SATHEE: Kinematics Question 515

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) $5 \sqrt{5} k m / h, t a n^{- 1} (1 / 2) S o f E$

B) $5 \sqrt{5} k m / h, t a n^{- 1} (1 / 2) E o f S$

C) $4 \sqrt{5} k m / h, t a n^{- 1} (1 / 2) S o f E$

D) $4 \sqrt{5} k m / h, t a n^{- 1} (1 / 2) E o f S$

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’, ${\vec{v}}_{A / B} = 5 j$

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, $v_{B} = \sqrt{10^{2} + 5^{2}} = 5 \sqrt{5} k m / h$

$\tan θ = \frac{5}{10} = \frac{1}{2}$

$θ = \tan^{- 1} (\frac{1}{2}) s$ of E

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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:

Answer:

Solution:

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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 $B$ .