Additional Math Help HERE ONLY...!

[vanibharutham]

At 0900 hours a ship sails from the point P with position vector (2i + 3j) km relative to an origin O.
The ship sails north-east with a speed of 15(2)^0.5 km h–1.

(i) Find, in terms of i and j, the velocity of the ship.

Ok, we need to be able to find a vector (xi + yj) whose magnitude is 15(2)^0.5 .
From the question we are given the ship sails north-east i.e. on a bearing of 45 degrees...hence giving us an isoceles triangle of 90,45,45. Because it is isoceles, we know that xi = yj
Therefore, just xi = xj

x² + x² = (15(2)^0.5 )²
x = 15

Therefore velcity of ship is 15i + 15j

(ii) Show that the ship will be at the point with position vector (24.5i + 25.5j) km at 1030 hours.

We are given that "At 0900 hours a ship sails from the point P with position vector (2i + 3j) km ". And we have also worked out that the ship is a velocity of (15i + 15j) km/h
In order to find the positiion of ship at 10:30
we know 10:30 is 1.5 hours after 9:00

Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
FINAL POSITION = (2i + 3j) + (15i + 15j)1.5
=> 2i + 3j + 22.5i + 22.5j
=> 24.5i + 25.5j

(iii) Find, in terms of i, j and t, the position of the ship t hours after leaving P.

Using the same formula as in the last part
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
Therefore in terms of i,j, and t

=> (2i + 3j) + (15i + 15j)t
=> 2i + 15ti + 3j + 15tj
=> (2+15t)i + (3+15t)j

At the same time as the ship leaves P, a submarine leaves the point Q with position vector
(47i – 27j) km. The submarine proceeds with a speed of 25 km h–1 due north to meet the ship.

(iv) Find, in terms of i and j, the velocity of the ship relative to the submarine.

The submarine travels due north... therefore it has a velocity of a (25j)km/h
The velocity of ship relative to submarine
=> 15i + 15j - 25j
=> 15i - 10j

(v) Find the position vector of the point where the submarine meets the ship.

In order for the ship and their submarine to intersect, their final positions must equal
Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
INITIAL POSITION + (VELOCITY * TIME) =  INITIAL POSITION + (VELOCITY * TIME)
(2+15t)i + (3+15t)j = (47i - 27j) + (25j)t
(2+15t)i + (3+15t)j = (47)i + (25t - 27)j

Take the i's sepeartely first

2 + 15t = 47
15t = 45
t = 3

According to this they should meet when t = 3 hours

We have to find it for j as well, and if j equals to 3, we know it will intercept

3+15t = 25t - 27
30 = 10t
t = 3

Yes therefore they meet when t = 3 hours

To find the position vector at t = 3 hours, simply substitute
(2+15t)i + (3+15t)j
(2 + 45)i + (3 + 45)j
=> 47i + 48j

[me...myself...n...i]

i attached the paper. question 6.....can anyone help me....

[vanibharutham]

same type of question as the other... you need to be able to break down the big question into small usable bits...

The diagram shows a large rectangular television screen in which one corner is taken as the origin O
and i and j are unit vectors along two of the edges. In a game, an alien spacecraft appears at the point
A with position vector 12j cm and moves across the screen with velocity (40i + 15j) cm per second. A
player fires a missile from a point B; the missile is fired 0.5 seconds after the spacecraft appears on the
screen. The point B has position vector 46i cm and the velocity of the missile is (ki +30j) cm per second,
where k is a constant. Given that the missile hits the spacecraft,

(i) show that the spacecraft moved across the screen for 1.8 seconds before impact,
(ii) find the value of k.

i) First lets identify our position vectors for each of our two things on the screen after t seconds....

