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

Answer

To solve the problem, we will use the Pythagorean theorem and related rates. Let's break it down step by step.

### Step 1: Understand the Problem
We have a ladder of length 2 m leaning against a wall. The top of the ladder is sliding down at a rate of 25 cm/sec. We need to find the rate at which the bottom of the ladder is sliding away from the wall when the top of the ladder is 1 m above the ground.

### Step 2: Set Up the Variables
Let:
- \( x \) = the distance from the wall to the bottom of the ladder (in meters)
- \( y \) = the height of the top of the ladder above the ground (in meters)
- The length of the ladder is constant: \( L = 2 \) m

### Step 3: Apply the Pythagorean Theorem
Since the ladder, wall, and ground form a right triangle, we can use the Pythagorean theorem:
\[ x^2 + y^2 = L^2 \]
Substituting \( L = 2 \):
\[ x^2 + y^2 = 4 \]

### Step 4: Differentiate with Respect to Time
We 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 have:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Dividing by 2:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

### Step 5: Substitute Known Values
We know that \( y = 1 \) m (the height of the ladder at the moment we are considering) and \( \frac{dy}{dt} = -25 \) cm/sec (which is -0.25 m/sec). We need to find \( x \) when \( y = 1 \):
\[ x^2 + 1^2 = 4 \implies x^2 = 3 \implies x = \sqrt{3} \text{ m} \]

Now substitute \( y = 1 \) m and \( \frac{dy}{dt} = -0.25 \) m/sec into the differentiated equation:
\[ \sqrt{3} \frac{dx}{dt} + 1 \cdot (-0.25) = 0 \]

### Step 6: Solve for \( \frac{dx}{dt} \)
Rearranging gives:
\[ \sqrt{3} \frac{dx}{dt} = 0.25 \]
\[ \frac{dx}{dt} = \frac{0.25}{\sqrt{3}} \text{ m/sec} \]

### Step 7: Convert to cm/sec
To convert \( \frac{dx}{dt} \) from m/sec to cm/sec, multiply by 100:
\[ \frac{dx}{dt} = \frac{0.25 \times 100}{\sqrt{3}} = \frac{25}{\sqrt{3}} \text{ cm/sec} \]

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