A convex, rearview mirror of focal length 20cm, is fitted in a car. A second car 2 m broad and 1.6m high is 6m away from the first car and overtakes the first car at a relative speed of `15 ms^(-1)`, then the speed of the first car is

Options

A. `0.016 ms^(-1)`
B. `0.257 ms^(-1)`
C. `0.162 ms^(-1)`
D. `0.0073 ms^(-1)`

Answer

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

### Step 1: Understand the problem setup
We have a convex rearview mirror with a focal length (F) of 20 cm. There are two cars: the first car (which has the mirror) and the second car (which is overtaking the first car). The second car is 6 m away from the first car and is moving at a relative speed of 15 m/s.

### Step 2: Convert units
Since the focal length is given in centimeters, we should convert all measurements to the same unit. The distance between the two cars is 6 m, which is equal to 600 cm.

### Step 3: Identify object distance (u)
For mirrors, the object distance (u) is taken as negative in the mirror formula. Therefore, we have:
\[ u = -600 \, \text{cm} \]

### Step 4: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{F} \]
Where:
- \( v \) = image distance
- \( u \) = object distance
- \( F \) = focal length

Rearranging the formula to find \( v \):
\[ \frac{1}{v} = \frac{1}{F} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{v} = \frac{1}{20} - \frac{1}{-600} \]
\[ \frac{1}{v} = \frac{1}{20} + \frac{1}{600} \]

### Step 5: Find a common denominator and calculate \( v \)
The common denominator for 20 and 600 is 600. Thus:
\[ \frac{1}{20} = \frac{30}{600} \]
So,
\[ \frac{1}{v} = \frac{30}{600} + \frac{1}{600} = \frac{31}{600} \]
Now, taking the reciprocal gives:
\[ v = \frac{600}{31} \, \text{cm} \]

### Step 6: Differentiate to find the velocity relationship
To find the relationship between the velocities, we differentiate the mirror formula with respect to time \( t \):
\[ \frac{d}{dt}\left(\frac{1}{v}\right) = \frac{d}{dt}\left(\frac{1}{F} - \frac{1}{u}\right) \]
This leads to:
\[ -\frac{1}{v^2} \frac{dv}{dt} = 0 - \left(-\frac{1}{u^2} \frac{du}{dt}\right) \]
Thus,
\[ \frac{dv}{dt} = \frac{v^2}{u^2} \frac{du}{dt} \]

### Step 7: Substitute values
Given that the relative speed of the second car (which is \( \frac{du}{dt} \)) is 15 m/s, we need to convert this to cm/s:
\[ \frac{du}{dt} = 15 \times 100 = 1500 \, \text{cm/s} \]
Now substituting \( v \) and \( u \):
\[ \frac{dv}{dt} = \frac{\left(\frac{600}{31}\right)^2}{(-600)^2} \cdot 1500 \]

### Step 8: Calculate \( \frac{dv}{dt} \)
Calculating this gives:
\[ \frac{dv}{dt} = \frac{360000}{961} \cdot \frac{1500}{360000} \]
This simplifies to:
\[ \frac{dv}{dt} = \frac{1500}{961} \approx 1.56 \, \text{cm/s} \]

### Step 9: Find the speed of the first car
Since the speed of the first car is equal to the speed of the image, we have:
\[ v_{\text{first car}} = \frac{1500}{961} \approx 1.56 \, \text{cm/s} \]
Converting this back to m/s:
\[ v_{\text{first car}} \approx 0.0156 \, \text{m/s} \]

### Final Answer
The speed of the first car is approximately **0.016 m/s**.