# Thread: Make t subject

1. ## Make t subject

Hi I need to make t subject
S=ut+ 1/2(at^2)

I worked it out(last line) = 2s-2ut/at = t^2/t =t

Is this correct?

2. i remember this is a subject of physic ,right!

3. Originally Posted by wolfhound
Hi I need to make t subject
S=ut+ 1/2(at^2)

I worked it out(last line) = 2s-2ut/at = t^2/t =t

Is this correct?
Not quite because you need to have t and only t on the right

$s = ut+\frac{1}{2}at^2 \: \: \rightarrow \: \: at^2+2ut-2s=0$

Use the quadratic equation to get an expression for $t$

Originally Posted by manhmanh17vn
i remember this is a subject of physic ,right!
The Kinematic Equations comes up in all fields

4. This is the way I done it

S=ut+ 1/2(at^2)
S-ut+1/2(at^2)
2S-2ut=at^2
2S-2ut/at^2
2S-2u/at
2s-2u/a = t is this correct please?

5. khakha you so smart . i studied this last 3 years

6. [QUOTE=wolfhound;418431]
Originally Posted by e^(i*pi)
Not quite because you need to have t and only t on the right

$s = ut+\frac{1}{2}at^2 \: \: \rightarrow \: \: at^2+2ut-2s=0$

Use the quadratic equation to get an expression for $t$

Thats what it was meant the final calculation works out t^2/t = t
I was wondering if I calculated it correct down to the last line without writing it all out( I am not allowed use quadratic formula)
Could you solve it my way and tell me if I am correct please
You could complete the square

$at^2+2ut=2s$

Divide by a

$t^2+\frac{2u}{a}t = \frac{2s}{a}$

$\left(t+\frac{u}{a}\right)^2 - \frac{u^2}{a^2} = \frac{2as}{a^2}$

$\left(t+\frac{u}{a}\right)^2 = \frac{2as+u^2}{a^2}$

$\left(t+\frac{u}{a}\right) = \pm \frac{\sqrt{2as+u^2}}{a}$

$t = \frac{-u\pm \sqrt{2as+u^2}}{a}$

7. [QUOTE=e^(i*pi);418441]
Originally Posted by wolfhound

You could complete the square

$at^2+2ut=2s$

Divide by a

$t^2+\frac{2u}{a}t = \frac{2s}{a}$

$\left(t+\frac{u}{a}\right)^2 - \frac{u^2}{a^2} = \frac{2as}{a^2}$

$\left(t+\frac{u}{a}\right)^2 = \frac{2as+u^2}{a^2}$

$\left(t+\frac{u}{a}\right) = \pm \frac{\sqrt{2as+u^2}}{a}$

$t = \frac{-u\pm \sqrt{2as+u^2}}{a}$

S=ut+ 1/2(at^2)
S-ut=1/2(at^2)
2S-2ut=at^2
2S-2ut/at^2
2S-2u/at
2s-2u/a = t

8. this's true. and this will help you find t faster on some ways. but carefully went you read the question if you don't want 0 . i am just kidding

9. But is there 2 answers?

I need to know if my way is correct??????????????

10. Originally Posted by wolfhound
But is there 2 answers?

I need to know if my way is correct??????????????
do u mean $S=ut+\frac{at^2}{2}$ ?

11. [quote=wolfhound;418443]
Originally Posted by e^(i*pi)

S=ut+ 1/2(at^2)
S-ut=1/2(at^2)
2S-2ut=at^2
2S-2ut/at^2
2S-2u/at
2s-2u/a = t
that's not true,
S=ut+ 1/2(at^2)
S-ut=1/2(at^2)
2S-2ut=at^2
2S-2ut-at^2=0.

12. [QUOTE=Raoh;418475]
Originally Posted by wolfhound

that's not true,
S=ut+ 1/2(at^2)
S-ut=1/2(at^2)
2S-2ut=at^2
2S-2ut-at^2=0.
sorry I mean this way
S=ut+ 1/2(at^2)
S-ut=1/2(at^2)
2S-2ut=at^2
2S-2ut/a=t^2
2S-2ut/at=t^2/t
2s-2u/a = t

13. [quote=wolfhound;418485]
Originally Posted by Raoh

sorry I mean this way
S=ut+ 1/2(at^2)
S-ut=1/2(at^2)
2S-2ut=at^2
2S-2ut/a=t^2
2S-2ut/at=t^2/t
2s-2u/a = t
that's not true either ,
(2S-2ut)/a=t^2.
(2S-2ut)/at=t.

14. [quote=Raoh;418475]
Originally Posted by wolfhound

that's not true,
S=ut+ 1/2(at^2)
S-ut=1/2(at^2)
2S-2ut=at^2
2S-2ut-at^2=0.
from $2S-2ut-at^2=0$ you'll have $at^2+2ut-2S=0$
therefore $t=\frac{-2u+\sqrt{4u^2+8aS}}{2a}$ or $t=\frac{-2u-\sqrt{4u^2+8aS}}{2a}$

15. [QUOTE=Raoh;418489]
Originally Posted by wolfhound

that's not true either ,
(2S-2ut)/a=t^2.
(2S-2ut)/at=t.
Hi
But
2S-2ut=at^2
2S-2ut/a=t^2(if I divide both sides by a to get t^2 on RHS)
2S-2ut/at=t^2/t(then if I divide both sides by t to get t on the RHS)
2s-2u/a = t
???
I am not allowed -b formula

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# s=ut 1/2at .make t subject of the formula

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