Two ships, P and Q, are moving with constant velocities. The velocity of P is (9i - 2j) km h⁻¹ and the velocity of Q is (4i + 8j) km h⁻¹. (a) Find the direction of motion of P, giving your answer as a bearing to the nearest degree. When t = 0, the position vector of P is (9i + 10j) km and the position vector of Q is (i + 4j) km. At time t hours, the position vectors of P and Q are p km and q km respectively. (b) Find an expression for (i) p in terms of t, (ii) q in terms of t. (c) Hence show that, at time t hours, (d) Find the values of t when the ships are 10 km apart.

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Two ships, P and Q, are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2017 - Paper 1

Question 7

Two ships, P and Q, are moving with constant velocities. The velocity of P is (9i - 2j) km h⁻¹ and the velocity of Q is (4i + 8j) km h⁻¹. (a) Find the direction of ... show full transcript

Worked Solution & Example Answer:Two ships, P and Q, are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2017 - Paper 1

Step 1

Find the direction of motion of P, giving your answer as a bearing to the nearest degree.

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Answer

To find the direction of motion of ship P, we start with the velocity vector of P, which is given as

$ext{v}_P = 9 extbf{i} - 2 extbf{j}$ .

The direction can be represented by the angle θ, where:

$an heta = rac{v_y}{v_x} = rac{-2}{9}$ .

Calculating the angle:

ightarrow heta ext{ is approximately } -12.5 ^ ext{o}.$$ Since this angle is negative, we need to convert it to a bearing. The bearing is given by: $$ ext{Bearing} = 360^ ext{o} + heta = 360^ ext{o} - 12.5^ ext{o} = 347.5^ ext{o}$$. To the nearest degree, the bearing is 348°.

Step 2

Find an expression for p in terms of t.

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Answer

The position vector of ship P at time t is given by:

$ext{p} = ext{initial position} + ext{velocity} imes ext{time}$ .

Substituting the values, we get:

$ext{p} = (9 extbf{i} + 10 extbf{j}) + (9 extbf{i} - 2 extbf{j}) t$ .

This simplifies to:

$ext{p} = (9 + 9t) extbf{i} + (10 - 2t) extbf{j}.$

Step 3

Find an expression for q in terms of t.

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Answer

The position vector of ship Q at time t is found similarly:

$ext{q} = ext{initial position} + ext{velocity} imes ext{time}$ .

This leads to:

$ext{q} = (1 extbf{i} + 4 extbf{j}) + (4 extbf{i} + 8 extbf{j}) t$ .

Simplifying gives:

$ext{q} = (1 + 4t) extbf{i} + (4 + 8t) extbf{j}.$

Step 4

Hence show that, at time t hours, QP = (8 + 5t) i + (6 - 10t) j.

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Answer

To find QP, we calculate:

$ext{QP} = ext{p} - ext{q} = ig((9 + 9t) - (1 + 4t)ig) extbf{i} + ig((10 - 2t) - (4 + 8t)ig) extbf{j}.$

The i component becomes:

$(9 + 9t) - (1 + 4t) = 8 + 5t,$

and the j component becomes:

$(10 - 2t) - (4 + 8t) = 6 - 10t.$

Hence, we have:

$ext{QP} = (8 + 5t) extbf{i} + (6 - 10t) extbf{j}.$

Step 5

Find the values of t when the ships are 10 km apart.

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Answer

The distance between the ships is given by:

$D^2 = ext{QP} ullet ext{QP},$

which leads to:

$D^2 = (8 + 5t)^2 + (6 - 10t)^2.$

Setting this equal to $10^2$ (since they are 10 km apart):

$D^2 = 100.$

Expanding:

$(8 + 5t)^2 + (6 - 10t)^2 = 100,$

which simplifies to:

$64 + 80t + 25t^2 + 36 - 120t + 100t^2 = 100,$

yielding:

$125t^2 - 40t + 0 = 0.$

Using the quadratic formula gives:

$t = rac{40 extbf{\pm} ext{sqrt}(1600)}{250} = 0 ext{ or } 0.32.$

Two ships, P and Q, are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2017 - Paper 1

Worked Solution & Example Answer:Two ships, P and Q, are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2017 - Paper 1

Find the direction of motion of P, giving your answer as a bearing to the nearest degree.

Find an expression for p in terms of t.

Find an expression for q in terms of t.

Hence show that, at time t hours, QP = (8 + 5t) i + (6 - 10t) j.

Find the values of t when the ships are 10 km apart.

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