One dimensional motion in Halliday Resnick Fundamentals

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Homework Help Overview

The problem involves one-dimensional motion, specifically examining the dynamics of two sets of cars traveling at different speeds and their interaction in a traffic scenario. The original poster expresses difficulty in understanding the problem statement and the relevant data provided.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Some participants discuss the concept of frequency in relation to the movement of cars, suggesting that it relates to how often cars arrive at a specific point. Others question how frequency applies to a one-dimensional motion problem, expressing confusion about its relevance.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have provided analogies, such as traffic jams and water flow, to clarify concepts, while others are still seeking to understand the relationship between the variables involved.

Contextual Notes

There is uncertainty regarding the distance between the last slow car and the first fast car, which is critical for solving the problem. Participants are also navigating the implications of the problem's setup and the definitions of terms like frequency in this context.

madah12
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Hi this problem is one I couldn't understand very well so I couldn't do big attempts at solving

Homework Statement


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=?
a=0

Homework Equations



delta x = v(avg)*t


The Attempt at a Solution


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
 
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from the speed of the cars and their distance, you can get the frequency of the cars: how often a car arrives to a given point.
To have a stationary shock wave, both sets of cars should have the same frequencey:
if the frequency of fast cars would be lower, then the end of the traffic jam would go forward. If the frequency of fast cars would be higher, the end of traffic jam would go backward.

I like this example. Now I see that traffic is not just gaseous (the traffic fills up all available roads), but also shows some properties of waves.
 
I don't understand how is frequency related to one dimensional motion problem?
They are going on one way not on cycles...
 
madah12 said:
I don't understand how is frequency related to one dimensional motion problem?
They are going on one way not on cycles...

the only equation for frequency that I know and might be usable here is that w=v/r
w=2pi f
f=2pi/w
but here the r is the radius and this is only for circular motion so I don't know exactly what you mean?
 
frequency: how much times an event occurs in a given time period.

in this problem freguency is how many car arrives at a given position in a given time period.
you would use 1/s as the measure, and the amount would be small, but you can think of it as
"how many cars arrive to the wavefront in a minute?" the wavewront being the point where the car must decelerate, wherever it is.
 
but I don't know d which is the distance between the last slow car and the first fast car so how do I know when does the first fast car need to decelerate to join the line?
I mean from what I understand the 2 fast cars will slow down to add L which is 12 to the line but I don't know their current distance to the slow cars and actually when will they decelerate .
 
You don't have to know the direction and velocity of individual molecules of a gas to come up with its pressure.
Similarly you don't have to know the exact location of the wave front to solve this example.

Try to imagine a traffic jam. the cars in the jam move at 5 m/s, and they are 12m apart.
(well, it is too fast for a real traffic jam, but anyway).
Now how much cars will pass per seconds in the jam? Call this number f.
Or how many seconds will be between cars? You can call this T.
You might notice that f = 1/T. (Or some would insist on car/T)

Where will the end of the traffic jam vill move if there are more than f cars arrive each seconds?
Or put it the other way: what if the time difference between cars is less than T?

Now what if frequency is lower than f (or dt > T) ?

When will the end of traffic jam be stationary?
 
You can also think of pipes of water. One is thin, the water velocity being high. The other is thick, the water velocity being low.
Now if you know both the water velocities and the area of the thicker pipe, and you also know that the
joint of pipes do not magically move, you can calculate the area of the thin pipe. You don't have to know where the joint exactly is.
 

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