Ship A is moving at a constant speed of 40 km h⁻¹ in the direction α° North of East, as shown, where tan α = rac{3}{4}. Ship B is moving at a constant speed of 38 km h⁻¹ in the direction due South. Find (i) the velocity of ship A in terms of $ \mathbf{i} $ and $ \mathbf{j} $ (ii) the velocity of ship 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 B is positioned d km due east of ship A at 2 pm. The closest distance between ship A and ship B in the subsequent motion is 35 km. (iv) Show that d = 37 km.

Leaving Cert Applied Maths Relative Velocity: Ship A is moving at a constant s

Ship A is moving at a constant speed of 40 km h⁻¹ in the direction α° North of East, as shown, where tan α = rac{3}{4} - Leaving Cert Applied Maths - Question 2 - 2018

Question 2

Ship A is moving at a constant speed of 40 km h⁻¹ in the direction α° North of East, as shown, where tan α = rac{3}{4}. Ship B is moving at a constant speed of 38 ... show full transcript

Worked Solution & Example Answer:Ship A is moving at a constant speed of 40 km h⁻¹ in the direction α° North of East, as shown, where tan α = rac{3}{4} - Leaving Cert Applied Maths - Question 2 - 2018

Step 1

Find the velocity of ship A in terms of $ \mathbf{i} $ and $ \mathbf{j} $

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Answer

To find the velocity of ship A, we first need to determine the angle α from the given tangent:

$\tan(\alpha) = \frac{3}{4}$

Using the relationship in a right triangle, we can find:

$\alpha = \arctan\left(\frac{3}{4}\right)$

With the speed of ship A being 40 km/h, we can express its velocity as:

$\mathbf{V_A} = 40 \cos(\alpha) \mathbf{i} + 40 \sin(\alpha) \mathbf{j}$

Calculating the components:

(\cos(\alpha) = \frac{4}{5}) and (\sin(\alpha) = \frac{3}{5})

Thus:

$\mathbf{V_A} = 40 \left(\frac{4}{5}\right) \mathbf{i} + 40 \left(\frac{3}{5}\right) \mathbf{j} = 32 \mathbf{i} + 24 \mathbf{j}$

Step 2

Find the velocity of ship B in terms of $ \mathbf{i} $ and $ \mathbf{j} $

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Answer

Ship B is moving directly south at 38 km/h:

$\mathbf{V_B} = 0 \mathbf{i} - 38 \mathbf{j}$

Step 3

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

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Answer

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

$\mathbf{V_{AB}} = \mathbf{V_A} - \mathbf{V_B}$

Substituting the values we found:

$\mathbf{V_{AB}} = (32 \mathbf{i} + 24 \mathbf{j}) - (0 \mathbf{i} - 38 \mathbf{j}) = 32 \mathbf{i} + 62 \mathbf{j}$

Step 4

Show that d = 37 km

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Answer

Using the given distance relationship and the formula for distance:

The distance ( d ) is calculated using the sin function:

$d \sin(\theta) = 35$

We are given: $\tan(\theta) = \frac{70}{24}$

To find ( d ), first relate it to the known angle from the right triangle:

$\theta = \arctan\left(\frac{70}{24}\right)$

From the sine relationship, substituting the values will yield:

After calculations, we find:

$d = 37 \text{ km}$

Ship A is moving at a constant speed of 40 km h⁻¹ in the direction α° North of East, as shown, where tan α = rac{3}{4} - Leaving Cert Applied Maths - Question 2 - 2018

Worked Solution & Example Answer:Ship A is moving at a constant speed of 40 km h⁻¹ in the direction α° North of East, as shown, where tan α = rac{3}{4} - Leaving Cert Applied Maths - Question 2 - 2018

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

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

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

Show that d = 37 km

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