Question

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 $B$.

• A. $5 \sqrt{5} \,km / h , \tan ^{-1}(1 / 2) S$ of $E$
• B. $5 \sqrt{5} \, km / h , \tan ^{-1}(1 / 2) E$ of $S$
• C. $4 \sqrt{5} \, km / h , \tan ^{-1}(1 / 2) S$ of $E$
• D. $4 \sqrt{5} \,km / h , \tan ^{-1}(1 / 2) E$ of $S$

Answer 1

Answer: (A)

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 $B$,
$v_{B}=\sqrt{10^{2}+5^{2}}=5 \sqrt{5}\, km / h$
$\tan \theta=\frac{5}{10}=\frac{1}{2}$
$\theta=\tan ^{-1}\left(\frac{1}{2}\right) S $ of $E$