An Add Maths Vector Question

"In this question, i is a unit vector due east and j is a unit vector due north. 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 km h^–1.

i) Find, in terms of i and j, the velocity of the ship.     [2]

ii) Show that the ship will be at the point with position vector (24.5i + 25.5j) km at 1030 hours.       [1]

iii) Find, in terms of i, j and t, the position of the ship t hours after leaving P.                [2]

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.                  [2]

v) Find the position vector of the point where the submarine meets the ship.                [2]

                  CIE ADD MATHS summer 2008 paper 1 question 10"

The velocity of the ship:                                                                                        

North-east has a compass bearing (angle with the y-axis or North axis) of 045°. Therefore the angle with the x-axis (i.e., East axis) is also 45° and the velocity vector forms an isosceles right triangle.  Recall that the

 sine and cosine of 45° are both 1/√2. Keep this exact value: do not use the approximation 0.707.

Since ship and sub both start with ‘s’, I have somewhat arbitrarily decided to use for the names of the boats their respective starting points in the equations which follow. ‘P’ subscripts denote the ship; ‘Q’ subscripts denote the submarine.

V_P = 15 √2 (1/√2) i +  15 √2 (1/√2) j

V_P = 15  i +  15 j

The position of the ship t hours after leaving P:                                      

Displacement of the ship from origin, O  = ship’s original position + velocity * time

D_P = P +t v_P

D_P = (2 i + 3 j) + (15 t i + 15 t j)

D_P = (2 + 15 t) i + (3 + 15 t) j

The  ship will be at the point (24.5i + 25.5j) km at 1030 hours      

Note that I did part iii) first and then returned to part ii) as a special case.

At 1030, one and a half hours have elapsed since 0900, so substitute t = 1.5 into the equation above.

D_P = (2 + 22.5) i + (3 + 22.5) j

D_P =  24.5 i + 25.5 j

The velocity of the ship relative to the submarine.                                            

The velocity of P relative to Q is

v_P – v_Q = (15  i + 15  j) – (0 i + 25 j)

= 15  i – 10 j

The position vector of the point where the submarine meets the ship.                                                                                                                                                                                                               

For the ship to meet the sub, the two position vectors (or using the vocabulary of physics, their displacements) must be equal for the same value of time, t.

D_P = D_Q                         

When vectors are equal, all their components are equal. Thus, the coefficients of i and of j for both displacements are equal.

Equating the coefficients of i we get:

2 + 15 t  = 47 

15 t = 45                              

t = 3                                    

Likewise, equating the coefficients of j we get:

3 + 15 t = 25 t – 27

30= 10 t

t = 3

As the t values are equal, we can be sure the ship meets the sub at 1200 noon.  I.e., noon is 3 hours after the starting time of 0900. (As an aside; had the solutions produced different values for t, we would have concluded the system was inconsistent, i.e., that the ship and the sub did not meet.)

The ship and sub meet at:

D_P = D_Q =  47 i + (25(3) – 27) j

= 47 i + 48 j

An alternate method:

The velocity of submarine relative to itself is v_Q – v_Q  = 0. ( A bolded zero is a zero vector.)

Hence the relative position of Q remains as 47 i – 27 j.  We can 'pretend' that Q is not moving.  We do this all the time: we pretend the earth is not whizzing through space at thousands of km per second!  For the ship to meet the sub, its relative position must also be 47 i – 27 j.

We can calculate the Relative Position of P (with respect to Q) at time t using:

Relative Position of P  = the ship’s original position + Relative velocity * time

47 i – 27 j = (2 i + 3 j) + (15 t i – 10 t j)

47 i – 2 i – 15 t i = 3 j + 27 j – 10 t j

45 i– 15 t i = 30 j – 10 t j

(45 – 15 t) i = (30 – 10 t) j

Since i and j are in different directions this vector equation can only have the solution  0 = 0.

With scalars, the equation 1x = 1y implies x = y. However vector equations such as1i = 1j or 2i = 2j, declare an impossibility, since perceptible movement East is never equal to any amount of movement North. However we do allow a movement of 0 units East to equal a movement of 0 units North.

We require that (45 – 15 t) i = 0 i .  Hence t = 3.

We also require that (30 – 10 t) j = 0 j .  It is confirmed that t = 3.

(Had we found two different values of t, the system of equations would have been inconsistent-implying no meeting of the ship and sub. As the system is consistent the ship and the sub do meet.)

The actual (not relative) position of ship P at 12 noon is found is found just as we found its position at 1030.

Displacement of the ship P from origin, O  = ship’s original position + velocity * time

D_P = v_P

D_P = (2 i + 3 j )+ (15 t i + 15 t j)

D_P = (2 + 15 (3)) i + (3 + 15 (3)) j

D_P = 47 i + 48 j