A motor car is fitted with a convex driving mirror of focal length `20 cm`. A second motor car is `6 m` away from the driving mirror of the first car. Calculate (i) position of second car as seen in the first car mirror.
(ii) if the second car is overtaking the first car at a relative speed of `15 m//s`, how will its image be moving and in what direction ?

Explanation

To solve the problem step by step, we will use the mirror formula and analyze the motion of the second car relative to the first car.

### Given:
- Focal length of the convex mirror, \( f = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Distance of the second car from the first car, \( u = -6 \, \text{m} \) (negative because the object is in front of the mirror)

### (i) Calculate the position of the second car as seen in the first car's mirror.

1. **Use the mirror formula**:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]

2. **Substitute the values**:
\[ \frac{1}{v} = \frac{1}{0.2} - \frac{1}{-6} \]

3. **Calculate \( \frac{1}{f} \) and \( \frac{1}{u} \)**:
\[ \frac{1}{0.2} = 5 \quad \text{and} \quad \frac{1}{-6} = -\frac{1}{6} \]
Therefore:
\[ \frac{1}{v} = 5 + \frac{1}{6} \]

4. **Finding a common denominator**:
The common denominator for 5 and \( \frac{1}{6} \) is 6:
\[ \frac{1}{v} = \frac{30}{6} + \frac{1}{6} = \frac{31}{6} \]

5. **Calculate \( v \)**:
\[ v = \frac{6}{31} \, \text{m} \approx 0.1935 \, \text{m} \]

### Conclusion for (i):
The position of the second car as seen in the first car's mirror is approximately \( 0.1935 \, \text{m} \) behind the mirror.

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### (ii) Determine the motion of the image of the second car.

1. **Relative speed**:
The second car is overtaking the first car at a relative speed of \( 15 \, \text{m/s} \).

2. **Direction of motion**:
Since the second car is moving towards the first car (and thus towards the mirror), the image of the second car will appear to move away from the mirror.

3. **Calculate the speed of the image**:
The speed of the image \( v_i \) can be determined using the magnification formula:
\[ \text{Magnification} (m) = -\frac{v}{u} \]
The speed of the image is given by:
\[ v_i = m \cdot v_o \]
where \( v_o \) is the speed of the object (the second car).

4. **Substituting the values**:
From our previous calculation:
\[ m = -\frac{0.1935}{-6} = \frac{0.1935}{6} \]
Therefore:
\[ v_i = \left(\frac{0.1935}{6}\right) \cdot 15 \]

5. **Calculate the speed of the image**:
\[ v_i = \frac{0.1935 \times 15}{6} \approx 0.484 \, \text{m/s} \]

### Conclusion for (ii):
The image of the second car is moving away from the mirror at a speed of approximately \( 0.484 \, \text{m/s} \).

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