Ram moves in east direction at a speed of 6 m/s and Shyam moves `30^(@)` east of north at a speed of 6 m/s. The magnitude of their relative velocity is

Options

A. 3 m/s
B. 6 m/s
C. ` 6 sqrt(3) m//s `
D. `6 sqrt(2) m//s `

Answer

To solve the problem, we need to find the magnitude of the relative velocity between Ram and Shyam. Let's break down the steps:

### Step 1: Define the velocities of Ram and Shyam
- Ram moves east at a speed of 6 m/s. We can represent this as:
\[ \vec{V_R} = 6 \hat{i} \]
where \(\hat{i}\) is the unit vector in the east direction.

- Shyam moves at a speed of 6 m/s at an angle of \(30^\circ\) east of north. To find the components of Shyam's velocity, we need to resolve it into its x (east) and y (north) components:
\[ \vec{V_S} = 6 \cos(30^\circ) \hat{i} + 6 \sin(30^\circ) \hat{j} \]

### Step 2: Calculate the components of Shyam's velocity
Using the trigonometric values:
- \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)
- \(\sin(30^\circ) = \frac{1}{2}\)

Now substituting these values:
\[ \vec{V_S} = 6 \left(\frac{\sqrt{3}}{2}\right) \hat{i} + 6 \left(\frac{1}{2}\right) \hat{j} \]
\[ \vec{V_S} = 3\sqrt{3} \hat{i} + 3 \hat{j} \]

### Step 3: Find the relative velocity of Ram with respect to Shyam
The relative velocity \(\vec{V_{R/S}}\) is given by:
\[ \vec{V_{R/S}} = \vec{V_R} - \vec{V_S} \]
Substituting the values we have:
\[ \vec{V_{R/S}} = 6 \hat{i} - (3\sqrt{3} \hat{i} + 3 \hat{j}) \]
\[ \vec{V_{R/S}} = (6 - 3\sqrt{3}) \hat{i} - 3 \hat{j} \]

### Step 4: Calculate the magnitude of the relative velocity
To find the magnitude of the relative velocity, we use the formula:
\[ |\vec{V_{R/S}}| = \sqrt{(6 - 3\sqrt{3})^2 + (-3)^2} \]
Calculating each component:
1. \((6 - 3\sqrt{3})^2 = 36 - 36\sqrt{3} + 27 = 63 - 36\sqrt{3}\)
2. \((-3)^2 = 9\)

Now substituting these back into the magnitude formula:
\[ |\vec{V_{R/S}}| = \sqrt{(63 - 36\sqrt{3}) + 9} = \sqrt{72 - 36\sqrt{3}} \]

### Step 5: Simplify the expression
To simplify further, we can calculate the approximate value:
- \(36\sqrt{3} \approx 62.35\)
- Thus, \(72 - 62.35 \approx 9.65\)

Taking the square root:
\[ |\vec{V_{R/S}}| \approx \sqrt{9.65} \approx 3.1 \text{ m/s} \]

### Final Result
The magnitude of the relative velocity between Ram and Shyam is approximately \(3.1 \text{ m/s}\).

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