Kinematics Question 515

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 $ .

Options:

A)$ 5\sqrt{5}km/h,ta{{n}^{-1}}(1/2)SofE $

B)$ 5\sqrt{5}km/h,ta{{n}^{-1}}(1/2)EofS $

C) $ 4\sqrt{5}km/h,ta{{n}^{-1}}(1/2)SofE $

D) $ 4\sqrt{5}km/h,ta{{n}^{-1}}(1/2)EofS $

Show Answer

Answer:

Correct Answer: A

Solution:

[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}}=5j $

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}}( \frac{1}{2} )s $ of E



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