First the Alien Spacecraft:
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
=>  (12j) +  (40i + 15j)(t)
=> (40t)i + (12 + 15t)j

Then the missile,
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
=>  (46i) +  (ki +30j)(t-0.5)
=> (46 + k(t - 0.5))i + (30(t-0.5))j
t has to be t - 0.5 here because it left 0.5 seconds after the alien spacecraft, so it the alien spacecraft left at t, then the missile leaves at t - 0.5

Now make your positions equal each other

(40t)i + (12 + 15t)j = (46 + k(t - 0.5))i + (30(t-0.5))j

Take the i's and j's sepeartely

therefore,
12 + 15t = (30(t-0.5))
12 + 15t = 30t - 15
27 = 15t
t = 1.8

ii) (40t) = (46 + k(t - 0.5))
We know that t = 1.8 from last question
Therefore,

(40*1.8) = 46 + k(1.8-0.5)
72 = 46 + 1.3k
1.3k = 26
k = 20


hope that helps

[me...myself...n...i]

Thanks......

[archangel]

At 0900 hours a ship sails from the point P with position vector (2i + 3j) km relative to an origin O.
The ship sails north-east with a speed of 15(2)^0.5 km h–1.

(i) Find, in terms of i and j, the velocity of the ship.

Ok, we need to be able to find a vector (xi + yj) whose magnitude is 15(2)^0.5 .
From the question we are given the ship sails north-east i.e. on a bearing of 45 degrees...hence giving us an isoceles triangle of 90,45,45. Because it is isoceles, we know that xi = yj
Therefore, just xi = xj

x² + x² = (15(2)^0.5 )²
x = 15

Therefore velcity of ship is 15i + 15j

(ii) Show that the ship will be at the point with position vector (24.5i + 25.5j) km at 1030 hours.

We are given that "At 0900 hours a ship sails from the point P with position vector (2i + 3j) km ". And we have also worked out that the ship is a velocity of (15i + 15j) km/h
In order to find the positiion of ship at 10:30
we know 10:30 is 1.5 hours after 9:00

Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
FINAL POSITION = (2i + 3j) + (15i + 15j)1.5
=> 2i + 3j + 22.5i + 22.5j
=> 24.5i + 25.5j

(iii) Find, in terms of i, j and t, the position of the ship t hours after leaving P.

Using the same formula as in the last part
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
Therefore in terms of i,j, and t

=> (2i + 3j) + (15i + 15j)t
=> 2i + 15ti + 3j + 15tj
=> (2+15t)i + (3+15t)j

At the same time as the ship leaves P, a submarine leaves the point Q with position vector
(47i – 27j) km. The submarine proceeds with a speed of 25 km h–1 due north to meet the ship.

(iv) Find, in terms of i and j, the velocity of the ship relative to the submarine.

The submarine travels due north... therefore it has a velocity of a (25j)km/h
The velocity of ship relative to submarine
=> 15i + 15j - 25j
=> 15i - 10j

(v) Find the position vector of the point where the submarine meets the ship.

In order for the ship and their submarine to intersect, their final positions must equal
Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
INITIAL POSITION + (VELOCITY * TIME) =  INITIAL POSITION + (VELOCITY * TIME)
(2+15t)i + (3+15t)j = (47i - 27j) + (25j)t
(2+15t)i + (3+15t)j = (47)i + (25t - 27)j

Take the i's sepeartely first

2 + 15t = 47
15t = 45
t = 3

According to this they should meet when t = 3 hours

We have to find it for j as well, and if j equals to 3, we know it will intercept

3+15t = 25t - 27
30 = 10t
t = 3

Yes therefore they meet when t = 3 hours

To find the position vector at t = 3 hours, simply substitute
(2+15t)i + (3+15t)j
(2 + 45)i + (3 + 45)j
=> 47i + 48j

whoa. this looks freakishly familiar...
does it ring a bell in anyone? 

[ineedaz]

At 0900 hours a ship sails from the point P with position vector (2i + 3j) km relative to an origin O.
The ship sails north-east with a speed of 15(2)^0.5 km h–1.

(i) Find, in terms of i and j, the velocity of the ship.

