A `2 m` ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate `25cm//sec.`, then the rate `("in cm"//"sec".)` at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is `1 m` above the ground is :

Options

A. `(25)/(sqrt3)`
B. 25
C. `25sqrt3`
D. `(25)/(3)`

Explanation

To solve the problem step by step, we will use the Pythagorean theorem and related rates.

### Step 1: Set up the problem
Let:
- \( x \) = the distance from the wall to the bottom of the ladder (horizontal distance)
- \( y \) = the height of the top of the ladder above the ground (vertical distance)
- The length of the ladder is constant at \( L = 2 \) meters.

### Step 2: Write the relationship using the Pythagorean theorem
According to the Pythagorean theorem, we have:
\[ x^2 + y^2 = L^2 \]
Substituting \( L = 2 \):
\[ x^2 + y^2 = 2^2 \implies x^2 + y^2 = 4 \]

### Step 3: Differentiate with respect to time
We will differentiate both sides of the equation with respect to time \( t \):
\[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(4) \]
Using the chain rule, we get:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Dividing through by 2:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

### Step 4: Solve for \( \frac{dx}{dt} \)
Rearranging gives:
\[ x \frac{dx}{dt} = -y \frac{dy}{dt} \]
Thus:
\[ \frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt} \]

### Step 5: Substitute known values
We know:
- \( \frac{dy}{dt} = -25 \) cm/sec (the negative sign indicates that \( y \) is decreasing).
- We need to find \( \frac{dx}{dt} \) when \( y = 1 \) m (or 100 cm).

Using the Pythagorean theorem to find \( x \) when \( y = 1 \):
\[ x^2 + 1^2 = 4 \implies x^2 + 1 = 4 \implies x^2 = 3 \implies x = \sqrt{3} \text{ m} = \sqrt{3} \times 100 \text{ cm} \]

### Step 6: Calculate \( \frac{dx}{dt} \)
Now substituting \( y = 100 \) cm and \( x = \sqrt{3} \times 100 \) cm into the equation:
\[ \frac{dx}{dt} = -\frac{100}{\sqrt{3} \times 100} \times (-25) \]
This simplifies to:
\[ \frac{dx}{dt} = \frac{25}{\sqrt{3}} \text{ cm/sec} \]

### Final Answer
Thus, the rate at which the bottom of the ladder slides away from the wall when the top of the ladder is 1 m above the ground is:
\[ \frac{dx}{dt} = \frac{25}{\sqrt{3}} \text{ cm/sec} \]