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Ship A is positioned 204 km due south of lighthouse L - Leaving Cert Applied Maths - Question 2 - 2014
Question 2
Ship A is positioned 204 km due south of lighthouse L. A is moving at an angle α east of north at a constant speed of 58 km h⁻¹, where tan α = \frac{a}{b}. Ship B ... show full transcript
Worked Solution & Example Answer:Ship A is positioned 204 km due south of lighthouse L - Leaving Cert Applied Maths - Question 2 - 2014
Step 1
Find (i) the velocity of A in terms of \mathbf{i} and \mathbf{j}.
Answer
The velocity of Ship A, \mathbf{v_a}$, can be calculated by using the components of the velocity in the eastward and northward directions. Given the speed of 58 km h⁻¹ and the angle α:
[
\mathbf{v_a} = 58 \sin(\alpha) \mathbf{i} + 58 \cos(\alpha) \mathbf{j}
]
Substituting in the known values and using the relationship for tan α = \frac{a}{b}:
[
\mathbf{v_a} = 58 \sin(\alpha) \mathbf{i} + (40 + 42) \mathbf{j}
]
Step 2
Step 3
Find (iii) the velocity of A relative to B in terms of \mathbf{i} and \mathbf{j}.
Answer
The relative velocity \mathbf{v_{ab}} of Ship A with respect to Ship B is given by:
[ \mathbf{v_{ab}} = \mathbf{v_a} - \mathbf{v_b} ] Substituting the expressions: [ \mathbf{v_{ab}} = (40 \mathbf{i} + 42 \mathbf{j}) - (40 \mathbf{i} + 0 \mathbf{j}) ] Simplifying yields: [ \mathbf{v_{ab}} = 0 \mathbf{i} + 42 \mathbf{j} ]
Step 4
Step 5
Find (v) the distance from lighthouse L to the meeting point.
Answer
The distance (d) from the lighthouse L to the meeting point can be calculated using: [ |\mathbf{C}| = 40 \times t ] Substituting t we find: [ |\mathbf{C}| = 40 \times 17 = 680 \text{ km}. ] Moreover, the total distance is: [ |\mathbf{C}| = \sqrt{(510)^{2} + (680)^{2}} = 850 \text{ km}. ]