Simple equation , need walk through

[PhillyGixxer]

Ok i am just moving around variables and the book got a result i cannot understand.

I have v = Vo + a*t

I want to solve for t
so i first moved the v over

Giving me 0 = Vo-V + a* t

Then i want to divide by t
giving t = Vo-V + a Right ?

the book has
...V - Vo
t= ------
... a

How did they get the a on bottom ? What step am i missing/not understanding ?

 

[vsage]

dividing zero by any number would result in zero. Dividing through by t would result in 0 = (vo-V)/t + a.

 

[VietDao29]

No, to solve a linear equation like that you must first isolate the unknow(s), like this:
Say you want to solve for x in the equation ax + y = t
Now isolate ax, i.e subtract y from both sides of the equation:
ax = t - y
Now to solve for x, we divide both sides by a (assuming that a is not 0)
So we have:
$$x = \frac{t - y}{a}$$
Can you get this?
Your problem can be solved in the same way. Can you go from here? :)

 

[PhillyGixxer]

No, to solve a linear equation like that you must first isolate the unknow(s), like this:
Say you want to solve for x in the equation ax + y = t
Now isolate ax, i.e subtract y from both sides of the equation:
ax = t - y
Now to solve for x, we divide both sides by a (assuming that a is not 0)
So we have:
$$x = \frac{t - y}{a}$$
Can you get this?
Your problem can be solved in the same way. Can you go from here? :)

v= Vo + at isolate

v-Vo = at divide t

v-Vo
----- = t
a

Got it, thank you

 

[VietDao29]

v= Vo + at isolate

v-Vo = at divide t

v-Vo
----- = t
a

Got it, thank you

Yes, it looks good. However, it should be "divide both sides by a" (not t) :)

 

[PhillyGixxer]

If i have an equation like

tan = v^2 / r * g

and i want v^2

How do i solve for this ?

Also what are these methods call so i can search for them. I search for "linear equation " but didn't find what i was looking for.

 

[dav2008]

Your goal is to manipulate the equation to obtain v² to be on one side. Remember that you have to perform the same operation to both sides (or multiply one of the sides by 1) to keep the equation true.

So you want to isolate the v², right? What can you do to both sides to get rid of the rg on the right side?

I recommend googling for "algebra tutorial" (Or "elementary algebra tutorial" to be more specific). I'm surprised that your school is making you take a physics course before you have even learned elementary algebra.

This website might help you: http://home.sprynet.com/~smyrl/TUTOR.HTM (namely the equations tutorials)

 

[PhillyGixxer]

If you can find a site that has examples/tips of sub variables and moving them around please post it. I cannot find anything. I checked both with yahoo and google. You know of any ?

This is holding me back in my studying.

 

[HallsofIvy]

Best thing you can do is stop thinking about "moving around variables"- remove that awful phrase from your vocabulary altogether- and think of "doing the same thing to both sides of the equation".

To solve v= v₀+ at, for t, what's done to t? Another way of phrasing that is "suppose you were given t and asked to find v- what would you do". The answer is: first multiply by a, then add v₀. Since you want to solve for t, rather than find v, you want to "undo" what's been done to t- do the opposite of what you would do to find v. The opposite of "multiply by a" is "subtract a" and the opposite of "add v₀" is "subtract v₀". And don't forget to do those in the opposite order! (In the morning I have to put on my socks before I put on my shoes. In the evening, to "undo" that I have take off my shoes first, then take off my socks.) So starting from v= v₀+ at, first subtract v₀ from both sides: v- v₀= v₀+ at- v₀= at. From v- v₀= at, divide both sides by a:
(v- v₀)/a= at/a= t: t= (v- v_0[\sub])/a.

 

[arildno]

Philly:
It might be a helpful teqnique for you to INTRODUCE new sub-variables, in order to SIMPLIFY your original expression into a shape you DO know how to manipulate.
As an example, let us take your expression (as I read it):
$$\tan(\theta)=\frac{v^{2}}{r*g}$$
Now, we introduce auxiliary variables $a=v^{2}, b=r*g$
Thus, your original equation may be written as:
$$\tan(\theta)=\frac{a}{b}$$
This looks simpler than your original expression, right?
Now, you were to solve for "a" (that is v^2).

1. Multiply both sides with b:
$$b*\tan(\theta)=b*\frac{a}{b}$$

2. Now, on your right handside, remember that you ALWAYS may write a fraction as follows $\frac{a}{b}=\frac{1}{b}*a$ no matter what the numbers a and b are.
Thus, your right-hand side can always be re-written as:
[tex]b*\frac{a}{b}=b*\frac{1}{b}*a[/itex]
(And therefore, your EQUATION in 1 can be rewritten as $b*\tan(\theta)=b*\frac{1}{b}*a$

3. Furthermore, no matter what sort of number b is, you always have:
$$b*\frac{1}{b}=1$$
(That's basically the old rule "A number divided by itself equals 1"!)
Hence, you may rewrite your equation from 2. as:
$$b*\tan(\theta)=1*a$$

4. You also know that any number a multiplied with 1 equals itself, 1*a=a, thus, you have:
$$b*\tan(\theta)=a$$
Or, switching about:
$$a=b*\tan(\theta)$$

5. Now, re-introduce v^2 and r*g from the definitions of a and b, and you have:

$$v^{2}=r*g*\tan(\theta)$$

 

[arildno]

Best thing you can do is stop thinking about "moving around variables"- remove that awful phrase from your vocabulary altogether- and think of "doing the same thing to both sides of the equation".

I just have to emphatically agree with you here, HallsofIvy!

If there is one way maths is taught in schools that can be seen as the prime obstacle preventing people from understanding maths, it is that damn "move over and change signs rule". :yuck: :yuck:

I have worked at some time in the school system, and I see how damaging this rule is for ordinary persons perception of maths.
It is simply incomprehensible mumbo-jumbo to them.

However, unfortunately, once they have seen that horrid rule, they fiercely cling to it (even if, and in particular when, they use it incorrectly), because the eminently comprehensible rule "do the same on both sides" takes slightly more space to write down on a piece of paper.