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    I would suggest that
    \vec{AE}=\vec{AC}+\vec{CE}
    \vec{AE}=\vec{AB}+\vec{BE}
    However, \vec{CE}+\vec{BE}=?
    Does it help?

    (Original post by fablereader)
    I'm having trouble wtih a vector problem.

    'The diagram shows a trapezium ABCD, with AB parallel to DC and AB twice as long as DC.
    E is the mid-point of BC. AD-> = p and DC-> = q.
    Find, in terms of p and q

    a AB->
    b AC->
    c CD->
    d DB->
    e AE->
    f ED->'


    -> signifies it is a translation instead of a distance. Unfortunately, I can't put in the diagram. I've solved a-d (2q, p + q, -q, 2q - p respectively), but I've stalled on e. Can anyone help?
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    (Original post by depymak)
    I would suggest that
    \vec{AE}=\vec{AC}+\vec{CE}
    \vec{AE}=\vec{AB}+\vec{BE}
    However, \vec{CE}+\vec{BE}=?
    Does it help?
    Yes, thanks! I've gotten the answers now.
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    I'm having trouble with this problem. Could anyone help me with it?

    'A boat P, capable of 6 m s-1 in still water, travels from a point A around a course ABC which is an equilateral triangle with sides 500 m long. A uniform current of 4 m s-1 flows in the direction AB->. An identical boat Q starts from A at the same time and travels the other way (ACB) around the course. Which boat gets back to A first, and what is the margin of victory (in seconds)?'
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    I'm still having trouble with the question, and I'm having trouble with two more. Could anyone help with any?

    11)

    'A boat P, capable of 6 m s-1 in still water, travels from a point A around a course ABC which is an equilateral triangle with sides 500 m long. A uniform current of 4 m s-1 flows in the direction AB->. An identical boat Q starts from A at the same time and travels the other way (ACB) around the course. Which boat gets back to A first, and what is the margin of victory (in seconds)?'

    12)

    'A river is D m wide and flows at u m s-1. A man can swim at v m s-1 in still water, where v > u. Find the ratio of the shortest time it would take him to swim across the river and back, and the time it would take him to swim D m upstream and back.'

    14)

    'OABC is a tetrahedron. L, M and N are the mid-points of OA, OB and OC respectively. P, Q and R are the mid-points of BC, AC and AB respectively. OA-> = a, OB-> = b and OC-> = c. Use vector methods to show that the lines PL, QM and RN bisect each other.'
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    12) Am I mistaken in considering the man's velocity at still water as the relative to water velocity when he swims at a moving water?
    If not, the resultant velocity is the vector sum of v and u.
    Should you try a draft sketch?
    What do you think?
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    (Original post by depymak)
    12) Am I mistaken in considering the man's velocity at still water as the relative to water velocity when he swims at a moving water?
    If not, the resultant velocity is the vector sum of v and u.
    Should you try a draft sketch?
    What do you think?
    I'm not entirely sure what you mean by that. If you mean that he swims at v when the river is moving at u, no, I don't think so. I think he swims at v when u essentially equals 0, i.e. the water is completely still. And you're right about the resultant velocity, thanks! I'll try a draft sketch.
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    Ok, thanks, we are on the same page then.We have to sum vectors in three different cases.
    (Original post by fablereader)
    .....And you're right about the resultant velocity, thanks! I'll try a draft sketch.
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    I'm having difficulty understanding something. I've spent an hour on it and looked through two different textbooks, but it simply isn't making sense in my head. Can anyone help?

    The situation is that of an inclined plane with an object on it. This object is in equilibrium. m is its mass, g is its gravity. The inclined plane is at angle θ. They say that the component of the weight parallel to the plane is mgsin θ and the component of the weight perpendicular to the plane is mgcos θ. However, I do not understand how they get to this equation for a component. Here is an example they give:

    'A stone of mass 0.1 kg rests on a smooth plane inclined at 20 degrees to the horizontal. It is held in equilibrium by a force, F, acting parallel to the plane, as shown (cannot include diagram).

    Find the magnitude of the force F and the normal reaction, R, of the plane on the stone.

