Is it posible to solve for t?
teewan
I recently encountered this eqquation in Physics:
d = vt + 1/2at^2
(displacement equals initial velocity times time interval plus the product
of one half acceleration times t squared)
Is it posible to solve for t?
I can't seem to isolate it. Neither can my Algebra teacher.
Brad
In <01bbcad7$c948ee40$7541...@teewan.epix.net> "teewan"
well dude it looks the algebra teacher should turn in his (most likely
its a male) =)) teaching license. This is a very easy equation to solve
for any variable. We're going to have to use our best friend, the
square root. Here are 2 equations that you should blab your teacher
with so they'll never forget it. =) I won't go into teaching though,
the pay is too low for me & I wouldn't be able to buy a Lamborghini
Diablo VT on a teacher's salary. I really want to be a Wall Street
Trader. =) And now for those awesome physics equations. =)
t = SQRT( v^2 + 2 * a * d ) - v
--------------------------------
a
t= -1 * [ SQRT( v^2 + 2 * a * d ) + v ]
---------------------------------------
a
have fun!!! now your teacher will like you even more =)
Kcpjp
Teewan asked:
>I recently encountered this eqquation in Physics:
>
>d = vt + 1/2at^2
>
>(displacement equals initial velocity times time interval plus the
product
>of one half acceleration times t squared)
>
>Is it posible to solve for t?
>
>I can't seem to isolate it. Neither can my Algebra teacher.
>
Brad responded:
>
>t = SQRT( v^2 + 2 * a * d ) - v
> --------------------------------
> a
>
>
>t= -1 * [ SQRT( v^2 + 2 * a * d ) + v ]
> ---------------------------------------
> a
My response:
First of all I'm rather upset that a high school math teacher can't solve
a basic, in fact THE basic quadratic equation. Teewan, I hope you continue
to explore you math talents because I think that just by asking your
question here you indicate an interest and a hunger to understand. Keep it
up; math, IMHO is the basis for almost all significnat intellectual
growth. Here is my response to your "physics" question:
The standard quadratic polynomial is: ax^2 + bx + c = 0 and the two
solutions (since it is second order) are :
x = -b + SQRT(b^2 - 4*a*c)
-----------------------------------
2a
x = -b - SQRT(b^2 -4*a*c)
--------------------------------
2a
Now, transpose your original equation to:
(1/2)*t^2 + v*t - d = 0 This is now in the "standard" form of the
quadratic, so we substitute from the physics equation into the "standard
form for the solution and get the following two solutions for t. Use t
for x; (1/2)*a for a; v for b; and d for c:
t = -v + SQRT(v^2 + 2*a*d)
----------------------------------
a
t = -v - SQRT( v^2 + 2*a*d)
-----------------------------------
a
Same answer as Brad's. just looking a little different.
Hope this helps you - Good luck
Phil
Gary Scott Simon
In article <55mt27$h...@dfw-ixnews5.ix.netcom.com>, amer...@ix.netcom.com
(Brad) wrote:
>In <01bbcad7$c948ee40$7541...@teewan.epix.net> "teewan"
><tee...@epix.net> writes:
>>
>>I recently encountered this eqquation in Physics:
>>
>>d = vt + 1/2at^2
>>
>>(displacement equals initial velocity times time interval plus the
>product
>>of one half acceleration times t squared)
>>
>>Is it posible to solve for t?
Yes.
Equations of the form:
ax^2 + bx + c = 0
are called "quadratic" equations, and are very common in a first year
algebra course (perhaps because the solution of such equations is so
important in physics). The general solution to such equations (which you
can, and should, learn to derive for yourself) is:
x = {-b +/- [sqrt(b^2 - 4ac)]}/2a (Note: +/- means "plus or minus")
We can re-express your equation into quadratic form (with a variable
of "t", rather than "x") in a couple of steps.
