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*M1 -- Bearing*
Please see the images attached below for the questions. There are 4 in total (images for 1,2 and 3 only)
(1) From M1 Jan 04
Answer: 031 degrees
(2) From M1 Jun 04
Answer: 117 degrees
(3) From Solomon Paper I
Answer to (c) : 286 degrees
(4) From Edexcel Specimen Paper
7. Two cars A and B are moving on straight horizontal roads with constant velocities. The velocity of A is 20 ms^-1 due east, and the velocity of B is (10i + 10j) ms^-1, where i and j are unit vectors directed due east and due north respectively. Initially A is at the fixed origin O, and the position vector of B is 300i m relative to O. At time t seconds, the position vectors of A and B are r metres and s metres respectively.
(a) Find expressions for r and s in terms of t.
Answer: r = 20ti s = (300 + 10t)i + (10t)j
(b) Hence write down an expression for AB(arrow on top) in terms of t.
Answer: (300-10t)i + (10t)j
(c) Find the time when the bearing of B from A is 045 degrees.
Answer: t = 15
(d) Find the time when the cars are again 300m apart.
Answer: t = 30
Just the bold red bit (part (c)) in this question please...I understand the rest of it.
........................................................................................................
So as you can see above I have a problem with Bearing...lol
...could anyone please explain the concept to me...I know it's something to do with taking the angle from the North but I have never quite understood it properly.
So could someone please explain how you would deal with the above questions and any other bearing question in general in M1 (please add a short explanation with your answers to help me!)...also have you got any good links to just explain the basics of bearing? There isn't much of an explanation in the M1 text book!
I keep looking at markschemes for the above questions and still don't understand them!
Thx in advance!
(1) From M1 Jan 04
Answer: 031 degrees
(2) From M1 Jun 04
Answer: 117 degrees
(3) From Solomon Paper I
Answer to (c) : 286 degrees
(4) From Edexcel Specimen Paper
7. Two cars A and B are moving on straight horizontal roads with constant velocities. The velocity of A is 20 ms^-1 due east, and the velocity of B is (10i + 10j) ms^-1, where i and j are unit vectors directed due east and due north respectively. Initially A is at the fixed origin O, and the position vector of B is 300i m relative to O. At time t seconds, the position vectors of A and B are r metres and s metres respectively.
(a) Find expressions for r and s in terms of t.
Answer: r = 20ti s = (300 + 10t)i + (10t)j
(b) Hence write down an expression for AB(arrow on top) in terms of t.
Answer: (300-10t)i + (10t)j
(c) Find the time when the bearing of B from A is 045 degrees.
Answer: t = 15
(d) Find the time when the cars are again 300m apart.
Answer: t = 30
Just the bold red bit (part (c)) in this question please...I understand the rest of it.
........................................................................................................
So as you can see above I have a problem with Bearing...lol
...could anyone please explain the concept to me...I know it's something to do with taking the angle from the North but I have never quite understood it properly. So could someone please explain how you would deal with the above questions and any other bearing question in general in M1 (please add a short explanation with your answers to help me!)...also have you got any good links to just explain the basics of bearing? There isn't much of an explanation in the M1 text book!
I keep looking at markschemes for the above questions and still don't understand them!Thx in advance!
you question 1.) says...
i vectors go eastward
j vectors go northward
therefore, 000, is a vector with only a positive j component, 090 is a vector with only a positive i component, 180 is a vector with only a negative j component, and 270 is a vector with only a negative i component. this help at all?
i vectors go eastward
j vectors go northward
therefore, 000, is a vector with only a positive j component, 090 is a vector with only a positive i component, 180 is a vector with only a negative j component, and 270 is a vector with only a negative i component. this help at all?
1.) direction vector = velocity
tan(@) = opposite/adjacent = 5/3 = 1.4
tan(@) = 1.4 => @=59 (above horizontal, bearings from vertical implies bearing = 90-@)
bearing = 031
2.) direction vector = (7,-7.5)-(4,-6) => d = (3,-1.5)
tan(@) = opposite/adjacent = -(1.5/3)
tant(@) = -0.5 => @ = 27 (below horizontal, therefore bearing = 90 + @)
bearing = 117
3.) try it yourself, if you cant do, i'll work it out...
generally direction vectors...
(ai + bj) => bearing = 90 - (tan^-1)(b/a)
(ai - bj) => bearing = 90 + (tan^-1)(b/a)
(-ai - bj) => bearing = 270 - (tan^-1)(b/a)
(-ai + bj) => bearing = 270 + (tan^-1)(b/a)
tan(@) = opposite/adjacent = 5/3 = 1.4
tan(@) = 1.4 => @=59 (above horizontal, bearings from vertical implies bearing = 90-@)
bearing = 031
2.) direction vector = (7,-7.5)-(4,-6) => d = (3,-1.5)
tan(@) = opposite/adjacent = -(1.5/3)
tant(@) = -0.5 => @ = 27 (below horizontal, therefore bearing = 90 + @)
bearing = 117
3.) try it yourself, if you cant do, i'll work it out...
generally direction vectors...
(ai + bj) => bearing = 90 - (tan^-1)(b/a)
(ai - bj) => bearing = 90 + (tan^-1)(b/a)
(-ai - bj) => bearing = 270 - (tan^-1)(b/a)
(-ai + bj) => bearing = 270 + (tan^-1)(b/a)
To help you with the vector question on the M1 specimen paper you can download the worked solutions from Examsolutions.co.uk for free! There is also worked solutions available online for M1 June 2004, and M1 Jan 2005.
