Question

Suppose while sitting in a parked car, you notice a jogger approaching you in the side view mirror of $R=2m$ . If the jogger is running at a speed of $5m{{s}^{-1}}$ , how fast image of the jogger appear to move when the jogger is:
A. 39m
B. 29m
C. 19m
D. 9m

Answer 1

Hint: Define the process of image formation in a spherical mirror. The mirror used as side view mirrors in cars are all convex mirrors. Obtain the mirror equation. Find its derivative with respect to time t and then we can obtain the relation for the velocity of the object to the velocity of the image.

Complete answer: Given in the question that, the radius of curvature of the die view mirror is,

$R=2m$
So, the focal length of the mirror is, $f=\dfrac{R}{2}=\dfrac{2m}{2}=1m$
Given the velocity of the jogger is, $v=5m{{s}^{-1}}$
So, we can write, $\dfrac{du}{dt}=-5m{{s}^{-1}}$

Now, the mirror equation gives the relation between the object distance, image distance and the focal length of the mirror. We can mathematically express it as,

$\begin{align}
  & \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f} \\
 & v=\dfrac{fu}{u-f} \\
\end{align}$

Differentiating the above equation with respect to time on both sides, we get

$\begin{align}
  & \dfrac{dv}{dt}=\dfrac{f\dfrac{du}{dt}-\dfrac{du}{dt}\left( fu \right)}{{{\left( u-f \right)}^{2}}} \\
 & \dfrac{dv}{dt}=\dfrac{-{{f}^{2}}\dfrac{du}{dt}}{{{\left( u-f \right)}^{2}}} \\
 & \dfrac{dv}{dt}=\dfrac{-{{f}^{2}}}{{{\left( u-f \right)}^{2}}}\dfrac{du}{dt} \\
\end{align}$

Here, $\dfrac{dv}{dt}$ will be the velocity of the image on the mirror and $\dfrac{du}{dt}$ will be the velocity of the object.

A. when the jogger is 39 m away, the velocity of the jogger on the mirror will be,
$\dfrac{dv}{dt}=\dfrac{-{{1}^{2}}\times \left( -5 \right)}{{{\left( 39-1 \right)}^{2}}}=3.46\times {{10}^{-3}}m{{s}^{-1}}$

B. when the jogger is 29 m away, the velocity of the jogger on the mirror will be,
$\dfrac{dv}{dt}=\dfrac{-{{1}^{2}}\times \left( -5 \right)}{{{\left( 29-1 \right)}^{2}}}=6.38\times {{10}^{-3}}m{{s}^{-1}}$

C. when the jogger is 19 m away, the velocity of the jogger on the mirror will be,
$\dfrac{dv}{dt}=\dfrac{-{{1}^{2}}\times \left( -5 \right)}{{{\left( 19-1 \right)}^{2}}}=1.54\times {{10}^{-2}}m{{s}^{-1}}$

D. when the jogger is 9 m away, the velocity of the jogger on the mirror will be,
$\dfrac{dv}{dt}=\dfrac{-{{1}^{2}}\times \left( -5 \right)}{{{\left( 9-1 \right)}^{2}}}=7.8\times {{10}^{-2}}m{{s}^{-1}}$

Note:
Since, the mirror used is a convex mirror, the object distance will be always negative and the image distance will be always positive. Again, the focal length of the mirror will also be positive. Since, the object distance is negative, we have set the velocity of the object as negative.