Question

A ship $A$ sailing due east with a velocity of $10km/h$ happens to appear sailing due north with a velocity of $5km/h$, 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 $E$