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    (Original post by Marccs)
    Can anyone help me with this question on vectors
    Watch this video, it'll help you http://www.examsolutions.net/maths-r...tutorial-2.php
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    (Original post by Don Joiner)
    Dam I forgot; here:

    Attachment 545515


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    (Original post by metrize)
    They set west as the positive direction, you can set east as positivw direction and you'd get the same answer
    But I don't get how the velocity of p can be the same before and after


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    Can someone explain why it cannot tilt about B for part (C)? Im confused because I thought both strings were meant to be taut Name:  Screen Shot 2016-06-07 at 20.13.28.png
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    (Original post by KloppOClock)
    Okay tell me which part you dont understand out of these steps...

    r= (5-3t)i + (t-3)j

    When moving parallel, it has to be moving directly right or left, for example, (5i 0j)

    so the j component must equal zero

    t-3 = 0
    t=3 when moving parallel
    >>r= (5-3t)i + (t-3)j
    the velocity is given but how did you get the initial position.
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    (Original post by KloppOClock)
    the reaction at the other support (not c) is zero
    Thanks
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    (Original post by fpmaniac)
    >>r= (5-3t)i + (t-3)j
    the velocity is given but how did you get the initial position.
    well you dont need the original position, i just used that formula to find velocity instead of location

    i used velocity and acceleration(t) instead of location and velocity(t)

    m and m/s vs m/s and m/s/s
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    (Original post by Don Pedro K.)
    Nigel is that you??
    I dont think im the nigel ur thinking off
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    (Original post by Don Pedro K.)
    Watch this video, it'll help you http://www.examsolutions.net/maths-r...tutorial-2.php
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    (Original post by KloppOClock)
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    Das what I got also
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    (Original post by NigelK98)
    I dont think im the nigel ur thinking off
    Ahh okay haha sorry! It's just there's this guy I know with the surname beginning with K as well xD
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    (Original post by Louiseelg0rt)
    Can someone explain why it cannot tilt about B for part (C)? Im confused because I thought both strings were meant to be taut Name:  Screen Shot 2016-06-07 at 20.13.28.png
Views: 148
Size:  119.3 KBAttachment 545553545555
    The beam stays horizontal as said
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    (Original post by KloppOClock)
    if it says awrt then yes, if not, probably not.

    Did you use an aproximation in your method such as g?

    also if you wrote the answer as a fraction before you rounded it they should mark it correct and ignore subsequent working i think
    (Original post by Don Pedro K.)
    Question paper?
    I only asked this because they showed the answer in different format in different papers. I hope I attached the images right

    I did the paper with Q4c first (Jan 2011) and gave my answer as a fraction, 5/3. The mark scheme said this was fine (in fact the only answer I see there ) so I was like that's lucky for me. I did the paper with Q6c after (Jun 2011) and gave my answer as a fraction because of what I seen before, but THIS time it says you don't get a mark for 1/14, only for that in 2/3 s.f. So what do I do? Is it the question type, or do they actually allow 3 s.f. in the Jan 2011 paper but just didn't say.
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    (Original post by Don Pedro K.)
    This is because as soon as one particle moves, the other one moves with it since the string can't stretch!
    Gotcha, thanks! What about light? and what does a "rod" tell us?
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    (Original post by Don Pedro K.)
    Das what I got also
    But you've labeled both 4u and u as positive even though one is going east and the other west?


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    (Original post by Kamara7)
    I only asked this because they showed the answer in different format in different papers. I hope I attached the images right

    I did the paper with Q4c first (Jan 2011) and gave my answer as a fraction, 5/3. The mark scheme said this was fine (in fact the only answer I see there ) so I was like that's lucky for me. I did the paper with Q6c after (Jun 2011) and gave my answer as a fraction because of what I seen before, but THIS time it says you don't get a mark for 1/14, only for that in 2/3 s.f. So what do I do? Is it the question type, or do they actually allow 3 s.f. in the Jan 2011 paper but just didn't say.
    Wow yeah that is quite weird! Although, it may perhaps not be the full mark scheme, so it might allow 5/3 in 3.s.f?
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    (Original post by Don Pedro K.)
    Ahh okay haha sorry! It's just there's this guy I know with the surname beginning with K as well xD
    no worries
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    (Original post by Kamara7)
    I only asked this because they showed the answer in different format in different papers. I hope I attached the images right

    I did the paper with Q4c first (Jan 2011) and gave my answer as a fraction, 5/3. The mark scheme said this was fine (in fact the only answer I see there ) so I was like that's lucky for me. I did the paper with Q6c after (Jun 2011) and gave my answer as a fraction because of what I seen before, but THIS time it says you don't get a mark for 1/14, only for that in 2/3 s.f. So what do I do? Is it the question type, or do they actually allow 3 s.f. in the Jan 2011 paper but just didn't say.
    as i said before, if you use an aproxamation such as g in your answer, you have to round.

    In jan 2011 you were doing vectors which is fine.

    in june 2011 you assumed that gravity equals 9.8 so you must round to 2 or 3 sf.
    it says this on the front of the exam paper
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    (Original post by Don Joiner)
    But you've labeled both 4u and u as positive even though one is going east and the other west?


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    You can choose what direction you want to call positive
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    (Original post by thelegend99)
    Gotcha, thanks! What about light? and what does a "rod" tell us?
    Means that it won't bend or flex or anything funny like that ! Check this video out https://www.youtube.com/watch?v=T7XO768pack from about half a minute in !
 
 
 
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