# Derivation of Formulas

1. Jun 30, 2009

### jaimieee

1. The problem statement, all variables and given/known data

a) Given: 2ad - Vf2 - Vl2 . Find Vl?
b) Given: KE = 1/2 mv2 . Find V?
c) Given: a= (Vf - Vl)/t Find Vl

d) d= 1/2 gt 2 find t?
e) ac= (mv2)r, find M?

2. Relevant equations
sorry, i do not know what are relevant eqs.

3. The attempt at a solution

i have answeres a,b,c but i do not know the rest.

2) v2 = (2KE)/m
V= sqrt(2KE)/m

3) Vf = Vl + at
Vl = V - at

Thanks.

2. Jun 30, 2009

### hage567

Are you just trying to rearrange the equations? If you can do the first three, I don't see why you shouldn't be able to get the last two. Can you show what you've tried so far?

3. Jun 30, 2009

### jaimieee

yes!certainly. and sure. i already have answers from the problems tho im not sure if those are correct.:

2) v2 = (2KE)/m
V= sqrt(2KE)/m

3) Vf = Vl + at
Vl = V - at

4) d= (gt^2)/2
t^2= 2(gd)
t=sqrt2(gd)

5) m=v^2 - acr

:) hope you could correct if ive done it wrongly :D

4. Jun 30, 2009

### cepheid

Staff Emeritus
No, 4 and 5 do not look correct.

For 4, you are trying to isolate t on one side of the equation. The first step is to multiply both sides of the equation by 2, which you have done.

What is the second step, for getting rid of the g on the right hand side?

5. Jun 30, 2009

### jaimieee

yea i think that's wrong too. i'm a little bit confused with that. i was comparing #4 from the 2nd problem. but i think i wasn't able to get it. :D what do you think?

6. Jun 30, 2009

### cepheid

Staff Emeritus
Hi jaimieee,

As I've already said your solutions for 4 and 5 are wrong. I was trying to guide you through 4 step by step, but you didn't answer my question...

7. Jun 30, 2009

### jaimieee

oops. i'm sorry. hmm. do i need to apply rationalization? i don't know, i was lost on that part. T__T coz what i did is.

d= (gt^2)/2
then i transposed t^2 it became
-t^2 = [g(-d)]/2
then i divided both sides by -1 to eliminate the negative sign on t
so it became
(-t^2)/-1 = [g(-d)]/2/-1
so i got the reciprocal
t^2= -2(-gd) :D
then to eliminate the exponent i extracted a sqrt
t=sqrt-2(-gd)

that's what i did.

Last edited: Jun 30, 2009
8. Jun 30, 2009

### cepheid

Staff Emeritus
I meant this question here, specifically. How would you get rid of the g?

We started with:

d = ½gt2

Then we got rid of the 1/2 by multiplying BOTH sides of the equation by 2. The result was:

2d = gt2

Now, we need to get rid of the g so that the t will be left by itself on the right hand side. How would you do that?

9. Jun 30, 2009

### jaimieee

transpose g to the left side so the equation becomes:
2d-g=t^2 :D

10. Jun 30, 2009

### cepheid

Staff Emeritus
What do you mean, "transposed?" This step does not make any sense mathematically.

If you have an equation like:

this = that​

then if you are going to do something to that equation, you have to do it to BOTH sides of the equation. If you do something to only one side, then it will no longer be true that it is equal to the other side. Do you understand?

So, (for example) multiplying both sides of the equation by 2 would be allowed:

2*this = 2*that​

but clearly doing it to only one side would NOT be allowed:

2*this = that​

because this statement is clearly FALSE.

11. Jun 30, 2009

### jaimieee

oops.this equation "-t^2 = [g(-d)]/2" is quite confusing. i was confused with that too. hehe.sorry

so it has to be equal. hmm. so i won't be applying transposition? sorry but i don't get it.

right i need to get rid of the g so what i did is transpose it to the left which caused its sign to change so this is the equation i formed:

2d-g=t^2 then i'll extract a sqrt
so t = sqrt2d-g?

this is confusing me. sorry

12. Jun 30, 2009

### cepheid

Staff Emeritus
This equation is not true.