El Stevo
1.) direction vector = velocity
tan(@) = opposite/adjacent = 5/3 = 1.4
tan(@) = 1.4 => @=59 (above horizontal, bearings from vertical implies bearing = 90-@)
bearing = 031
2.) direction vector = (7,-7.5)-(4,-6) => d = (3,-1.5)
tan(@) = opposite/adjacent = -(1.5/3)
tant(@) = -0.5 => @ = 27 (below horizontal, therefore bearing = 90 + @)
bearing = 117
3.) try it yourself, if you cant do, i'll work it out...
generally direction vectors...
(ai + bj) => bearing = 90 - (tan^-1)(b/a)
(ai - bj) => bearing = 90 + (tan^-1)(b/a)
(-ai - bj) => bearing = 270 - (tan^-1)(b/a)
(-ai + bj) => bearing = 270 + (tan^-1)(b/a)
tan(@) = opposite/adjacent = 5/3 = 1.4
tan(@) = 1.4 => @=59 (above horizontal, bearings from vertical implies bearing = 90-@)
bearing = 031
2.) direction vector = (7,-7.5)-(4,-6) => d = (3,-1.5)
tan(@) = opposite/adjacent = -(1.5/3)
tant(@) = -0.5 => @ = 27 (below horizontal, therefore bearing = 90 + @)
bearing = 117
3.) try it yourself, if you cant do, i'll work it out...
generally direction vectors...
(ai + bj) => bearing = 90 - (tan^-1)(b/a)
(ai - bj) => bearing = 90 + (tan^-1)(b/a)
(-ai - bj) => bearing = 270 - (tan^-1)(b/a)
(-ai + bj) => bearing = 270 + (tan^-1)(b/a)
yes thank you so much for the general solutions...so much easier now...
but i still dont understand this:
El Stevo
you question 1.) says...
i vectors go eastward
j vectors go northward
therefore, 000, is a vector with only a positive j component, 090 is a vector with only a positive i component, 180 is a vector with only a negative j component, and 270 is a vector with only a negative i component.
i vectors go eastward
j vectors go northward
therefore, 000, is a vector with only a positive j component, 090 is a vector with only a positive i component, 180 is a vector with only a negative j component, and 270 is a vector with only a negative i component.
and I really dont understand (4)...like i said i looked at the markscheme and for (4) they just made the i component = j component...I can't understand why!
thx for ur help so far guys...really appreciate it...
The i and j are equal, because i something is at a 45 degree angle, opp/adj is equal to one (do tan 45 on a calc). If something equals 1, it has to be the same number (i.e. 2/2=1, 7/7=1). therefore, what's going across(i component, which is also adj) equals what's going up (j component, which is also opp).
Hope this helps..
Hope this helps..
Queen_A
The i and j are equal, because i something is at a 45 degree angle, opp/adj is equal to one (do tan 45 on a calc). If something equals 1, it has to be the same number (i.e. 2/2=1, 7/7=1). therefore, what's going across(i component, which is also adj) equals what's going up (j component, which is also opp).
Hope this helps..
Hope this helps..
ahh ok...
what about when it is 90 degrees from it? and 180? 270? etc etc
If it was 90 degrees from A dat wuld be parallel to the j vector, so the i vector would be equal to 0. I doubt they'l ask anything other than acute angles.
El Stevo
1.) direction vector = velocity
tan(@) = opposite/adjacent = 5/3 = 1.4
tan(@) = 1.4 => @=59 (above horizontal, bearings from vertical implies bearing = 90-@)
bearing = 031
2.) direction vector = (7,-7.5)-(4,-6) => d = (3,-1.5)
tan(@) = opposite/adjacent = -(1.5/3)
tant(@) = -0.5 => @ = 27 (below horizontal, therefore bearing = 90 + @)
bearing = 117
3.) try it yourself, if you cant do, i'll work it out...
generally direction vectors...
(ai + bj) => bearing = 90 - (tan^-1)(b/a)
(ai - bj) => bearing = 90 + (tan^-1)(b/a)
(-ai - bj) => bearing = 270 - (tan^-1)(b/a)
(-ai + bj) => bearing = 270 + (tan^-1)(b/a)
tan(@) = opposite/adjacent = 5/3 = 1.4
tan(@) = 1.4 => @=59 (above horizontal, bearings from vertical implies bearing = 90-@)
bearing = 031
2.) direction vector = (7,-7.5)-(4,-6) => d = (3,-1.5)
tan(@) = opposite/adjacent = -(1.5/3)
tant(@) = -0.5 => @ = 27 (below horizontal, therefore bearing = 90 + @)
bearing = 117
3.) try it yourself, if you cant do, i'll work it out...
generally direction vectors...
(ai + bj) => bearing = 90 - (tan^-1)(b/a)
(ai - bj) => bearing = 90 + (tan^-1)(b/a)
(-ai - bj) => bearing = 270 - (tan^-1)(b/a)
(-ai + bj) => bearing = 270 + (tan^-1)(b/a)
thx stevo those general solutions work well...
but i still dont quite understand bearing...for example when u worked out 1) and 2) how did you know it was above/below horizontal...and then how did you know whether to add/subtract 90 degrees etc...
can anyone please explain this...or if you know a website which explains bearing well...i can't seem to understand it at all!
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