A ship S is moving with constant velocity $(-2.5 extbf{i} + 6 extbf{j})$ km h$^{-1}$. At time 1200, the position vector of S relative to a fixed origin O is $(16 extbf{i} + 5 extbf{j})$ km. Find (a) the speed of S, (b) the bearing on which S is moving. The ship is heading directly towards a submerged rock R. A radar tracking station calculates that, if S continues on the same course with the same speed, it will hit R at the time 1500. (c) Find the position vector of R. The tracking station warns the ship’s captain of the situation. The captain maintains S on its course with the same speed until the time is 1400. He then changes course so that S moves now on that course at a constant speed of $5$ km h$^{-1}$. Assuming that S continues to move with this new constant velocity, d) an expression for the position vector of the ship t hours after 1400, e) the time when S will be due east of R, f) the distance of S from R at the time 1600.

A-Level Edexcel Maths Mechanics Constant Acceleration - 1D: A ship S is moving with constant

A ship S is moving with constant velocity $(-2.5 extbf{i} + 6 extbf{j})$ km h$^{-1}$ - Edexcel - A-Level Maths Mechanics - Question 7 - 2006 - Paper 1

Question 7

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A ship S is moving with constant velocity $(-2.5 extbf{i} + 6 extbf{j})$ km h$^{-1}$. At time 1200, the position vector of S relative to a fixed origin O is $(16 ext... show full transcript

Worked Solution & Example Answer:A ship S is moving with constant velocity $(-2.5 extbf{i} + 6 extbf{j})$ km h$^{-1}$ - Edexcel - A-Level Maths Mechanics - Question 7 - 2006 - Paper 1

Step 1

Find the speed of S

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Answer

To calculate the speed of S, use the formula for the magnitude of the velocity vector:

orm{(-2.5 extbf{i} + 6 extbf{j})} = rac{1}{h} ext{sqrt}{(-2.5)^2 + (6)^2}$$ $$= ext{sqrt}(6.25 + 36)$$ $$= ext{sqrt}(42.25)$$ $$= 6.5 ext{ km h}^{-1}$$

Step 2

Find the bearing on which S is moving

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Answer

The bearing can be calculated using the arctangent function:

$ext{Bearing} = 360 - ext{arctan} rac{(2.5)}{(6)}$ Calculating gives: $= 360 - 22.62 = 337^ ext{o}$

Step 3

Find the position vector of R

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Answer

To find the position vector of R at time 1500, calculate the position of S at that time:

At 1500: $s = (16 extbf{i} + 5 extbf{j}) + (3 ext{ h} imes (-2.5 extbf{i} + 6 extbf{j}))$ Calculate: $s = (16 - 7.5) extbf{i} + (5 + 18) extbf{j}$ $= 8.5 extbf{i} + 23 extbf{j}$ Hence R = $8.5 extbf{i} + 23 extbf{j}$ .

Step 4

an expression for the position vector of the ship t hours after 1400

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Answer

For the time after 1400, the position vector can be expressed as:

$s(t) = (11 extbf{i} + 17 extbf{j}) + (t ext{ hours}) imes (5 extbf{i})$ Therefore, $s(t) = (11 + 5t) extbf{i} + 17 extbf{j}$

Step 5

the time when S will be due east of R

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Answer

For S to be due east of R, the i-component of the position vector of S must equal the i-component of R. Setting: $11 + 5t = 8.5$ Solving gives: $5t = -2.5$ $t = -0.5$ So S will be due east of R at 1512 hours.

Step 6

the distance of S from R at the time 1600

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Answer

At 1600, find the position of S: $s(2) = (11 extbf{i} + 17 extbf{j}) + (2 ext{ hours}) imes (5 extbf{i})$ This gives: $s(2) = (11 + 10) extbf{i} + 17 extbf{j} = 21 extbf{i} + 17 extbf{j}$ Now calculate the distance from R:

orm{(21 - 8.5) extbf{i} + (17 - 23) extbf{j}}$$ $$= orm{12.5 extbf{i} - 6 extbf{j}} = ext{sqrt}(12.5^2 + (-6)^2) = ext{sqrt}(156.25 + 36) = ext{sqrt}(192.25) \\ ext{approximately } 4.72 ext{ km}$$

A ship S is moving with constant velocity $(-2.5 extbf{i} + 6 extbf{j})$ km h$^{-1}$ - Edexcel - A-Level Maths Mechanics - Question 7 - 2006 - Paper 1

Worked Solution & Example Answer:A ship S is moving with constant velocity $(-2.5 extbf{i} + 6 extbf{j})$ km h$^{-1}$ - Edexcel - A-Level Maths Mechanics - Question 7 - 2006 - Paper 1

Find the speed of S

Find the bearing on which S is moving

Find the position vector of R

an expression for the position vector of the ship t hours after 1400

the time when S will be due east of R

the distance of S from R at the time 1600

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