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
BETA
