# How to re-arrange S = ut + 1/2at² to make T subject?

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#21

(Original post by

s=ut + 1/2at^2

s/t=u+1/2at

2s/t=u+at

2s/t-u=at

2s-u/a= t^2

√(2s-u)/a= t

voila.

**Rooster523**)s=ut + 1/2at^2

s/t=u+1/2at

2s/t=u+at

2s/t-u=at

**2s-u= at^2**2s-u/a= t^2

√(2s-u)/a= t

voila.

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#22

(Original post by

This step is wrong. You should get 2s-ut= at^2

**03gshep**)This step is wrong. You should get 2s-ut= at^2

ahh well spotted

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#23

(Original post by

You can, using the formula like above.

**03gshep**)You can, using the formula like above.

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#25

woah I have no idea what you guys are on about I do it like this....

S = ut +1/2at(squared)

S - ut/0.5a = t?

or

S-1/2at(squared) divided by U?

One of these is correct Im sure.

S = ut +1/2at(squared)

S - ut/0.5a = t?

or

S-1/2at(squared) divided by U?

One of these is correct Im sure.

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#26

(Original post by

woah I have no idea what you guys are on about I do it like this....

S = ut +1/2at(squared)

S - ut/0.5a = t?

or

S-1/2at(squared) divided by U?

One of these is correct Im sure.

**Advanced Subsidiary**)woah I have no idea what you guys are on about I do it like this....

S = ut +1/2at(squared)

S - ut/0.5a = t?

or

S-1/2at(squared) divided by U?

One of these is correct Im sure.

The question was make t the

**subject**.

So although the following are true they are not the solution to the problem.

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#27

well then the if none of those are correct, then t can't be the subject in this equation. That's the way I was taught to re-arrange equations.

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Here's the Solution given by Steve

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#30

I am amazed how many physics students do not know how to solve a quadratic equation...

BTW time can be negative, in many cases you will expect two results, for example throwing a ball in the air, it will be at ground level exactly twice. It all depends when you define your t=0. Even if you say time can't be negative it is an error to neglect the - root because the solution corresponding to that one can still be in positive time (for example if your u is negative)

BTW time can be negative, in many cases you will expect two results, for example throwing a ball in the air, it will be at ground level exactly twice. It all depends when you define your t=0. Even if you say time can't be negative it is an error to neglect the - root because the solution corresponding to that one can still be in positive time (for example if your u is negative)

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(Original post by

I am amazed how many physics students do not know how to solve a quadratic equation...

BTW time can be negative, in many cases you will expect two results, for example throwing a ball in the air, it will be at ground level exactly twice. It all depends when you define your t=0. Even if you say time can't be negative it is an error to neglect the - root because the solution corresponding to that one can still be in positive time (for example if your u is negative)

**MikeyW09**)I am amazed how many physics students do not know how to solve a quadratic equation...

BTW time can be negative, in many cases you will expect two results, for example throwing a ball in the air, it will be at ground level exactly twice. It all depends when you define your t=0. Even if you say time can't be negative it is an error to neglect the - root because the solution corresponding to that one can still be in positive time (for example if your u is negative)

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#32

(Original post by

I think I know how to Solve Quadratics. And BTW you're doing PHD and I am only a AS Student.

**jasbirsingh**)I think I know how to Solve Quadratics. And BTW you're doing PHD and I am only a AS Student.

Right. Once you know how to solve a quadratic equation, you don't really get much better at it over time, in fact you almost certainly solve them more often than a phd student.

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(Original post by

Then why make the thread

Right. Once you know how to solve a quadratic equation, you don't really get much better at it over time, in fact you almost certainly solve them more often than a phd student.

**TableChair**)Then why make the thread

Right. Once you know how to solve a quadratic equation, you don't really get much better at it over time, in fact you almost certainly solve them more often than a phd student.

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#34

(Original post by

Because I am writing program in Pascal which solves the SUVAT equations that's why I needed the re-arranged version.

**jasbirsingh**)Because I am writing program in Pascal which solves the SUVAT equations that's why I needed the re-arranged version.

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(Original post by

But you just said you could rearrange it yourself

**TableChair**)But you just said you could rearrange it yourself

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#36

(Original post by

hence its a quadritic and you can use the formula

end of.

**cambo211**)hence its a quadritic and you can use the formula

end of.

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#37

(Original post by

I said I can use this equation to solve the questions by putting numbers first..and then solving it

**jasbirsingh**)I said I can use this equation to solve the questions by putting numbers first..and then solving it

My comment wasn't directed at you, it was at the other people who posted in this thread and got it wrong.

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#38

(Original post by

I said I can use this equation to solve the questions by putting numbers first..and then solving it ( which is much easier)...and I tried re-arranging it but couldn't...I need the re-arranged version to make a program (SUVAT Calculator).

**jasbirsingh**)I said I can use this equation to solve the questions by putting numbers first..and then solving it ( which is much easier)...and I tried re-arranging it but couldn't...I need the re-arranged version to make a program (SUVAT Calculator).

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#39

Wrong on some counts.

Step 3: is 2s/t = 2u + at (because u is a separate entity so the 2 has to apply to it on its own) but the more misleading assumption really is in step 5. 2s/t-u=at, but multiplying all by t gives 2s-ut=at2 and not 2s-u=at2 as above. So, the t still persists with the u, and has not gone away at all.

Step 3: is 2s/t = 2u + at (because u is a separate entity so the 2 has to apply to it on its own) but the more misleading assumption really is in step 5. 2s/t-u=at, but multiplying all by t gives 2s-ut=at2 and not 2s-u=at2 as above. So, the t still persists with the u, and has not gone away at all.

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#40

(Original post by

Wrong on some counts.

Step 3: is 2s/t = 2u + at (because u is a separate entity so the 2 has to apply to it on its own) but the more misleading assumption really is in step 5. 2s/t-u=at, but multiplying all by t gives 2s-ut=at2 and not 2s-u=at2 as above. So, the t still persists with the u, and has not gone away at all.

**yomex234**)Wrong on some counts.

Step 3: is 2s/t = 2u + at (because u is a separate entity so the 2 has to apply to it on its own) but the more misleading assumption really is in step 5. 2s/t-u=at, but multiplying all by t gives 2s-ut=at2 and not 2s-u=at2 as above. So, the t still persists with the u, and has not gone away at all.

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