A car is fitted with a convex side-view mirror of focal length $$20 \, \text{cm}$$. A second car $$2.8 \, \text{m}$$ behind the first car is overtaking the first car at relative speed of $$15 \, \text{m/s}$$. The speed of the image of the second car as seen in the mirror of the first one is:

To find the speed of the image in a convex mirror, we can use the relationship between object velocity and image velocity derived from the mirror formula.

1. Identify the given values:

   Focal length of the convex mirror, $$f = +20\text{ cm} = +0.2\text{ m}$$ (positive for a convex mirror)

   Object distance, $$u = -2.8\text{ m}$$ (negative according to sign convention)

   Relative speed of the object (second car), $$v_o = 15\text{ m/s}$$

2. Use the mirror formula to find the image distance ($$v$$):$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

Substitute the values into the formula:

$$\frac{1}{0.2} = \frac{1}{v} + \frac{1}{-2.8}$$

$$5 = \frac{1}{v} - \frac{1}{2.8}$$

$$\frac{1}{v} = 5 + \frac{1}{2.8}$$

$$\frac{1}{v} = 5 + \frac{10}{28} = 5 + \frac{5}{14}$$

$$\frac{1}{v} = \frac{70 + 5}{14} = \frac{75}{14}$$

$$v = \frac{14}{75}\text{ m}$$

1. Relate the velocity of the image ($v_i$) to the velocity of the object ($$v_o$$):

   Differentiating the mirror formula with respect to time gives the relation for the speed of the image along the principal axis:

   $$v_i = -\left(\frac{v}{u}\right)^2 v_o$$

Since we only need the magnitude of the speed:

$$|v_i| = \left(\frac{v}{u}\right)^2 |v_o|$$

Substitute the values of $$v$$, $$u$$, and $$v_o$$:

$$|v_i| = \left(\frac{14/75}{-2.8}\right)^2 \times 15$$

Simplify the fraction inside the square:

$$\frac{14/75}{2.8} = \frac{14}{75 \times 2.8} = \frac{14}{210} = \frac{1}{15}$$

Now, calculate the speed of the image:

$$|v_i| = \left(\frac{1}{15}\right)^2 \times 15$$

$$|v_i| = \frac{1}{225} \times 15$$

$$|v_i| = \frac{1}{15}\text{ m/s}$$