This problem involves finding the speed of an image formed by a convex mirror when the object is moving relative to the mirror. Let's break it down step by step.

Step 1: Understand the Mirror and Sign Convention

The mirror is convex, so its focal length (f) is positive. Given f = 20 cm. Convert to meters for consistency: f = 0.2 m.

We use the Cartesian sign convention: distances measured opposite to the direction of incident light are negative. For a convex mirror:

• Focal length (f) is positive.
• Object distance (u) is negative (since object is in front of the mirror).
• Image distance (v) is positive (since image is behind the mirror, virtual).

Step 2: Write the Mirror Formula

The mirror formula is: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

For our convex mirror, f = +0.2 m. The object (second car) is 2.8 m behind the first car (which has the mirror). Since the object is in front of the mirror, u = -2.8 m.

Step 3: Find the Image Position

Plug values into the mirror formula to find v: $\frac{1}{0.2}=\frac{1}{v}+\frac{1}{-2.8}$

Simplify: $5=\frac{1}{v}-\frac{1}{2.8}$ $\frac{1}{v}=5+\frac{1}{2.8}=\frac{14}{2.8}+\frac{1}{2.8}=\frac{15}{2.8}$ $v=\frac{2.8}{15}\approx 0.1867m$

Since v is positive, the image is virtual and behind the mirror.

Step 4: Relate Object and Image Speeds

To find the speed of the image, we differentiate the mirror formula with respect to time (t).

Start with: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Since f is constant, differentiate both sides: $0=-\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}$

Let $\frac{dv}{dt}=v_i$ (speed of image) and $\frac{du}{dt}=u_o$ (speed of object). Note: Since u is negative and object is moving towards the mirror (decreasing |u|), $u_o$ is positive. Rearranging: $-\frac{1}{v^2}{v}_i=\frac{1}{u^2}{u}_o$

So, $v_i=-\frac{v^2}{u^2}{u}_o$

Step 5: Plug in the Values

Given: $u=-2.8m$ , $v=\frac{2.8}{15}m$ , $u_o=15m/s$ (relative speed, object approaching mirror).

Calculate: $v_i=-\frac{{\left(\frac{2.8}{15}\right)}^2}{{(-2.8)}^2}\times 15$

Simplify: $v_i=-\frac{{2.8}^2}{{15}^2}\times \frac{1}{{2.8}^2}\times 15=-\frac{1}{{15}^2}\times 15=-\frac{1}{15}m/s$

The negative sign indicates that the image is moving in the opposite direction to the object's motion. Since the object is moving towards the mirror, the image (virtual) also moves towards the mirror but slower. The magnitude is $\frac{1}{15}m/s$.

Final Answer

The speed of the image is $\frac{1}{15}m/s$.

Related Topics and Formulae

Mirror Formula: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ , where f is focal length, v is image distance, u is object distance. Sign convention is crucial.

Differentiation of Mirror Formula: Used to find rate of change of image position with respect to time. $v_i=-\frac{v^2}{u^2}{u}_o$

Convex Mirror Properties: Always forms virtual, erect, and diminished images. Focal length is positive, image distance is positive.