#17: The Beep.

The Question

I was recently driving back to Northborough from Rochester, NY after the holidays when a thought hit me. After rubbing the bruise the thought gave me I began to think. Say one person in a car crosses the MA border from NY on I-90 and beeps his/her horn. The next person closest to them on I-90 then beeps there horn in some sort of automotive relay race. This continues for however long it takes until “The Beep” reaches Boston. Now stemming from this thought I had several quandries; 1: How long would it take “The Beep” to reach Boston from the Ma/NY border? 2: How many cars would it take to necsesitate “The Beep”? and finally 3: Would the fact that the cars are all moving towards boston speed up “The Beep” in any measurable amount?
Hope you had a good holiday shaun and i can’t wait to hear your answer!


Comment by Kurt — 12/30/2004 @ 1:42 pm

The Answer

I had to call in the big guns on this one. Many thanks to Jen (of Shamus and Jen) for the mathematical support.

Here are the FACTS:

1. The Mass Pike is 138 miles long from the NY border to its end in Boston.
2. Sound travels at 740 miles per hour in 30 degree air.
3. The cars are moving at an average of 70 miles per hour.
4. At 70 MPH, the safe distance between you and the next car is 7 times the length of your car.
5. The average car is 9 feet in length.
6. There are 5280 feet in a mile.

Here are the ASSUMPTIONS:

1. The location of the horn is in the back of the FIRST car.
2. The number of cars is fixed (moving area of interest).

Here is the MATH:

5280 ft * 138 miles = 728640 ft.
10120 cars to get through.
(9 *7) = 63 ft between cars.
63 ft / 5280 = 0.011932 miles between cars.
740 MPH (speed of sound) – 70 MPH (speed of car) = 670 MPH.
1 hour / 670 MPH * 0.011932 = 0.000018 hrs * 3600 seconds in an hour = 0.064111 seconds.
0.085 seconds to get sound from back of car to front.

So the final equation is:

(10120 cars) * (0.085 seconds) + (10120) * (0.064111 seconds) = 1509 seconds.

1509 seconds = 25.15 minutes.

So, to answer your questions:

1. It would take 25.15 minutes for “The Beep” to get to Boston.
2. The number of cars needed to actualize “The Beep” would be 10120.
3. The speed the cars are traveling does NOT increase the speed of the beep. Rather the opposite actually.


8 thoughts on “#17: The Beep.

  • 2/8/2005 at 6:14 pm

    Shaun, I have two concerns.

    First, I am slightly confused by your statement that the speed of the car slows down the progress of the beep. It is my understanding that sound will always travel at the speed of sound, no speeding up, no slowing down (provided the sound is traveling through the same density material). The speed does change the pitch, because the wavelengths change in size.
    Please see my paint drawing:
    But I do not believe it would influence the speed.

    Secondly, there’s another issue I have with the car speed. Imagine this: You have all your cars laid out between NY and Boston. If none of them were moving, there would be a total of 10120 cars (as you stated) and assuming the rest of the math is correct and the speed of sound used was 670mph it would indeed take 25 minutes. However, when they are all moving at a set 70 mph, the situation is different. As soon as the first car beeps the cars at the Boston end of the line start getting off the Mass Pike. so really, you’d have to remove (25 min = .4 hr)
    (.4)*70mph = 28 miles worth of cars.

    In the 25 minutes that the beep takes to travel, the line of cars would have moved 28 miles, and the resulting time required for the beep’s travel would change! So, I think in order to properly predict the time it takes for the beep to travel, you’d have to use limits to determine the correct number of cars. This would make me believe that the speed of the cars actually DOES impact how long it takes for the Beep to reach Boston from NY.

    AND, with car numbers so high, you may want to consider multiplying the time it takes for the average human to physically react to a stimulus (which I believe is 650 milliseconds).

    Despite my concerns, I still think you did a Bang up job answering this question. Bravo!

    -mike d.

  • 2/8/2005 at 9:24 pm

    I’m more than happy to address your concerns (although my pathetic mathematical skills will no doubt be evident).

