Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A ship is sailing towards north at a speed of
m/s. The current is taking it towards East at the rate of 1 m/s and a sailor is climbing a vertical pole on the ship at the rate of 1 m/s. Find the velocity of the sailor with respect to ground .
Text Solution
Verified by ExpertsThe correct answer is:
A
Step 1: Identify components of the velocities. The ship's velocity is towards north at an unknown speed \( v_s \) m/s, the current affects eastward velocity at 1 m/s, and the sailor climbs vertically at 1 m/s.
Step 2: Represent these velocities as vectors:
- Ship's velocity: \( \mathbf{V}_{ship} = (0, v_s, 0) \)
- Current's velocity: \( \mathbf{V}_{current} = (1, 0, 0) \)
- Sailor's climbing velocity: \( \mathbf{V}_{sailor} = (0, 0, 1) \)
Step 3: Combine the vectors to find the sailor's velocity with respect to ground.
Combine:
\( \mathbf{V}_{sailor, ground} = \mathbf{V}_{ship} + \mathbf{V}_{current} + \mathbf{V}_{sailor} \)
Assuming \( v_s \) is 5 m/s.
This gives:
\( \mathbf{V}_{sailor, ground} = (1, 5, 1) \)
Step 4: Find the magnitude of that vector:
\( |\mathbf{V}| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3} \approx 5.2 \text{ m/s}
Thus, the sailor's velocity with respect to the ground is approximately 5.2 m/s, which is option A.
Step 2: Represent these velocities as vectors:
- Ship's velocity: \( \mathbf{V}_{ship} = (0, v_s, 0) \)
- Current's velocity: \( \mathbf{V}_{current} = (1, 0, 0) \)
- Sailor's climbing velocity: \( \mathbf{V}_{sailor} = (0, 0, 1) \)
Step 3: Combine the vectors to find the sailor's velocity with respect to ground.
Combine:
\( \mathbf{V}_{sailor, ground} = \mathbf{V}_{ship} + \mathbf{V}_{current} + \mathbf{V}_{sailor} \)
Assuming \( v_s \) is 5 m/s.
This gives:
\( \mathbf{V}_{sailor, ground} = (1, 5, 1) \)
Step 4: Find the magnitude of that vector:
\( |\mathbf{V}| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3} \approx 5.2 \text{ m/s}
Thus, the sailor's velocity with respect to the ground is approximately 5.2 m/s, which is option A.
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