A point object is moving on the principal axis of a concave mirror of focal length `24 cm` towards the mirror. When it is at a distance of `60 cm` from the mirror, its velocity is . `9 sec// cm` What is the velocity of the image at that instant

Options

A. `5 cm//sec` towards the mirror
B. `4 cm//sec` towards the mirror
C. `4 cm//sec` away from the mirror
D. `9 cm//sec` away from the mirror

Answer

To solve the problem, we will use the mirror formula and the concept of related rates in calculus. Let's break it down step by step:

### Step 1: Identify the given values
- Focal length of the concave mirror, \( f = -24 \) cm (negative because it is a concave mirror)
- Object distance, \( u = -60 \) cm (negative as per the sign convention)
- Velocity of the object, \( \frac{du}{dt} = -9 \) cm/s (negative because the object is moving towards the mirror)

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{-24} = \frac{1}{v} + \frac{1}{-60} \]

### Step 3: Solve for \( v \)
Rearranging the equation:
\[ \frac{1}{v} = \frac{1}{-24} + \frac{1}{60} \]
Finding a common denominator (which is 120):
\[ \frac{1}{v} = \frac{-5}{120} + \frac{2}{120} = \frac{-3}{120} = \frac{-1}{40} \]
Thus,
\[ v = -40 \text{ cm} \]

### Step 4: Differentiate the mirror formula
Now we differentiate the mirror formula with respect to time \( t \):
\[ \frac{d}{dt}\left(\frac{1}{f}\right) = \frac{d}{dt}\left(\frac{1}{v}\right) + \frac{d}{dt}\left(\frac{1}{u}\right) \]
Since \( f \) is constant, \( \frac{d}{dt}\left(\frac{1}{f}\right) = 0 \):
\[ 0 = -\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} \]

### Step 5: Substitute known values
Substituting \( v = -40 \) cm, \( u = -60 \) cm, and \( \frac{du}{dt} = -9 \) cm/s:
\[ 0 = -\frac{1}{(-40)^2} \frac{dv}{dt} - \frac{1}{(-60)^2} (-9) \]
This simplifies to:
\[ 0 = -\frac{1}{1600} \frac{dv}{dt} + \frac{9}{3600} \]

### Step 6: Solve for \( \frac{dv}{dt} \)
Rearranging gives:
\[ \frac{1}{1600} \frac{dv}{dt} = \frac{9}{3600} \]
Multiplying both sides by 1600:
\[ \frac{dv}{dt} = \frac{9 \times 1600}{3600} \]
Calculating:
\[ \frac{dv}{dt} = \frac{14400}{3600} = 4 \text{ cm/s} \]

### Final Answer
The velocity of the image at that instant is \( 4 \) cm/s.

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