Really struggling with M1

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

Posted from TSR Mobile

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?


Posted from TSR Mobile

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)?


Posted from TSR Mobile

So if ship P was: r= ( 4i + 10j) + ( 2i- 6j)t
I just do (2i-6j)/(i-3j)?
What about if it asks me find the time when?


Posted from TSR Mobile

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


Posted from TSR Mobile

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

Posted from TSR Mobile

May you tell me a question you've seen on a past paper please so I can understand it with a context?

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

Posted from TSR Mobile

okay I'll have a go at it, thankyou!!

okay I'll have a go at it, thankyou!!

How's it going?

Posted from TSR Mobile

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