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Ship A is travelling east 30° north at a constant speed of 14 km h⁻¹ - Leaving Cert Applied Maths - Question 2 - 2015
Question 2
Ship A is travelling east 30° north at a constant speed of 14 km h⁻¹. Ship B is travelling due north at a constant speed of 20 km h⁻¹. B is positioned 10 km due ea... show full transcript
Worked Solution & Example Answer:Ship A is travelling east 30° north at a constant speed of 14 km h⁻¹ - Leaving Cert Applied Maths - Question 2 - 2015
Step 1
Express the velocity of A and the velocity of B in terms of \( \mathbf{i} \) and \( \mathbf{j} \)
Answer
To express the velocities, we decompose the speeds of the ships into their components:
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For Ship A:
- The eastward (( \mathbf{i} )) component: ( v_A^x = 14 \cos(30°) = 14 \cdot \frac{\sqrt{3}}{2} = 7\sqrt{3} ) km h⁻¹.
- The northward (( \mathbf{j} )) component: ( v_A^y = 14 \sin(30°) = 14 \cdot \frac{1}{2} = 7 ) km h⁻¹.
Therefore, the velocity of A is: ( \mathbf{v_A} = 7\sqrt{3} \mathbf{i} + 7 \mathbf{j} ) km h⁻¹.
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For Ship B:
- The velocity of B is directed straight north, so: ( \mathbf{v_B} = 0 \mathbf{i} + 20 \mathbf{j} ) km h⁻¹.
Step 2
Find the velocity of A relative to B in terms of \( \mathbf{i} \) and \( \mathbf{j} \)
Answer
The velocity of A relative to B is calculated by subtracting the velocity of B from A:
( \mathbf{v_{AB}} = \mathbf{v_A} - \mathbf{v_B} = (7\sqrt{3} \mathbf{i} + 7 \mathbf{j}) - (0 \mathbf{i} + 20 \mathbf{j}) )
Thus: ( \mathbf{v_{AB}} = 7\sqrt{3} \mathbf{i} - 13 \mathbf{j} ) km h⁻¹.
Step 3
Calculate the shortest distance between the ships in their subsequent motion
Answer
To find the shortest distance between the ships, we consider their positions over time and use the relative velocity:
- The relative speed between the ships in the ( \mathbf{i} ) direction is constant at ( 7\sqrt{3} ) km h⁻¹.
- The initial separation in the ( \mathbf{j} ) direction (north) is ( 10 ) km.
Using the depth projection, the angle ( \alpha ) can be calculated: ( \alpha = \tan^{-1}\left(\frac{13}{7\sqrt{3}}\right) \approx 47° ).
- The distance ( d ) can then be calculated from the position over time: ( d = 10 \sin(\alpha) = 10 \sin(47°) \approx 7.3 ) km.
Step 4
Find the distance between the ships one hour after the instant that they were closest together
Answer
The distance one hour later can be computed using the velocities:
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The magnitude of the relative velocity is: ( |\mathbf{v_{AB}}| = \sqrt{(7\sqrt{3})^2 + (-13)^2} = \sqrt{(21 + 169)} = \sqrt{190} \approx 17.776 ) km h⁻¹.
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Therefore, the distance after one hour: ( d_1 = 17.776 \cdot 1 = 17.776 ) km.
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Finally, we can find this distance using the Pythagorean theorem: ( d = \sqrt{(7.3)^2 + (17.776)^2} \approx 19.2 ) km.