A ladder placed on a smooth floor slips. If at a given instant the velocity with which the ladder is slipping on the floor is `v_1`, and the velocity of that part of ladder which is touching the wall is `v_1`, then the velocity of the centre of the ladder at that instant is:

Options

A. `v_1`
B. `v_2`
C. `(v_1+v_2)/2`
D. `sqrt(v_1^2+v_2^2)/2`

Explanation

To find the velocity of the center of the ladder at the instant when it is slipping, we can follow these steps:

### Step 1: Define the coordinates of the ladder
Assume the ladder makes contact with the wall at point (0, y) and with the floor at point (x, 0). The length of the ladder is L. The coordinates of the center of the ladder can be expressed as:
\[ \text{Center coordinates} = \left(\frac{x}{2}, \frac{y}{2}\right) \]

### Step 2: Establish the velocities of the points
Let:
- \( v_1 \): the velocity of the point in contact with the floor (point at (x, 0)).
- \( v_2 \): the velocity of the point in contact with the wall (point at (0, y)).

Since the ladder is slipping, we have:
\[ \frac{dx}{dt} = v_1 \quad \text{(for the point on the floor)} \]
\[ -\frac{dy}{dt} = v_2 \quad \text{(for the point on the wall)} \]

### Step 3: Find the velocity of the center of the ladder
The position vector of the center of the ladder is given by:
\[ \mathbf{R} = \left(\frac{x}{2}, \frac{y}{2}\right) \]
To find the velocity of the center, differentiate the position vector with respect to time:
\[ \mathbf{V}_{\text{center}} = \frac{d\mathbf{R}}{dt} = \left(\frac{1}{2} \frac{dx}{dt}, \frac{1}{2} \frac{dy}{dt}\right) \]
Substituting the values of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):
\[ \mathbf{V}_{\text{center}} = \left(\frac{1}{2} v_1, \frac{1}{2} (-v_2)\right) = \left(\frac{1}{2} v_1, -\frac{1}{2} v_2\right) \]

### Step 4: Calculate the magnitude of the velocity of the center
The magnitude of the velocity of the center can be calculated using the Pythagorean theorem:
\[ |\mathbf{V}_{\text{center}}| = \sqrt{\left(\frac{1}{2} v_1\right)^2 + \left(-\frac{1}{2} v_2\right)^2} \]
This simplifies to:
\[ |\mathbf{V}_{\text{center}}| = \frac{1}{2} \sqrt{v_1^2 + v_2^2} \]

### Final Result
Thus, the velocity of the center of the ladder at that instant is:
\[ \frac{1}{2} \sqrt{v_1^2 + v_2^2} \]

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