Math Is Fun Forum

well i should honestly say that my formula was derived by the same theory, and inadequate

No air resistance, RFcomes only from in between four wheels and road.

adding air resistance would make RF more complicated. air resistance would be proportional to the speed when low, to square of it when high.

i haven't got that kind of facts, if you like, you can check out papers, published experiment results, etc

ignoring direction, consider velocity as speed  (presented as v).

Net a= dv/dt=v'   Newton's  Law,
Net a= P/(vm) - RF/m -mathsyperson's equation
define P/m=k k is a constant to v
in a friction process, friction force RF= u (Pressing Force) = u m g, where u is determined by the friction surface, m is the car mass, g is acceleration due to gravity
thus RF/m = u g , regardless of mass or speed. define u g= p

v'=dv/dt=k/v-p (former formula doesn't conclude -p here cuz i was not thinking clearly then )

v'-k/v=-p

for simple linear air resistance

v'= k/v -p - av ( backward acceleration due to air)

Who can solve that?

to make more real, net acceleration is fixed at first for some time when v is too low.

Perhaps we can use a software to solve it, i think you are good at it, sam

EA    Seconds to 200km/hr
10    12.5
v=200 t=10

v= k log(1+t)  <v velocity t time elapsed> , ideally.

Thank you, i will do it

midhole spays furtherst, if the bottle is placed on ground.

the position of x should be a function of t. x = p(t)
- can we use a cartesian co-ords? A water particle's position is
(x,y)= (x(t),y(t)) , in which x is the horizontal distance from bottle, y is the height it has fallen

define d= depth

x= √(2dg) t  (1)
y= g t^2 /2   (2)

solving 2 for t gets
t= √(2y/g)

hence x can be expressed  an explicit function of y as
x=2 √(dy)

if you place a full bottle on the ground, a hole at 1/n height of bottle and another hole at 1-1/n height will have their water stream intersect on the ground.

but the hole at the mid point will spray water further, with x=2d , the same as bottle height, when y= d = 1/2 bottle height

((1-t^2)/t^1/2) = 1/t^1/2 - t^2/t^1/2 =t^-1/2 -t^3/2

t^a/t^b=t^a-b and 1/t^a=t^-a

Both velocity and acceleration are vectors. All vectors can be decomponented into a sum of inter-perpendicular vectors. perpendicular means they don't interfere each other. in this case, a vector can be decomponented to a horizontal one and a verticle one, (eg Sqrt(2gh) i and 0j, where i and j stand for horizantal and verticle unit vector) and the verticle component of the acceleration does not influence the horizontal componet , like g doesn't interfere horizontal velocity.

Thus a motion containing accelerations and velocities and displacements can be decomponented into 3 dimensions.
here 2-d is enough.

horizontally, no a, so v constantly √(2gh) thus  displacement sx=v t, where sx stands for horizontal displacement.
vertically, a=g, v0=0 so vt = at = gt, displacement sy=1/2(v0+vt)t = 1/2 a t^2 = 1/2 g t^2
now we know after a certain time t, sy=d,
solve t out, t=√(2d/g)
sx=v t = √(2gh) √(2d/g)= 2√(hd)
sx=2√(hd), when sy=d
t is not contained in this formula, and it applies to  water particle escaping from any holes. it states its horizontal displacement after having fallen distance d.

uh o, it seems a lot of difficulty to go further, lazy person like i would like to pass it onto next solver...

Do you agree real numbers are defined as variables, and they are mobiles?

I could give a example that approaching is not being even in real number system.

What's the difference between [0,1) and [0,1]?

Ho Odg!!

Q15 is especially hard

I think it's time to make up some agreement and brief the points.

1-0.999...=infinitesmall by k/10^n type
1/9-0.111...=infinitesmall by k/10^n type
1/infinitesmall is Exactly Stable 0,so that  .999...=Exactly1 , 0.111...=Exactly 1/9, or else you admit infinitesmall isn't  as  SAME as 0, THOUGH "CLOSE"
between 2 real numbers exists a real number.

By the way, no easy excuses, "you cannot simply apply finite to infinite" is one of them, it will be considered an excuse, not one explaination. Because the question "Why?" "How?" will challenge it.

