NOVA physics: chapter 1, question 5. proportional question, doesn't make sense.

[ibeatupnerds]

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You are given this equation: E=0.2(Density)(Area)(Distance)(v^2)
If just her area is reduced by 20%, how much further can she drive and still use the same amount of energy?

I get the explanation in the back of the book: (1) solve for Distance. Distance = [(0.2)(Energy)] / [(density)(Area)(v^2)]
(2) if a reduced by 20%, then A is multiplied by 0.8 (assuming her original was 1)
(3) so Distance = (0.8)^(-1)=1.25 thus she has 25% increase [had to use a calculator to confirm this, not sure how to do -1 exponents]


yeah, ok, I get that.

BUT WHY
isn't it 20% further. I mean if her area decreases by 20% than why wouldn't her distance increase by 20% since everything else is constant.


Can someone tell me why I can't use the second way and maybe even give me an example when to use the first way and when to use the second.

 

[loveoforganic]

1) That's not the equation I would get solving for distance

distance = E / [(0.2)(density)(area)(v^2)]

2) good

3) x^-1 = 1/x

Finally, the way you've written it, it should be 20% further, unless I'm missing something. Otherwise, you've written it down wrong or the book is wrong.

 

[ibeatupnerds]

steps (1) (2) and (3) was the books explanation. which makes sense (at least if you plug in the numbers)

the 20% is what I guessed (which was wrong, according to the book).

I don't get why you'd do it the books way and get 25% and why not my way which leads to 20%

 

[Catburr]

Why is it that a 20% decrease in one variable isn't directly "made up for" by a 20% increase in another?

Think about it this way:

I have an equation that says W = YZ. If Y decreases by 20%, then we now have 0.8(YZ). But we find out that we can boost Z to bring us back to 1(YZ). What do we multiply by to get back to 1(YZ)? Not 1.2, because 0.8 x 1.2 = 0.96.

To find the necessary multiplier to get us back to 1, you just divide: 1/0.8 = 1.25, which means a 25% increase is required to recover from a 20% decrease. (1.25-1)*100% = 25%

Similarly, to recover from a 30% decrease, we'd look at 1/0.7 = 1.43, so we need a 43% increase.

It becomes more intuitive, at least to me, once you get into the whole number multipliers - for example, a 50% decrease is easy to understand: 1/0.5 = 2, so we need a (2-1)*100% = 100% increase. We lost half, so we multiply by 2. And so on, 75% decrease means we lost 3/4, have only 1/4 left, so we multiply by 4 to make up for it.

This kind of math is commonly used when we dilute samples to run an assay. Don't feel bad about finding it counter-intuitive...my Ph.D. holding boss messes it up quite frequently. 🙄

Hope this was vaguely helpful.

 

[loveoforganic]

d'oh, good stuff!

 

[ibeatupnerds]

yeah, that helps. I guess I should of thought about it bit more. It's a bit different using percentages because it changes the final product (versus basic v=x/t, if someone goes twice the velocity, than they go the twice the distance in the same time), first time I've encountered percentages. Thanks!

 

[Catburr]

yeah, that helps. I guess I should of thought about it bit more. It's a bit different using percentages because it changes the final product (versus basic v=x/t, if someone goes twice the velocity, than they go the twice the distance in the same time), first time I've encountered percentages. Thanks!

No problem. In all likelihood I'm missing your point, but I want to point out that whether you are using decimals, or fractions, or percentages, the math is actually the same, as long as you correctly convert between them. Using a different notation shouldn't change your final product, unless I'm missing something. 2/5 = 0.4 = 40%, for example.

Yes, if you go twice as fast, you go twice the distance if you travel for the same amount of time:

Percentages: v --> 200% v, so x --> 200% x, if t stays at 100% t

Whole numbers: v --> 2v, so x --> 2x if t stays at 1t