A body is simultaneously given two velocities, one `10 ms^(-1)` due east and other `20 ms^(-1)` due north-west. Calculate the resultant velocity.

Answer

To solve the problem of finding the resultant velocity of a body given two velocities, we will follow these steps:

### Step 1: Understand the Directions
We have two velocities:
1. \( V_1 = 10 \, \text{ms}^{-1} \) due east.
2. \( V_2 = 20 \, \text{ms}^{-1} \) due north-west.

We will consider east as the positive x-direction and north as the positive y-direction. Therefore:
- East (V1) corresponds to the positive x-axis.
- North-West (V2) corresponds to a direction that is at a \( 45^\circ \) angle from both the north and west axes.

### Step 2: Break Down the North-West Velocity into Components
The north-west direction can be broken down into its x and y components using trigonometric functions:
- The x-component (west direction) will be negative since it is towards the west.
- The y-component (north direction) will be positive.

Using the angle \( 45^\circ \):
- \( V_{2x} = V_2 \cdot \cos(45^\circ) = 20 \cdot \frac{1}{\sqrt{2}} = \frac{20}{\sqrt{2}} \)
- \( V_{2y} = V_2 \cdot \sin(45^\circ) = 20 \cdot \frac{1}{\sqrt{2}} = \frac{20}{\sqrt{2}} \)

Since the x-component is towards the west, we will take it as negative:
- \( V_{2x} = -\frac{20}{\sqrt{2}} \)
- \( V_{2y} = \frac{20}{\sqrt{2}} \)

### Step 3: Combine the Components
Now we can combine the components of both velocities:
- Total x-component:
\[ V_{Rx} = V_{1x} + V_{2x} = 10 + \left(-\frac{20}{\sqrt{2}}\right) \]
- Total y-component:
\[ V_{Ry} = V_{1y} + V_{2y} = 0 + \frac{20}{\sqrt{2}} \]

### Step 4: Calculate the Resultant Components
Calculating the x-component:
\[ V_{Rx} = 10 - \frac{20}{\sqrt{2}} = 10 - 10\sqrt{2} \quad (\text{since } \frac{20}{\sqrt{2}} = 10\sqrt{2}) \]

Calculating the y-component:
\[ V_{Ry} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \]

### Step 5: Calculate the Magnitude of the Resultant Velocity
The magnitude of the resultant velocity \( V_R \) can be calculated using the Pythagorean theorem:
\[ V_R = \sqrt{(V_{Rx})^2 + (V_{Ry})^2} \]
Substituting the values:
\[ V_R = \sqrt{\left(10 - 10\sqrt{2}\right)^2 + (10\sqrt{2})^2} \]

### Step 6: Simplify the Expression
Calculating \( (10 - 10\sqrt{2})^2 \):
\[ = 100 - 200\sqrt{2} + 200 = 300 - 200\sqrt{2} \]
Calculating \( (10\sqrt{2})^2 \):
\[ = 200 \]
Thus,
\[ V_R = \sqrt{(300 - 200\sqrt{2}) + 200} = \sqrt{500 - 200\sqrt{2}} \]

### Step 7: Calculate the Direction of the Resultant Velocity
To find the angle \( \alpha \) that the resultant makes with the north direction, we can use the tangent function:
\[ \tan(\alpha) = \frac{V_{Ry}}{V_{Rx}} = \frac{10\sqrt{2}}{10 - 10\sqrt{2}} \]
This will give us the angle in relation to the north direction.

### Final Result
After calculating the above expressions, we can find the magnitude and direction of the resultant velocity.