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    I need help with this question.

    A particle is moving along a straight line OA with constant acceleration 2ms-2. At O the particle is moving towards A with speed 5.5ms-1. The distance OA is 20m. Find the time the particle takes to move from O to A.

    S=20 V=5.5 A=2 T=?

    I used the formula s=vt-1/2at^2 but I got the wrong answer.

    Any help?
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    How can the particle be moving at 5.5ms-1 at OA? Since OA is a distance over which it has non-zero acceleration. Are you sure it's not just at O?
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    Well, all I can suggest is that you try again. I have an answer; what did you get and what should the answer be, before I say? :p:
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    v is not 5.5, its 'u'
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    (Original post by ForGreatJustice)
    How can the particle be moving at 5.5ms-1 at OA? Since OA is a distance over which it has non-zero acceleration. Are you sure it's not just at O?
    I'm assuming it means the speed at A is 5.5ms-1
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    (Original post by ForGreatJustice)
    How can the particle be moving at 5.5ms-1 at OA? Since OA is a distance over which it has non-zero acceleration. Are you sure it's not just at O?
    That was a typo, i've edited the question.

    (Original post by DarkWhite)
    I have an answer; what did you get and what should the answer be, before I say?
    The answer is
    Spoiler:
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    2.5
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    (Original post by Mist786)
    That was a typo, i've edited the question.



    The answer is
    Spoiler:
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    2.5
    Ok, in which case remember that v is the final velocity, but you're only given the initial velocity u, for which there is a (very similar) formula for which will work
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    (Original post by ForGreatJustice)
    Ok, in which case remember that v is the final velocity, but you're only given the initial velocity u, for which there is a (very similar) formula for which will work
    I still get the wrong answer when I substitute the values, I think i'm rearrange it wrong. Any help? I'll rep
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    s=ut+1/2at^2 then if you substitute the values in, you get a quadratic in t, which gives the answer.

    If you still get the wrong anwser, post your working.
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    (Original post by ForGreatJustice)
    s=ut+1/2at^2 then if you substitute the values in, you get a quadratic in t, which gives the answer.

    If you still get the wrong anwser, post your working.
    20=5.5t-1/2*2*t^2
    20=5.5t-t^2
    20=t(5.5+t)
    20/t=5.5+t

    ^^That's wrong .. idk
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    ok...you're using the wrong equation, we want s=ut + 1/2at^2

    putting the numbers in to this gives 20=5.5t+t^2

    t^2+5.5t-20=0

    which can be either factorised (easiest if you can spot it), completing the square (messy in this case) or the quadratic formula.

    You will get 2 values for t, but clearly one will be absurd in this context, and the other is the answer you are looking for.
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    You need to use two of the formulae - here goes.

    S= 20 U = 5.5 V = ? A= 2 T= ?

    V^2=U^2+2AS
    Therefore V = Sqrt(30.25+80) = 10.5

    Then plug into V= U+AT (Which we'll rearrange to (V-U)/A = T ) and you have (10.5-5.5)/2 = 2.5 = T
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    (Original post by Longorefisher)
    You need to use two of the formulae - here goes.

    S= 20 U = 5.5 V = ? A= 2 T= ?

    V^2=U^2+2AS
    Therefore V = Sqrt(30.25+80) = 10.5

    Then plug into V= U+AT (Which we'll rearrange to (V-U)/A = T ) and you have (10.5-5.5)/2 = 2.5 = T
    you can, but the one simple formula works.
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    (Original post by ForGreatJustice)
    ok...you're using the wrong equation, we want s=ut + 1/2at^2

    putting the numbers in to this gives 20=5.5t+t^2

    t^2+5.5t-20=0

    which can be either factorised (easiest if you can spot it), completing the square (messy in this case) or the quadratic formula.

    You will get 2 values for t, but clearly one will be absurd in this context, and the other is the answer you are looking for.
    I still get the wrong answer when I do the quadratic formula and I have no idea how to do completing the square.
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    (Original post by ForGreatJustice)
    you can, but the one simple formula works.
    Yeah, I would agree although I just ran the quadratic formula on it and it's not coming out at the answer I'd expect.
    Edit - calc error. Using the wrong co-efficent of x^2 - I do agree with you now.
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    you shouldn't do, but the quadratic factorises to (t-2.5)(t+8)=0 which gives the answer

    just tried the quadratic formula, that works too

    out of curiosity what did you get from the quadratic formula?
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    (Original post by ForGreatJustice)
    you shouldn't do, but the quadratic factorises to (t-2.5)(t+8)=0 which gives the answer

    just tried the quadratic formula, that works too

    out of curiosity what did you get from the quadratic formula?
    The quadratic formula gave me -45.5 and -40 :rolleyes:

    Don't laugh at me
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    t=\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    where at^2+bt+c=0

    So you just put a=1, b=5.5, and c=-20 in :p:

    are you just putting the whole load into a calculator? since if you don't bracket up right, things go wrong
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    (Original post by ForGreatJustice)
    t=\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    where at^2+bt+c=0

    So you just put a=1, b=5.5, and c=-20 in :p:

    are you just putting the whole load into a calculator? since if you don't bracket up right, things go wrong
    lol I realised my mistake

    Thanks again
 
 
 
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