    Resolving parallel to the plane:

    F - 0.1gsin 20 = 0
    F = 0.1gsin 20
    = 0.335 N

    Resolving perpendicular to the plane:

    R - 0.1gcos 20 = 0
    R = 0.1gcos 20
    = 0.921 N'
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    (Original post by fablereader)
    I'm having difficulty understanding something. I've spent an hour on it and looked through two different textbooks, but it simply isn't making sense in my head. Can anyone help?

    inclined plane is at angle θ. They say that the component of the weight parallel to the plane is mgsin θ and the component of the weight perpendicular to the plane is mgcos θ. However, I do not understand how they get to this equation for a component. Here is an example they give:


    The blue vertical line is your weight veritcally. The blue perpendicular line is the reaction force. The red line is the extension of the reaction force. Can you see how the vertical blue lines forms a right angle triangle? This means I can then say the angle I've marked in red (angle between slope and weight) is 90-theta. Now, since the parallel and perpendicular of the slope is perpendicular, I can then say the other angle I've marked in red is 90-(90-theta) = theta.

    Now, if I want to resolve perpendicular to the surface, I get: R acting perpendicular "away" from the surface and mg\cos \theta acting perpendicular "into" the surface.

    Hence, since the particle is on the surface and not levitating, it must mean that these two forces cancel each other out. i.e: my net perpendicular force is R - mg\cos \theta = 0. Re-arranging this gets me R = mg\cos \theta. My reaction.

    In this example, there's no force acting down the plane apart from the component of weight, so the force acting down the slope is mg \cos (90 - \theta) = mg\sin \theta. This acts down the slope.

    'A stone of mass 0.1 kg rests on a smooth plane inclined at 20 degrees to the horizontal. It is held in equilibrium by a force, F, acting parallel to the plane, as shown (cannot include diagram).

    Find the magnitude of the force F and the normal reaction, R, of the plane on the stone.
    Here, we resolve perpendicular. It's obviously not levitating, so the force perpendicular to the surface "outwards" (reaction) must equal the force into the surface perpendicular to it (perpendicular component of weight), hence: R - mg\cos 20^{\circ} = 0. Hence R = mg \cos 20^{\circ}.

    We can ignore the force F here because we are resolving perpendicularly (we resolve perpendicularly outwards) and the force acts parallel, so the two components are mutually perpendicular.

    Now, we resolve parallel. The weight obviously acts "down" the plane and the force acts "up" the plane, since it's in equilibrium, these two forces cancel each other out.

    Resolve either parallel up or parallel down the slope, either worse. I'll go with up. This gets us F - mg\sin 20^{\circ} = 0.

    Was that any clearer?
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    (Original post by Zacken)

    Was that any clearer?
    I'm afraid not. I'm not entirely sure if I made myself clear on what I was confused about (my fault, bad wording). What I'm confused about is how mgcos θ can be the force equal to (R). Am I making sense?
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    (Original post by fablereader)
    I'm afraid not. I'm not entirely sure if I made myself clear on what I was confused about (my fault, bad wording). What I'm confused about is how mgcos θ can be the force equal to (R). Am I making sense?
    From my diagram, can you see that the force acting away from the surface is R the reaction? And that the angle between the vertical (weight) and perpendicular is  \theta? i.e: the angle between the blue line and red line?

    Now resolving the blue line into the component along the red line gets us mg \cos \theta. But now obviously the reaction force away from the plane has to equal the force "into" the plane so as to keep the ball on the surface.

    Hence, the net force acting perpendicularly is R - mg\cos \theta = 0.
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    (Original post by fablereader)
    I'm afraid not. I'm not entirely sure if I made myself clear on what I was confused about (my fault, bad wording). What I'm confused about is how mgcos θ can be the force equal to (R). Am I making sense?
    If I gave you this diagram:

    And I say that the blue line is B and the red line is R, how would you find the component of B acting in the direction of R?
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    (Original post by Zacken)
    If I gave you this diagram:

    And I say that the blue line is B and the red line is R, how would you find the component of B acting in the direction of R?
    I'm still not understanding this (particularly the 'acting in direction of' part) but this is my gut reaction:

    Form a right-angle triangle, then use:

    cos θ = adjacent/hypotenuse

    multiply both sides by hypotenuse:

    hypotenuse(cos θ) = adjacent

    divide both sides by cos θ

    hypotenuse = adjacent/cos θ
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    (Original post by fablereader)
    I'm still not understanding this (particularly the 'acting in direction of' part) but this is my gut reaction
    Watch this and this. It should hopefully clear up everything.
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    (Original post by Zacken)
    Watch this and this. It should hopefully clear up everything.
    OK, I think I get it now. Thanks!
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    (Original post by fablereader)
    OK, I think I get it now. Thanks!
    Phew!
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    This isn't necessarily going with the title, but I've just done a past paper and marked it and I have a few questions (Please help):

    How necessary is it for my answer to be to 2-3 significant figures? One of my answers was 117.6 (exact answer), but they seem to specifically penalize that.

