A ship is sailing towards the north at a speed of 1.25 m/s. The current is taking it towards the east at the rate of 1 m/s and a sailor is climbing a vertical pole on the ship at the rate of 0.5 m/s. Find the velocity of the sailor in space.

Explanation

To find the velocity of the sailor in space, we will analyze the problem step by step.

### Step 1: Define the Directions
We will define our coordinate system:
- Let the **i** direction represent the east.
- Let the **j** direction represent the north.
- Let the **k** direction represent the vertical (upward).

### Step 2: Determine the Velocity Components
1. The ship is sailing towards the north at a speed of **1.25 m/s**. Therefore, the velocity component in the **j** direction is:
\[ \vec{V}_{\text{ship}} = 1.25 \, \hat{j} \]

2. The current is taking the ship towards the east at a speed of **1 m/s**. Therefore, the velocity component in the **i** direction is:
\[ \vec{V}_{\text{current}} = 1 \, \hat{i} \]

3. The sailor is climbing a vertical pole on the ship at a rate of **0.5 m/s**. Therefore, the velocity component in the **k** direction is:
\[ \vec{V}_{\text{sailor}} = 0.5 \, \hat{k} \]

### Step 3: Combine the Velocity Components
Now, we can combine these components to find the total velocity of the sailor in space:
\[ \vec{V}_{\text{sailor in space}} = \vec{V}_{\text{current}} + \vec{V}_{\text{ship}} + \vec{V}_{\text{sailor}} = 1 \, \hat{i} + 1.25 \, \hat{j} + 0.5 \, \hat{k} \]

### Step 4: Calculate the Magnitude of the Velocity
To find the magnitude of the velocity vector, we use the formula:
\[ |\vec{V}| = \sqrt{(V_x)^2 + (V_y)^2 + (V_z)^2} \]
Substituting the values:
\[ |\vec{V}| = \sqrt{(1)^2 + (1.25)^2 + (0.5)^2} \]
Calculating each term:
- \( (1)^2 = 1 \)
- \( (1.25)^2 = 1.5625 \)
- \( (0.5)^2 = 0.25 \)

Adding these values:
\[ |\vec{V}| = \sqrt{1 + 1.5625 + 0.25} = \sqrt{2.8125} \]

### Step 5: Final Calculation
Now, we calculate the square root:
\[ |\vec{V}| \approx 1.677 \, \text{m/s} \]

### Conclusion
The velocity of the sailor in space is approximately **1.677 m/s**.

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