A motor boat is racing towards North at `25 km//h` and the water current in that region is `10 km//h` in the direction of `60 ^@` East of South. Find the resultant velocity of the boat.

Options

A. 11 km `h^(-1)`
B. 22 km `h^(-1)`
C. 33 km `h^(-1)`
D. 44 km `h^(-1)`

Explanation

To solve the problem of finding the resultant velocity of the motorboat racing towards the north while considering the water current, we can follow these steps:

### Step 1: Identify the velocities
- The velocity of the motorboat (V_boat) is 25 km/h towards the North.
- The velocity of the water current (V_current) is 10 km/h at an angle of 60° East of South.

### Step 2: Break down the current's velocity into components
To find the resultant velocity, we need to break down the current's velocity into its North and East components.

- The South component of the current can be calculated as:
\[ V_{current, South} = V_{current} \cdot \cos(60°) = 10 \cdot \frac{1}{2} = 5 \text{ km/h} \]

- The East component of the current can be calculated as:
\[ V_{current, East} = V_{current} \cdot \sin(60°) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ km/h} \approx 8.66 \text{ km/h} \]

### Step 3: Determine the effective North velocity
Since the current is directed South, it will reduce the effective North velocity of the boat:
\[ V_{effective, North} = V_{boat} - V_{current, South} = 25 - 5 = 20 \text{ km/h} \]

### Step 4: Combine the components to find the resultant velocity
Now we have:
- North component: 20 km/h
- East component: \(5\sqrt{3}\) km/h

We can find the resultant velocity (R) using the Pythagorean theorem:
\[ R = \sqrt{(V_{effective, North})^2 + (V_{current, East})^2} \]
Substituting the values:
\[ R = \sqrt{(20)^2 + (5\sqrt{3})^2} = \sqrt{400 + 75} = \sqrt{475} \approx 21.91 \text{ km/h} \]

### Step 5: Determine the direction of the resultant velocity
To find the angle θ of the resultant velocity with respect to the North direction, we use:
\[ \tan(\theta) = \frac{V_{current, East}}{V_{effective, North}} = \frac{5\sqrt{3}}{20} \]
Calculating θ:
\[ \theta = \tan^{-1}\left(\frac{5\sqrt{3}}{20}\right) = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \approx 15.9° \]

### Final Result
The resultant velocity of the boat is approximately 21.91 km/h at an angle of about 15.9° East of North.

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