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# Really struggling with M1 watch

1. Hello I have M1 edexcel in a few weeks and haven't done it for ages. I seem to be getting confused on the same concepts as I did last year. I got a C in M1 last year so this year really want to get 80+ Need an A overall in maths for uni

Some of the things I am confused about
Which way do I label Tension? I seem to just memorise it rather than understand it.
In vector questions, to work out when Ship P is due North West of Q, what is the procedure to work out these questions? I know the magnitude is parallel but what do I equate what to?
There are a few more things probably but these are what come in mind for now!thanks
2. (Original post by frozo123)
Hello I have M1 edexcel in a few weeks and haven't done it for ages. I seem to be getting confused on the same concepts as I did last year. I got a C in M1 last year so this year really want to get 80+ Need an A overall in maths for uni

Some of the things I am confused about
Which way do I label Tension? I seem to just memorise it rather than understand it.
In vector questions, to work out when Ship P is due North West of Q, what is the procedure to work out these questions? I know the magnitude is parallel but what do I equate what to?
There are a few more things probably but these are what come in mind for now!thanks
For tension always remember that tension can only ever pull an object, which should allow you to work out the directions of tensions. I'll elaborate a bit on your other question in the morning when I wake up (sorry I'm very tired right now) so if you have any others post them as well and I'll have a look for you

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3. (Original post by samb1234)
For tension always remember that tension can only ever pull an object, which should allow you to work out the directions of tensions. I'll elaborate a bit on your other question in the morning when I wake up (sorry I'm very tired right now) so if you have any others post them as well and I'll have a look for you

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https://2802a3b1a650824d2586fd3336bd...%20Edexcel.pdf
Last part of Question 6 and 5d) I thought the magnitude of the position vector would give you the distance?
And yeah idk how to do those parallel questions 😩

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4. Well, assuming you're talking about (c) Find the time when(i) Q is due north of P,(ii) Q is north-west of P.

i) When Q is due north of P, draw a diagram. Q is directly above P. Therefore, their i components (x on a graph) can be equalled to each other and solved for t using the expressions you worked out in b.

ii) When Q is north-west of P the bearing of vector PQ will be 315 degrees. So, tan(x) = 45

x = 1

therefore vertical component of PQ/horizontal component of PQ = 1. Solve for t.
5. (Original post by Mattematics)
Well, assuming you're talking about (c) Find the time when(i) Q is due north of P,(ii) Q is north-west of P.

i) When Q is due north of P, draw a diagram. Q is directly above P. Therefore, their i components (x on a graph) can be equalled to each other and solved for t using the expressions you worked out in b.

ii) When Q is north-west of P the bearing of vector PQ will be 315 degrees. So, tan(x) = 45

x = 1

therefore vertical component of PQ/horizontal component of PQ = 1. Solve for t.
Sorry don't understand what you did after the bearing bit? could I use a magnitude ( mu) And then do mu( -I +j) instead?

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6. (Original post by frozo123)
Sorry don't understand what you did after the bearing bit? could I use a magnitude ( mu) And then do mu( -I +j) instead?

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Sorry, could have been clearer about that. If one is north west of the other, their vertical and horizontal components have to be the same. That's because the angle when they're north west is 45 degrees, and to calculate the angle you do tan^-1(vertical/horizontal)

We can say that as its 45 degrees, it's going to be tan45 which equals one.

Therefore, vertical/horizontal of the vector PQ = 1

So vertical = horizontal, or in other words

-1 + t (horizontal component of PQ) = -3 + 7t (vertical component).
7. (Original post by Mattematics)
Sorry, could have been clearer about that. If one is north west of the other, their vertical and horizontal components have to be the same. That's because the angle when they're north west is 45 degrees, and to calculate the angle you do tan^-1(vertical/horizontal)

We can say that as its 45 degrees, it's going to be tan45 which equals one.

Therefore, vertical/horizontal of the vector PQ = 1

So vertical = horizontal, or in other words

-1 + t (horizontal component of PQ) = -3 + 7t (vertical component).
Alright thankyou I'll have another go at similar questions tomorrow, what would you do if it asked to worked out when it's travelling parallel to (i-3j)?

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8. (Original post by frozo123)
Alright thankyou I'll have another go at similar questions tomorrow, what would you do if it asked to worked out when it's travelling parallel to (i-3j)?

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If they're travelling parallel to each other their velocities will be multiples of each other. Divide the velocities by each other to get the ratio.
9. So if ship P was: r= ( 4i + 10j) + ( 2i- 6j)t
I just do (2i-6j)/(i-3j)?

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10. (Original post by frozo123)
So if ship P was: r= ( 4i + 10j) + ( 2i- 6j)t
I just do (2i-6j)/(i-3j)?

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Firstly you should really sort your notation out. Much easier to write (4+2t)i +(10-6t)j. If it is parallel it is a multiple of the other one so you can set up a simultaneous equation with the coefficients: 4+2t = n(1) and 10-6t = n(-3) you can then solve for t

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11. (Original post by samb1234)
Firstly you should really sort your notation out. Much easier to write (4+2t)i +(10-6t)j. If it is parallel it is a multiple of the other one so you can set up a simultaneous equation with the coefficients: 4+2t = n(1) and 10-6t = n(-3) you can then solve for t

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However this method wouldn't work if one ship is NE of another ship, would you have to find the the vector between the ships and then equate it to n(i + j) in this case?
12. (Original post by frozo123)
However this method wouldn't work if one ship is NE of another ship, would you have to find the the vector between the ships and then equate it to n(i + j) in this case?
No that method is for when its velocity vector is parallel to something. For position vectors you need to look at the I and j the compass point refers to. Eg north west could be represented by -i +j. You would then have to find the position such that the change in i between the two points = - the distance in j between the two points

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13. (Original post by samb1234)
No that method is for when its velocity vector is parallel to something. For position vectors you need to look at the I and j the compass point refers to. Eg north west could be represented by -i +j. You would then have to find the position such that the change in i between the two points = - the distance in j between the two points

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May you tell me a question you've seen on a past paper please so I can understand it with a context?
14. (Original post by frozo123)
May you tell me a question you've seen on a past paper please so I can understand it with a context?
Do you want me to do a question?

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15. (Original post by frozo123)
May you tell me a question you've seen on a past paper please so I can understand it with a context?
OK look at January 2013 q6 try it and I'll do a full solution

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16. (Original post by samb1234)
OK look at January 2013 q6 try it and I'll do a full solution

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okay I'll have a go at it, thankyou!!
17. (Original post by frozo123)
okay I'll have a go at it, thankyou!!
Let me know when you're done/need help

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18. (Original post by frozo123)
okay I'll have a go at it, thankyou!!
How's it going?

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19. (Original post by samb1234)
How's it going?

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yeah I couldn't do it I just think I don't understand vectors in M1
couldn't do the 5 mark one at the end
20. (Original post by frozo123)
yeah I couldn't do it I just think I don't understand vectors in M1
couldn't do the 5 mark one at the end
OK before I give you the full solution I'll try and lead you through it as it will be more useful than me just telling you how to do it. What have you done on the last question so far?

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