First, subtract d from both sides:
0 = vt + (1/2)at^2 - d
Next, re-arrange the terms on the right-hand side of the equation:
0 = (1/2)at^2 + vt - d
0 = (a/2)t^2 + vt - d
Applying the general solution to quadratic equations in this case,
where the first coefficient (i.e. the "a" in the general form) is a/2,
the second (i.e. the "b" in the general form) is v and the third (i.e.
the "c" in the general form) is -d:
t = {-v +/- [sqrt(v^2 + 2ad]}/a
>>I can't seem to isolate it. Neither can my Algebra teacher.
>well dude it looks the algebra teacher should turn in his (most likely
>its a male) =)) teaching license. This is a very easy equation to solve
>for any variable. We're going to have to use our best friend, the
>square root. Here are 2 equations that you should blab your teacher
>with so they'll never forget it. =) I won't go into teaching though,
>the pay is too low for me & I wouldn't be able to buy a Lamborghini
>Diablo VT on a teacher's salary. I really want to be a Wall Street
>Trader. =)
Although I agree that teewan's teacher must have been having a truly
bad day, or shouldn't be teaching algebra, I'ma bit upset by Brad's
attitide toward teaching. Those of use who understand the math behind,
for example, imputing the volatility of a futures cntract from options
prices should be committed to restoring the status of teachers in society.
Brad
In <garscosi-051...@ip15.an17.new-york4.ny.psi.net>
I'm very much into options though. Selling options is the greatest way
to make money. Like for example, let's take Novell which has a share
price of around $10. If I spend $10,000 on Novell, I get a 1000 shares.
That means I can write 10 contracts. I would choose a December call at
15. Now if you're following the market, you would know that novell
never increases or decreases more than 3/8 in a day. So, it's 99.89425%
likely that Novell will not reach 15 by December. So basically its
almost a guaranntee profit, therefore anyone would be stupid to buy the
option, but there's some suckers in America who would. =) Now that's
the real way to make money. SELL options, DO NOT BUY options. =)
teach
"teewan" <tee...@epix.net> wrote:
>I recently encountered this eqquation in Physics:
>d = vt + 1/2at^2
>(displacement equals initial velocity times time interval plus the product
>of one half acceleration times t squared)
>Is it posible to solve for t?
>I can't seem to isolate it. Neither can my Algebra teacher.
This is quadratic in "t"
In fact it is a commonly known equation of motion.
Re-arranging:
at^2 + 2vt - 2d = 0
t = {-2v +- SQRT[(vt)^2 - 4*a*(-2d)]}/2a
You take it from there.
David.
Dan Larsen
You betcha. Ever heard of the quadratic? ax^2+bx+c=0 solves to ((-b(+ or
-) Sqr(b^2-4ac))/(2a)=x
teewan <tee...@epix.net> wrote in article
<01bbcad7$c948ee40$7541...@teewan.epix.net>...
Robin E. Brown
teewan wrote:
>
> I recently encountered this eqquation in Physics:
>
> d = vt + 1/2at^2
>
> (displacement equals initial velocity times time interval plus the product
> of one half acceleration times t squared)
>
> Is it posible to solve for t?
>
nice : t=(sqrt(v^2 +2ad) - v)/a or (-sqrt(v^2 +2ad) - v)/a. Obviously, a
can not equal 0. I personally think you'd be better off substituting
known values of v, a and d into the original formula and solving the
resulting equation via the quadratic formula. Also, in most practical
applications, negative time should be disregarded.
Hope this helps -
Robin
Gary Scott Simon
In article <55o4tg$m...@dfw-ixnews2.ix.netcom.com>, amer...@ix.netcom.com
(Brad) wrote:
>I'm very much into options though. Selling options is the greatest way
>to make money. Like for example, let's take Novell which has a share
>price of around $10. If I spend $10,000 on Novell, I get a 1000 shares.
>That means I can write 10 contracts. I would choose a December call at
>15. Now if you're following the market, you would know that novell
>never increases or decreases more than 3/8 in a day. So, it's 99.89425%
>likely that Novell will not reach 15 by December.
