Question

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

Answer 1

Step 1: Given

1. Focal length of the convex mirror is $20cm$ which is fitted on the first car

1. Second car $2.8m$ behind the first car
2. Second car overtake the fast car at a relative speed of $15m/s$

Step 2: Formula used

1. $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ where $u$ is the distance of the object from the mirror, $v$ is the distance of the image from the mirror and $f$ is the focal length

Step 3: Find the the distance of the second car

Here, we are aware that the focal length is $f=+20cm$. Our object distance, denoted by $u$, where $u=-280cm$, is $2.8m$, or behind the first car. The first vehicle is moving at a $15m/s$ relative speed.

So $\frac{du}{dt}=15m/s$

Now, we utilize the mirror formula to determine the image distance; specifically, the second car's distance is indicated by the symbol $v$, and $v$ may be calculated as follows:

$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ \Rightarrow \frac{1}{v}=\frac{u-f}{uf} \\ \Rightarrow \frac{1}{v}=\frac{-280-20}{\left(-280\right)·\left(20\right)} \\ \Rightarrow v=\frac{280}{15}cm \end{gathered}$$

Here, $u$ is the distance of the first car, and $f$ is the focal length of the second car’s side view mirror.

Step 4: Find the speed of the image of the second car as seen in the mirror of the first one

We have

$\frac{du}{dt}=15m/s$

$$\begin{gathered} v=\frac{280}{15}cm \\ u=280cm \end{gathered}$$

The second car's speed will now be determined by differentiating the mirror formula with regard to time in the manner shown below:

$\frac{d}{dt}\left(\frac{1}{u}\right)+\frac{d}{dt}\left(\frac{1}{v}\right)=\frac{d}{dt}\left(\frac{1}{f}\right)$

But the focal length of the mirror remains constant; hence its derivative will be zero.

$$\begin{gathered} \frac{du}{dt}\left(-\frac{1}{u^2}\right)+\frac{dv}{dt}\left(-\frac{1}{v^2}\right)=0 \\ \Rightarrow \frac{du}{dt}\left(-\frac{1}{u^2}\right)=\frac{dv}{dt}\left(\frac{1}{v^2}\right) \\ \Rightarrow \frac{dv}{dt}=-\frac{v^2}{u^2}\frac{du}{dt} \\ \Rightarrow \frac{dv}{dt}=-\frac{{\left({\displaystyle \frac{280}{15}}\right)}^2}{{\left(280\right)}^2}15 \\ \Rightarrow \frac{dv}{dt}=-\frac{1}{15}m/s \end{gathered}$$

Hence the required answer is $-\frac{1}{15}m/s$