Two bodies A and B are at position `(2, 3)m` amd `(5, 7)m` respectively. These two are moving with constant velocity `3 ms^(-1)` along `-ve` y axis and `4ms^(-1)` along `+ve` x axis respectively. Minimum separation between them during their motion is

Options

A. `5 m`
B. `3 m`
C. `4 m`
D. `10 m`

Explanation

To find the minimum separation between the two bodies A and B during their motion, we can follow these steps:

### Step 1: Identify Initial Positions
The initial positions of the two bodies are given as:
- Body A is at position \( (2, 3) \, \text{m} \)
- Body B is at position \( (5, 7) \, \text{m} \)

### Step 2: Determine the Velocities
The velocities of the bodies are:
- Body A moves with a velocity of \( 3 \, \text{m/s} \) in the negative y-direction. This means its velocity vector can be represented as \( \vec{v_A} = (0, -3) \, \text{m/s} \).
- Body B moves with a velocity of \( 4 \, \text{m/s} \) in the positive x-direction. This means its velocity vector can be represented as \( \vec{v_B} = (4, 0) \, \text{m/s} \).

### Step 3: Write the Equations of Motion
The positions of the bodies as functions of time \( t \) can be expressed as:
- For Body A:
\[ \text{Position of A} = (2, 3 - 3t) \]
- For Body B:
\[ \text{Position of B} = (5 + 4t, 7) \]

### Step 4: Calculate the Distance Between the Bodies
The distance \( d \) between the two bodies at any time \( t \) is given by the formula:
\[ d(t) = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \]
Substituting the positions of A and B:
\[ d(t) = \sqrt{(5 + 4t - 2)^2 + (7 - (3 - 3t))^2} \]
This simplifies to:
\[ d(t) = \sqrt{(3 + 4t)^2 + (4 + 3t)^2} \]

### Step 5: Expand and Simplify
Expanding the squares:
\[ d(t) = \sqrt{(3 + 4t)^2 + (4 + 3t)^2} = \sqrt{(9 + 24t + 16t^2) + (16 + 24t + 9t^2)} \]
Combining like terms:
\[ d(t) = \sqrt{25 + 48t + 25t^2} \]

### Step 6: Find the Minimum Distance
To find the minimum distance, we can minimize the expression \( d(t) \). Since \( d(t) \) is a quadratic function in terms of \( t \), we can find the vertex of the parabola given by:
\[ d(t) = \sqrt{25t^2 + 48t + 25} \]
The minimum occurs at:
\[ t = -\frac{b}{2a} = -\frac{48}{2 \times 25} = -\frac{48}{50} = -0.96 \]
Since \( t \) cannot be negative in this context, we evaluate \( d(t) \) at \( t = 0 \):
\[ d(0) = \sqrt{25} = 5 \, \text{m} \]

### Conclusion
The minimum separation between the two bodies during their motion is \( 5 \, \text{m} \).

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