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    I understand how to do the question but I don't understand why the reaction at D (Q) isn't perpendicular to the pole. Can someone explain why?

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    (Original post by ozmo19)
    I understand how to do the question but I don't understand why the reaction at D (Q) isn't perpendicular to the pole. Can someone explain why?

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    I'm uncertain as to why you think D (Q) is a reaction force? It's just a random force, not a reaction one. Hence there's no reason to believe it's perpendicular to the pole.
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    (Original post by Zacken)
    I'm uncertain as to why you think D (Q) is a reaction force? It's just a random force, not a reaction one. Hence there's no reason to believe it's perpendicular to the pole.
    Ahh, I misread the question when it said P was at a right angle, I assumed they meant Q was as well. Thanks.
    I have another question. In 6(iii) when finding the integral of v to find an expression for s the mark scheme states that the constant cancels for the times given. I don't understand this as Ive never encountered it. Can you explain this?

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    (Original post by ozmo19)
    Ahh, I misread the question when it said P was at a right angle, I assumed the meant Q was as well. Thanks.
    I have another question. In 6(iii) when finding the integral of v to find an expression for s the marks scheme states that the constant cancels for the times given. I don't understand this as Ifve never encountered it. Can you explain this?

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    It's always the same when you are evaluating between limits. Say you were doing [x^2 +7x +c] with bounds 4 and 0. You would get (4^2 +28 +c) -(0^2+0+c) and as you can see the c cancels
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    (Original post by ozmo19)
    In 6(iii) when finding the integral of v to find an expression for s the marks scheme states that the constant cancels for the times given. I don't understand this as Ive never encountered it. Can you explain this?

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    If you've ever integrated between limits, you should be aware of this :/

    If not ~ then you should find that the constant will cancel anyway.
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    (Original post by ozmo19)
    Ahh, I misread the question when it said P was at a right angle, I assumed the meant Q was as well. Thanks.
    I have another question. In 6(iii) when finding the integral of v to find an expression for s the marks scheme states that the constant cancels for the times given. I don't understand this as Ive never encountered it. Can you explain this?

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    Are you familiar with the idea that displacement is the area under the velocity-time graph?

    Are you familiar with the idea that area under a graph f(x) between x=a and x=b is \int_a^b f(x) \, \mathrm{d}x?

    So, the displacement here is \int_3^5 v \, \mathrm{d}t = \cdots , where the definite integral has no constant of integration. i.e: you just integrate and then do integral(upper limit) - integral(lower limit).

    The reason why this is so is, let's assume f(x) integrates to F(x) + c, then:

    \displaystyle 

\begin{equation*}\int_a^b f(x) \, \mathrm{d}x = \bigg[F(x) + c\bigg]_a^b = (F(b) + c) - (F(a) +c) = F(b) - F(a) + c - c = F(b) - F(a)\end{equation*}
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    (Original post by samb1234)
    It's always the same when you are evaluating between limits. Say you were doing [x^2 +7x +c] with bounds 4 and 0. You would get (4^2 +28 +c) -(0^2+0+c) and as you can see the c cancels
    (Original post by Alexion)
    If you've ever integrated between limits, you should be aware of this :/

    If not ~ then you should find that the constant will cancel anyway.
    The mark scheme didn't use limits in the integral. Im aware you don't add a constant when using limits.
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    (Original post by ozmo19)
    The mark scheme didn't use limits in the integral. Im aware you don't add a constant when using limits.
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    yes it does. They haven't put the limits on the integral sign but what are they doing lol - they're finding the distance at the two times and subtracting them, i.e. evaluating between limits
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    (Original post by ozmo19)
    The mark scheme didn't use limits in the integral. Im aware you don't add a constant when using limits.
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    The mark scheme is effectively just doing it the long way. Using limits is a very valid way to answer the question.

    The way it does it effectively just breaks down the process. But if you were to do it without removing the constants, you'd get two expressions t1 + d and t2 + d so when you subtract them, they would cancel out, just like when using limits.
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    (Original post by ozmo19)
    The mark scheme didn't use limits in the integral. Im aware you don't add a constant when using limits.
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    They are integrating between limits, just in a stupid and potentially confusing way.
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    (Original post by Zacken)
    They are integrating between limits, just in a stupid and potentially confusing way.
    Do i need to integrate between 3 and 4 and then between 4 and 5 then add the result since it stops when t=4 and possibly changes direction?

    Thanks
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    (Original post by ozmo19)
    Do i need to integrate between 3 and 4 and then between 4 and 5 then add the result since it stops when t=4 and possibly changes direction?

    Thanks
    If it changes direction, then yes.
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    (Original post by Zacken)
    x
    Why is there actually no reaction in this question????

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    (Original post by ozmo19)
    Why is there actually no reaction in this question????

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    Who says there isn't one?
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    (Original post by Zacken)
    Who says there isn't one?
    I redraw a diagram and did the question with one but the mark scheme doesn't include a reaction
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    (Original post by ozmo19)
    I redraw a diagram and did the question with one but the mark scheme doesn't include a reaction
    Link the markscheme?
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    (Original post by Zacken)
    Link the markscheme?
    EDIT: the ms doesn't make use of a reaction force to calculate either answers. Just to clarify, there is one and I should include one on the diagram in the exam?
    Thanks
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    (Original post by ozmo19)
    EDIT: the ms doesn't make use of a reaction force to calculate either answers. Just to clarify, there is one and I should include one on the diagram in the exam?
    Thanks
    MS doesnt make use of a reaction force, because you don't need it to do the question. If, in the exam, you are asked to draw out the diagram marking all forces, then include the normal reaction, yes.
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    (Original post by ozmo19)
    EDIT: the ms doesn't make use of a reaction force to calculate either answers. Just to clarify, there is one and I should include one on the diagram in the exam?
    Thanks
    There is one and if asked to show all forces you wpuld need to include it. It's not required for this question because you need only resolve parallel.
 
 
 
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