Ship A is positioned 80 km south of ship B. A is moving north-east at a constant speed of $30\sqrt{2}$ km h$^{-1}$. B is moving due west at a constant speed of 15 km h$^{-1}$. Find (i) the velocity of A in terms of $ extbf{i}$ and $ extbf{j}$ (ii) the velocity of B in terms of $ extbf{i}$ and $ extbf{j}$ (iii) the velocity of A relative to B in terms of $ extbf{i}$ and $ extbf{j}$ (iv) the shortest distance between A and B in the subsequent motion.

Leaving Cert Applied Maths Relative Velocity: Ship A is positioned 80 km south

Ship A is positioned 80 km south of ship B - Leaving Cert Applied Maths - Question 2 - 2012

Question 2

Ship A is positioned 80 km south of ship B. A is moving north-east at a constant speed of $30\sqrt{2}$ km h$^{-1}$. B is moving due west at a constant speed of 15 km... show full transcript

Worked Solution & Example Answer:Ship A is positioned 80 km south of ship B - Leaving Cert Applied Maths - Question 2 - 2012

Step 1

the velocity of A in terms of $\textbf{i}$ and $\textbf{j}$

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Answer

The velocity of ship A, moving north-east at a speed of $30\sqrt{2}$ km h $^{-1}$ , can be broken down into its components:

$\vec{v_A} = 30\sqrt{2}\sin(45)\, \textbf{i} + 30\sqrt{2}\cos(45)\, \textbf{j}$

Calculating the components:

$\vec{v_A} = 30\sqrt{2}\cdot\frac{\sqrt{2}}{2}\, \textbf{i} + 30\sqrt{2}\cdot\frac{\sqrt{2}}{2}\, \textbf{j}$

Thus,

$\vec{v_A} = 30\, \textbf{i} + 30\, \textbf{j}$

Step 2

the velocity of B in terms of $\textbf{i}$ and $\textbf{j}$

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Answer

The velocity of ship B, moving due west at a speed of 15 km h $^{-1}$ , is given by:

$\vec{v_B} = -15\, \textbf{j}$

Step 3

the velocity of A relative to B in terms of $\textbf{i}$ and $\textbf{j}$

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Answer

The velocity of A relative to B can be calculated using:

$\vec{v_{AB}} = \vec{v_A} - \vec{v_B}$

Substituting the velocities:

$\vec{v_{AB}} = (30\, \textbf{i} + 30\, \textbf{j}) - (-15\, \textbf{j})$

This simplifies to:

$\vec{v_{AB}} = 30\, \textbf{i} + 45\, \textbf{j}$

Step 4

the shortest distance between A and B in the subsequent motion

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Answer

To find the shortest distance, we will use the angle and the distance formula. The angle $\theta$ between the velocities can be calculated using:

$\theta = \tan^{-1}\left(\frac{30}{15}\right) = 33.69^\circ$

The shortest distance can then be calculated using the initial distance (80 km) and the angle:

$d = 80 \cdot \cos(33.69) \approx 66.56 \text{ km}$

Ship A is positioned 80 km south of ship B - Leaving Cert Applied Maths - Question 2 - 2012

Worked Solution & Example Answer:Ship A is positioned 80 km south of ship B - Leaving Cert Applied Maths - Question 2 - 2012

the velocity of A in terms of $\textbf{i}$ and $\textbf{j}$

the velocity of B in terms of $\textbf{i}$ and $\textbf{j}$

the velocity of A relative to B in terms of $\textbf{i}$ and $\textbf{j}$

the shortest distance between A and B in the subsequent motion

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