A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8m behind the first car is overtaking the first car at a relative speed of 15 `m/s`. The speed of the image of the second car as seen in the mrror of the first one is:

Options

A. `(1)/(15)(m)/(s)`
B. `10(m)/(s)`
C. `15(m)/(s)`
D. `(1)/(10)(m)/(s)`

Explanation

To solve the problem, we will use the mirror formula and the concept of relative speed. The steps are as follows:

### Step 1: Understand the given data
- Focal length of the convex mirror, \( F = +20 \, \text{cm} = +0.2 \, \text{m} \)
- Distance of the second car behind the first car, \( U = -2.8 \, \text{m} \) (negative because the object is in front of the mirror)
- Relative speed of the second car with respect to the first car, \( \frac{dU}{dt} = -15 \, \text{m/s} \) (negative because the second car is moving towards the first car)

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{V} + \frac{1}{U} = \frac{1}{F} \]
where \( V \) is the image distance.

### Step 3: Differentiate the mirror formula
Differentiating the mirror formula with respect to time \( t \):
\[ -\frac{1}{V^2} \frac{dV}{dt} - \frac{1}{U^2} \frac{dU}{dt} = 0 \]
This can be rearranged to find \( \frac{dV}{dt} \):
\[ \frac{dV}{dt} = -\frac{V^2}{U^2} \frac{dU}{dt} \]

### Step 4: Find \( V \) using the mirror formula
First, we need to find \( V \) using the mirror formula:
\[ \frac{1}{V} = \frac{1}{F} - \frac{1}{U} \]
Substituting the values:
\[ \frac{1}{V} = \frac{1}{0.2} - \frac{1}{-2.8} \]
Calculating each term:
\[ \frac{1}{0.2} = 5 \quad \text{and} \quad \frac{1}{-2.8} \approx -0.3571 \]
Thus,
\[ \frac{1}{V} \approx 5 + 0.3571 = 5.3571 \]
So,
\[ V \approx \frac{1}{5.3571} \approx 0.186 \, \text{m} \]

### Step 5: Substitute values into the differentiated equation
Now substitute \( V \), \( U \), and \( \frac{dU}{dt} \) into the differentiated equation:
\[ \frac{dV}{dt} = -\frac{(0.186)^2}{(-2.8)^2} \cdot (-15) \]
Calculating \( (0.186)^2 \approx 0.034596 \) and \( (-2.8)^2 = 7.84 \):
\[ \frac{dV}{dt} = \frac{0.034596}{7.84} \cdot 15 \]
Calculating further:
\[ \frac{dV}{dt} \approx \frac{0.034596 \times 15}{7.84} \approx \frac{0.51894}{7.84} \approx 0.0663 \, \text{m/s} \]

### Final Answer
The speed of the image of the second car as seen in the mirror of the first car is approximately \( 0.0663 \, \text{m/s} \).

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