A motor car A is travelling with a velocity of 20 m/s in the north-west direction and another motor car B is travelling with a velocity of 15 m/s in the north-east direction. The magnitude of relative velocity of B with respect to A is

Options

A. `25 m//s`
B. `15 m//s`
C. `20 m//s`
D. `35 m//s`

Answer

To find the magnitude of the relative velocity of motor car B with respect to motor car A, we can follow these steps:

### Step 1: Define the velocities of cars A and B
- Car A is traveling at a velocity of 20 m/s in the north-west direction.
- Car B is traveling at a velocity of 15 m/s in the north-east direction.

### Step 2: Break down the velocities into components
- The north-west direction makes an angle of 45 degrees with both the north and west axes.
- The north-east direction also makes an angle of 45 degrees with both the north and east axes.

Using trigonometric functions, we can express the velocities in terms of their x (east-west) and y (north-south) components.

For car A:
- \( V_A = 20 \, \text{m/s} \)
- \( V_{Ax} = -20 \cos(45^\circ) = -20 \times \frac{1}{\sqrt{2}} = -\frac{20}{\sqrt{2}} \, \text{m/s} \) (westward)
- \( V_{Ay} = 20 \sin(45^\circ) = 20 \times \frac{1}{\sqrt{2}} = \frac{20}{\sqrt{2}} \, \text{m/s} \) (northward)

For car B:
- \( V_B = 15 \, \text{m/s} \)
- \( V_{Bx} = 15 \cos(45^\circ) = 15 \times \frac{1}{\sqrt{2}} = \frac{15}{\sqrt{2}} \, \text{m/s} \) (eastward)
- \( V_{By} = 15 \sin(45^\circ) = 15 \times \frac{1}{\sqrt{2}} = \frac{15}{\sqrt{2}} \, \text{m/s} \) (northward)

### Step 3: Calculate the relative velocity of B with respect to A
The relative velocity \( V_{BA} \) is given by:
\[ V_{BA} = V_B - V_A \]

Calculating the components:
- \( V_{BAx} = V_{Bx} - V_{Ax} = \frac{15}{\sqrt{2}} - \left(-\frac{20}{\sqrt{2}}\right) = \frac{15 + 20}{\sqrt{2}} = \frac{35}{\sqrt{2}} \, \text{m/s} \)
- \( V_{BAy} = V_{By} - V_{Ay} = \frac{15}{\sqrt{2}} - \frac{20}{\sqrt{2}} = \frac{15 - 20}{\sqrt{2}} = -\frac{5}{\sqrt{2}} \, \text{m/s} \)

### Step 4: Calculate the magnitude of the relative velocity
The magnitude of the relative velocity \( |V_{BA}| \) is given by:
\[ |V_{BA}| = \sqrt{(V_{BAx})^2 + (V_{BAy})^2} \]

Substituting the values:
\[ |V_{BA}| = \sqrt{\left(\frac{35}{\sqrt{2}}\right)^2 + \left(-\frac{5}{\sqrt{2}}\right)^2} \]
\[ = \sqrt{\frac{1225}{2} + \frac{25}{2}} \]
\[ = \sqrt{\frac{1250}{2}} \]
\[ = \sqrt{625} \]
\[ = 25 \, \text{m/s} \]

### Final Answer
The magnitude of the relative velocity of B with respect to A is **25 m/s**.