    I tend to immediately put in 9.8 as g, but they don't seem to do that until much later. Is that significant?

    When I draw a graph, do I have to make some kind of marker line to mark significant points (i.e. a dotted line for a break between two abrupt changes of acceleration on an acceleration-time graph)?

    And finally, one of my methods is different from the method they specify in the mark scheme. Not only is it different, but I'm not sure what it's talking about. Would mine still count if it leads to the correct answer and what is their solution mean?

    5. A car accelerates uniformly from rest for 20 seconds. It moves at constant speed v m s-1 for the next 40 seconds and then decelerates uniformly for 10 seconds until it comes to rest.

    (a) For the motion of the car, sketch
    (i) a speed-time graph,
    (ii) an acceleration-time graph

    Given that the total distance moved by the car is 880 m,
    (b) find the value of v.

    (This is my solution for (b))

    area under line in speed-time graph (A) = distance
    A = 880
    A - (v * 40) + (1/2 (20 * v)) + (1/2 (10 * v))
    880 = 40v + 10v + 5v
    880 = 55v
    v = 16

    answer:
    16 ms-1

    (This is their notes on the question)
    ((7 * 40)/2) * v = 880
    v = 880 * (2/110) = 16
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    (Original post by fablereader)
    How necessary is it for my answer to be to 2-3 significant figures? One of my answers was 117.6 (exact answer), but they seem to specifically penalize that.
    Depends on the question, if it's a question where you have used the approximation g=9.8 then you must give your answer to 2-3 s.f because otherwise you are claiming your answer is more accurate that it could possibly be given the nature of your approximation of g = 9.8 to s.f

    If you do a question where you don't use such an approximation, then it's fine for you to give an exact answer or an answer to however many s.f you want, but the best practice is to give it to 3 s.f.

    I tend to immediately put in 9.8 as g, but they don't seem to do that until much later. Is that significant?
    Not at all, that's fine. Put it in whenever you want.

    When I draw a graph, do I have to make some kind of marker line to mark significant points (i.e. a dotted line for a break between two abrupt changes of acceleration on an acceleration-time graph)?
    Not really but it's helpful for both you and the marker so I would recommend it.

    And finally, one of my methods is different from the method they specify in the mark scheme. Not only is it different, but I'm not sure what it's talking about. Would mine still count if it leads to the correct answer and what is their solution mean?
    Yes, certainly. You would get full marks as long as it's correct.

    5. A car accelerates uniformly from rest for 20 seconds. It moves at constant speed v m s-1 for the next 40 seconds and then decelerates uniformly for 10 seconds until it comes to rest.

    (a) For the motion of the car, sketch
    (i) a speed-time graph,
    (ii) an acceleration-time graph

    Given that the total distance moved by the car is 880 m,
    (b) find the value of v.

    (This is my solution for (b))

    area under line in speed-time graph (A) = distance
    A = 880
    A - (v * 40) + (1/2 (20 * v)) + (1/2 (10 * v))
    880 = 40v + 10v + 5v
    880 = 55v
    v = 16

    answer:
    16 ms-1

    (This is their notes on the question)
    ((7 * 40)/2) * v = 880
    v = 880 * (2/110) = 16
    They've done the a same thing - it just looks like they spotted that they could find the area under the curve in a much easier way than you did (looks like they noted the shape under the curve was a triangle?).

    You'd get full marks; do not that the markscheme tends to be less verbose than an actual candidate answer.
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    (Original post by Zacken)
    Depends on the question, if it's a question where you have used the approximation g=9.8 then you must give your answer to 2-3 s.f because otherwise you are claiming your answer is more accurate that it could possibly be given the nature of your approximation of g = 9.8 to s.f

    If you do a question where you don't use such an approximation, then it's fine for you to give an exact answer or an answer to however many s.f you want, but the best practice is to give it to 3 s.f.



    Not at all, that's fine. Put it in whenever you want.



    Not really but it's helpful for both you and the marker so I would recommend it.



    Yes, certainly. You would get full marks as long as it's correct.



    They've done the a same thing - it just looks like they spotted that they could find the area under the curve in a much easier way than you did (looks like they noted the shape under the curve was a triangle?).

    You'd get full marks; do not that the markscheme tends to be less verbose than an actual candidate answer.
    Thank you very much!
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    (Original post by fablereader)
    Thank you very much!
    No worries!
 
 
 
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