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A ship A sailing due east with a velocity of 10 km/h happens to appear sailing due north with a velocity of 5 km/h , to a person, sitting in a moving ship B. Determine the velocity (absolute) of ship B.

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$5 \sqrt{5} km / h, \tan^{- 1} (1 / 2) S of E$

$5 \sqrt{5} km / h, \tan^{- 1} (1 / 2) E of S$

$4 \sqrt{5} km / h, \tan^{- 1} (1 / 2) S of E$

$4 \sqrt{5} km / h, \tan^{- 1} (1 / 2) E of S$

answer is A.

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Detailed Solution

Here we are given velocity of ‘A’, ${\vec{v}}_{A} = 10 \hat{i}$

Velocity of 'A' w.r.t ‘B’, ${\vec{v}}_{A / B} = 5 \hat{j}$

$Now {\vec{v}}_{A / B} = {\vec{v}}_{A} - {\vec{v}}_{B}$

$5 \hat{j} = 10 \hat{i} - {\vec{v}}_{B} \Rightarrow {\vec{v}}_{B} = 10 \hat{i} - 5 \hat{j}$

Hence velocity of B,

$\begin{array}{l} v_{B} = \sqrt{10^{2} + 5^{2}} = 5 \sqrt{5} km / h \\ \tan θ = \frac{5}{10} = \frac{1}{2} \\ θ = \tan^{- 1} (\frac{1}{2}) S of E \end{array}$

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