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HallsofIvy

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quasar987

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The key idea is that F, m and a represent numbers, like 1, 2, 3¼, pi³,...etc. And for any number 'n', there exists another number 'm' such that n*m = 1. This number m is of course, the inverse of n: m = 1/n.

So if for exemple, we want to isolate m in F=ma, we want to multiply both sides by the inverse of a, like so

[tex]F=ma \Leftrightarrow \frac{1}{a}F=m\frac{a}{a} \Leftrightarrow \frac{F}{a} = m[/tex]

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t=(Vf-Vi)/a if I wanted to find "a" I would multiply both sides by a, yes? but that would leave me with at=Vf-Vi right? and "a" would not be isolated on one side of the =. what am I doing wrong?HallsofIvy said:You "undo" whatever is done to a. In F= ma, a is multiplied by m. To "undo" that, you do the opposite of "multiply by m"- youdivide by m. Dividing both sides of the equation by m, F/m= (ma)/m= a so a= F/m.

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quasar987

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Continue. Divide both sides by t.

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TD

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a) An eqaution remains equivalent when you add the same number to both sides (this may be negative if you wish to 'substract').

Example, we want a out of: [tex]a + b = c \Leftrightarrow a + b - b = c - b \Leftrightarrow a = c - b[/tex]

b) An eqaution remains equivalent when you multiply both sides with the same factor ([tex] \ne 0[/tex])

Example, we want a out of: [itex]a\frac{b}{c} = d \Leftrightarrow a\frac{b}{c}\frac{c}{b} = d\frac{c}{b} \Leftrightarrow a = \frac{{dc}}{b}

[/itex]

That's all you need here

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thanks everyone

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EnumaElish

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An eqaution remains equivalent when you multiply both sides with the same factor, including 0.TD said:b) An eqaution remains equivalent when you multiply both sides with the same factor ([tex] \ne 0[/tex])

Generally, an eqaution remains equivalent when you apply an invertible (strictly increasing or strictly decreasing) function to both sides:

a = b implies Log(a) = Log(b);

a = sqrt(b) implies a

In general, if a = f(b) and g is the inverse of f, then g(a) = g(f(b)) = b.

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TD

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In that case, you have to exclude 0, no?

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EnumaElish

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You have to pay attention to the parantheses.Serj said:

a = (V2 - V1)/t

at = V2 - V1

at + V1 = V2

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EnumaElish

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I see what you're saying. I am splitting hairs when I point out that strictly speaking, a = b is preserved under multiplication with zero, which is not covered under your (or, the)definition of equivalent equations.TD said:

In that case, you have to exclude 0, no?

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why did you out everything except V2 on the other side of the = instead of putting V2 were "a" was and put "a" on the other side?EnumaElish said:You have to pay attention to the parantheses.

a = (V2 - V1)/t

at = V2 - V1

at + V1 = V2

why did you multiply "t" and "a"?

if "t" was the numerator, would you still multiply it with "a"?

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The the next year in Phys12 I sit down at my desk the first day and written on my desk is the little triangle cheat method and someone commented below it "morons use triangles". I always got a kick out of that.

How old are you and what grade are you in? I didn't take a physics class until grade 10 and we were doing stuff like this in grade 7/8. I find it odd that you would be taking a physics class without knowing these algebra fundamentals.

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why did you multiply "t" and "a"?

if "t" was the numerator, would you still multiply it with "a"?

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EnumaElish

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Serj said:why did you out everything except V2 on the other side of the = instead of putting V2 were "a" was and put "a" on the other side?

why did you multiply "t" and "a"?

if "t" was the numerator, would you still multiply it with "a"?

a = (V2 - V1)/t

(V2 - V1)/t = a

t(V2 - V1)/t = ta

V2 - V1 = ta

V2 - V1 + V1 = ta + V1

V2 = V1 + ta

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d/V=Vt/V+1/2 at^2

d/V=t +1/2 at^2

2d/V=t +2*1/2 at^2

2 d/V=t + at^2

(2 d/V)/a=t + at^2/a

(suare root of)(2 d/V)/a=t + (square root of)t^2

(suare root of)(2 d/V)/a=t +t

((suare root of)(2 d/V)/a)/2=2t/2

((suare root of)(2 d/V)/a)/2=t

What did I do wrong? how do I fix it so Im not dividing

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And is Vt one variable or two?

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VietDao29

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You are wrong at the second line.Serj said:...d/V=Vt/V+1/2 at^2

d/V=t +1/2 at^2

You have:

[tex]a = b + c[/tex]

[tex]\Leftrightarrow \frac{a}{d} = \frac{b + c}{d} = \frac{b}{d} + \frac{c}{d}[/tex]

We divide both sides by d <> 0.