Ok, we need to be able to find a vector (xi + yj) whose magnitude is 15(2)^0.5 .
From the question we are given the ship sails north-east i.e. on a bearing of 45 degrees...hence giving us an isoceles triangle of 90,45,45. Because it is isoceles, we know that xi = yj
Therefore, just xi = xj

x² + x² = (15(2)^0.5 )²
x = 15

Therefore velcity of ship is 15i + 15j

(ii) Show that the ship will be at the point with position vector (24.5i + 25.5j) km at 1030 hours.

We are given that "At 0900 hours a ship sails from the point P with position vector (2i + 3j) km ". And we have also worked out that the ship is a velocity of (15i + 15j) km/h
In order to find the positiion of ship at 10:30
we know 10:30 is 1.5 hours after 9:00

Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
FINAL POSITION = (2i + 3j) + (15i + 15j)1.5
=> 2i + 3j + 22.5i + 22.5j
=> 24.5i + 25.5j

(iii) Find, in terms of i, j and t, the position of the ship t hours after leaving P.

Using the same formula as in the last part
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
Therefore in terms of i,j, and t

=> (2i + 3j) + (15i + 15j)t
=> 2i + 15ti + 3j + 15tj
=> (2+15t)i + (3+15t)j

At the same time as the ship leaves P, a submarine leaves the point Q with position vector
(47i – 27j) km. The submarine proceeds with a speed of 25 km h–1 due north to meet the ship.

(iv) Find, in terms of i and j, the velocity of the ship relative to the submarine.

The submarine travels due north... therefore it has a velocity of a (25j)km/h
The velocity of ship relative to submarine
=> 15i + 15j - 25j
=> 15i - 10j

(v) Find the position vector of the point where the submarine meets the ship.

In order for the ship and their submarine to intersect, their final positions must equal
Therefore
FINAL POSITION = INITIAL POSITION + (VELOCITY * TIME)
INITIAL POSITION + (VELOCITY * TIME) =  INITIAL POSITION + (VELOCITY * TIME)
(2+15t)i + (3+15t)j = (47i - 27j) + (25j)t
(2+15t)i + (3+15t)j = (47)i + (25t - 27)j

Take the i's sepeartely first

2 + 15t = 47
15t = 45
t = 3

According to this they should meet when t = 3 hours

We have to find it for j as well, and if j equals to 3, we know it will intercept

3+15t = 25t - 27
30 = 10t
t = 3

Yes therefore they meet when t = 3 hours

To find the position vector at t = 3 hours, simply substitute
(2+15t)i + (3+15t)j
(2 + 45)i + (3 + 45)j
=> 47i + 48j

whoa. this looks freakishly familiar...
does it ring a bell in anyone? 

indeed!!! 

[vanibharutham]

you guys owe me lol

[Anonymous]

What answer did you get for the vector q?

[vanibharutham]

cant remember exactly, but i think they interecepted at 13:30

[ineedaz]

yea i got 1330 aswell... thnk the vector was 30 something i and 40 somethng j
SIMPLEST paper ive done so far... hope paper 2 stays that way...
PREDICTIONS FOR PAPER 2 anyone??
- permutations and combinations
- matrices
- sets
- functions
- more calculus (DUHH!!) perhaps approximate change, more area, probably another kinematics question(which will not be optional),
- simple vectors
- trigonometry equations
- graph sketching
- inequalities
- binomial expansions
don't forget to study what came in paper 1 cause they may bring it again!!! for instance, i think they might bring circular measure again cause the one in paper 1 was tooooo easy!!

[sweetsh]

I think it will be easier than Paper 1 right?

[Anonymous]

was the relative velocity like 2i + 4j?

And was the position vector for part ii) 16i + 28j?

[ineedaz]

was the relative velocity like 2i + 4j?

And was the position vector for part ii) 16i + 28j?

yup thats what i got!!

[ineedaz]

I think it will be easier than Paper 1 right?

let us hope so... but i wouldn't bet on it!! 

[sweetsh]

Paper 1 got some hard questions, the ones you predicted are pretty easy.