It is not true because this is not a valid step. The reason is because what "transposing" something from the right hand side to the left hand side really means is subtracting it from both sides. For example:

a = b + c​

If I "move the c" to the left hand side, what I am really doing is subtracting c from BOTH sides of the equation like so:

a - c = b + c - c​

which results in

a - c = b​

I did something to both sides of the equation. That's why this step is allowed. I cannot emphasize this enough:

If you do something to an equation, you have to do it to both sides of the equation

It is important that you understand WHY this is a rule. Do you understand why this is a rule? If you don't do it to both sides, then it won't be an equation anymore, because one side will no longer be equal to the other.

For the sake of argument, what would happen if we "transposed" a g to the left-hand side? Well, as I explained above, "transposing the g to the left hand side" really means, "subtracting g from both sides." Doing that would result in the following:

2d - g = gt2 - g ​

So, transposing (i.e. subtracting) is not helpful because it doesn't get rid of the g on the right hand side.

Last edited: Jun 30, 2009
13. Jun 30, 2009

### jaimieee

so ya. i get it. it needs to be equal so:

2d - g = gt^2 -g
and on the right side g-g is cancelled
the eq will be
t^2=2d-g
to eliminate the exponent, ill extract a square root

t=sqrt 2d-g?

so is it correct? :D:D:D:D

14. Jun 30, 2009

### cepheid

Staff Emeritus
No, the two g's do NOT cancel. Whoever assigned you these problems assumed that you understood basic algebra. Now, I don't wish to sound mean, but it is clear from what I have quoted above that you do not understand basic algebra. I will make one last effort to explain this, but I have to say that after that if you still do not understand, I cannot help you. It is not reasonable to expect somebody on this online forum to teach you algebra.

First of all, saying that the g's cancel is like saying that:

ab - a = b​

This is most emphatically NOT true.

ab - a IS NOT EQUAL TO b​

This should be obvious, but if you don't understand, then at least try plugging in example numbers so that you can see what I am saying is true

a = 3
b = 7

ab - a = 3*7 - 3 = 21 - 3 = 18 WHICH IS NOT EQUAL TO 7

What I was trying to say in my previous post is that subtracting g from both sides will NOT help. Look at the right hand side. The right hand side is multiplied by g. If something is multiplied by g, what operation could I do to that something to get rid of the g (hint: it would be something that "undos" the multiplication).

15. Jun 30, 2009

### jaimieee

LOL. 2d - g = gt^2 -g < ya i agree with that. my bad. the g's on the right side should not be cancelled. but on our algebra that's the way we were taught. We have Adv. Algebra and Analytic Geometry and I really don't have a problem with that. But the one you're telling me made me confused coz it's really different. But Thank you, i really appreciate and i'm learning.

t=sqrt 2d-g? is it wrong? :D

16. Jun 30, 2009

### Staff: Mentor

If that's the way you were taught, your teacher was incompetent. Starting with

d = 1/2 * g * t2

how would you get the t2 factor all by itself. Notice that the operations on the right side are multiplications, so to undo those multiplications you should not be thinking of doing subtraction. What operation is the opposite of multiplication?

17. Jul 1, 2009

### jaimieee

hi Cepheid, I was wrong and i was able to get your point. We were taught about that transposition thingy but not at that case. Sorry, my bad. But I was able too see my mistake and I got a perfect score in my homework.

d= (gt^2)/2
my solution is:
i multiplied both sides by two to eliminate the denominator on the right side so it became:
2d=gt^2
then to eliminate g i divided G to both sides:
(2d)/g=(gt^2)/g
so the G on the right side is cancelled
(2d)/g=t^2
then extracting square root:
t=sqrt(2d)/g]
that's my answer and i got a perfect score so i believe it's correct :D.

on the 5th problem:
ac= (mv^2)/r, find M?

what i did is is multiplied both sides by r to eliminate the r on the right side so:
r(ac)=mv^2

then to eliminate mv^2 i divided both sides by mv^2 so:
[r(ac)]/v^2=[mv^2]/v^2
so i cancelled mv^2 on the right:
m=(r[ac])/v^2

Thank you very much Cepheid! Efforts are so much appreciated. ^^ cheers.

Division :D . Well my bad, I was wrong sorry for giving a false info. anyway i was able to get everything. thanks too. :)

18. Jul 1, 2009

### cepheid

Staff Emeritus
jaimieee,

It looks good to me! And you are very welcome. I think that you have now understood what I was trying to say.

Regards,

cepheid