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

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10 years ago
#21
(Original post by 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.
This step is wrong. You should get 2s-ut= at^2
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10 years ago
#22
(Original post by 03gshep)
This step is wrong. You should get 2s-ut= at^2

ahh well spotted
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10 years ago
#23
(Original post by 03gshep)
You can, using the formula like above.
You can't fully because there is .
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10 years ago
#24
(Original post by lolhenry)
You can't fully because there is .
hence its a quadritic and you can use the formula

end of.
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10 years ago
#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.
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10 years ago
#26
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.
Neither are correct

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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10 years ago
#27
(Original post by steve2005)
Neither are correct

The question was make t the subject.

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

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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10 years ago
#28
It's this:

t= (-u +or- sqroot(2sa+u^2))/a

End of.
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#29
(Original post by steve2005)
Neither are correct

The question was make t the subject.

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

The Solution you sent me perfectly worked.. Thank You .
Here's the Solution given by Steve

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10 years ago
#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)
1
#31
(Original post by 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)
I think I know how to Solve Quadratics. And BTW you're doing PHD and I am only a AS Student.
1
10 years ago
#32
(Original post by 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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#33
(Original post by TableChair)

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.
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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10 years ago
#34
(Original post by jasbirsingh)
Because I am writing program in Pascal which solves the SUVAT equations that's why I needed the re-arranged version.
But you just said you could rearrange it yourself
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#35
(Original post by TableChair)
But you just said you could rearrange it yourself
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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10 years ago
#36
(Original post by cambo211)
hence its a quadritic and you can use the formula

end of.
lol ok (Y)
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10 years ago
#37
(Original post by jasbirsingh)
I said I can use this equation to solve the questions by putting numbers first..and then solving it
I don't get what you can't do, if you know how to solve quadratic equations? Even if you can't complete the square, the quadratic formula is your answer right there......

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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10 years ago
#38
(Original post by 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).
It's exactly the same. The letters represent numbers...
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5 years ago
#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.
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5 years ago
#40
(Original post by 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.
Duudde this was 4 years ago...
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