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

Explanation

To solve the problem step by step, we will use the mirror formula and the concept of relative speed.

### 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 overtaking the first car, \( \frac{du}{dt} = 15 \, \text{m/s} \)

### Step 2: Use the mirror formula
The mirror formula for a convex mirror is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where:
- \( f \) is the focal length
- \( v \) is the image distance
- \( u \) is the object distance

### Step 3: Substitute the values into the mirror formula
Substituting the known values into the mirror formula:
\[ \frac{1}{0.2} = \frac{1}{v} + \frac{1}{-2.8} \]
This simplifies to:
\[ 5 = \frac{1}{v} - \frac{1}{2.8} \]

### Step 4: Solve for \( \frac{1}{v} \)
Rearranging gives:
\[ \frac{1}{v} = 5 + \frac{1}{2.8} \]
Calculating \( \frac{1}{2.8} \):
\[ \frac{1}{2.8} \approx 0.3571 \]
Thus:
\[ \frac{1}{v} \approx 5 + 0.3571 = 5.3571 \]
Now, taking the reciprocal to find \( v \):
\[ v \approx \frac{1}{5.3571} \approx 0.186 \, \text{m} \]

### Step 5: Differentiate with respect to time
Now, we need to find the speed of the image. We differentiate the mirror formula with respect to time \( t \):
\[ 0 = -\frac{1}{u^2} \frac{du}{dt} - \frac{1}{v^2} \frac{dv}{dt} \]
Rearranging gives:
\[ \frac{dv}{dt} = -\frac{v^2}{u^2} \frac{du}{dt} \]

### Step 6: Substitute the known values
Substituting \( u = -2.8 \, \text{m} \), \( \frac{du}{dt} = 15 \, \text{m/s} \), and \( v \approx 0.186 \, \text{m} \):
\[ \frac{dv}{dt} = -\frac{(0.186)^2}{(-2.8)^2} \cdot 15 \]
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:
\[ \frac{dv}{dt} \approx -\frac{0.034596 \cdot 15}{7.84} \approx -0.066 \, \text{m/s} \]

### Step 7: Conclusion
The speed of the image of the second car as seen in the mirror of the first one is approximately \( -0.066 \, \text{m/s} \). The negative sign indicates that the image is moving in the opposite direction to the object.