IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => IGCSE/ GCSE => Math => Topic started by: Ghost Of Highbury on June 30, 2009, 12:24:34 pm

Title: Add math doubt...astar plz help
Post by: Ghost Of Highbury on June 30, 2009, 12:24:34 pm: m/j 2008 paper 1 additional math
Q 10
Title: Re: Add math doubt...astar plz help
Post by: astarmathsandphysics on July 01, 2009, 10:44:17 am: I cant do this until i get to a pc. Someone else can have a go.
Title: Re: Add math doubt...astar plz help
Post by: astarmathsandphysics on July 02, 2009, 10:51:01 am: I tried to download the paper to answer the question but it may be corrupted.
Title: Re: Add math doubt...astar plz help
Post by: Ghost Of Highbury on July 02, 2009, 11:38:44 am: find it attached
plzz help
i'm really stuck
Title: Re: Add math doubt...astar plz help
Post by: Ghost Of Highbury on July 02, 2009, 06:58:25 pm: astar u dere
sir plzz help
i uploaded the question too..
Title: Re: Add math doubt...astar plz help
Post by: Theguitarist on July 02, 2009, 08:11:41 pm: Ok, I will answer this question.

(i) If u look closely it says that the ship sails 'north east'. hence its unit vector north east will be (i+j)/ square root of 2. And since velocity has both speed and direction, its velocity = 15* square root of 2 * the unit vector which = 15i + 15j

(ii) Since we have established that the velocity in the stated direction is 15i + 15j, the distance it has traveled at 10:30 can be found by multiplying 1.5 into the velocity. which = 22.5i + 22.5j. But since they have asked for the position vector, it will be equal to 22.5i + 22.5j + 2i + 3j = 24.5i + 25.5j

(iii) Similarly for part three; the distance travelled t hours after leaving = (15t)i + (15t)j. Hence in terms of 't' the position will be = (15t + 2)i + (15t + 3)j

(iv) In this subquestion they have simply asked for the velocity of the ship relative to the submarine (Vssu) = Velocity of ship - Velocity of submarine =
(15i + 15j) - (25j). NOTE: 25j is because it is only travelling north. = 15i -10j

Answer to final part in the next post
Title: Re: Add math doubt...astar plz help
Post by: Theguitarist on July 02, 2009, 08:20:18 pm: Assuming t = the time when the two bodies meet.

After 't' hours the position vector of the ship = (2 + 15t)i + (3+ 15t)j

After 't' hours the position vector of the submarine = 47i + (-27+ 25t)j.

At their point of collision, the position vectors should be the same. Hence to find 't' we equate: 47 = 2 + 15t. Therefore, t = 3 hours

Hence final position vector can be found by substituting t in any one of the position vectors which should be 47i + 48j.

Hope that helped!
Title: Re: Add math doubt...astar plz help
Post by: astarmathsandphysics on July 02, 2009, 09:07:46 pm: I will try it tomorrow. I cant download it to my phone and open it.
Title: Re: Add math doubt...astar plz help
Post by: astarmathsandphysics on July 03, 2009, 07:34:38 am: Thamks guitarman
Title: Re: Add math doubt...astar plz help
Post by: Ghost Of Highbury on July 03, 2009, 11:41:42 am: why will it be i+j/root 2
didnt get the first part itself?
Title: Re: Add math doubt...astar plz help
Post by: astarmathsandphysics on July 03, 2009, 01:53:56 pm: Resolving vertically: 1cos45=1/ $sqrt(2)$ ie j/ $sqrt(2)$

resolving horizontally 1cos45=1/ $sqrt(2)$ ie i/ $sqrt(2)$

unit vector is i/ $sqrt(2)$ + j/ $sqrt(2)$ =(i+j)/ $sqrt(2)$
Title: Re: Add math doubt...astar plz help
Post by: Ghost Of Highbury on July 03, 2009, 04:17:33 pm: could u plz provide me with a digram regarding its movement
does it move from O to P and then further parllel to the x-axis

wht does relative to the orgin mean?
Title: Re: Add math doubt...astar plz help
Post by: Ghost Of Highbury on July 03, 2009, 04:19:16 pm: Quote from: astarmathsandphysics on July 03, 2009, 01:53:56 pm
Resolving vertically: 1cos45=1/ $sqrt(2)$ ie j/ $sqrt(2)$

resolving horizontally 1cos45=1/ $sqrt(2)$ ie i/ $sqrt(2)$

unit vector is i/ $sqrt(2)$ + j/ $sqrt(2)$ =(i+j)/ $sqrt(2)$
which rule have u applied
how come cos 45
where is 45 degree
plz explain clearly
i am totlly blank abt this question
Title: Re: Add math doubt...astar plz help
Post by: Theguitarist on July 04, 2009, 07:51:53 pm: i think astar explained it in a slightly complicated manner but u get (i+j)/ root 2 by simply finding the unit vector :P which is the vector (i+j) divided by its magnitude which is root 2 (found by pythagoras of '1')
Title: Re: Add math doubt...astar plz help
Post by: Ghost Of Highbury on July 07, 2009, 07:36:12 pm: what are the steps to find a unit vector??
Title: Re: Add math doubt...astar plz help
Post by: Theguitarist on July 08, 2009, 07:12:34 am: the definition of a unit vector is essentially one which has a magnitude of 1. so if i had a vector i+j, the unit vector would be (i+j) / root 2. Root 2 is obtained by doing a square root of 1 squared + 1 squared (because it is 1i+1j i.e i+j). so it is basically the vector itself divided by its magnitude.
Title: Re: Add math doubt...astar plz help
Post by: Ghost Of Highbury on July 09, 2009, 03:48:48 pm: hey thanks the guitarist
appreciated
Title: Re: Add math doubt...astar plz help
Post by: astarmathsandphysics on August 28, 2009, 10:44:14 pm: A(7,2) C(1,4) are two vertices of a square ABCD
(a)find the equation of the diagonol BD
(b) Find the coordinates of B and of D

a) gradient of AC=m=(4-2)/(1-7)=-1/3 Now use y-y₁=m(x-x₁)

y-2=-1/3(x-7) so y=-1/3x+13/3

AC=(-6,2) and BD is at right angles so BD=(2.6) so from centre to B=(1,3) and to D=(-1,-3). Ccentre is at midpoint of AC=1/2(7+1,2+4)=(4,3) so B=(1,3)+(4,3)=(5,6) and C=(-1,-3)+(4,3)=(3,0)