that is the answer to c).

This finally IS correct. :-) Many apologies for making an error previously.

Cheers,
Daniel

---------- FOLLOW-UP ----------

QUESTION: I am really sorry but I think I messed up something wrong and want you elaborate on what it is because I got d = 48 and that is through this :(I am going to write everything I thought of after your explanation)
I choose initial position of slow car =0
so initial position of fast car = -(d+l)
from what I understood from you
total distance traveled by fast car (s)= v(fast)*t=d+L
total distance travel by slow car = L=v(slow)*t
t=L/v(slow) = 12/5 S
s=v(fast)*t=d+L
25*12/5 = d+L
60=d+L --->>>60=d+12 --->>>>d=48m
Did I make an algebriac mistake somewhere Or I totally misunderstood you?

Answer
Hi,

You are right, I made a mess. I should have corrected part a) as well. The bumper-to-bumper separation is indeed 48 m. For some reason I had re-defined d to mean the fast car spacing in the front-to-front sense. For the part b) I was numerically correct in the first place, but that was caused by the fact that 2.5m/s is half of 5m/s, which "intuitively" corresponds to the two-fold increase in the separation of the fast cars.

The numerical solution of part b) changes - the analytic part really should be correct this time. v(tail) = 5m/s - 20m/s/8 = 2.5m/s. This is the same number as I gave you the first time, but here it has been derived correctly.

Cheers,
Daniel