A driver stops his car at a red light. The car is fitted with side view mirror of focal length 10 m. An ambulance is approaching the car at a constant speed of `16 ms^(-1)`. The speed of image of the ambulance, as seen by the driver in the side view mirror, when it is at distance of 50 m from the mirror will be

Answer

To solve the problem, we need to find the speed of the image of the ambulance as seen by the driver in the side view mirror. We will use the mirror formula and the concept of relative velocity.

### Step-by-Step Solution:

1. **Identify the type of mirror**:
The side view mirror in the car is a **convex mirror**.

2. **Set the given values**:
- Focal length of the mirror, \( f = +10 \, \text{m} \) (positive for convex mirror).
- Distance of the object (ambulance) from the mirror, \( u = -50 \, \text{m} \) (negative as per sign convention).
- Speed of the ambulance, \( v_o = 16 \, \text{m/s} \) (approaching the mirror).

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

4. **Substitute the values**:
\[ \frac{1}{v} = \frac{1}{10} - \frac{1}{-50} \]
\[ \frac{1}{v} = \frac{1}{10} + \frac{1}{50} \]
To add these fractions, find a common denominator (which is 50):
\[ \frac{1}{v} = \frac{5}{50} + \frac{1}{50} = \frac{6}{50} \]
Therefore,
\[ v = \frac{50}{6} = \frac{25}{3} \, \text{m} \approx 8.33 \, \text{m} \]

5. **Calculate the speed of the image**:
The speed of the image \( v_i \) can be found using the formula:
\[ v_i = -\left(\frac{v^2}{u^2}\right) v_o \]
Here, \( v \) is the image distance we found, and \( u \) is the object distance.

6. **Substituting the values**:
\[ v_i = -\left(\frac{(\frac{25}{3})^2}{(-50)^2}\right) \cdot 16 \]
Calculating \( \left(\frac{25}{3}\right)^2 = \frac{625}{9} \) and \( (-50)^2 = 2500 \):
\[ v_i = -\left(\frac{625/9}{2500}\right) \cdot 16 \]
Simplifying:
\[ v_i = -\left(\frac{625 \cdot 16}{9 \cdot 2500}\right) \]
\[ v_i = -\left(\frac{10000}{22500}\right) = -\left(\frac{4}{9}\right) \, \text{m/s} \approx -0.44 \, \text{m/s} \]

7. **Conclusion**:
The speed of the image of the ambulance, as seen by the driver, is approximately **-0.44 m/s**. The negative sign indicates that the image is moving in the opposite direction to the approach of the ambulance.