I would suggest that
However,
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 midpoint 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 ad (2q, p + q, q, 2q  p respectively), but I've stalled on e. Can anyone help?

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 27032016 12:17

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 27032016 12:48

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 27032016 23:11
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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 30032016 20:15
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 midpoints of OA, OB and OC respectively. P, Q and R are the midpoints 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.'Last edited by fablereader; 30032016 at 22:00. 
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 31032016 08:50
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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 31032016 09:23
(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? 
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 31032016 10:21
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.Last edited by depymak; 31032016 at 10:23. 
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 17042016 20:03
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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 17042016 20:20
(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 90theta. 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(90theta) = theta.
Now, if I want to resolve perpendicular to the surface, I get: acting perpendicular "away" from the surface and 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 . Rearranging this gets me . 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 . 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.
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 .
Was that any clearer?Last edited by Zacken; 17042016 at 20:31. 
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 17042016 20:31

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 17042016 20:34
(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?
Now resolving the blue line into the component along the red line gets us . 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 .Last edited by Zacken; 17042016 at 20:43. 
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 17042016 20:38
(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?
And I say that the blue line is and the red line is , how would you find the component of acting in the direction of ? 
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 17042016 20:43
(Original post by Zacken)
If I gave you this diagram:
And I say that the blue line is and the red line is , how would you find the component of acting in the direction of ?
Form a rightangle 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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 17042016 20:52
(Original post by fablereader)
I'm still not understanding this (particularly the 'acting in direction of' part) but this is my gut reaction 
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 17042016 21:17

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 17042016 21:19
(Original post by fablereader)
OK, I think I get it now. Thanks! 
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 05052016 21:55
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 23 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 accelerationtime 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 speedtime graph,
(ii) an accelerationtime 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 speedtime 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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 05052016 22:22
(Original post by fablereader)
How necessary is it for my answer to be to 23 significant figures? One of my answers was 117.6 (exact answer), but they seem to specifically penalize that.
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?
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 accelerationtime 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 speedtime graph,
(ii) an accelerationtime 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 speedtime 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
You'd get full marks; do not that the markscheme tends to be less verbose than an actual candidate answer. 
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 05052016 22:32
(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 23 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. 
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 05052016 22:36
(Original post by fablereader)
Thank you very much!
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