And it's very unlikely that I would lose twelve spins in a row
betting a row (at 2-1 odds) at a roulette wheel. So, if I keep doubling
my bet on every spin, the odds are that I'll be a winner. Unfortunately,
on those relatively rare occasions that I am a loser, I'll lose much
bigger than I won when I was a winner.
>So basically its
>almost a guaranntee profit, therefore anyone would be stupid to buy the
>option, but there's some suckers in America who would. =) Now that's
>the real way to make money. SELL options, DO NOT BUY options. =)
People buy options (and buy and sell futures contracts on
commodities) for a wide variety of reason, not always directly related to
an expectation of profit on a particular options transaction. Those who
sell option are, in reality, selling insurance, and had better take care
that the "premium" they charge is sufficient to cover their losses in the
unlikely even they have to "pay off".
Brad
In <garscosi-051...@news.pipeline.com> gars...@pipeline.com
First I'd stick to selling stock options. Once I make lots of money
from it, I'd go into selling futures and commodities options. It's such
an exciting field. =)
Gary Scott Simon
In article <55p7lv$a...@dfw-ixnews9.ix.netcom.com>, amer...@ix.netcom.com
(Brad) wrote:
It's very exciting, because everything I wrote about risk and
insurance applies equally well to stock options. On those relatively
rare occasions that options get into the money, the losses sustained by
options sellers are often far larger than the proceeds "pocketed" when
options expire without value.
Brad
In <garscosi-061...@news.pipeline.com> gars...@pipeline.com
hehe, we would both make a great money making team hehe
Tunes
Use the quadratic formula:
x = (-b+- sqrt(b^2 - 4ac)) / 2a
d = vt + 1/2at^2
1/2at^2 + vt - d = 0
a = 1/2a
b = v
c = -d
t = (-v +- sqrt(v^2 - 4(1/2a)(-d))) / (2)(1/2a)
have your math teacher e-mail me with any questions
B.
teewan <tee...@epix.net> wrote in article
<01bbcad7$c948ee40$7541...@teewan.epix.net>...
the oatman
>
> d = vt + 1/2at^2
>
> (displacement equals initial velocity times time interval plus the
> product
> of one half acceleration times t squared)
>
> Is it posible to solve for t?
>
> I can't seem to isolate it. Neither can my Algebra teacher.
>
where x={-b+sqrt(b^2-4ac)}/2a
or x={-b-sqrt(b^2-4ac)}/2a
First, set the equation equal to 0.
.5at^2+vt-d=0
let a=.5a
let b=v
let c=-d
so t={-v+sqrt(v^2-4(.5a)(-d))}/(2(.5a))
or t={-v-sqrt(v^2-4(.5a)(-d))}/(2(.5a))
simplified, this is
t={-v+sqrt(v^2+2ad)}/a
or
t={-v+sqrt(v^2+2ad)}/a
exclude negative answers.
good luck
Marty Kulinski
I agree that Tunes has the correct answer except that the answer is not in
its most simplest form. That would be:
t = (-v +- sqrt(v^2 + 2ad)) / a
Tunes <lu...@bigfoot.com> wrote in article
<01bbcdad$e6ee6500$f8304010@lunas2>...
> Use the quadratic formula:
>
>
> x = (-b+- sqrt(b^2 - 4ac)) / 2a
>
>
>
> d = vt + 1/2at^2
>
> 1/2at^2 + vt - d = 0
>
>
>
> a = 1/2a
>
> b = v
>
> c = -d
>
>
> t = (-v +- sqrt(v^2 - 4(1/2a)(-d))) / (2)(1/2a)
>
> have your math teacher e-mail me with any questions
>
> B.
>
> teewan <tee...@epix.net> wrote in article
> <01bbcad7$c948ee40$7541...@teewan.epix.net>...
> > I recently encountered this eqquation in Physics:
> >
