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Ship A is moving at a constant speed of 34 km h⁻¹ in the direction east α° north, where tan α = 8/15 - Leaving Cert Applied Maths - Question 2 - 2017
Question 2
Ship A is moving at a constant speed of 34 km h⁻¹ in the direction east α° north, where tan α = 8/15. Ship B is moving at a constant speed of 26 km h⁻¹ in the direc... show full transcript
Worked Solution & Example Answer:Ship A is moving at a constant speed of 34 km h⁻¹ in the direction east α° north, where tan α = 8/15 - Leaving Cert Applied Maths - Question 2 - 2017
Step 1
(i) the velocity of ship A in terms of \( \hat{i} \) and \( \hat{j} \)
Answer
To find the velocity of ship A, we resolve its speed into horizontal and vertical components based on the angle α:
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The horizontal component: [ v_{A,x} = 34 \cos(\alpha) = 34 \cos\left(\tan^{-1}\left(\frac{8}{15}\right)\right) ] Using the relation, ( \cos(\alpha) = \frac{15}{\sqrt{8^2 + 15^2}} = \frac{15}{\sqrt{289}} = \frac{15}{17} ), we get: [ v_{A,x} = 34 \cdot \frac{15}{17} = 30 \text{ km h}^{-1} ]
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The vertical component: [ v_{A,y} = 34 \sin(\alpha) = 34 \sin\left(\tan^{-1}\left(\frac{8}{15}\right)\right) ] Using the relation, ( \sin(\alpha) = \frac{8}{17} ), we get: [ v_{A,y} = 34 \cdot \frac{8}{17} = 16 \text{ km h}^{-1} ]
Thus, the velocity of ship A is: [ \vec{v}_A = 30 \hat{i} + 16 \hat{j} \text{ km h}^{-1} ]
Step 2
(ii) the velocity of ship B in terms of \( \hat{i} \) and \( \hat{j} \)
Answer
For ship B, we again resolve its speed into components based on the angle β:
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The horizontal component (west is negative): [ v_{B,x} = -26 \cos(\beta) = -26 \cos\left(\tan^{-1}\left(\frac{12}{5}\right)\right) ] Using the relation, ( \cos(\beta) = \frac{5}{\sqrt{12^2 + 5^2}} = \frac{5}{\sqrt{169}} = \frac{5}{13} ), we get: [ v_{B,x} = -26 \cdot \frac{5}{13} = -10 \text{ km h}^{-1} ]
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The vertical component (south is negative): [ v_{B,y} = -26 \sin(\beta) = -26 \sin\left(\tan^{-1}\left(\frac{12}{5}\right)\right) ] Using the relation, ( \sin(\beta) = \frac{12}{13} ), we get: [ v_{B,y} = -26 \cdot \frac{12}{13} = -24 \text{ km h}^{-1} ]
Thus, the velocity of ship B is: [ \vec{v}_B = -10 \hat{i} - 24 \hat{j} \text{ km h}^{-1} ]
Step 3
(iii) the velocity of A relative to B in terms of \( \hat{i} \) and \( \hat{j} \)
Answer
The velocity of ship A relative to ship B can be calculated as: [ \vec{v}_{A \text{ relative to } B} = \vec{v}_A - \vec{v}B ] Thus: [ \vec{v}{A \text{ relative to } B} = (30 \hat{i} + 16 \hat{j}) - (-10 \hat{i} - 24 \hat{j}) ] [ = (30 + 10)\hat{i} + (16 + 24)\hat{j} = 40 \hat{i} + 40 \hat{j} ]
Step 4
Step 5
(v) the distance from lighthouse L to the meeting point
Answer
To find the distance from lighthouse L to the meeting point after time t:
We calculate the distance traveled by ship A: [ d = 34 \cdot t = 34 \cdot 13 = 442 \text{ km} ]
Now, we find the straight-line distance from L: Using Pythagorean theorem: [ d = \sqrt{(520)^2 + (442)^2} ] [ = \sqrt{270400 + 195364} = \sqrt{465764} \approx 245.28 \text{ km} ]