Ship A is 126 km due west of ship B. A is moving at a constant speed of 50 km h⁻¹ in the direction east α north where \( \tan \alpha = \frac{24}{7} \). B is moving due north at a constant speed of 48 km h⁻¹. Find (i) the velocity of A in terms of \( \mathbf{i} \) and \( \mathbf{j} \) (ii) the velocity of B in terms of \( \mathbf{i} \) and \( \mathbf{j} \) (iii) the velocity of A relative to B in terms of \( \mathbf{i} \) and \( \mathbf{j} \). Ship A intercepts ship B after t hours. Find (iv) the value of t (v) the distance each ship travels in this time t.

Leaving Cert Applied Maths Relative Velocity: Ship A is 126 km due west of shi

Ship A is 126 km due west of ship B - Leaving Cert Applied Maths - Question 2 - 2011

Question 2

Ship A is 126 km due west of ship B. A is moving at a constant speed of 50 km h⁻¹ in the direction east α north where \( \tan \alpha = \frac{24}{7} \). B is moving ... show full transcript

Worked Solution & Example Answer:Ship A is 126 km due west of ship B - Leaving Cert Applied Maths - Question 2 - 2011

Step 1

Find (i) the velocity of A in terms of \( \mathbf{i} \) and \( \mathbf{j} \)

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Answer

To find the velocity of ship A, we express its motion in terms of its components. Given ( \alpha = \tan^{-1}(\frac{24}{7}) ), we split the velocity into its components:

Velocity of A: [ \mathbf{v_A} = 50 \cos(\alpha) \mathbf{i} + 50 \sin(\alpha) \mathbf{j} ] Calculating: [ \mathbf{v_A} = 50 \cdot \frac{7}{25} \mathbf{i} + 50 \cdot \frac{24}{25} \mathbf{j} = 14 \mathbf{i} + 48 \mathbf{j} ]

Step 2

Find (ii) the velocity of B in terms of \( \mathbf{i} \) and \( \mathbf{j} \)

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Answer

Ship B is moving purely north, therefore its velocity can be expressed as:

Velocity of B: [ \mathbf{v_B} = 0 \mathbf{i} + 48 \mathbf{j} ]

Step 3

Find (iii) the velocity of A relative to B in terms of \( \mathbf{i} \) and \( \mathbf{j} \)

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Answer

To find the velocity of A relative to B, we subtract the velocity of B from the velocity of A:

[ \mathbf{v_{AB}} = \mathbf{v_A} - \mathbf{v_B} = (14 \mathbf{i} + 48 \mathbf{j}) - (0 \mathbf{i} + 48 \mathbf{j}) = 14 \mathbf{i} + 0 \mathbf{j} ]

Step 4

Find (iv) the value of t

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Answer

Given that ship A is 126 km away from ship B and moves towards it:

Distance = velocity × time [ 126 = 14t ] Solving for t: [ t = \frac{126}{14} = 9 \text{ hours} ]

Step 5

Find (v) the distance each ship travels in this time t

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Answer

Calculating the distance for each ship in 9 hours:

For ship A: [ S_A = 50 \times 9 = 450 \text{ km} ]

For ship B: [ S_B = 48 \times 9 = 432 \text{ km} ]

Ship A is 126 km due west of ship B - Leaving Cert Applied Maths - Question 2 - 2011

Worked Solution & Example Answer:Ship A is 126 km due west of ship B - Leaving Cert Applied Maths - Question 2 - 2011

Find (i) the velocity of A in terms of \( \mathbf{i} \) and \( \mathbf{j} \)

Find (ii) the velocity of B in terms of \( \mathbf{i} \) and \( \mathbf{j} \)

Find (iii) the velocity of A relative to B in terms of \( \mathbf{i} \) and \( \mathbf{j} \)

Find (iv) the value of t

Find (v) the distance each ship travels in this time t

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