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Q.

Ship A is travelling with a velocity of 5 kmh^–1 due east. The second ship is heading 30° east of north. What should be the speed of second ship if it is to remain always due north with respect to the first ship?

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a

$8 {kmh}^{- 1}$

b

$9 {kmh}^{- 1}$

c

$7 {kmh}^{- 1}$

d

$10 {kmh}^{- 1}$

answer is A.

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Detailed Solution

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$\begin{array}{l} {\vec{V}}_{A} = 5 \hat{i} km / h \\ {\vec{V}}_{B} = Vsin 30^{\circ} \hat{i} + Vcos 30^{\circ} \hat{j} \\ = (\frac{V}{2} \hat{i} + \frac{\sqrt{3} V}{2} \hat{j}) \\ ∴ {\vec{V}}_{BA} = {\vec{V}}_{B} - {\vec{V}}_{A} = (\frac{V}{2} - 5) \hat{i} + \frac{\sqrt{3} V}{2} \hat{j} \\ ∴ (\frac{V}{2} - 5) = 0 \Rightarrow V = 10 km / h . \end{array}$

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