Question

A ladder AB, of length 13m, has one end A on a levelled horizontal ground and the other end B resting against a vertical wall. If the end A begins to slip away from the wall with constant speed 0.25 m/s, and the end B slips down the wall, then the speed of the end B, when B has reached a height of 5m above the ground, is

• A. 0.6 m/s
• B. 0.5 m/s
• C. 0.45 m/s
• D. 0.4 m/s
• E. 0.35 m/s

Hint:

The ratio of speeds is inversely proportional to the distances: \(\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}\). Since \(x > y\), the speed of B will be greater than the speed of A.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a related rates problem. The ladder, the wall, and the ground form a right-angled triangle.
As the ladder slips, the base and height of the triangle change, but the hypotenuse (the ladder) remains constant.

Step 2: Key Formula or Approach:
1. Pythagoras Theorem: \(x^2 + y^2 = L^2\), where \(x\) is the distance of A from the wall, \(y\) is the height of B, and \(L\) is the length of the ladder.
2. Differentiate with respect to time \(t\): \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\).

Step 3: Detailed Explanation:
Given:
Length of ladder \(L = 13\) m.
Speed of A moving away from the wall: \(\frac{dx}{dt} = 0.25\) m/s.
We need to find the speed of B (\(|\frac{dy}{dt}|\)) when height \(y = 5\) m.
First, find the distance \(x\) when \(y = 5\):
\[ x^2 + 5^2 = 13^2 \implies x^2 = 169 - 25 = 144 \implies x = 12 \text{ m} \]
Now, differentiate the relationship \(x^2 + y^2 = 13^2\):
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \implies x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]
Substitute the known values:
\[ (12)(0.25) + (5) \frac{dy}{dt} = 0 \]
\[ 3 + 5 \frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -0.6 \text{ m/s} \]
The speed is the magnitude of the rate of change, so speed = \(0.6\) m/s.

Step 4: Final Answer:
The speed of end B is 0.6 m/s.