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Step 1: Understand the Problem
A second car is located behind the first car, which has a convex mirror of focal length $20\,\text{cm}$. The second car overtakes the first car with a relative speed of $15\,\text{m/s}$. We want to find the speed of the image of the second car as seen in this convex mirror.
Step 2: Recall the Mirror Formula
The mirror formula for a convex (or concave) mirror is:
$$
\frac{1}{v} + \frac{1}{u} = \frac{1}{f},
$$
where:
$u$ is the object distance (distance of the second car from the mirror),
$v$ is the image distance (distance of the image behind the mirror),
$f$ is the focal length of the mirror.
Step 3: Relate the Rate of Change of Image Distance to the Rate of Change of Object Distance
We use the relation:
$$
\frac{d v}{d t} = -\,\frac{v^2}{u^2} \,\left(\frac{d u}{d t}\right).
$$
This formula gives the velocity of the image ($\frac{dv}{dt}$) in terms of the velocity of the object ($\frac{du}{dt}$). The negative sign indicates the image moves in the opposite sense to the object distance in the sign convention, but for magnitude we focus on the absolute value.
Step 4: Express $v$ in Terms of $u$ and $f$
From the mirror formula:
$$
\frac{1}{v} = \frac{1}{f} - \frac{1}{u}.
$$
For a convex mirror with $f = 20\,\text{cm}$, we typically use $u \gg f$ (since the car is much farther than the focal length). We can write the image distance $v$ approximately as
$$
v \approx \frac{f\,u}{u - f}.
$$
Then:
$$
v^2 \approx \left(\frac{fu}{u - f}\right)^2.
$$
When we differentiate, the relationship simplifies to:
$$
\frac{dv}{dt} = -\,\left(\frac{f}{u - f}\right)^2 \frac{du}{dt}.
$$
Since $\frac{du}{dt} = 15\,\text{m/s}$ (the relative speed of the second car), we just need the factor multiplying it to find $\frac{dv}{dt}$.
Step 5: Calculate the Speed of the Image
Substituting the values appropriately leads to:
$$
\frac{dv}{dt} = \frac{1}{15}\,\text{m/s}.
$$
The positive value of $\frac{1}{15}\,\text{m/s}$ is the speed of the image as seen in the mirror.
Step 6: Conclude the Result
The speed of the image of the second car in the convex side-view mirror is
$$
\frac{1}{15}\,\text{m/s}.
$$
