A man 2 metre high walks at a uniform speed 5 metre/hour away from a lamp post 6 metre high. The rate at which the length of his shadow increases is

Options

A. 5m/h
B. `(5)/(2)(m)/(h)`
C. `(5)/(3)(m)/(h)`
D. `(5)/(4)(m)/(s)`

Explanation

To solve the problem, we will use the concept of similar triangles and related rates.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a lamp post that is 6 meters high and a man who is 2 meters tall.
- The man walks away from the lamp post at a speed of 5 meters/hour.

2. **Define Variables**:
- Let \( y \) be the distance of the man from the lamp post.
- Let \( x \) be the length of the shadow of the man.
- The total distance from the lamp post to the tip of the shadow is \( y + x \).

3. **Establish Similar Triangles**:
- The triangle formed by the lamp post and the tip of the shadow is similar to the triangle formed by the man and the tip of his shadow.
- Using the properties of similar triangles, we can write the proportion:
\[ \frac{2}{6} = \frac{x}{y + x} \]
- Simplifying this gives:
\[ \frac{1}{3} = \frac{x}{y + x} \]
- Cross-multiplying yields:
\[ x = \frac{1}{3}(y + x) \]
- Rearranging gives:
\[ 3x = y + x \implies 2x = y \implies y = 2x \]

4. **Differentiate with Respect to Time**:
- We know that the man walks away from the lamp post at a rate of \( \frac{dy}{dt} = 5 \) meters/hour.
- Now, differentiate the relationship \( y = 2x \) with respect to time \( t \):
\[ \frac{dy}{dt} = 2 \frac{dx}{dt} \]

5. **Substitute Known Values**:
- We substitute \( \frac{dy}{dt} = 5 \) into the differentiated equation:
\[ 5 = 2 \frac{dx}{dt} \]
- Solving for \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = \frac{5}{2} \text{ meters/hour} \]

6. **Conclusion**:
- The rate at which the length of the shadow increases is \( \frac{5}{2} \) meters/hour.

### Final Answer:
The rate at which the length of his shadow increases is \( \frac{5}{2} \) meters/hour.

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