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[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given with respect to a fixed origin.] A ship S is moving with constant velocity (–12i + 7.5j) km h⁻¹ - Edexcel - A-Level Maths Mechanics - Question 6 - 2012 - Paper 1
Question 6
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[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given with respect to a fixed origin.] A ship S i... show full transcript
Worked Solution & Example Answer:[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given with respect to a fixed origin.] A ship S is moving with constant velocity (–12i + 7.5j) km h⁻¹ - Edexcel - A-Level Maths Mechanics - Question 6 - 2012 - Paper 1
Step 1
Find the direction in which S is moving, giving your answer as a bearing.
Answer
To find the direction, we compute the angle using the tangent ratio based on the velocity components. The velocity vector is given as ( \mathbf{v} = -12i + 7.5j ). The bearing is calculated as follows:
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Calculate the angle ( \theta ) from the components: [ \theta = \arctan\left(\frac{7.5}{-12}\right) ]
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This will give an angle in the second quadrant since the x-component is negative. Compute the angle: [ \theta = \arctan\left(\frac{7.5}{12}\right)] which equals approximately ( 32° ) (using a calculator).
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To find the bearing, add 180° to this result: [ \text{Bearing} = 180° + 32° = 212° ] Therefore, the bearing is ( 212° ).
Step 2
Step 3
Find the distance of S from B when t = 3.
Answer
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Substitute ( t = 3 ) into the position vector ( s ): [ s = (40 - 12 \times 3)i + (-6 + 7.5 \times 3)j = (40 - 36)i + (-6 + 22.5)j = 4i + 16.5j ]
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The position vector of beacon B is ( (7i + 12.5j) ).
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Find the vector from S to B: [ SB = B - S = (7i + 12.5j) - (4i + 16.5j) = (3i - 4j) ]
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Calculate the distance: [ |SB| = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ km} ]
Step 4
Find the distance of S from B when S is due north of B.
Answer
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For S to be due north of B, their i-components must be equal. The position vector of S is ( (40 - 12t)i + (-6 + 7.5t)j ) and B is positioned at ( (7i + 12.5j) ). Thus, we set: [ 40 - 12t = 7 \implies 12t = 33 \implies t = \frac{33}{12} \approx 2.75 ]
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Substitute ( t = \frac{33}{12} ) back into ( s ): [ s = (40 - 12 \times \frac{33}{12})i + (-6 + 7.5 \times \frac{33}{12})j ] Simplifying: [ s = (40 - 33)i + (-6 + 20.625)j = 7i + 14.625j ]
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Now calculate the distance from S to B: [ SB = B - S = (7i + 12.5j) - (7i + 14.625j) = 0i - 2.125j ]
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The distance is: [ |SB| = \sqrt{(0)^2 + (-2.125)^2} = 2.125 \text{ km} ]