- #1

dark_exodus

I need to rearrange the following formula to get a(Acceleration):

s = ut-1/2at

The last t is squared

Any help

My maths is a little rusty.

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- Thread starter dark_exodus
- Start date

- #1

dark_exodus

I need to rearrange the following formula to get a(Acceleration):

s = ut-1/2at

The last t is squared

Any help

My maths is a little rusty.

- #2

- 933

- 0

The correct formula is

s=(v*t)+((1/2)*a*t*t)

Then

a=(s-(v*t))/((1/2)*t*t)

s=(v*t)+((1/2)*a*t*t)

Then

a=(s-(v*t))/((1/2)*t*t)

- #3

dark_exodus

Could you give a break down of how you got that answer?

- #4

Integral

Staff Emeritus

Science Advisor

Gold Member

- 7,212

- 56

s= ut + at^{2}/2

s - ut = at^{2}/2

2(s - ut) = at^{2}

a = 2(s - ut))/(t^{2})

s - ut = at

2(s - ut) = at

a = 2(s - ut))/(t

- #5

HallsofIvy

Science Advisor

Homework Helper

- 41,847

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You have the formula s = ut-1/2at

a= something.

You do that using two basic concepts: (1) For everything that is alread "done" to a, do the opposite (2) Anything you do to one side of the equation you must do to the other.

So:

s= ut- (1/2) at

s- ut= ut- (1/2) at

Now the only problem is that a is multiplied by -(1/2) t

So, divide both sides of the equation by -(1/2) t

That gives (s- ut)/((-1/2)t

a= -2(s-ut)/t

- #6

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- #7

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hiya, can u help me rearrange the following equation 2 get R² -

f = (1/2πC)*√(R¹+R²/R¹R²R³)

thanx

- #8

- 695

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You're going to have to put in more parantheses in that expression in the radical... Do you mean:

[tex]R^1 + \frac{R^2}{R^1R^2R^3}[/tex]

or:

[tex]R^1 + \frac{R^2}{R^1}R^2R^3[/tex]

or perhaps:

[tex]\frac{R^1 + R^2}{R^1R^2R^3}[/tex]

?

[tex]R^1 + \frac{R^2}{R^1R^2R^3}[/tex]

or:

[tex]R^1 + \frac{R^2}{R^1}R^2R^3[/tex]

or perhaps:

[tex]\frac{R^1 + R^2}{R^1R^2R^3}[/tex]

?

Last edited:

- #9

- 2

- 0

f = 1/(2πC)*√(R¹+R²

------

R¹R²R³)

plz try and help 2 rearrange 4 R² = equation

thanx

------

R¹R²R³)

plz try and help 2 rearrange 4 R² = equation

thanx

- #10

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 966

That's even worse!

I'm going to assume that "f = (1/2πC)*√(R¹+R²/R¹R²R³)" means

[tex]f= (\frac{1}{2}\pi C)\sqrt{\frac{R_1+R_2}{R_1R_2R_3}}[/tex]

Notice that I have also changed to sub-scripts rather than super-scripts since I tend to confuse those with exponents (I am assuming they are NOT exponents!).

First thing you do is divide both sides by that number outside the square root to get

[tex]\frac{2f}{\pi C}= \sqrt{\frac{R_1+R_2}{R_1R_2R_3}}[/tex]

Now get rid of that square root by squaring both sides:

[tex]\frac{4f^2}{\pi^2C^2}= \frac{R_1+R_2}{R_1R_2R_3}[/tex]

Multiply on both sides by R_{1}R_{2}R_{3} so we don't hav e that fraction to worry about:

[tex](R_1R_2R_3)\frac{4f^2}{\pi^2C^2}= R_1+ R_2[/tex]

Subtract R_{2} from both sides so that we have the quantity we are solving for on the left:

[tex](R_1R_2R_3)\frac{4f^2}{\pi^2C^2}- R_2= R_1[/tex]

and, since there is an "R_{2}" in each term, factor that out:

[tex]R_2((R_1R_3)\frac{4f^2}{\pi^2C^2}- 1)= R_1[/tex]

Finally, isolate R_{2} by dividing both sides of the equation by everything on the left except R_{2}:

[tex]R_2= \frac{R_1(\pi^2C^2-1)}{R_1R_3\pi^2C^2}[/tex]

You are welcome to go back to superscripts now if that was the way the problem was given.

I'm going to assume that "f = (1/2πC)*√(R¹+R²/R¹R²R³)" means

[tex]f= (\frac{1}{2}\pi C)\sqrt{\frac{R_1+R_2}{R_1R_2R_3}}[/tex]

Notice that I have also changed to sub-scripts rather than super-scripts since I tend to confuse those with exponents (I am assuming they are NOT exponents!).

First thing you do is divide both sides by that number outside the square root to get

[tex]\frac{2f}{\pi C}= \sqrt{\frac{R_1+R_2}{R_1R_2R_3}}[/tex]

Now get rid of that square root by squaring both sides:

[tex]\frac{4f^2}{\pi^2C^2}= \frac{R_1+R_2}{R_1R_2R_3}[/tex]

Multiply on both sides by R

[tex](R_1R_2R_3)\frac{4f^2}{\pi^2C^2}= R_1+ R_2[/tex]

Subtract R

[tex](R_1R_2R_3)\frac{4f^2}{\pi^2C^2}- R_2= R_1[/tex]

and, since there is an "R

[tex]R_2((R_1R_3)\frac{4f^2}{\pi^2C^2}- 1)= R_1[/tex]

Finally, isolate R

[tex]R_2= \frac{R_1(\pi^2C^2-1)}{R_1R_3\pi^2C^2}[/tex]

You are welcome to go back to superscripts now if that was the way the problem was given.

Last edited by a moderator:

- #11

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Worked it out thnx.

Last edited:

- #12

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- #13

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[tex] T = 2\pi \sqrt (\frac{\ell}g)[/tex]

Heres what I've done.

[tex] T = 2\pi \sqrt (\frac{\ell}g)[/tex]

Divide both sides by [tex]2\pi[/tex]

[tex]\frac{T}{2\pi} = \sqrt (\frac{\ell}g)[/tex]

Remove the root by squaring both sides

[tex]\frac{T^2}{4\pi^2} = \frac{\ell}g[/tex]

This is where I'm a bit confused. Does the rearrangement finish as :

[tex]\ell[/tex] divided by [tex](\frac{T^2}{4\pi^2})[/tex] = [tex]g[/tex] ?

I've got the figures for [tex]T[/tex], [tex]\ell[/tex] and [tex]g[/tex]. If I use that final rearrangement I can get closer to the answer for [tex]g[/tex] than any other, but my answer is slightly larger than the given answer.

I would appreciate any help at all, and I apologise for any mistakes in my coding. Its the first time I've ever used Tex.

- #14

- 445

- 5

Hi I need help rearranging s=ut+1/2at2 to find t

any help would be appreciated

Do you mean t

If not, simply isolate the ut term, then divide by u. Otherwise, perhaps take s from both sides and you have a quadratic in t, which you can solve using standard methods.

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