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Math Help - Make t subject

  1. #1
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    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?
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  2. #2
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    i remember this is a subject of physic ,right!
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  3. #3
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    Quote Originally Posted by wolfhound View Post
    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

    Quote Originally Posted by manhmanh17vn View Post
    i remember this is a subject of physic ,right!
    The Kinematic Equations comes up in all fields
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  4. #4
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    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?
    Last edited by wolfhound; December 5th 2009 at 09:45 AM.
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  5. #5
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    khakha you so smart . i studied this last 3 years
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  6. #6
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    [QUOTE=wolfhound;418431]
    Quote Originally Posted by e^(i*pi) View Post
    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}
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  7. #7
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    [QUOTE=e^(i*pi);418441]
    Quote Originally Posted by wolfhound View Post

    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}
    what about this way

    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
    Last edited by wolfhound; December 5th 2009 at 10:07 AM.
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  8. #8
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    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
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  9. #9
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    But is there 2 answers?

    I need to know if my way is correct??????????????
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  10. #10
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    Smile

    Quote Originally Posted by wolfhound View Post
    But is there 2 answers?

    I need to know if my way is correct??????????????
    do u mean S=ut+\frac{at^2}{2} ?
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  11. #11
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    Smile

    [quote=wolfhound;418443]
    Quote Originally Posted by e^(i*pi) View Post

    what about this way

    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.
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  12. #12
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    [QUOTE=Raoh;418475]
    Quote Originally Posted by wolfhound View Post

    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
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  13. #13
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    [quote=wolfhound;418485]
    Quote Originally Posted by Raoh View Post

    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.
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  14. #14
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    [quote=Raoh;418475]
    Quote Originally Posted by wolfhound View Post

    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}
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  15. #15
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    [QUOTE=Raoh;418489]
    Quote Originally Posted by wolfhound View Post

    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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