Anyway, to find t from:

[tex]d = vt + \frac{1}{2}a t ^ 2[/tex]

[tex]\Leftrightarrow \frac{a}{2}t ^ 2 + vt - d = 0[/tex]

t is the unknown.

Can you solve:

[tex]\alpha x ^ 2 + \beta x + \gamma = 0[/tex]

for x?

[tex]\alpha , \beta , \gamma[/tex] are already known.

Viet Dao,

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x^2=(-bx-y)/a

square root both sides

x= square root of (-bx-y)/a

That's really messy and I don't know if thats how you would do it.

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Moonbear

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First, you made an error in your first step. If you divide by V, you have to divide EVERYTHING by V, so your second equation would be:Serj said:

d/V=Vt/V+1/2 at^2

d/V=t +1/2 at^2

2d/V=t +2*1/2 at^2

2 d/V=t + at^2

(2 d/V)/a=t + at^2/a

(suare root of)(2 d/V)/a=t + (square root of)t^2

(suare root of)(2 d/V)/a=t +t

((suare root of)(2 d/V)/a)/2=2t/2

((suare root of)(2 d/V)/a)/2=t

What did I do wrong? how do I fix it so Im not dividingfractions

d/V=Vt/V + 1/2(at^2)/V

However, to solve for t in this case, you can't just rearrange the equation. Now you need to go back through some of the other kinematic equations you have learned so far and use those to substitute equivalent terms in this equation so you end up isolating t.

At this point, you need to pay very careful attention that V in this equation is actually Vo, or the initial velocity in case your book uses a different way to symbolize that.

Since this is pretty complicated, I'll get you started, but then you're going to have to look through your notes and see if you can find another equation that helps substitute terms where I leave off. Something else I want to point out from reading what's been done in this thread previously is to remember to pay attention to the order of operations and watch what terms are enclosed in parentheses. In case you've forgotten the order of operations, remember the acronym PEMDAS for Parentheses, Exponents, Multiply, Divide, Add, Subtract. If you aren't careful of doing everything in this order, you'll introduce errors.

1) d=Vot+1/2at^2 (your starting equation)

2) d=(Vo + 1/2at)t (factored out t)

3) d=2/2(Vo + 1/2at)t (multiplied the right side of the equation by 2/2; this is not an intuitive step, which is why I wanted to walk you through to this point. You can always multiply by 1 on one side of an equation without changing it's value, and sometimes that means using a fraction like 2/2...back when I was learning this stuff, figuring out when to use that particular trick was really difficult and sometimes it's a matter of staring at the other equations you have to use long enough until one of them jumps out as being a likely candidate if only you had a different denominator somewhere)

4) d=((2Vo + at)/2)t (factored out the 2/2)

5) d= ((Vo + Vo + at)/2)t (substituted Vo + Vo for 2Vo; basic arithmetic)

Okay, now I think this should have gotten you started well on your way. Look through your other equations and see if there's something that allows you to eliminate a term that includes t so you can then rearrange the equation to finish solving for the other t.

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HallsofIvy

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[tex]d= Vt- \frac{1}{2}at^2[/tex] is a quadratic equation: you can't solve it for t by just "rearranging" the equation. With some specific values of V and a, you might be able to factor but for the general equation you need to use the "quadratic formula". Certainly if you've taken two years of algebra you should know that formula:

If [tex]ax^2+ bx+ c= 0[/tex] then [tex]x= \frac{-b \pm\sqrt{b^2- 4ac}}{2a}[/tex].

Here [tex] \frac{a}{2}t^2- Vt+ d= 0[/tex] so [tex]t= \frac{V\pm\sqrt{V^2- 2ad}}{a}[/tex].

If [tex]ax^2+ bx+ c= 0[/tex] then [tex]x= \frac{-b \pm\sqrt{b^2- 4ac}}{2a}[/tex].

Here [tex] \frac{a}{2}t^2- Vt+ d= 0[/tex] so [tex]t= \frac{V\pm\sqrt{V^2- 2ad}}{a}[/tex].

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Moonbear

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Oh, but take note that Halls inverted a sign in his starting equation. It doesn't change how you use the formula, just try your hand at your own starting equation (maybe he did it on purpose so you don't just copy down his answer but have to try to work it out for yourself).

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HallsofIvy

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Oh, that's a good excuse! What really happening is, since this is clearly a "motion" problem with acceleration a, I was thinking of a falling object with a= -g and miswrote it.

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Moonbear

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http://www.purplemath.com/modules/sqrquad.htm

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