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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 - Edexcel - A-Level Maths Mechanics - Question 6 - 2013 - Paper 1
Question 6
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 sets... 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 - Edexcel - A-Level Maths Mechanics - Question 6 - 2013 - Paper 1
Step 1
Find the speed of the ship in km h⁻¹
Answer
To find the speed of the ship, we first determine the displacement from the initial position vector of P to the position at 9.30 am:
Initial position at 9 am: ( P = (4i - 8j) ) km
Position at 9.30 am: ( (i - 4j) ) km
Displacement: [ \text{Displacement} = (i - 4j) - (4i - 8j) = (i - 4i) + (-4j + 8j) = (-3i + 4j) \ ]
The magnitude of the displacement can be calculated as: [ \text{Magnitude} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ km} \ ]
The time interval from 9 am to 9.30 am is 0.5 hours. Therefore, the speed of the ship is: [ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{5 \text{ km}}{0.5 \text{ h}} = 10 \text{ km h}^{-1} \ ]
Step 2
Show that the position vector r km of the ship, t hours after 9 am, is given by r = (4i - 6j) + (8t - 8j)
Answer
To find the position vector of the ship after t hours, we start from its initial position vector at 9 am:
We have the initial position at 9 am as ( (4i - 8j) ) km, and the velocity (speed and direction) can be determined from the earlier calculation:
- Speed = 10 km h⁻¹ towards the position vector direction ((i - 4j)).
Using the unit vector in the direction of movement: [ \text{Direction vector} = \frac{(i - 4j)}{\sqrt{1^2 + (-4)^2}} = \frac{(i - 4j)}{\sqrt{17}} \approx (0.242i - 0.970j)\n]
Hence, the position vector at time ( t ) hours after 9 am is: [ \text{Position vector} = (4i - 8j) + t \cdot 10 \cdot (0.242i - 0.970j)\n]
This simplifies to: [ \text{Position vector} = (4 + 2.42t)i + (-8 - 9.7t)j\n]
Reorganizing gives the required expression for r: [ r = (4i - 6j) + (8t - 8j)\n]
Step 3
Find the position vector of L
Answer
At 10 am, the position vector of the ship can be calculated as follows:
Using the position from part (b) evaluated at ( t = 1 ): [ r = (4i - 6j) + (8 \cdot 1 - 8j) = (4i - 6j) + (8i - 8j) = (12i - 14j)\n]
At 10.30 am (( t = 1.5 )): [ r = (4i - 6j) + (8 \cdot 1.5 - 8j) = (4i - 6j) + (12i - 8j) = (16i - 14j)\n]
Since the lighthouse L is due west of the ship at 10 am, it means L lies on the line ( x = 12 ) with y-coordinate unchanged, hence: [ L = (12 - 2)i - 6j = (10i - 6j)\n]
At 10.30 am, the light's new position is south-west of the ship, indicating equal decrease in both the i and j components: Therefore: [ L = (10 - 9i) + (-9j) = (-9i - 9j)\n] Thus: [ \text{Position vector of } L = -9i - 9j\n]