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When a train moves at x miles an hour the total pressure on the 
front of the locomotive due to air pressure (y lb.) is given by 
the following:

x   0       20      40       60       80          100
y   0     79,5     318    715,5     1272       1987,5

Is there a simple formula connecting x and y? If yes, 
what is it? --Sawyer, "Mathematician's Delight"

I can't see what's the formula here.

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4 Answers 4

up vote 3 down vote accepted

$f(40)=79.5\cdot 4,\;\;f(60)=79.5\cdot9,\;\;f(80)=79.5\cdot16\cdots$ suggests $f(x)=79.5\cdot \left(\dfrac{x}{20}\right)^2 $

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Instead of giving you the plain answer, which has already been given, I will try to let you see how i did it

First notice that the 0.5 occurs one time, then it doesn't the other one, then it does again etc. 79.5 -- 318 -- 715.5 etc

One can guess that the 0.5 occurs when you divide for a power of two that is one higher the highest power of two in the numerator (e.g. you do 4/8, 16/32 etc)

If you look at the power of two, you find that when appears the 0.5 the highest power of two is 4; so you may guess that you divide by 8 when computing the number.

Then multiply by 8 all the numbers (to see what were the factor of the numerator before you divide by 8) and you find out that the factor 159 is always present.

In fact, you get $636 = 4*159, 2544 = 16*159, 5724=36*159, 10176=64*159, 15900=100*159 $

Nice.

So remember we know $y = 159/8*Something$.

Get rid of the 159 (divide everything by 159) to get

20 --> 4 | 40 --> 16 | 60 --> 36 | 80 --> 64 | 100 --> 100

It's easy to see that the pattern here is (x/10)^2

putting all together we get

$y = 159/800x^2$

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If you plot the data, it looks parabolic. I would start by seeing if you can find $k$ so that you can represent it as $y=kx^2$. This works well.

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$$159/800(x)^2$$ So $159/800(40^2)=318$ etc

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Maybe this can be edited to avoid the 1/ab ambiguity (i.e., does 1/ab = 1/(ab) or 1/ab = (1/a)b?). –  Douglas S. Stones Oct 17 '12 at 2:57

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