    Your first objection is to the overall speed decrease. Think of it this way. If we were running at the same speed, and I threw a bottle of yellow mustard at your head, the bottle would start at the point I was at when I threw it. But you’d be running away from that point, so it would actually have to travel farther than the distance between us to hit you and dump yellow mustard all over your nice new shirt. Of course, that doesn’t explain away your objection, but it makes me feel better to throw mustard at your head.

    For your second objection, simply re-read assumption #2.

    I did find a similar figure for the human reaction time, but we wanted to keep the scenario as simple as possible (given my obvious lack of math skills and the fact that the question was asked by Kurt, who has the same love of math that I do).

    Thanks for the insightful comments!

  • 2/9/2005 at 3:35 pm

    You know what we can do. We can compare shauns answer with Google’s Answer. Google has a question answering feature that if you give them say, 5 bucks they answer a question. More money for difficult questions. I say that we try to raise money to put towards this and have the good people at Google answer this questions then we can compare answers. I have one dollar for this investiture.

  • 2/18/2005 at 2:35 pm

    I think there is one other obvious flaw in the initial assumptions and problem setup. When have you EVER seen all the people on the MA Pike driving the “safe distance” from other cars?! I think you should include the Masshole factor, which dictates that for every 1 safe driver, there are approximately 5.5 unsafe drivers, and 0.5 grandmothers (I looked that fact up in my handy “make-up-anything-you-like-and-pass-it-off-as-reality” guide). I believe the unsafe drivers would be tailgating at approximately 1-2 carlengths MAX, and the grandmas would be travelling at 40-50 mph tops. If you figure this in, I think the rate at which “the beep” moves will be less linear, and thus the total “sound flux” would be the integral of the velocity profile across the “sound transport area”. Also, since the fast lane would approximate a car/air interface, at that boundary you could assume the viscous force is essentially zero. To the right of the slow lane (aka granma lane) is the breakdown lane, where cars do not move, and would give you the so called “no-slip condition”. With those boundary conditions in place, you could then solve for the constants of integration and arrive at the appropriate velocity profile. If you then divide by the “sound tranport” area, you would obtain the average velocity of the sound, which could then be used with the distances in question to arrive at the time necessary. Of course, you could always make the “plug flow” assumption if you feel the traffic flow (and therfore sound flow) is very turbulent and well developed, in which case the velocity profile would simplify and you could use just inlet and outlet flows and forces to arrive at the velocity. But in either case, you still have to assume the cars are Newtonian. If not, you’ll have to use a numerical solving package such as Mathematica or the like to figure out the partial differential equations that arrive when you combine the momentum equation with the equation of motion. I think this is the right way to go with solving the problem, and I expect a full answer derived from first principles by morning, including all assumptions with an appropriately labeled diagram, and a computer simulation of the whole shebang! Yep, I’m just that bored at work right now… and also working on my Transport Phenomena classwork.

  • 2/19/2005 at 7:53 am

    My main complaint here, is that the average speed from the Mass/NY border is probably about 70mph as stated in the answer. However, the average speed of any driver on the Mass Pike inside of rt. 495 is less than a tortoise during rush hour (rush hour defined as the hours between 5am and 9pm).

    So, considering that traffic density generally inceases as you go from east to west, and thus, the average speed of the drivers decrease… is it safe the say that the beep will move faster as it approaches Boston?

  • 2/19/2005 at 7:54 am

    oh yeah…I read Google Answers alot when I’m bored, here’s another dollar to get experts on the question!

  • 2/21/2005 at 9:27 pm

    i HAVE A NOTHER REBUTTAL Shaun. While driving back to Northborough from Rochester again this weekend, there was a sign just over the border of NY/MA and it said that the distance to Boston was 132 miles. That should alter your calculations. I just want to get this as accurate as possible.

    Becky, I spoke to Shaun about this in person, and we decided to ignore ruch hour and all that, kinda going for an ideal situation with the amount of cars on the Pike. I hope I helped answer that for you Becky, and free up some of your time to get to other questions Shaun.

  • 2/22/2005 at 4:42 pm

    Kurt – a quick comment. Just because there is 132 miles to Boston, doesn’t mean the Mass Pike is 132 miles long. In fact, all that means is that there is 132 miles to the Boston City Limits. There is still 6 miles of Pikeage inside of Boston! Hoo Ha!


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