INFINITE is not a CARD GAME, with whatever set-up rules in mathematists' imaginary world.
It should be LOGICALLY CONSISTENT .
Next time i will explain it in more detail.

FORMER PROOF of
               What the Quotient and the Remainder of the dividend 1/9 are
        as soon as  the Nth digit after the decimal point of the Quotient Comes Out

Hypothesis:(H(n)) the quotient is 0.11...1 (n 1s) and the remainder is 0.0...0(n 0s so far)1=1/10^n

Proof:
1) n=1

    0.1
   ____
9) 1.0
       9
  _____
       1------show the remainder is 0.1

H(1) is true

2) If H(k) is true, when k>=1, deduct H(k+1)
the quotient is 0.11...1(k 1s) and the remainder 0.00...(altogether k 0s)1, continue the division
            kk+1 th digit after decimal point
            | |
   0.11...11
   _______
9)1.0
   ...
   ...
           __
           10
             9
            __
           *10
              9
            __
              1---the remainder is 0.00..0(altogether k+1 0s)1
* the remainder in H(k) is 0.00...(altogether k 0s)1 with the form 1 corresponding the nth digit after the decimal point in the quotient. To further compute, add 0 on its right.

When H(k) is true, H(k+1) is also true

Conculusion)  for n>=1, H(n) is true.

Proof of the error of the quotient to 1/9 is 1/900...0(n 0s)when the former has n digits after decimal point---
1/9-0.11...1(n 1s) =

10^n/(9*10^n) - 99...9/(9*10^n)

= 1/(9*10^n)=1/900...0(n 0s)

0.11...1 still not a rational number, the only rational way up to now is to express it as a series
{S(n)}={0.1, 0.11, 0.111, ...},
Limit[S(n)]=1/9 when n->infinitity
since the error has been derived, proof is omitted to the following
Error |S(n)-1/9|=1/(9*10^n) forever shrinks when n->infinity
Limit|S(n)-1/9|=0 when n->infinity

another expression for above is

S(n)->1/9, when n->infinitity
|S(n)-1/9|->0, when n->infinitity

Clearly |S(n)-1/9|=1/(9*10^n) is the infinitesmall in this case. it's a function of integers. it's not ZERO. it approaches to it. Any infinitesmall is a VARIABLE induced by n, x or some other varible.

If you say infinitesmall is zero here inorder to say 0.111... =1/9 Berkeley's Attack is far more than enough to tear the logic of calculus and limit theory.
if infinitesmall is zero ds/dt has a denomirator 0, no sense
if infinitesmall is not zero 0.111... never equal to 1/9, and realtime velocity can never be approximate:
V(t)=ds/dt=(k(t+dt)²-kt²)/dt=k+kdt kdt?

"And what are these fluxions? The velocities of evanescent increments. And what are these same evanescent increments? They are neither finite quantities, nor quantities infinitely small, nor yet nothing. May we not call them ghosts of departed quantities? "
   
    ---The analyst: or a discourse addressed to an infidel mathematician, written by Berkeley, published in 1734.
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Berkeley.html

the former proof implies it's not controversial to rewrite 0.111... as {S(n)}, since it's a series covergent to 1/9, there always exists another series convergent more rapidly to 1/9.

Almost all theroems in real numbers come from one origin, Cantor's definition of real numbers.
"There always exist a number between two numbers" is saying real numbers are thick. rational numbers are thick, too.

But can you even say 0.99... or 0.11 a real number? (the first time i doubted this)

"This division will never be exact.  We will keep getting  090909...  To round it off to three places, we have calculated the digit in the fourth place.  It is 9.  Therefore, we will add 1 to the previous digit:

0.09090.091

One sometime sees,  1 ÷ 11 = 0.090909...  The three dots, called ellipsis, mean "and so on for as far as you please according to the indicated pattern, or according to the rule."

!!!!This means that 1 ÷ 11 cannot be expressed exactly as a decimal. !!!! However, we may approximate it to as many decimal places as we please by following the pattern  090909."

http://www.themathpage.com/ARITH/divide-whole-numbers-2.htm

between two !!!! 's(added by me) intrigued my doubt.

Maybe i should reread some books before exactly illustrating my point of view on Cantor's definition